Welcome to Modern
Abstract Algebra
MAM3011F, 3AL
Ready to exercise your brain?
Contents
1 Preliminaries
1
1.1 Well-Ordering Axiom. Induction. Strong Induction. . . . . . . .
1
1.2
Division Algorithm. Gcd’s. Primes . . . . . . . . . . . . . . . . .
6
1.3
Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.4
Integers modulo n . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
1.5
Permutation groups . . . . . . . . . . . . . . . . . . . . . . . . . . 15
1.6
1.5.1
Cycles. Disjoint permutations . . . . . . . . . . . . . . . . 17
1.5.2
The Alternating group . . . . . . . . . . . . . . . . . . . . 20
Cyclic groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
1.6.1
Maps between groups . . . . . . . . . . . . . . . . . . . . . 32
2 Dihedral groups. Lagrange’s Theorem
2.1
35
Dihedral groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
2.2 The centre of the octic group . . . . . . . . . . . . . . . . . . . . . 37
2.3
Cosets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
2.3.1
2.4
Some important properties of cosets . . . . . . . . . . . . 39
Lagrange’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 41
3 Isomorphism Theorems
45
3.1
Normal subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
3.2
Factor groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
3.3 The Correspondence Theorem . . . . . . . . . . . . . . . . . . . . 51
3.3.1
Maximal normal subgroups . . . . . . . . . . . . . . . . . 56
3.4 The First Isomorphism Theorem . . . . . . . . . . . . . . . . . . 57
3.5 The Second Isomorphism Theorem . . . . . . . . . . . . . . . . . 62
3.6 The Third Isomorphism Theorem . . . . . . . . . . . . . . . . . . 64
iii
iv
CONTENTS
4 Cauchy’s Theorem
67
4.1
Group Action on a Set . . . . . . . . . . . . . . . . . . . . . . . . 67
4.2
Conjugation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
4.3
Cauchy’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
5 Sylow Theorems
79
5.1 The Sylow Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . 79
5.2 Applications of the Sylow Theorems . . . . . . . . . . . . . . . . 84
6 Finite Abelian groups
87
6.1
Direct Sums and Products . . . . . . . . . . . . . . . . . . . . . . 87
6.2
Fundamental Theorem of Finite Abelian Groups . . . . . . . . . 91
6.2.1
6.3
Constructing all abelian finite groups . . . . . . . . . . . 91
Proof of the Fundamental Theorem of Finite Abelian Groups . 96
Chapter 1
Preliminaries
In this first chapter, we will revisit some foundational concepts and results.
Although some of these might be familiar to you, the goal here is to refresh
your mind, make sure you know these if you didn’t and set you up for a great
semester!
1.1
Well-Ordering Axiom. Induction. Strong
Induction.
Let us begin by recalling the meaning of “proving by induction”. Consider
the following sequence of integers:
1 = 1,
(S)
1 + 3 = 4,
1 + 3 + 5 = 9,
1 + 3 + 5 + 7 = 16,
⋮
If we look at carefully both left and right side of (S), then we will realise that
these follow a pattern. More precisely:
1
2
J. Sánchez-Ortega
Left side
Right side
Sum of the first n odd integers
12 , 22 , 32 , . . . , n2
Keeping in mind that the n-th odd integer is 2n − 1, it is natural to ask
ourselves whether
?
1 + 3 + 5 + . . . + (2n − 1) = n2
(pn )
But how can we prove the validity of the statement (pn )?
Principle of Mathematical Induction
Let pn be a property or statement defined for each n ∈ N. Suppose that
the following two conditions hold.
(1) Basis Step: p1 is true.
(2) Inductive Step: if pk is true, then pk+1 is true. (k ≥ 1)
Then the property pn is true for every n ∈ N.
Definition 1.1. The proof of a result that applies the Principle Mathematical
Induction is called a proof by induction. The assumption that pk is true is
called the induction hypothesis.
Example 1.2. Prove by induction that
1 + 3 + 5 + . . . + (2n − 1) = n2 , for all n ≥ 1
Solution: Let pn be the property 1 + 3 + 5 + . . . + (2n − 1) = n2 . It is clear that
the base step, (the validity of p1 ) holds since 1 = 12 . Assume that pk is true
and let us show that pk+1 is also true. The induction hypothesis tells us that
1 + 3 + 5 + . . . + (2k − 1) = k 2 . Using this we obtain that
1 + 3 + 5 + . . . + (2k − 1) + (2k + 1) = k 2 + 2k + 1 = (k + 1)2 ,
proving the validity of pk+1 and the induction principle applies to get that pn
is true for all n ≥ 1.
Notice that we have started the Principle of Mathematical Induction at 1
but we can indeed replace it by any integer m. More concretely:
1. Preliminaries
3
General Principle of Mathematical Induction
If m ∈ N and pm , pm+1 , . . . are statements such that
(1) Base Step: pm is true.
(2) Inductive Step: if pk is true, then pk+1 is true, where k ≥ m.
Then pn is true for each n ≥ m.
To prove the General Principle of Mathematical Induction, apply the
Principle of Mathematical Induction to the following family of statements:
tn = pm+n−1 ,
n ≥ 1.
Sometimes, to prove that pk+1 holds, it is not enough to assume that pk is
true. In those situations, we will need to apply a variant of the Principle of
Mathematical Induction called Strong Induction.
Strong Induction
Let m ∈ N and for each n ≥ m, let pn be a statement. If the following
conditions are satisfied:
(1) pm is true.
(2) If k ≥ m and all of pm , pm+1 , . . . , pk are true, then pk+1 is also
true.
Then pn is true for every n ≥ m.
Proof. Given n ≥ m, we define the following statement:
tn ∶
pm , pm+1 , . . . , pn are true
Note that tm is true by (1). Assume that tk is true for some k ≥ m, and let us
show that tk+1 is so. Note that the validity of tk means that pm , pm+1 , . . . , pk
are true. Then from (2) we get that pk+1 is true, so tk+1 is true, as desired.
An application of the General Principle of Mathematical Induction concludes
the proof.
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J. Sánchez-Ortega
Recall that p ∈ N is a prime number (or, simply prime) if p ≠ 1 and the
only natural numbers divisors of p are 1 and p.
Example 1.3. Use Strong Induction to show that every integer n ≥ 2 is a
product of (one or more) primes.
Solution: For n ≥ 2, consider the statement
pn ∶ n is a product of primes.
It is clear that p2 is true, since 2 is a prime. Take k ≥ 2 and assume that
p2 , p3 , . . . , pk are true, i.e., 2, 3, . . . , k are all products of primes. In order to
apply Strong Induction, we need to show that k + 1 is a product of primes.
We distinguish two cases:
C1 k + 1 is a prime. In this case we are done! :)
C2 k + 1 = ab, for some 2 ≤ a, b ≤ k.
In the second case, applying the Strong Induction hypothesis, we get that
a and b are product of primes, so is ab = k + 1.
We close this section recalling a very useful property of the natural numbers
N, which turns out to be equivalent to Induction and Strong Induction.
Well-Ordering Axiom/ Least Integer Principle
If X is a nonempty subset of N, then there exists a ∈ X such that a ≤ x
for all x ∈ X.
Theorem 1.4. The following are equivalent:
(i) Well-Ordering;
(ii) Induction;
(iii) Strong Induction.
1. Preliminaries
5
Proof. (i) ⇒ (ii)
Let p1 , p2 , . . . be statements such that p1 is true and pk ⇒ pk+1 for every
k ≥ 1. We have to show that pn is true for every n ≥ 1. To do so we define the
set X = {n ≥ 1 ∣ pn is false}, that consists on all the natural numbers n such
that pn is false. Our goal is to show that X = ∅. Assume, on the contrary,
that X ≠ ∅. Then by the Well-Ordering Principle X has a smallest element,
say m. Notice that m > 1 since p1 is true, so m − 1 ≥ 1. Notice that m − 1 ∉ X
because m is the least element of X. Thus pm−1 is true and so is pm (because
pm−1 ⇒ pm ). But this is impossible because pm is false, since m ∈ X. Thus
X = ∅ and hence pn is true for all n ≥ 1.
(ii) ⇒ (iii) has been proved.
(iii) ⇒ (i)
Let X be a nonempty subset of N and let us show that X contains a least
element. For n ≥ 1, we define the statement
pn ∶ if n ∈ X then X has a least element.
Notice that p1 is true; in fact, if 1 ∈ X, then 1 is the least element of X. Take
k ≥ 1 and assume that p1 , . . . , pk are true and let us show that pk+1 is also true.
Suppose then that k + 1 ∈ X and let us prove that X has a least element. We
consider two cases:
C1 k + 1 is the least element of X. Then there is nothing to prove.
C2 There exists m ∈ X such that m < k + 1. In this case, then pm is true,
and since m ∈ X, this implies that X has a least element. In any case,
we have proved that if k + 1 ∈ X, then X has a least element, that is,
pk+1 is true.
Applying Strong Induction we can conclude that pn is true for all n ≥ 1.
Keeping in mind that X is nonempty, we get that X has a least element.
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J. Sánchez-Ortega
1.2
Division Algorithm. Gcd’s. Primes
Using the Well-Ordering Axiom one can prove the existence of the quotient
and remainder in the integer long division. In algebra, this result is known as
the Division Algorithm.
Division Algorithm
Let n and d ≥ 1 be integers. Then there exist unique integers q and r,
called quotient and remainder , respectively, satisfying that n = qd + r
and 0 ≤ r < d.
Proof. We divide the proof into two parts:
Part 1: Existence of quotient and remainder.
We claim that the set X = {n − td ∣ t ∈ Z, n − td ≥ 0} is non-empty. In fact:
• If n ≥ 0, then n ∈ X since n = n − 0 ⋅ d ≥ 0.
• If n < 0, then n − nd ∈ X since n − nd = n(1 − d) ≥ 0.
Hence X is non-empty, and the Well-Ordering Axiom applies to get that
X has a least element r. Then r ≥ 0 and r = n − qd for some q ∈ Z. Thus,
n = qd + r. It remains to show that r < d. Suppose on the contrary that
r ≥ d. Then r − d ∈ X since r − d ≥ 0 and r − d = n − qd − d = n − (q + 1)d. This
contradicts the minimality of r (because r − d < r) and so r < d.
Part 2: Uniqueness of quotient and remainder.
Suppose that we can also write n = q ′ d + r′ for some integers q ′ and r′ such
that 0 ≤ r′ < d. We can assume (without loss of generality) that r ≥ r′ . Then
subtracting both expressions we obtain that 0 = (q − q ′ )d + r − r′ , which implies
that (q ′ − q)d = r − r′ . On the other hand, we know that 0 ≤ r − r′ ≤ r < d, and
q ′ − q is an integer. So necessarily r = r′ and q = q ′ .
Example 1.5. When dividing −17 by 5 we obtain a quotient of −4 and a
remainder of 3, since −17 = (−4) ⋅ 5 + 3. While the quotient and remainder
when dividing 4187 by 129 are 32 and 59, respectively. This means that
4187 = 32 ⋅ 129 + 59.
1. Preliminaries
7
Let n, m, d be integers with d ≠ 0. Recall that n is said divisible by d, or
that d is a divisor of n, if there exists an integer q such that n = qd. We
write d∣n . We also say that n is a multiple of d. If d does not divide n, we
write d ∣/ n . We call d a common divisor of m and n if d∣m and d∣n.
There is a distinguished common divisor, known as, the greatest common
divisor, or gcd for short, that satisfy a very important property. We recall
these now.
Definition 1.6. Let m and n be integers, not both zero. An integer d is
called the greatest common divisor of m and n, written d = gcd(m, n)
if it satisfies the following:
(1) d ≥ 1.
(2) d∣m and d∣n.
(3) k∣m and k∣n ⇒ k∣d.
We say that m and n are coprime or relatively prime if gcd(m, n) = 1.
Bézout’s Lemma/ Bézout’s Identity
Let m and n be two integers, not both zero. Then d = gcd(m, n) exists
and d = xm + yn for some integers x and y.
Proof. Let X = {sm + tn ∣ s, t ∈ Z and sm + tn ≥ 1}. Note that X is nonempty, since m2 + n2 ∈ X. By the Well-Ordering Axiom we can find a least
element in X, say d. Then d ≥ 1 and d = xm + yn for integers x and y; if k∣m
and k∣n, then k∣(xm + yn) = d. This shows that any common divisor of m and
n is a divisor of d. It remains to show that d is a common divisor of m and
n. Here, we will show that d∣n. One can show that d∣m similarly. Apply the
Division Algorithm to find q, r ∈ Z such that n = qd + r and 0 ≤ r < d. Then
r = n − qd = n − q(xm + yn) = (−qx)m + (1 − qy)n,
which says that r is a linear combination of m and n. If r > 0 we would have
that r ∈ X, which contradicts the minimality of d in X. Thus r = 0.
Bézout’s Lemma is quite handy when proving statements in group theory.
An important consequence is the following:
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J. Sánchez-Ortega
Corollary 1.7. Two integers m and n are relatively prime if and only if
there exist integers x and y such that xm + yn = 1. In particular, any two
consecutive integers k and k + 1 are relatively prime.
Proof. If m and n are relatively prime, then the result follows by Bézout’s
Lemma. Suppose now that there exist integers x and y such that xm + yn = 1,
and let d = gcd(m, n). Then d∣(xm + yn), since d is a common divisor of m
and n. But xm + yn = 1, so d∣1 and since d ≥ 1 (by the definition of the gcd)
we have that d = 1, proving that m and n are relatively prime.
I am sure you learnt at the school that any integer n ≥ 2 can be factorised
into primes in an unique way: n = pn1 1 pn2 2 . . . pnr r , where the pi are all the
distinct prime divisors of n, and ni ≥ 1 for each i. In algebra, this fact is
known as the Prime Factorisation Theorem and can be proved using Strong
Induction and the Well-Ordering Axiom. To prove it, we first need to prove a
few results about divisibility and prime numbers.
Theorem 1.8. Let m and n be relatively prime integers. Then the following
assertions hold for any integer k:
(1) m∣k, n∣k ⇒ mn∣k.
(2) m∣kn ⇒ m∣k.
(3) Euclid’s Lemma. If p∣mn where m and n are integers, then either p∣m
or p∣n.
Proof. By Bézout’s Lemma, we have 1 = xm + yn, for x and y integers.
(1) If k = qm and k = pn for some integers p and q, then
k = 1 ⋅ k = (xm + yn)k = xmk + ynk = xm(pn) + yn(qm) = (xp + yq)mn.
Hence mn∣k, proving (1).
(2) Let nk = qm, where q is an integer. Then
k = 1 ⋅ k = (xm + yn)k = xmk + ynk = xmk + y(qm) = (xk + yq)m,
which shows that m∣k, as desired.
1. Preliminaries
9
(3) Let d = gcd(m, p). Then d∣p, which implies that either d = 1 or d = p
(because p is a prime). If d = p, then p∣m; otherwise if d = 1, then m and p are
coprime, and (2) applies to get that p∣n.
Prime Factorisation Theorem
(1) Every integer n ≥ 2 is a product of (one or more)
primes.
(2) The factorization is unique up to the order of the
factors. That is, if n = p1 p2 . . . pr = q1 q2 . . . qs , where all the pi
and qj are primes, then r = s and the qj can be relabelled so that
pi = qi for all i = 1, 2, . . . , r.
Proof. (1) was proved in Example 1.3.
(2) Assume, on the contrary, that (2) fails, which means that there exits an
integer ≥ 2 admitting two different factorisations into primes. Then by the
Well-Ordering Axiom we can choose m ≥ 2 to be the smallest integer admitting
two different factorisations into primes:
m = p 1 p2 . . . p r = q1 q 2 . . . q s .
Notice that m is not a prime (otherwise, it would only admit a prime factorisation; the trivial one) and so r ≥ 2, s ≥ 2. We have p1 ∣q1 q2 . . . qs , so p1 ∣qj (for
some j) by Euclid’s Lemma. By relabelling the qj , we may assume that p1 ∣q1 ,
which yields that p1 = q1 because both are primes, so
m
= p2 . . . p r = q2 . . . q s ,
p1
is an integer (smaller than m) that admits two distinct factorisations into
primes. It contradicts the choice of m, and therefore proves (2).
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J. Sánchez-Ortega
1.3
Groups
Definitions 1.9. A group (G, ⋅) is a set G equipped with a binary operation
⋅ ∶ G × G → G satisfying the following axioms:
(G1) The associative law holds: a ⋅ (b ⋅ c) = (a ⋅ b) ⋅ c for every a, b, c ∈ G.
(G2) There is a unity 1 for G. Then 1 ⋅ a = a ⋅ 1 = a for all a ∈ G.
(G3) Every element of G has an inverse in G. In other words, for every a ∈ G
there is an element b ∈ G such that a ⋅ b = b ⋅ a = 1
The group G is called abelian if, in addition, it satisfies
(G4) The operation ⋅ is commutative: a ⋅ b = b ⋅ a holds for every a, b ∈ G.
If G is finite, the number ∣G∣ is called the order of G.
Although we write most groups multiplicatively, many important groups
are written additively. Then the unity element is denoted 0 and is called zero,
and the inverse of x is denoted −x and is called the negative of x.
A subset H of a group G is called a subgroup of G if H is itself a group
using the operation of G.
Examples 1.10.
1. The sets Z, Q, R, C are all abelian groups under addition. In each
case the identity element is 0, and the inverse of a is −a. Each of them
is a subgroup of the larger ones.
2. Let R denote Z, Q, R, or C. Then the set U (Mn (R), ⋅) of invertible
n × n matrices over R, that is, the matrices A ∈ Mn (R) such that det(A)
has a mutiplicative inverse in R, is a group called the general linear
group of degree n over R, denoted GL(n, R). More precisely:
If R = Q, R, or C then GL(n, R) = {A ∈ Mn (R) ∣ det(A) ≠ 0}
Gl(n, Z) = {A ∈ Mn (Z) ∣ det(A) = ±1}
The special linear group is the following subgroup of GL(n, R):
SL(n, R) = {A ∈ Mn (R) ∣ det(A) = 1}
Recall that to find out whether a given set of a group was a subset we
used the Subgroup Test, which can be illustrated as follows:
1. Preliminaries
11
G
1G
h1
h
hh1
h−1
H
h−1
1
Subgroup Test
A subset H of a group G is a subgroup of G if and only if
the following three conditions are satisfied.
(1) 1G ∈ H, where 1G is the identity element of G.
(2) If h ∈ H and h1 ∈ H, then hh1 ∈ H.
(3) If h ∈ H, then h−1 ∈ H. Here h−1 denotes the inverse of h in G.
In this case, H has the same unity as G, and if h ∈ H, its inverse in
H is the same as its inverse in G.
Example 1.11. Let n ≥ 0 be an integer. Then the set of all multiples of n,
namely, nZ = {nk ∣ k ∈ Z} is a subgroup of the additive group (Z, +).
In fact, the identity element of Z is 0, and 0 = n ⋅ 0 ∈ nZ. If a and b are in nZ,
then a = nk and b = nm, for some k, m ∈ Z, and
a + b = nk + nm = n(k + m) ∈ nZ,
−a = −(nk) = n(−k) ∈ nZ.
From the Subgroup Test we get that nZ is a subgroup of Z. Recall that all
these are all the subgroups of Z.
Example 1.12. Recall that the centre Z(G) of a group G is the set of
elements of G that commutes with every element of G; that is:
Z(G) = {z ∈ G ∣ zg = gz for all g ∈ G}.
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J. Sánchez-Ortega
Notice that G is abelian if and only if Z(G) = G. To show that Z(G) is
indeed a subgroup of G we use the Subgroup Test. It is clear that 1 ∈ Z(G).
For z, z ′ ∈ Z(G) we need to prove that zz ′ ∈ Z(G) and z −1 ∈ Z(G). To do so,
take any g ∈ G and check that zz ′ and z −1 commute with g of G; in fact:
g(zz ′ ) = (gz)z ′
z∈Z(G)
=
(zg)z ′ = z(gz ′ )
z ′ ∈Z(G)
=
z(z ′ g) = (zz ′ )g,
which shows that zz ′ ∈ Z(G). Now, since z ∈ Z(G) we have that
zg = gz ⇒ z −1 (zg) = z −1 (gz) ⇒ g = (z −1 g)z ⇒ gz −1 = z −1 g,
proving that z −1 ∈ Z(G).
When manipulating groups, the exponential and cancellation laws are very
useful. We will these a lot this year, so make sure you remember them!
Exponent and Cancellation Laws
Let G be a group and f, g, h ∈ G.
(1) g n g m = g n+m for all n, m ∈ Z.
(2) (g n )m = g nm for all n, m ∈ Z.
(3) If gh = hg, then (gh)n = g n hn for all n ∈ Z.
(4) Left cancellation Law: if gh = gf , then h = f .
(5) Right cancellation Law: if hg = f g, then h = f .
Unless otherwise is specified, when dealing with groups, we adopt a multiplicative notation for their operation. Recall that if the operation in a group
G is written additively, for example in (Z, +), the unity is denoted 0 and
the inverse of g ∈ G is denoted −g. The exponent g n (in multiplicative
notation) becomes ng. The exponents laws look like:
Multiplicative notation
Additive notation
g n+m = g n g m
(g n )m = g nm
(n + m)g = ng + mg
m(ng) = (mn)g
(gh)n = g n hn
if g and h commute
n(g + h) = ng + nh
1. Preliminaries
1.4
13
Integers modulo n
Given an integer n ≥ 2, one can define on the set of integers Z an equivalence
relation ≡, called congruence modulo n as follows:
Two integers a and b are said to be congruent modulo n if n∣(a − b). In
such a case, we write a ≡ b (mod n) .
Example 1.13. 2 ≡ 5 (mod 3),
21 ≡ 16 (mod 5),
−4 ≡ 2 (mod 6).
The equivalence class [a]n or [a] of an integer a with respect to congruence
modulo n is called its residue class modulo n, while a is said to be a
generator of [a]n . That is,
[a]n = {x ∈ Z ∣ x ≡ a (mod n)} .
Notice that we that x ∈ [a]n , or equivalently, x ≡ a (mod n) if and only
if the remainder when dividing x by n is a . In fact:
x ∈ [a]n ⇔ x ≡ a (mod n) ⇔ n∣(x − a) ⇔ x = a + qn, for q ∈ Z.
This means that [a]n consists on the integers whose remainder is a
when divided by n.
Two residue classes [a]n and [b]n are equal if and only if a ≡ b (mod n) .
In fact, if a ≡ b (mod n) and x ∈ [a]n , then x ≡ a (mod n). Using that
congruence modulo n is an equivalence relation we obtain that x ≡ b (mod n),
which means that x ∈ [b]n . This proves that [a]n ⊆ [b]n ; the other containment
can be shown in a similar way. The converse trivially holds.
The set consisting of all the residue classes modulo n is denoted by
Zn , and is called the set of integers modulo n. In what follows, to ease
the notation, we write ā to refer to [a]n . Using the Division Algorithm we
can prove that the order of Zn is n. More precisely, Zn = {0̄, 1̄, . . . , n − 1} .
We can define an addition (respectively, multiplication) on the set of
integers Zn modulo n by adding (respectively, multiplying) the generators of
the corresponding residue classes. More precisely, for ā, b̄ ∈ Zn we define:
ā + b̄ = a + b
and
ā ⋅ b̄ = ab
For instance, in Z6 we have that 3̄ + 5̄ = 3 + 5 = 8̄ = 2̄, since 8 ≡ 2 (mod 6).
depend on the generator of the residue classes. For example, 9̄ = 3̄ in Z6 and
9̄ + 5̄ = 9 + 5 = 14 = 2̄, since 14 ≡ 2 (mod 6),
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J. Sánchez-Ortega
that is, we obtain the same result, as expected. Some properties of the modular
operations (that is, the addition and multiplication in Zn ) are collected below.
Theorem 1.14. Let n ≥ 2 be a fixed modulus and a, b, c ∈ Z. Then the
following hold in Zn .
(1) ā + b̄ = b̄ + ā
and
āb̄ = b̄ā.
(2) ā + (b̄ + c̄) = (ā + b̄) + c̄ and
(3) ā + 0̄ = ā and
ā(b̄c̄) = (āb̄)c̄.
ā1̄ = ā . We call 0̄ the zero of Zn , while 1̄ is said to be
the unity of Zn .
(4) ā + −a = 0̄ . We usually write −ā to denote −a.
(5) ā(b̄ + c̄) = āb̄ + āc̄.
Remark 1.15. There are several differences between the arithmetic of Zn
and that of Z. For example:
• 0 and 1 are the only integers satisfying that k 2 = k, while in Z6 the
elements 3̄ and 4̄ have that property:
(3̄)2 = 9̄ = 3̄, since 9 ≡ 3 (mod 6),
(4̄)2 = 16 = 4̄, since 16 ≡ 4 (mod 6),
• the cancellation law holds in Z: if ab = ac and a ≠ 0, then b = c. In
general, we do not have cancellation in Zn . For instance, in Z6 we
have that 4̄ ⋅ 2̄ = 4̄ ⋅ 5̄, where 4̄ ≠ 0̄ and 2̄ ≠ 5̄.
• If ab = 0 in Z, then either a = 0 or b = 0. This need not hold in Zn . For
instance, in Z6 we have that 2̄ ⋅ 3̄ = 0̄, and 2̄ ≠ 0̄, 3̄ ≠ 0̄.
From Theorem 1.14 and Remark 1.15 we deduce that (Zn , +, 0̄) is an
abelian group of order n, while (Zn , ⋅, 1̄) is NOT a group , in general.
Moreover, recall that ā ∈ Zn has a multiplicative inverse in Zn if and
only if a and n are relatively prime, that is, gcd(a, n) = 1. In particular,
if p is prime, then (Zp , ⋅, 1̄) is an abelian group of order p.
1. Preliminaries
1.5
15
Permutation groups
A permutation of a set X is a bijection α ∶ X → X. The set SX of all
permutations on X is a group under composition, called a permutation
group. Notice that the identity element in SX is the identity map IdX .
For a given integer n ≥ 1, we write Xn = {1, 2, . . . , n} ; the permutation
group Sn is called the symmetric group of degree n. In what follows, we
will recall some features of this group. We begin by introducing First a more
convenient way to write down permutations in Sn . This notation is known as
the Matrix-Type Notation.
Matrix-Type Notation. We write σ ∈ Sn as a matrix as follows:
• in the top row we list the elements 1, 2, . . . , n of Xn ;
• in the bottom row we put the numbers after shuffling, that is,
σ(1), σ(2), . . . , σ(n).
2
...
n ⎞
⎛ 1
1
2
...
n
↓
↓ ⎟
σ=(
)=⎜ ↓
σ(1) σ(2) . . . σ(n)
⎝σ(1) σ(2) . . . σ(n)⎠
In other words, the elements of Xn are listed in the top row, and beneath
each element of Xn is its image.
Example 1.16. The permutation σ ∈ S4 defined by σ(1) = 3, σ(2) = 1,
1 2 3 4
σ(3) = 4, σ(4) = 2 in matrix notation is σ = (
).
3 1 4 2
Theorem 1.17. The symmetric group Sn of degree n has ∣Sn ∣ = n! elements.
Proof. Since permutations in Sn are bijective maps Xn → Xn , we have that
two permutations σ and τ in Sn are equal if and only if σ(k) = τ (k) for every
k ∈ Xn . On the other hand, in order to construct a permutation
1
2
...
n
σ=(
)
σ(1) σ(2) . . . σ(n)
we have to choose the numbers σ(1), σ(2), . . . , σ(n) from Xn , so that they
are all distinct. Hence we have n choices for σ(1), then n − 1 choices for
σ(2), then n − 2 choices for σ(3), and so on. Thus, in total σ can be built in
n(n − 1) . . . 2 ⋅ 1 = n! different ways, which shows that ∣Sn ∣ = n!
16
J. Sánchez-Ortega
Recall that the group operation in Sn is the composition of maps. Given
γ, σ ∈ Sn , to find their composition γ ○ σ we first apply σ and then γ:
(γ ○ σ)(k) = γ(σ(k)),
for all k ∈ Xn .
We write γσ to denote γ ○ σ, and refer to it as the product of the permutations γ and σ. It is way more practical to compute γσ using the matrix
expressions of γ and σ. In fact, to compute γσ we stack σ (which is the
permutation on the right-most position) on top of γ and follow the direction
of the arrows like shown in the example below.
1 2 3
1 2 3
Example 1.18. Find γσ, for σ = (
) and γ = (
).
3 1 2
1 3 2
Solution: We stack σ on top of γ:
σ=
γ=
1
2
3
3
1
2
1
2
3
1
3
2
1 2 3
We have obtained that γσ = (
).
2 1 3
The identity permutation ε in Sn is the identity map: ε(k) = k for every
1 2 ... n
k ∈ Xn . We write it as ε = (
)
1 2 ... n
The inverse of σ ∈ Sn is the (unique) map σ −1 ∶ Xn → Xn such that
σ(σ −1 (k)) = k, σ −1 (σ(k)) = k . These equations tell us that each of σ and
σ −1 reverses the action of the other. So, we can easily obtain the matrix
expression of σ −1 from the matrix of σ by reading up:
2
...
n ⎞
⎛ 1
↑
↑ ⎟
σ=⎜ ↑
⎝σ(1) σ(2) . . . σ(n)⎠
1. Preliminaries
17
1 2 3 4 5 6 7 8
For instance, the inverse permutation of σ = (
) ∈ S8
4 1 8 3 2 5 6 7
can be found by looking at the second row of σ and reading up to its top row:
⎛1 2 3 4 5 6 7 8⎞
σ = ⎜↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑⎟
⎝4 1 8 3 2 5 6 7⎠
From here we obtain that
σ −1 = (
1 2 3 4 5 6 7 8
).
2 5 4 1 6 7 8 3
We check below that σ −1 σ = ε. Similarly, one can check that σσ −1 = ε.
σ=
σ −1 =
1.5.1
1
2
3
4
5
6
7
8
4
1
8
3
2
5
6
7
1
2
3
4
5
6
7
8
2
5
4
1
6
7
8
3
Cycles. Disjoint permutations
An element k ∈ Xn is said to be fixed by a permutation σ ∈ Sn if σ(k) = k .
Otherwise if σ(k) ≠ k , we say that k is moved by σ. Two permutations
σ and τ are disjoint if there is no element of Xn moved by both. For
1 2 3 4
1 2 3 4
example, the permutations σ = (
) and τ = (
) are disjoint
3 2 1 4
1 4 3 2
permutations, since σ moves 1 and 3, while τ moves 2 and 4.
Algebraists (or mathematicians, in general) like separating (or breaking)
complex things (algebraic structures, in our case) into their simplest component parts. This intellectual “divide and conquer” helps us to understand
complicated processes and solve difficult problems. We can apply this approach to permutations and decompose any permutation into more simple
ones, called cycles. For instance, consider the permutation
1 2 3 4 5 6
σ=(
) ∈ S6
4 6 3 2 5 1
18
J. Sánchez-Ortega
and look at carefully how σ moves the elements of X6 = {1, 2, 3, 4, 5, 6}:
G1
'
6h
4
3e
5e
2
In other words, σ fixes 3 and 5, and moves 1, 2, 4, 6 in a cyclic way. Due to
this reason, we call σ a cycle, and write it in a more compact way called the
cycle notation: σ = (1 4 2 6) . Notice that in the cycle notation
- we have only listed the elements that are moved by sigma;
- each element is moved to its right neighbor, except the last
element, which “cycles around” to the first.
In a more general way, given k1 , k2 , . . . , kr distinct elements of Xn . If σ ∈ Sn
fixes the remaining elements of Xn (if any) and
σ(k1 ) = k2 ,
σ(k2 ) = k3 , . . . , σ(kr−1 ) = kr ,
σ(kr ) = k1 ,
then σ is called an r-cycle; r is called the length r of σ. We express σ as
σ = (k1 k2 . . . kr ),
which means that
k1 g
/ k2
/ ...
/ kr−1
/ kr
Notice that the only cycle of length 1 is the identity permutation,
while a 2-cycle (k1 k2 ) interchanges k1 and k2 , and fixes everything else; 2cycles are called transpositions. Using 3-cycles and transpositions, we can
easily list all the elements in S3 , namely:
S3 = {ε, (1 2), (1 3), (2 3), (1 2 3), (1 3 2)}
One can prove that the inverse of a cycle is a cycle:
(k1 k2 . . . kr−1 kr ) = (kr kr−1 . . . k2 k1 )
1. Preliminaries
19
Cycle Decomposition Theorem
Every permutation σ ≠ ε (in Sn ) can be factorised as a product of
(one or more) disjoint cycles of length at least 2. Such factorisation is
unique up to the order of the factors.
Proof. Part 1: Existence
Let σ ∈ Sn with σ ≠ ε. We proceed by induction on n ≥ 2. If n = 2, then
S2 = {ε, (1 2)} and σ = (1 2) is already a cycle (it is a transposition).
Assume that the result is true for permutations in Sn−1 and let σ ∈ Sn . If
σ(n) = n, we can view σ as an element of Sn−1 and by the induction hypothesis,
we are done. We can then assume that σ(n) ≠ n, and let k = σ −1 (n). Then
σ(k) = σ(σ −1 (n)) = n, and so k ≠ n, because we are assuming that σ(n) ≠ n.
So, we can consider the transposition γ = (k n), and the permutation τ = σγ.
From γ 2 = ε we obtain that σ = τ γ. Moreover, τ (n) = σγ(n) = σ(k) = n, and
so τ ∈ Sn−1 . Applying the induction hypothesis we obtain that τ is a product
of disjoint cycles. At this point, we distinguish two cases:
• Case 1: τ (k) = k.
Then τ and γ are disjoint permutations since τ (n) = n, and γ = (k n).
We are done in this case because σ = τ (k n) and τ is a product of disjoint
cycles.
• Case 2: τ (k) ≠ k.
Then k is moved by (exactly one) cycle factor of τ . Notice that τ
is a product of disjoint cycles by the inductive hypothesis. These
considerations allow us to write τ = µ(k k1 k2 . . . kr ), where µ is a
product of disjoint cycles fixing k, k1 , . . . , kr , and n since τ (n) = n. Thus
σ = τ γ = µ(k k1 k2 . . . kr )(k n) = µ(k n k1 k2 . . . kr ) is a product of
disjoint cycles, as desired.
Part 2: Uniqueness
Suppose that σ = γa . . . γ2 γ1 = δb . . . δ2 δ1 are two factorizations of σ into
disjoint cycles. We need to show that a = b and γi = δi for all i = 1 . . . a (after
possible reordering).
20
J. Sánchez-Ortega
We proceed by induction on t ∶= max(a, b). If t = 1 there is nothing to
prove since σ = γ1 = δ1 . Otherwise, assume that the result is true for t − 1
and let σ moves m. Then m occurs in exactly one γi and exactly one δj . By
reordering the factors (if necessary) we can assume that m occurs in γ1 and
in δ1 , so we can write:
γ1 = (m k2 . . . kr ),
δ1 = (m ℓ2 . . . ℓs ),
Assume, for example, that r ≤ s. Then:
k2 = σ(m) = ℓ2 ,
k3 = σ(k2 ) = σ(ℓ2 ) = ℓ3 ,
⋮
kr = σ(kr−1 ) = σ(ℓr−1 ) = ℓr .
Now, if r < s then m = k1 = σ(kr ) = σ(ℓr ) = ℓr+1 , which is impossible since δ1
is a cycle. Therefore r = s and γ1 = δ1 . Write µ = γ1 = δ1 . Then
σµ−1 = γa . . . γ2 = δb . . . δ2 ,
is a product of a − 1 (b − 1) disjoint cycles, and so the hypothesis induction
yields that a = b, and (after possible reordering) γi = δi for i = 2, 3, . . . , a, which
finishes the proof.
1.5.2
The Alternating group
Here, we focus our attention on transpositions. Recall that a transposition
δ is a 2-cycle, so it has the form δ = (m n) (where m ≠ n) and satisfies
δ 2 = ε,
δ −1 = δ . Transpositions are quite important since every permu-
tation is a product of transpositions. This follows from the Cycle
Decomposition Theorem together with the fact that every r-cycle is a product
of r − 1 transpositions:
(k1
k2 . . . kr ) = (k1
k2 )(k2
k3 ) . . . (kr−2
kr−1 )(kr−1
kr )
Remark 1.19. In contrast to the factorisation into cycles, factorisations
into transpositions are not unique. For example:
(1 2)(2 4)(4 5) = (1 2 4 5) = (1 5)(1 4)(1 2).
1. Preliminaries
21
Although the factorisation into transpositions of a permutation σ is far from
being unique, it has a very nice property: the numbers of transpositions
into any two factorisations of σ into transpositions have the same
parity. To be more precise, if σ admits a factorisation having an even
(respectively, odd) number of transpositions, then any other factorisation of σ
into transpositions must have an even (respectively, odd) number of factors.
This is known as the Parity Theorem.
Parity Theorem
If a permutation σ admits two factorisations
σ = γr . . . γ2 γ1 = µs . . . µ2 µ1 ,
where each γi and µj is a transposition, then both r and s are even
or both are odd.
The Parity Theorem allows us to introduce the following notions:
Definitions 1.20. A permutation σ is called even or odd if it can be written as a product of an even or odd number of transpositions. The set of all even
permutations in Sn is denoted An , and is called the alternating group of
degree n .
1 2 3 4 5 6 7 8 9
Example 1.21. Factorise the permutation σ = (
)
5 4 6 1 7 8 2 9 3
into a product of transpositions and find its parity.
Solution: We first factorise σ into a product of disjoint cycles:
σ = (1 5 7 2 4)(3 6 8 9).
Notice that σ is the product of a 5-cycle (which is even) and a 4-cycle (which
is odd), and so σ is odd (even + odd = odd). To finish, we factorise of the
5-cycle and the 4-cycle above into transpositions:
(1 5 7 2 4) = (1 5)(5 7)(7 2)(2 4),
(3 6 8 9) = (3 6)(6 8)(8 9)
From here, we obtain that:
σ = (1 5)(5 7)(7 2)(2 4)(3 6)(6 8)(8 9).
22
J. Sánchez-Ortega
To prove the Parity Theorem we need a preliminary result:
Lemma 1.22. The identity permutation ε can not be written as the product
of an odd number of transpositions.
Proof. Suppose that the identity permutation ε is a product of transpositions
ε = γ1 . . . γr , and let us show that r is even. Notice that r ≥ 2 (that is, we
can not have only one transposition) because a transposition can not be the
identity permutation. We proceed by (strong) induction on r:
If r = 2, then there is nothing to prove. Suppose then that r > 2 and the
result true for all factorisations of the identity permutation ε onto a product
of transpositions with < r factors. Let ε = γ1 γ2 . . . γr−1 γr , where the γj ’s are
all transpositions.
We first look at γr−1 and γr . If γr = (a b) (where a ≠ b), then we have four
possibilities for γr−1 :
• γr−1 = (b a) = (a b)
In this case, γr−1 γr = ε, so we have that ε = γ1 . . . γr−2 . By the inductive
hypothesis, we obtain that r − 2 is even, which implies that r is even.
• γr−1 = (a c), for c ≠ a and c ≠ b
Then γr−1 γr = (a c)(a b) = (a b)(b c), and so ε = γ1 . . . γr−2 (a b)(b c).
• γr−1 = (b c), for c ≠ a and c ≠ b
Then γr−1 γr = (b c)(a b) = (a c)(b c), and so ε = γ1 . . . γr−2 (a c)(b c).
• γr−1 = (c d), for c ≠ d, and both c and d different from a, b
In this case, γr−1 and γr commute, since they are disjoint transpositions.
We obtain that ε = γ1 . . . γr−2 (a b)(c d).
Notice that in any of the last three cases, we have replaced the product
γr−1 γr by the product of other two transpositions αr−1 and αr such that
a appears in αr−1 but not in αr . Repeating this process, at some point
we will have to land in the first case, and the result will follow by the
inductive hypothesis. Otherwise, we would obtain a product of r transpositions
1. Preliminaries
23
β1 β2 . . . βr such that β1 β2 . . . βr = ε, and a appears only in β1 . But this is
impossible since such product does not fix a, while ε permutations does.
We are now in a position to prove the Parity Theorem.
Proof. Suppose that a permutation σ admits two factorizations into transpositions:
σ = γn . . . γ2 γ1 = µm . . . µ2 µ1 . We have to show that both n and m
are either even or odd. Keeping in mind that µ−1
j = µj for all j, we obtain that
ε = µ1 µ2 . . . µm γn . . . γ2 γ1 .
An application of Lemma 1.22 tells us that m + n is even, which implies that
n and m are even (respectively, odd), as desired.
We finish this section with a few important properties of the alternating
group An .
Theorem 1.23. Let n ≥ 2. Then An is group (a subgroup of Sn ). More
precisely:
(1) The identity permutation ε is in An .
(2) If σ and τ are in An , then σ −1 and στ are in An .
(3) ∣An ∣ = 21 n!
Proof. (1) Notice that ε is obviously even since, for example, ε = (1 2)(1 2).
(2) Suppose that σ and τ are even permutations, which means that σ and τ
are products of an even number of transpositions. Let
σ = γ1 γ2 . . . γr ,
τ = δ1 δ2 . . . δ s ,
where r and s are even, and the γi and δj are transpositions. Then
στ = γ1 γ2 . . . γr δ1 δ2 . . . δs ,
and since r + s is even, we have that στ ∈ An . We prove now that σ −1 ∈ An ,
which follows from the fact that σ −1 = γr . . . γ2 γ1 ; in fact, we have that
(γ1 γ2 . . . γr−1 γr )(γr γr−1 . . . γ2 γ1 ) = γ1 γ2 . . . γr−1 γ 2 r γr−1 . . . γ2 γ1
= γ1 γ2 . . . γr−2 γ 2 r−1 γr−2 . . . γ2 γ1 = . . . = ε,
24
J. Sánchez-Ortega
and (γr γr−1 . . . γ2 γ1 )(γ1 γ2 . . . γr−1 γr ) = ε, which finishes the proof of (2).
(3) Let On denote the set of odd permutations in Sn . Then Sn = An ∪ On and
An ∩ On = ∅ by the Parity Theorem. Since n! = ∣Sn ∣ = ∣An ∣ + ∣On ∣, it is enough
to show that ∣An ∣ = ∣On ∣; in other words, the number of even permutations
coincide with the number of odd permutations. To prove that ∣An ∣ = ∣On ∣, we
will provide a bijective map f ∶ An → On .
Let γ = (1 2) and define f (σ) = γσ, for all σ ∈ An . Observe that f is
well-defined, namely, the permutation γσ is odd (γ is odd and σ is even). The
fact that γ 2 = ε implies that f is a bijection. In fact:
• f is one-to-one: f (σ1 ) = f (σ2 ) ⇒ γσ1 = γσ2 ⇒ γ 2 σ1 = γ 2 σ2 ⇒ σ1 = σ2
• f is surjective: Given τ ∈ On , then σ ′ = γτ ∈ An and
f (σ ′ ) = γσ ′ = γ(γτ ) = γ 2 τ = τ.
Thus f is a bijection, and therefore ∣An ∣ = ∣On ∣.
1.6
Cyclic groups
Last year, you learnt about a very important class of abelian groups, called
cyclic groups. Here we briefly recall the definition of a cyclic group and
some important results. We begin by recalling the definition of the order of
an element in a group.
Definition 1.24. The order of an element g in a group G is defined as the
smallest positive integer n such that g n = 1 . If such n does not exist, that
is, g k ≠ 1 for all k ≥ 1, then we say that g has infinite order. We write o(g) ,
or ∣g∣ , to denote the order of g.
Notice that in additive notation, o(g) = n means that n is the smallest
positive integer such that ng = 0.
Remark 1.25. Given g ∈ G of order n. Using the Division Algorithm, one
can prove if g ℓ = 1 for some ℓ ≥ 1, then n∣ℓ.
1. Preliminaries
25
Example 1.26. In the additive group Z4 of integers modulo 4, we have that
o(2̄) = 2 and o(3̄) = 4 since
2 ⋅ 2̄ = 2̄ + 2̄ = 2 + 2 = 4̄ = 0̄,
2 ⋅ 3̄ = 3̄ + 3̄ = 3 + 3 = 6̄ = 2̄,
3 ⋅ 3̄ = 3̄ + 3̄ + 3̄ = 2̄ + 3̄ = 2 + 3 = 5̄ = 1̄,
4 ⋅ 3̄ = 3̄ + 3̄ + 3̄ + 3̄ = 1̄ + 3̄ = 1 + 3 = 4̄ = 0̄.
Definition 1.27. Let g be an element of a group G. The subset of G consisting
of all the powers of g, namely, ⟨g⟩ = {g k ∣ k ∈ Z} is a subgroup of G (this
follows from the exponent laws), called the cyclic subgroup of G generated
by g. Moreover, ⟨g⟩ is indeed the smallest subgroup of G containing the
element g. Notice that the order of g coincides with the order of ⟨g⟩.
We say that G is a cyclic group if there exists g ∈ G such that G = ⟨g⟩ .
Notice that if g has order n, then G is a cyclic group of order n; if g has
order infinity, we say that G is a cyclic group of order infinity.
Remark 1.28. If the operation in the group G is written additively, the
cyclic subgroup generated by g consists on the multiples of g:
⟨g⟩ = {kg ∣ k ∈ Z} = Zg.
Remember that in additive notation “powers” are replaced by “multiples”.
Examples 1.29.
1. The cyclic subgroup generated by σ ∈ S3 is ⟨σ⟩ = {ε, σ, σ 2 }.
2. The cyclic subgroup generated by 2 ∈ Z consists on the even numbers.
3. The group (Z, +) is cyclic and 1 and -1 are the only generators.
If k ∈ Z we have k = k ⋅ 1 ∈ ⟨1⟩ and k = (−k) ⋅ (−1) ∈ ⟨−1⟩. Thus
Z = ⟨1⟩ = ⟨−1⟩. Moreover, if n is an integer different from 1 and −1 then
nZ ≠ Z since, for example, n + 1 ∉ nZ.
4. The group (Zn , +) is cyclic with generator 1̄ and order n.
Recall that Zn = {0̄, 1̄, . . . , n − 1}. Given k̄ ∈ Zn we have that k̄ = k 1̄ and
so k̄ ∈ ⟨1̄⟩. It follows that Zn = ⟨1̄⟩, as desired.
26
J. Sánchez-Ortega
5. The (multiplicative) group Z∗9 = {1̄, 2̄, 4̄, 5̄, 7̄, 8̄} consisting on the elements of Z9 having a multiplicative inverse is a cyclic group. In fact,
Z∗9 = ⟨2̄⟩ because
2̄0 = 1̄, 2̄1 = 2̄, 2̄2 = 4̄, 2̄3 = 8̄, 2̄4 = 16 = 7̄, 2̄5 = 14 = 5̄, 2̄6 = 10 = 9̄,
which shows that all the elements of Z∗9 are powers of 2̄.
Remark 1.30. Using the Exponent Laws one can show that every cyclic
group is abelian. Notice that the converse DOES NOT hold: not every
abelian group is cyclic. For example, Z2 × Z2 is abelian but not cyclic, since
all its elements have order 2. This group is known as the Klein group.
Notice that a cyclic group can have more than one generator. For
instance, consider the additive cyclic group Z12 = {0̄, 1̄, 2̄, . . . , 11}, which we
know is generated by 1̄. We can easily prove that 5̄ also generates Z12 , that is
to say that Z12 = ⟨5̄⟩ = {k 5̄ ∣ k ∈ Z}.
To prove that 5̄ is a generator of Z12 , we need to show that any element
x̄ ∈ Z12 can be written as x̄ = 5̄ + 5̄ + . . .(k + 5̄ = k 5̄ for some k ∈ Z. In fact:
0̄ = 0 ⋅ 5̄,
1̄ = 5 ⋅ 5̄,
since 25 = 1̄, because 25 ≡ 1 (mod 12),
2̄ = 10 ⋅ 5̄,
since 50 = 2̄, because 50 ≡ 2 (mod 12),
3̄ = 3 ⋅ 5̄,
since 15 = 3̄, because 15 ≡ 3 (mod 12),
4̄ = 8 ⋅ 5̄,
since 40 = 3̄, because 40 ≡ 4 (mod 12),
⋮
In the next result, we provide a necessary and sufficient condition for an
element of a (finite) cyclic group to be a generator.
Theorem 1.31. Let G = ⟨g⟩ be a cyclic group with generator g. Then:
(1) Every subgroup of G is cyclic.
(2) Suppose that o(g) = n. Then G = ⟨g k ⟩ if and only if gcd(k, n) = 1.
1. Preliminaries
27
Proof. (1) Suppose that H is a subgroup of G. If H = {1}, then we are
done because H = ⟨1⟩ is cyclic. Otherwise, let g k ∈ H, k ≠ 0. We can assume
(without loss of generality) that k > 0, since g −k = (g k )−1 ∈ H because H is a
subgroup of G. By the Well-Ordering Principle we can choose m to be the
smallest positive integer such that g m ∈ H. We claim that ⟨g m ⟩ = H. In fact,
it is clear that ⟨g m ⟩ ⊆ H, since g m ∈ H and H is a subgroup of G. For the
other containment, take g s ∈ H and apply the Division Algorithm to write
s = qm + r, with 0 ≤ r < m. Our goal is to show that r = 0; if r ≠ 0, then
g r = (g m )−q g s ∈ H, by the exponent laws. But it contradicts the minimality of
m. Therefore r = 0 and g s = (g m )q ∈ ⟨g m ⟩, as desired.
(2) Suppose first that g k is a generator, that is, G = ⟨g k ⟩. Then, in particular,
g ∈ ⟨g k ⟩. Thus g = (g k )m = g km , for some m ∈ Z. Theorem 2.5 implies that
n∣(1 − km), and so 1 − km = qn for q ∈ Z; that is, 1 = km + qn, which implies
that gcd(k, n) = 1. Conversely, if gcd(k, n) = 1, then 1 = xk + yn for some
integers x and y. Hence:
g = g 1 = g xk+yn = (g k )x (g n )y = (g k )x (1)y = (g k )x ∈ ⟨g k ⟩,
which implies that G = ⟨g k ⟩.
Example 1.32. The generators of Z12 are the elements k̄ with gcd(k, 12) = 1,
namely, 1̄, 5̄, 7̄, and 11.
Next, we recall the Fundamental Theorem of Finite Cycle Groups, which
gives a precise description of all subgroups of a finite cyclic group G.
It tells us how many subgroups G has and how to determine them. To be more
precise: G has a unique cyclic subgroup of order k for every divisor
k of n and those are the only subgroups of G. In other words: the
subgroups of G are in one-to-one correspondence with the positive
divisors of n.
28
J. Sánchez-Ortega
Fundamental Theorem of Finite Cyclic Groups
Let G = ⟨g⟩ be a cyclic group of order n.
(1) If H is a subgroup of G, then H = ⟨g d ⟩ for some d∣n.
(2) If H is any subgroup of G with order ∣H∣ = k, then k∣n.
(3) If k∣n, then ⟨g n/k ⟩ is the unique subgroup of G of order k.
In particular, the order of any subgroup of G divides the order of G.
This tells us that cyclic groups satisfy Lagrange’s Theorem. To prove the
Fundamental Theorem of Finite Cyclic Groups, we need a preliminary result:
Lemma 1.33. Let g be an element of a group G of order n. If d∣n then the
order of the element g d is n/d.
Proof. Let k = nd . Then (g d )k = g dk = g n = 1. It remains to show that k is
the smallest positive integer with that property. Suppose that (g d )r = 1 for
some r ≥ 1. Then g dr = 1, which implies that n∣dr since o(g) = n. We then can
write that dr = qn, for some q ≥ 1. But then dr = qn = q(dk) = d(qk), and so
r = qk, which implies that r ≥ k, as desired.
Proof of the Fundamental Theorem of Finite Cyclic Groups.
(1) The result is clear for ∣H∣ = 1. Otherwise, from Theorem 1.31 (1) we get
that H = ⟨g m ⟩ for some m > 0. Let d = gcd(m, n). Then d∣n and we only need
to show that H = ⟨g d ⟩.
From d∣m, we have that m = qd for some q ∈ Z. Thus g m = (g d )q ∈ ⟨g d ⟩ and
therefore H = ⟨g m ⟩ ⊆ ⟨g d ⟩. On the other hand, by Bézout’s Lemma we have
that d = xm + yn, where x, y ∈ Z, and so
g d = g xm+yn = (g m )x (g n )y = (g m )x (1)y = (g m )x ∈ ⟨g m ⟩ = H ⇒ ⟨g d ⟩ ⊆ H,
finishing the proof of (1).
(2) Let H be a subgroup of G of order k. From (1) we obtain that H = ⟨g d ⟩, for
some d∣n. If k = ∣H∣, then by Lemma 1.33 we have that k = ∣H∣ = o(g d ) = n/d,
which shows that k∣n.
1. Preliminaries
29
(3) Suppose that K is any subgroup of G of order k. By (1) we have that
K = ⟨g m ⟩ where m∣n. Then from Lemma 1.33 we obtain that
k = ∣K∣ = o(g m ) =
n
n
n
⇒ m = ⇒ K = ⟨g k ⟩,
m
k
which finishes the proof.
We close the section with a few applications of the Fundamental Theorem
of Finite Cyclic Groups.
Corollary 1.34. Let G = ⟨g⟩ be a cyclic group of order n. Then for any
positive integer m we have that ⟨g m ⟩ = ⟨g d ⟩ , where d = gcd(m, n) . Moreover,
the order of g m is n/d.
Proof. Take m an arbitrary positive integer and let d = gcd(m, n). Proceeding like in the proof of (1) in the Fundamental Theorem of Finite Cyclic
Groups, we have that ⟨g m ⟩ = ⟨g d ⟩. On the other hand, we know that
o(g m ) = ∣⟨g m ⟩∣ = ∣⟨g d ⟩∣ = o(g d ) =
n
,
d
by Lemma 1.33.
Example 1.35. In the (additive) cyclic group Z30 , find the order of 18.
Solution: Notice that Z30 = ⟨1̄⟩ and 18 = 18 ⋅ 1̄. We will use Corollary 1.34.
In this case, n = 30 and m = 18. Since gcd(30, 18) = 6, we have that ⟨18⟩ = ⟨6̄⟩
and the order of 18 in Z30 is 30
6 = 5. In other words, the subgroup generated
by 18 in Z30 has 5 elements: ⟨18⟩ = ⟨6̄⟩ = {0̄, 6̄, 12, 18, 24}, because in Z30 we
have that 0 ⋅ 6̄ = 0̄, 1 ⋅ 6̄ = 6̄, 2 ⋅ 6̄ = 12, 3 ⋅ 6̄ = 18, 4 ⋅ 6̄ = 24, 5 ⋅ 6̄ = 30 = 0̄.
The Fundamental Theorem of Finite Cyclic Groups can be used to find
all the subgroups of a finite cyclic group and draw the corresponding lattice
diagram. To do so, notice that ⟨g m ⟩ ⊆ ⟨g k ⟩ ⇔ k∣m
Example 1.36. Find all subgroups of the multiplicative cyclic group
C12 and draw its lattice diagram.
Write C12 = ⟨g⟩, where o(g) = 12. Since the positive divisors of 12 are 1, 2,
3, 4, 6, 12, the Fundamental Theorem of Finite Cyclic Groups allows us to
30
J. Sánchez-Ortega
conclude that C12 has exactly 6 subgroups. To determine the unique subgroup
of each of these orders, we use Lemma 1.33. Clearly, {1} = ⟨1⟩ = ⟨g 12 ⟩.
12
⇒ d = 6;
d
12
Order 3: 3 = o(g d ) =
⇒ d = 4;
d
12
⇒ d = 3;
Order 4: 4 = o(g d ) =
d
12
Order 6: 6 = o(g d ) =
⇒ d = 2;
6
Order 2: 2 = o(g d ) =
the subgroup of order 2 is ⟨g 6 ⟩;
the subgroup of order 3 is ⟨g 4 ⟩;
the subgroup of order 4 is ⟨g 3 ⟩;
the subgroup of order 6 is ⟨g 2 ⟩;
To picture the diagram, notice that
⟨g 6 ⟩ ⊆ ⟨g 2 ⟩ and also ⟨g 6 ⟩ ⊆ ⟨g 3 ⟩
⟨g 4 ⟩ ⊆ ⟨g 2 ⟩
C12
⟨g 2 ⟩
⟨g 4 ⟩
⟨g 3 ⟩
⟨g 6 ⟩
⟨1⟩
Example 1.37. Find all subgroups of the additive cyclic group Z100
and draw the lattice diagram.
Since the positive divisors of 100 are 1, 2, 4, 5, 10, 20, 25, 50 and 100, by
the Fundamental Theorem of Finite Cyclic Groups we can assert that Z100 has
exactly 9 subgroups of orders the mentioned divisors of 100. Clearly, {0̄} = ⟨0̄⟩.
To determine the remaining subgroups, we use Lemma 1.33 in its additive
version:
2 = o(d1̄) =
100
⇒ d = 50; the subgroup of order 2 is ⟨50⟩ = 50Z100 ;
d
1. Preliminaries
31
4 = o(d1̄) =
100
⇒ d = 25; the subgroup of order 4 is ⟨25⟩ = 25Z100 ;
d
5 = o(d1̄) =
100
⇒ d = 20; the subgroup of order 5 is ⟨20⟩ = 20Z100 ;
d
10 = o(d1̄) =
100
⇒ d = 10; the subgroup of order 10 is ⟨10⟩ = 10Z100 ;
d
20 = o(d1̄) =
100
⇒ d = 5; the subgroup of order 20 is ⟨5̄⟩ = 5Z100 ;
d
25 = o(d1̄) =
100
⇒ d = 4; the subgroup of order 25 is ⟨4̄⟩ = 4Z100 ;
d
50 = o(d1̄) =
100
⇒ d = 2; the subgroup of order 50 is ⟨2̄⟩ = 2Z100 ;
d
To picture the diagram we use that
mZ ⊆ kZ ⇔ k∣m
In our case, we have the following containments:
20Z100 ⊆ 4Z100 ⊆ 2Z100
and
20Z100 ⊆ 10Z100 ⊆ 5Z100
50Z100 ⊆ 10Z100 ⊆ 2Z100
and
50Z100 ⊆ 25Z100 ⊆ 5Z100
Z100
2Z100
4Z100
5Z100
10Z100
20Z100
25Z100
50Z100
⟨0̄⟩
32
J. Sánchez-Ortega
1.6.1
Maps between groups
The main idea behind two groups G1 and G2 being isomorphic is that these
two groups are the “same” (from an algebraic point of view) except that their
elements may have different labels. The group G1 becomes G2 after relabelling
its elements; the map which does the relabelling is precisely an isomorphism
from G1 to G2 . More formal definitions follow.
Definitions 1.38. Let G1 and G2 be two groups. A mapping φ ∶ G1 → G2
is called a group homomorphism if φ(a ⋅1 b) = φ(a) ⋅2 φ(b) , for all a, b ∈
G1 . A monomorphism is an injective homomorphism, while a surjective
homomorphism is called an epimorphism. Homomorphisms which are
bijectives are said to be isomorphisms. When an isomorphism exists from
G1 to G2 we say that G1 is isomorphic to G2 , and we write G1 ≅ G2 .
Isomorphisms from a group to itself are called automorphisms.
The set consisting of all automorphisms of G, denoted Aut(G) , is a group
under composition called the group of automorphisms of G.
We close this review chapter with an important result about cyclic groups.
Isomorphism of Cyclic Groups
(1) For every positive integer n, every cyclic group of order n is
isomorphic to Zn . Thus, any two cyclic groups of order n are
isomorphic.
(2) Every infinite cyclic group is isomorphic to Z. Therefore
any two cyclic groups of order infinity are isomorphic.
Proof. (1) Let Cn = ⟨a⟩ be a (multiplicative) cyclic group of order n. Then
⟨a⟩ = {1, a, a2 , . . . , an−1 }.
Comparing this group with the additive cyclic group Zn = {0̄, 1̄, 2̄, . . . , n − 1},
⟨a⟩ = {a0 ,
a1 ,
a2 ,
Õ
×
×
×
Ö
Õ
×
×
×
Ö
Õ
×
×
×
Ö
{0̄,
1̄,
2̄,
Zn =
...,
an−1 }
Õ
×
×
×
Ö
...,
n − 1}
1. Preliminaries
33
one realises that the map φ ∶ Zn → ⟨a⟩ given by φ(i) = ai is a bijective map.
But this map also satisfies that
φ(i + j) = ai+j = ai aj = φ(i)φ(j),
which shows that φ is an isomorphism from Zn to ⟨a⟩.
(2) Suppose that ⟨a⟩ is a cyclic group of order infinity. Then
⟨a⟩ = {. . . , a−2 , a−1 , 1, a, a2 , . . .}.
Proceeding like in (1) we compare this group with Z:
⟨a⟩ = {. . . ,
Z=
{. . . ,
a−2 ,
a−1 , 1,
a1
a2 ,
Õ
×
×
×
Ö
Õ
×
×
×
Ö
Õ
×
×
×
Ö
Õ
×
×
×
Ö
Õ
×
×
×
Ö
−2,
−1,
0,
1,
2,
. . .}
. . .}
Then it is clear that the map φ ∶ Z → ⟨a⟩ given by φ(i) = ai is the desired
isomorphism.
Chapter 2
Dihedral groups. Lagrange’s
Theorem
In this chapter, we introduce a very important type of groups, and we revisit
one of the most important results for finite groups: Lagrange’s Theorem.
2.1
Dihedral groups
Let n ≥ 2 be an integer, the dihedral group Dn is the group of order
2n presented as Dn = {1, a, a2 , . . . , an−1 , b, ba, ba2 , . . . , ban−1 } , where o(a) = n ,
o(b) = 2 , and aba = b .
Notice that D3 is isomorphic to S3 because D3 = {1, a, a2 , b, ba, ba2 }, o(a) =
3, o(b) = 2, and aba = b. In S3 make a = (1 2 3) and b = (1 2).
If n = 2, then D2 is the Klein group because D2 = {1, a, b, ab}, o(a) = 2,
o(b) = 2, and aba = b. From o(a) = 2 we obtain that a−1 = a, which implies
that D2 is abelian (ba = a−1 b = ab).
Notice that a and b are the generators of Dn ; this means that any other
element of Dn can be written as a product of powers of a and b. We can
compute any product in Dn using that o(a) = n, o(b) = 2 and aba = b.
Moreover, from aba = b and using induction we can obtain that ak bak = b for
all k ∈ Z. Using that an = 1 we get that ak b = ba−k = ban−k for all k ∈ Z. It
implies o(bak )2 = 1 for all k ∈ Z since
(bak )2 = (bak )(bak ) = b(ak b)ak = b(ban−k )ak = b2 an = 1
Using these new formulas we can fill in the Cayley table for Dn . For example,
35
36
J. Sánchez-Ortega
the Cayley table for D4 , which is called the octic group, is as follows:
D4
1
a
a2
a3
b
ba
ba2
ba3
1
1
a
a2
a3
b
ba
ba2
ba3
a2
a2
a3
1
a
ba2
ba3
b
ba
a
a
a2
a3
1
ba
ba2
ba3
b
a3
a3
1
a
a2
ba3
b
ba
ba2
b
b
ba3
ba2
ba
1
a3
a2
a
ba
ba
b
ba3
ba2
a
1
a3
a2
ba2
ba2
ba
b
ba3
a2
a
1
a3
ba3
ba3
ba2
ba
b
a3
a2
a
1
For n ≥ 3, the dihedral group Dn can be seen as the group of motions of a
regular n-gon, which is a plane figure with n vertices evenly placed on a circle.
Here, by a motion we mean a permutation of its vertices that can be
performed by either a rotation or a symmetry. Such these motions bring the
polygon back to itself. For example, consider a square (as given below) whose
vertices are labelled as 1, 2, 3 and 4 and let us think about all possible ways
in which such a square can be repositioned without altering its shape and its
position in space.
2
3
1
4
We can rotate the square counterclockwise by an angle of π/2, π and 3π/2
because a rotation of 2π with gives us the same position of the labelling in
the vertices. Notice that there are also four different symmetries: flipping
about a horizontal (respectively, vertical) axis, and about the main diagonal
(respectively, the other diagonal); see the figures below.
2
3
2
3
2
3
1
4
1
4
R0
1
4
Rπ/2
3
4
2
1
2. Dihedral groups. Lagrange’s Theorem
2
3
37
4
1
2
3
3
2
1
4
1
4
2
3
Rπ
1
4
2
3
R3π/2
Sh
1
4
2
3
4
2
4
3
3
2
4
1
2
1
3
4
Sv
2
3
1
4
4
3
2
3
Smd
1
1
Sd
1
2
1
4
Notice that there are no more possible ways to reposition our square, since
its position after performing a motion is completely determined by the position
and orientation, either face up or face down, of any of the vertices. There
are exactly four different locations and two orientations for a given vertex,
which gives a total of eight motions. Now, let us rename the elements of these
motions as 1 ∶= R0 , a ∶= Rπ/2 and b ∶= Sh . Then Rπ = a2 , R3π/2 = a3 , a4 = 1,
Smd = ba, Sv = ba2 and Sd = ba3 , and we have obtained the octic group.
2.2
The centre of the octic group
Consider the octic group
D4 = {1, a, a2 , a3 , b, ba, ba2 , ba3 }, where o(a) = 4, o(b) = 2, aba = b.
Here, we will show that its centre is Z(D4 ) = {1, a2 }. Notice that we need to
show that the only elements of D4 which commute with any other element in
D4 are 1 and a2 . It is clear that 1 commutes with all the elements in D4 .
We will use that ak b = ba−k = ba4−k for all k ∈ Z. Let n ∈ {0, 1, 2, 3} and let
us show that ban ∉ Z(D4 ). From
a(ban ) = (ab)an = (ba3 )an = ban+3 ≠ ban+1 = (ban )a,
38
J. Sánchez-Ortega
we can see that ban does not commute with a, which implies that ban ∉ Z(D4 )
for n ∈ {0, 1, 2, 3}. On the other hand, from
ab = ba3 ≠ ba,
a3 b = ba ≠ ba3
we conclude that a, a3 ∉ Z(D4 ). We have now proved that Z(D4 ) ⊆ {1, a2 }.
But notice that a2 commutes with both generators a and b (a2 a = a3 = aa2 ,
a2 b = ba2 ), which says that a2 ∈ Z(D4 ) and so Z(D4 ) = {1, a2 }.
2.3
Cosets
To prove Lagrange’s Theorem, we first need to introduce the required machinery,
namely, cosets. The notion of a coset was introduced by Galois in 1930 but
was G. A. Miller the one who name them cosets.
Definition 2.1. Let H be a subgroup of a group G and a ∈ G. The set
Ha = {ha ∣ h ∈ H} is called the right coset of H in G generated by a.
Similarly, aH = {ah ∣ h ∈ H} is the left coset of H in G generated by a.
Observe that in additive notation the left and right cosets are:
H + a = {h + a ∣ h ∈ H},
a + H = {a + h ∣ h ∈ H}
It is important to notice that every coset in a group G is a subset of G.
Cosets are not elements in G!
Example 2.2. Find all the left cosets of H ∶= {ε, (1 3)} in G ∶= S3 .
Solution: Recall that S3 = {ε, (1 2), (1 3), (2 3), (1 2 3), (1 3 2)}. You
can easily check that
εH = H = (1 3)H,
(1 2)H = {(1 2), (1 3 2)} = (1 3 2)H,
(2 3)H = {(2 3), (1 2 3)} = (1 2 3)H.
Example 2.3. Find all the right cosets of 4Z in (the additive group) Z.
Solution: Given a ∈ Z the right coset generated by a is 4Z + a. But
4Z + a = a + 4Z = ā in Z4 .
2. Dihedral groups. Lagrange’s Theorem
39
We only have four right cosets, namely, the four residue classes modulo 4:
4Z + 0 = {. . . , −12, −8, −4, 0, 4, 8, 12, . . .}
4Z + 1 = {. . . , −11, −7, −3, 1, 5, 9, 13, . . .}
4Z + 2 = {. . . , −10, −6, −2, 2, 6, 10, 14, . . .}
4Z + 3 = {. . . , −9, −5, −1, 3, 7, 11, 15, . . .}
Example 2.4. Consider the octic group D4 = {1, a, a2 , a3 , b, ba, ba2 , ba3 },
where o(a) = 4, o(b) = 2 and aba = b. Find the cosets of Z(D4 ) = {1, a2 }.
Solution: Let us write Z = Z(D4 ) for short. Notice that left and right
cosets coincide because Z its the centre and recall that elements in the centre
commutes with every element in the group.
Z1 = Z
Za = {a, a3 }
Za2 = {a2 , a4 = 1} = Z
Za3 = {a3 , a5 = a} = Za
Zb = {b, a2 b = ba2 } = {b, ba2 }
Z(ba) = {ba, a2 (ba)} = {ba, ba3 }
Z(ba2 ) = {ba2 , a2 (ba2 )} = {ba2 , ba4 = b} = {ba2 , b} = Zb
Z(ba3 ) = {ba3 , a2 (ba3 )} = {ba3 , ba5 } = {ba3 , ba} = Z(ba)
Thus, the different cosets are Z, Za, Zb and Z(ba).
2.3.1
Some important properties of cosets
Let a ∈ G and H a subgroup of G. Then:
1. H1 = H = 1H, which means that H is a right and left coset of itself.
2. It is obvious that a ∈ Ha and a ∈ aH since a = 1 ⋅ a = a ⋅ 1 for 1 ∈ H.
3. If G is abelian, then Ha = aH for all a ∈ G and all subgroups H of G.
4. 3 might not be true if G is not abelian.
5. If a ∈ Hb, then Ha = Hb.
40
J. Sánchez-Ortega
6. Ha = Hb if and only if ab−1 ∈ H.
7. Ha = H if and only if a ∈ H.
8. There is a bijection from H to Ha. In particular, if H is finite, then any
coset Ha has the same number of elements as H and therefore all the
cosets have the same number of elements.
9. The family of all the cosets of Ha, as a ranges over G, is a partition of
G. It means that G = ∪a∈G Ha, and Ha ≠ Hb implies Ha ∩ Hb = ∅.
G
H1 = H
Hc
Ha
Hb
Observe that 6 and 7 in additive notation become
H + a = H + b ⇔ a − b ∈ H;
H +a=H ⇔a∈H
For instance, 6Z + 7 = 6Z + 1 since 7 − 1 = 6 ∈ 6Z.
Proof. Properties 1 , 2 and 3 trivially hold.
4. Take G = S3 and H = {ε, (1 3)}. Then one can easily check that
(1 2)H = {(1 2), (1 3 2)} ≠ {(1 2), (1 2 3)} = H(1 2)
5. To show that the cosets Ha and Hb are equal, we must show that
they are equal sets. Suppose that a ∈ Hb. Then there exists h ∈ H such that
a = hb. If x ∈ Ha, then x = h1 a for some h1 ∈ H. From a = hb we get that
x = h1 a = h1 (hb) = (hh1 )b ∈ Hb. This proves that Ha ⊆ Hb. The proof of the
containment Hb ⊆ Ha is similar (use now that b = h−1 a).
6. If ab−1 ∈ H, then ab−1 = h and so a = hb ∈ Hb. An application of 5
gives that Ha = Hb. If Ha = Hb, then a ∈ Hb so a = hb for some h ∈ H. This
gives that ab−1 = h ∈ H, as desired.
7 follows from 6 by taking b = 1.
2. Dihedral groups. Lagrange’s Theorem
41
8. Define ψ ∶ H → Ha by ψ(h) = ha. Clearly, ψ is surjective because every
element of Ha is of the form ha for some h ∈ H and so ha = ψ(h). Lastly, if
ψ(h1 ) = ψ(h2 ), then h1 a = h2 a and applying the cancellation law in G we get
that h1 = h2 , which proves that ψ is injective.
9. We first need to show that any two cosets, say Ha and Hb, are either
disjoint or equal. If they are disjoint, then we are done. Otherwise, let
x ∈ Ha ∩ Hb. Then x = h1 a = h2 b for some h1 , h2 ∈ H. Thus solving for a, we
obtain a = (h−1
1 h2 )b ∈ Hb. It follows from 5 that Ha = Hb.
Next, we need to check that G is the union of all the cosets. In order to
do so, we must show that every element g ∈ G is in one of the cosets of H.
But this clearly holds, since g = 1 ⋅ g ∈ Hg.
Thus, the family of all the cosets of H is a partition of G.
2.4
Lagrange’s Theorem
Now we are in a position to prove Lagrange’s Theorem.
Lagrange’s Theorem
Let H be a subgroup of a finite group G. Then the order of G is a
multiple of the order of H.
Proof. Let Ha1 , Ha2 , . . . , Hak be the distinct cosets of H in G. Then
G = Ha1 ∪ Ha2 ∪ . . . ∪ Hak , which is a disjoint union by Property 9 . Next,
using Property 8 we have that ∣Hai ∣ = ∣H∣ for each i, so
∣G∣ = ∣Ha1 ∣ + . . . + ∣Hak ∣ = ∣H∣ + . . . + ∣H∣ = k∣H∣,
which shows that the order of G is a multiple of the order of H, as desired.
A few applications of Lagrange’s Theorem follow.
Corollary 2.5. The order of any element of a finite group divides the order
of the group.
Proof. Let G be a finite group and g ∈ G. By definition the order o(g) of g
coincides with the order of the cyclic subgroup ⟨g⟩ of G generated by g. The
result now follows by applying Lagrange’s Theorem to ⟨g⟩ and G.
42
J. Sánchez-Ortega
Corollary 2.6. Every group of prime order is cyclic.
Proof. Let p be a prime and G a group of order p. Take g ∈ G with g ≠ 1
and consider the subgroup ⟨g⟩ of G, which is non-trivial because g ≠ 1. By
Lagrange’s Theorem we have that ∣⟨g⟩∣∣p, and so ∣⟨g⟩∣ = p, that is, G = ⟨g⟩.
Last year, you saw that every group of order 4 = 22 is either cyclic or
isomorphic to the Klein group. Recall that all the elements of the Klein group
have order 2. We can now use Lagrange’s Theorem to prove an analogous
result for an arbitrary prime p.
Corollary 2.7. If ∣G∣ = p2 where p is a prime, then either G is cyclic or
g p = 1 for every element g in G.
Proof. Assume that G is not cyclic, and let g ∈ G. Write H = ⟨g⟩ for the
cyclic subgroup of G generated by g. Then by Corollary 2.5 the order of H,
namely, o(g) = ∣H∣ divides the order of G, i.e., p2 . From here we derive that
o(g) = 1, p, or p2 . But o(g) ≠ p2 because G is not cyclic, so o(g) is either 1 or
p. In any case, g p = 1, concluding the proof.
Let G be a group and H a subgroup of G, the index of H in G is the
number of cosets of H in G. We denote it by ∣G ∶ H∣ . If G is finite, from the
proof of Lagrange’s Theorem, we know that the number of elements in G is
equal to the number of elements in H, multiplied by the number of cosets of
H in G. In other words:
∣G ∶ H∣ =
order of G ∣G∣
=
order of H ∣H∣
Note that H can be of finite index in G even if both H and G are
infinite. For example, H = 4Z has index 4 in G = Z and both 4Z and Z are
infinite.
Remark 2.8. The converse of Lagrange’s Theorem is NOT TRUE in
general! Consider, for example, the alternating group A4 , that is, the subgroup
of S4 consisting on all the even permutations. The order of A4 is 12 and A4
does not contain a subgroup of order 6. In fact:
A4 = {ε, (1 2 3), (1 3 2), (1 2 4), (1 4 2), (1 3 4), (1 4 3), (2 3 4), (2 4 3),
2. Dihedral groups. Lagrange’s Theorem
43
(1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}
If A4 contains a subgroup H of order 6, then ε ∈ H and H must contain either
(1 2)(3 4), or (1 3)(2 4) or (1 4)(2 3). This is because if H contains one of
the 3-cycles (a b c), then (a b c)−1 = (a c b) ∈ H. So, H can not consists only
on 3-cycles and the identity, since that will not produce an even number of
elements. Thus, H must contain ε, some 3-cycle (a b c), its inverse (a c b) and
a product of (disjoint) transpositions, say (a b)(c d). Since we are assuming
that H is a subgroup, then the following two products must elements of H:
(a b c)(a b)(c d) = (b c d)
(a b)(c d)(a b c) = (a c d)
But then (b c d)−1 = (b d c) and (a c d)−1 = (a d c) and ∣H∣ ≥ 7. From here, we
can conclude that A4 does not have a subgroup of order 6.
Chapter 3
Isomorphism Theorems
This chapter is devoted to three very important theorems in algebra: the
Isomorphism Theorems. The main ingredient is a special type of subgroups,
called normal subgroups. We begin this chapter studying these.
3.1
Normal subgroups
Let H be a subgroup of a group G and a ∈ G, we have already seen that
the left and right cosets aH and Ha might be different. We say that H is
a normal subgroup of G, or that H is normal in G, if gH = Hg for all
g ∈ G. We write H G.
Examples 3.1.
1. Let G be a group. Then {1} G and G G since g{1} = {g} = {1}g,
gG = G = Gg, for all g ∈ G. The centre Z(G) of G is always a normal
subgroup of G.
2. Every subgroup of an abelian group is normal.
3. Let G = S3 = {ε, σ, σ 2 , τ, τ σ, τ σ 2 }, where σ 3 = τ 2 = ε and στ σ = τ . Write
H = {ε, τ } and K = {ε, σ, σ 2 }. Then one can check that K S3 but H
is not normal in S3 . In fact:
Hσ = {hσ ∣ h ∈ H} = {εσ, τ σ} = {σ, τ σ},
σH = {σh ∣ h ∈ H} = {σε, στ } = {σ, τ σ 2 },
45
46
J. Sánchez-Ortega
since στ = σ 2 τ σ = σ 3 τ σ 2 = τ σ 2 . Thus Hσ ≠ σH, and H is not normal in
S3 . On the other hand, K S3 because
αK = K = Kα, for all α ∈ K,
τ K = {τ, τ σ, τ σ 2 } = Kτ,
τ σK = {τ σ, τ σ 2 , τ } = Kτ σ,
τ σ 2 K = {τ σ 2 , τ, τ σ} = Kτ σ 2 .
The definition of a normal subgroup does not provide us a practical way to
check whether a given subgroup is normal, since it requires the computation
of all left and right cosets of such subgroup. In what follows, we will learn a
more convenient method to test whether a subgroup is normal.
Normality Test
The following conditions are equivalent for a subgroup of a group G.
(1) H is normal in G.
(2) gHg −1 ⊆ H for all g ∈ G.
(3) gHg −1 = H for all g ∈ G.
Proof. (1) ⇒ (2). Let x ∈ gHg −1 , say x = ghg −1 for some h ∈ H. Then
gh ∈ gH = Hg by (1), and so gh = h1 g for some h1 ∈ H. But then x = ghg −1 =
h1 gg −1 = h1 ∈ H, which proves (2).
(2) ⇒ (3). For g ∈ G, apply (2) to g −1 to get that g −1 H(g −1 )−1 = g −1 Hg ⊆ H.
From here we obtain that
(2)
H = 1H1 = (gg −1 )H(gg −1 ) = g(g −1 Hg)g −1 ⊆ gHg −1 ⊆ H ⇒ H = gHg −1 ,
proving (3).
(3) ⇒ (1). For g ∈ G, we have gH = gH1 = gH(g −1 g) = (gHg −1 )g = Hg,
and (1) follows.
3. Isomorphism Theorems
47
A very important application of the Normality Test is the following:
Proposition 3.2. Let φ ∶ G → G′ be a group homomorphism. Then the
kernel of φ, defined as, ker φ = {k ∈ G ∣ φ(k) = 1G′ } is a normal subgroup of
G. Moreover, φ is a monomorphism if and only if ker φ = {1G }.
Proof. We first need to check that ker φ is a subgroup of G. To do so, we
use the Subgroup Test: for k, k ′ ∈ ker φ, using that φ is a group homomorphism
we have that
φ(kk ′ ) = φ(k)φ(k ′ )
φ(k −1 ) = φ(k)−1
k,k′ ∈ ker φ
k∈ ker φ
=
=
1G′ 1G′ = 1G′
1G′
which imply that kk ′ , k −1 ∈ ker φ, and so ker φ is a subgroup of G. Next, we
show that ker φ is normal in G. To do so, we check that condition (2) in
the Normality Test is satisfied. Take g ∈ G and k ∈ ker φ and let us show
that gkg −1 ∈ ker φ, that is, φ(gkg −1 ) = 1G′ . In fact, using that φ is a group
homomorphism and that k ∈ ker φ we have:
φ(gkg −1 ) = φ(g)φ(k)φ(g −1 ) = φ(g)φ(g −1 )φ(g)φ(g)−1 ,
which shows that gkg −1 ∈ ker φ, as required.
To finish, we prove the “if and only if” condition. Suppose first that φ is a
monomorphism. Since φ is a group homomorphism, we have that φ(1G ) = 1G′ .
For k ∈ ker φ we have that φ(k) = 1G′ = φ(1G ), which implies k = 1G since φ is
injective. This shows that ker(φ) = {1G }.
Conversely, assume that ker(φ) = {1G }. Since we are assuming that φ is
a group homomorphism, we only need to check that φ is injective. To do
so, suppose that g1 , g2 ∈ G are such that φ(g1 ) = φ(g2 ), and let us show that
g1 = g2 . From φ(g1 ) = φ(g2 ) multiplying by the inverse of φ(g2 ), which is,
φ(g2 )−1 = φ(g2−1 ), we obtain that
1G′ = φ(g1 )φ(g2 )−1 = φ(g1 )φ(g2−1 ) = φ(g1 g2−1 ) ⇒ g1 g2−1 ∈ ker(φ) = {1G },
which implies that g1 g2−1 = 1G , and so g1 = g2 . This shows that φ is injective,
finishing the proof.
48
J. Sánchez-Ortega
Example 3.3. The special linear group SL(2, R) is a normal subgroup of
the general linear group GL(2, R).
Solution: Recall that
G ∶= GL(2, R) = {A ∈ M2 (R) ∶ det(A) ≠ 0},
H ∶= SL(2, R) = {A ∈ M2 (R) ∶ det(A) = 1}.
We will prove that H satisfies (2) in the Normality Test. Given A ∈ G, we
need to show that AHA−1 ⊆ H. If B ∈ H, then
det(ABA−1 ) = det(A) det(B)
1
= det(B) = 1 ⇒ ABA−1 ∈ H,
det(A)
as desired.
3.2
Factor groups
Let G be a group and H a subgroup of G. We can combine cosets, via the
coset operation as follows: the coset of a, operated with the coset of b is
defined to be the coset of ab. In symbols: Ha ⋅ Hb = H(ab) . This definition
seems to be quite innocent but it is not that simple! The difficulty arises
from the fact that the result of operate two cosets Ha and Hb might not be
uniquely defined. Recall that Ha may be the same coset as Hc (it happens if
c ∈ Ha), and similarly, Hb may be the same coset as Hd. Therefore, Ha ⋅ Hb
should be the same as Hc ⋅ Hd. Unfortunately, it may happen that H(ab) is
not the same coset as H(cd). Graphically,
≠
=
=
Ha ⋅ Hb = H(ab)
Hc ⋅ Hd = H(cd)
For example, if G = S3 and H = {ε, (1 2)}, then
H(1 3) = {(1 3), (1 3 2)} = H(1 3 2)
H(2 3) = {(2 3), (1 2 3)} = H(1 2 3)
and yet H(1 3)(2 3) = H(1 3 2) ≠ H = Hε = H(1 3 2)(1 2 3).
Therefore, coset multiplication does not work as an operation on the cosets
of H = {ε, (1 2)} in S3 / Why not? What did it go wrong? Well ...
3. Isomorphism Theorems
49
Although H is a subgroup, H is not a normal subgroup of S3 . So then
the next questions are “what if H were a normal subgroup?”“Would coset
operation work?” Good news! ,, Check out the following theorem!
Theorem 3.4. Let H be a normal subgroup of a group G. If Ha = Hc and
Hb = Hd then H(ab) = H(cd).
Proof. Assume that Ha = Hc and Hb = Hd. Then a ∈ Hc and b ∈ Hd, and
we can write a = h1 c and b = h2 d for some h1 , h2 ∈ H. Thus:
ab = (h1 c)(h2 d) = h1 (ch2 )d
(3.1)
But ch2 ∈ cH = Hc because H is normal, and so ch2 = h3 c for some h3 ∈ H.
Returning to (3.1) we get that ab = h1 (ch2 )d = h1 (h3 c)d = (h1 h3 )(cd) ∈ H(cd).
An application of Cosets Property 5 gives that H(ab) = H(cd).
Let G be a group and H a normal subgroup of G. Let G/H be the set
consisting of all the cosets of H in G. Put it this way, if Ha, Hb, Hc, . . . are
the cosets of H in G, then G/H = {Ha, Hb, Hc, . . .} = {Hg ∣ g ∈ G}. We have
just proved that coset operation is a valid operation on G/H. In our next
result, we prove that G/H is indeed a group under coset operation.
Theorem 3.5. Let G be a group and H a normal subgroup of G. Then G/H
is a group under coset operation: Ha ⋅ Hb = H(ab), for all a, b ∈ G.
Proof. 1. Coset operation is associative:
Ha ⋅ (Hb ⋅ Hc) = Ha ⋅ H(bc) = H(a(bc)) = H((ab)c) = H(ab) ⋅ Hc =
= (Ha ⋅ Hb) ⋅ Hc
2. The unity for G/H is H:
Ha ⋅ H = Ha ⋅ H1 = H(a1) = H(1a) = H1 ⋅ Ha = H ⋅ Ha
3. The inverse of Ha is Ha−1 :
Ha ⋅ Ha−1 = H(aa−1 ) = H1 = H,
Ha−1 ⋅ Ha = H(a−1 a) = H1 = H
50
J. Sánchez-Ortega
The group G/H is said to be the factor group , or quotient group of
G by H. Notice that the order of G/H is number of distinct cosets of H in G,
that is, the index of H in G: ∣G/H∣ = ∣G ∶ H∣ . If G is finite, then from the
∣G∣
proof of Lagrange’s Theorem we can deduce that ∣G/H∣ = ∣H∣
. An important
map is the coset map π ∶ G → G/H given by π(a) = Ha . The coset map is
a group epimorphism with kernel ker π = H .
Example 3.6. Let Z be the additive group of the integers and ⟨6⟩ = 6Z the
cyclic subgroup of Z consisting of all the multiples of 6. Notice that 6Z is
a normal subgroup of Z, since Z is abelian. Therefore, we can consider the
factor group Z/6Z, which elements are the cosets of 6Z in Z, namely:
6Z + 0 = {. . . , −18, −12, −6, 0, 6, 12, 18, . . .}
6Z + 1 = {. . . , −17, −11, −5, 1, 7, 13, 19, . . .}
6Z + 2 = {. . . , −16, −10, −4, 2, 8, 14, 20, . . .}
6Z + 3 = {. . . , −15, −9, −3, 3, 9, 15, 21, . . .}
6Z + 4 = {. . . , −14, −8, −2, 4, 10, 16, 22, . . .}
6Z + 5 = {. . . , −13, −7, −1, 5, 11, 17, 23, . . .}
These are all the different cosets of 6Z. Observe that 6Z + 6 = 6Z + 0 (since
6 ∈ 6Z + 0), 6Z + 7 = 6Z + 1 (since 7 ∈ 6Z + 1), 6Z + 8 = 6Z + 2, etc.
Next, the operation in Z is +, so we use + for the coset operation:
(6Z + a) + (6Z + b) = 6Z + (a + b).
To ease our notation, we write the coset in the following shorter form:
0̄ = 6Z + 0 1̄ = 6Z + 1 2̄ = 6Z + 2 3̄ = 6Z + 3 4̄ = 6Z + 4 5̄ = 6Z + 5
Then Z/6Z = {0̄, 1̄, 2̄, 3̄, 4̄, 5̄}, and its Cayley table is:
+
0̄
1̄
2̄
3̄
4̄
5̄
0̄
0̄
1̄
2̄
3̄
4̄
5̄
1̄
1̄
2̄
3̄
4̄
5̄
0̄
2̄
2̄
3̄
4̄
5̄
0̄
1̄
3̄
3̄
4̄
5̄
0̄
1̄
2̄
4̄
4̄
5̄
0̄
1̄
2̄
3̄
5̄
5̄
0̄
1̄
2̄
3̄
4̄
3. Isomorphism Theorems
51
Notice that the factor group Z/6Z is isomorphic to Z6 . In general, Z/nZ ≅ Zn .
Example 3.7. Consider the octic group D4 = {1, a, a2 , a3 , b, ba, ba2 , ba3 },
where o(a) = 4, o(b) = 2 and aba = b. Find the Cayley table of the factor group
D4 /Z(D4 ) and conclude that D4 /Z(D4 ) is isomorphic to the Klein group K4 .
Solution: Write Z = Z(D4 ) for short. From Example 2.4 we know that
D4 /Z = {Z, Za, Zb, Z(ba)}. Keeping in mind how multiplication works in
factor groups and the Cayley table of D4 displayed in §2.1 we obtain that
D4 /Z
Z
Za
Zb
Z(ba)
Z
Z
Za
Zb
Z(ba)
Za
Za
Z
Z(ba)
Zb
Zb
Zb
Z(ba)
Z
Za
Z(ba)
Z(ba)
Zb
Za
Z
From the table we can see that D4 /Z is not a cyclic group and has order 4, so
D4 /Z ≅ K4 . Remember that (up to isomorphisms) there are only two groups
of order 4: the cyclic group C4 and the Klein group K4 .
A few important properties of factor groups follow. We will prove
these in our tut sessions.
Theorem 3.8. Let H be a normal subgroup of a group G.
(1) If G is abelian, then G/H is abelian.
(2) If G = ⟨g⟩ is cyclic, then G/H is also cyclic: G/H = ⟨Hg⟩.
(3) If H ⊆ Z(G) satisfies that G/H is cyclic, then G is abelian.
3.3
The Correspondence Theorem
The goal in this section is to study the subgroups of a factor group G/K.
We will show that there is a correspondence between the set of subgroups
of G that contains K and the subgroups of G/K. Moreover, if we are able
to determine all the subgroups in one of the sets, then we will be also able
to easily compute the subgroups in the other set. To motivate and also
understand how this correspondence works, we take K G and H a subgroup
52
J. Sánchez-Ortega
of G such that K ⊆ H. Notice that K H so it makes sense to consider the
factor group H/K, defined as H/K = {Kh ∣ h ∈ H}. It is obvious that H/K is
a subgroup of G/K (remember how the coset operation works; do not forget
that H is a subgroup of G). Here, we will prove that all the subgroups of
G/K arises as this way. We start with a very simple but useful result.
Lemma 3.9. Let K ⊆ H ⊆ G be groups with K G. If K ⊆ H1 ⊆ G, then
H ⊆ H1 if and only if H/K ⊆ H1 /K.
Proof. It is clear that H ⊆ H1 implies H/K ⊆ H1 /K. Conversely, assume
that H/K ⊆ H1 /K and take h ∈ H. Then Kh ∈ H/K ⊆ H1 /K, so we can find
h1 ∈ H1 such that Kh = Kh1 . This is equivalent to say that hh−1
1 ∈ K ⊆ H1 .
Therefore h = (hh−1
1 )h1 ∈ H1 .
Correspondence Theorem
Let K G be groups and define a map
Θ ∶ {H ∣ H is a subgroup of G, K ⊆ H} → {H ∣ H is a subgroup of G/K}
by Θ(H) = H/K. Then:
(1) If H is a subgroup of G/K, then H = H/K where
H = {g ∈ G ∣ Kg ∈ H}.
(2) The map Θ is a bijection.
Let H, H1 and H2 be subgroups of G containing K. Then:
(3) Θ preserves containment: H ⊆ H1 if and only if H/K ⊆ H1 /K.
(4) Θ preserves normality: H H1 if and only if H/K H1 /K.
(5) Θ preserves intersections: (H1 /K) ∩ (H2 /K) = (H1 ∩ H2 )/K.
(6) Θ preserves products: if H1 H2 is a subgroup of G, then
(H1 /K)(H2 /K) = (H1 H2 )/K.
Proof. (1). Let H be a subgroup of G/K, and define H = {g ∈ G ∣ Kg ∈ H}.
Notice that K ⊆ H since for any k ∈ K we have that Kk = K ∈ H because H is
3. Isomorphism Theorems
53
a subgroup of G/K. Moreover, we claim that H is a subgroup of G. In fact,
if h, h′ ∈ H, then Kh, Kh′ ∈ H and so
K(hh′ ) = (Kh)(Kh′ ) ∈ H,
Kh−1 = (Kh)−1 ∈ H
Thus hh′ , h−1 ∈ H and the Subgroup Test allows us to conclude that H is
?
a subgroup of G. We can then consider H/K = H. Notice that H ⊆ H/K
because Kg ∈ H implies (by definition of H) that g ∈ H, and so Kg ∈ H/K.
Conversely, if Kg ∈ H/K, then Kg = Kh for some h ∈ H. From K ⊆ H we
obtain that g ∈ Kh ⊆ Hh ⊆ H and so Kg ∈ H. It shows that H/K ⊆ H and
hence H/K = H, as required.
(2). Notice that Θ is surjective by (1). If θ(H) = θ(H1 ), that is, H/K =
H1 /K. Then from Lemma 3.9 we can conclude that H = H1 . It shows that Θ
is injective.
(3) follows from Lemma 3.9.
(4). Assume that H/K H1 /K and let us show that H H1 . Given h ∈ H
and h1 ∈ H1 , using the Normality Test it is enough to prove that h1 hh−1
1 ∈ H.
Using that H/K H1 /K we obtain that
−1
K(h1 hh−1
∈ (Kh1 )(H/K)(Kh1 )−1 ⊆ H/K
1 ) = (Kh1 )(Kh)(Kh1 )
−1
′
′
′
Thus K(h1 hh−1
1 ) = Kh for some h ∈ H. So, h1 hh1 ∈ Kh ⊆ H because K ⊆ H.
Hence H H1 .
Conversely, let H H1 . Then (Kh1 )(Kh)(Kh1 )−1 = K(h1 hh−1
1 ) ∈ H/K
for all h1 ∈ H1 and h ∈ H, because h1 hh−1
1 ∈ H. This shows that
(Kh1 )(H/K)(Kh1 )−1 ⊆ H/K ⇒ H/K H1 /K,
as desired.
(5) and (6) will be proved in our tut sessions.
Remark 3.10. Since the bijective correspondence Θ preserves inclusion, the
lattice diagram of the subgroups of G/K has the same form as the lattice
diagram of all subgroups of G containing K. In particular:
G ↔ G/K
54
J. Sánchez-Ortega
K ↔ K/K = {K} − the trivial subgroup of G/K
Example 3.11. Let C12 = ⟨a⟩ be the cyclic group of order 12, and let
K = ⟨a6 ⟩ and K1 = ⟨a4 ⟩. Draw the lattice diagram of all subgroups
of C12 , and use the correspondence theorem to obtain the lattice
diagram of all subgroups of C12 /K and C12 /K1 .
Solution: To determine the subgroups of C12 , we use the Fundamental
Theorem of Finite Cyclic Groups. The subgroups of C12 are given by the
positive divisors of 12, namely, 1, 2, 3, 4, 6, 12. The Fundamental Theorem of
Finite Cyclic Groups allows us to conclude that C12 has exactly 6 subgroups.
To determine the unique subgroup of each of these orders, remember that
o(ad ) = n/d, provided d∣n. We have that {1} = ⟨1⟩ = ⟨a12 ⟩, and
12
⇒ d = 6;
d
12
Order 3: 3 = o(ad ) =
⇒ d = 4;
d
12
Order 4: 4 = o(ad ) =
⇒ d = 3;
d
12
⇒ d = 2;
Order 6: 6 = o(ad ) =
6
Order 2: 2 = o(ad ) =
the subgroup of order 2 is ⟨a6 ⟩;
the subgroup of order 3 is ⟨a4 ⟩;
the subgroup of order 4 is ⟨a3 ⟩;
the subgroup of order 6 is ⟨a2 ⟩;
To draw the lattice diagram remember that ⟨am ⟩ ⊆ ⟨ak ⟩ ⇔ k∣m
C12
⟨a2 ⟩
K1 = ⟨a4 ⟩
⟨a3 ⟩
K = ⟨a6 ⟩
⟨1⟩
To obtain the lattice diagram of C12 /K we just need to look at the
subgroups H of C12 that contains K (upper left square) and consider H/K.
Similarly, to picture the lattice diagram of C12 /K1 , we focus our attention on
3. Isomorphism Theorems
55
the subgroups N of C12 containing K1 , and we consider N /K1 . Notice that
in this case N ∈ {K1 , ⟨a2 ⟩, C12 }.
C12 /K
⟨a2 ⟩/K
C12 /K1
⟨a3 ⟩/K
⟨a2 ⟩/K1
{K}
{K1 }
Example 3.12. Consider the octic group D4 = {1, a, a2 , a3 , b, ba, ba2 , ba3 },
where o(a) = 4, o(b) = 2 and aba = b. For Z = Z(D4 ) = {1, a2 }, determine all
the subgroups H of D4 such that Z ⊆ H ⊆ D4 .
Solution: From Example 3.7 we know that D4 /Z = {Z, Za, Zb, Z(ba)} ≅
K4 , the Klein group. Hence, the subgroups of D4 /Z are
{Z},
H1 = {Z, Za},
H2 = {Z, Zb},
H3 = {Z, Z(ba)},
D4 /Z
The lattice diagram of D4 /Z is the following:
D4 /Z
H1
H2
H3
{Z}
The Correspondence Theorem tells us that for i ∈ {1, 2, 3} we can find a unique
subgroup Hi of D4 such that Z ⊆ Hi and Hi = Hi /Z. More precisely, from the
Correspondence Theorem (1) we know that
H1 = {g ∈ D4 ∣ Zg ∈ H1 },
H2 = {g ∈ D4 ∣ Zg ∈ H2 },
H3 = {g ∈ D4 ∣ Zg ∈ H3 }
On the other hand from Example 2.4 we know that
Za2 = Z1 = Z,
Za3 = Za,
Z(ba2 ) = Zb,
Z(ba3 ) = Z(ba),
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J. Sánchez-Ortega
which yields that
H1 = {1, a, a2 , a3 },
H2 = {1, b, a2 , ba2 },
H3 = {1, ba, a2 , ba3 }
The Correspondence Theorem asserts that (besides D4 and Z) these are the
only subgroups H of D4 such that Z ⊆ H ⊆ D4 , and that the lattice diagram
of these subgroups has exactly the same shape that the lattice diagram of
D4 /Z. Moreover, the subgroups H1 , H2 and H3 are normal in D4 since H1 ,
H2 and H3 are normal in D4 /Z.
D4
H1
H2
H3
Z
3.3.1
Maximal normal subgroups
An important special case of the Correspondence Theorem describes when a
factor group is simple.
Definitions 3.13. A group G is said to be simple if G ≠ {1} and its only
normal subgroups are {1} and G. A normal subgroup K ≠ G is called a
maximal normal subgroup if the only normal subgroups H of G satisfying
that K ⊆ H ⊆ G are H = K and H = G.
Theorem 3.14. Let G be a group and K G with K ≠ G. Then K is a
maximal normal subgroup if and only if G/K is simple.
Proof. By definition the factor group G/K is simple if and only if its only
normal subgroups are the trivial subgroup K/K = {K} and the whole group
G/K. An application of the Correspondence Theorem tells us that the only
normal subgroups H of G such that K ⊆ H ⊆ G are H = K and H = G, which
finishes the proof.
3. Isomorphism Theorems
57
Example 3.15. Find all the maximal normal subgroups of Z.
Solution: Let us start by noticing that all the subgroups of Z are normal
subgroups because Z is abelian. Moreover, we know that Z is a cyclic group
so all its subgroups are cyclic, as well. This implies that all the subgroups
of Z are of the form nZ, where n ∈ Z. Theorem 3.14 tells us that finding
the maximal normal subgroups of Z is equivalent to determine the normal
subgroups of Z such that the corresponding factor group is simple. In other
words, we need to find when the factor group Z/nZ is simple. From Z/nZ ≅ Zn
we obtain that the order of Z/nZ is n. Now, from Lagrange’s Theorem the
order of any subgroup of Z/nZ divides n. So, Z/nZ will have the trivial
subgroup nZ/nZ = {nZ} and the whole group Z/nZ as normal subgroups if
and only if n = p is a prime. Thus, the maximal normal subgroups of Z are of
the form pZ with p a prime.
3.4
The First Isomorphism Theorem
Let G and G′ be groups. If there exists an epimorphism from G onto G′ , we
say that G′ is a homomorphic image of G. Using this terminology, the
canonical projection tells us that any factor group G/H is a homomorphic
image of G. This means that when we construct factor groups of G, we are,
in fact, constructing homomorphic images of G.
Therefore, the factor group construction gives us a way of manufacturing
homomorphic images of any group G. In this section, we will show that this is
indeed a way of manufacturing all the homomorphic images of G. To be more
precise, we will prove that every homomorphic image of G is isomorphic
to a factor group of G. As a consequence, G′ is a homomorphic image of
G if and only if G′ is (or is isomorphic to) a factor group of G. Let us start
by proving a simple theorem.
Theorem 3.16. Let φ ∶ G → G′ be a group homomorphism with kernel K.
Then φ(a) = φ(b) if and only if Ka = Kb. In other words, any two elements a
and b in G have the same image under φ if and only if they are in the same
coset of K.
58
J. Sánchez-Ortega
Proof. If a and b are two elements in G, then
φ(a) = φ(b) ⇔ 1 = φ(a)φ(b)−1 = φ(a)φ(b−1 ) = φ(ab−1 ) ⇔ ab−1 ∈ K
Applying Cosets Property 6 we get
φ(a) = φ(b) ⇔ ab−1 ∈ K ⇔ Ka = Kb,
as desired.
Given φ ∶ G → G′ an epimorphism with kernel K. Theorem 3.16 tells us
that all the elements in a coset of K in G, say Kx, have the same image under
φ, and, conversely, elements which have the same image under φ are in the
same coset of K.
K = K1G
φ
1G′
1G
φ(a) = φ(b)
Ka = Kb
a
b
Kx
φ(x)
x
G′
G
We have then built a bijective mapping which matches cosets of K, namely,
elements of the factor group G/K, with elements in G′ . To be more precise,
the coset Kx is matched with the element φ(x) ∈ G′ . It remains to check that
this correspondence is indeed an isomorphism.
3. Isomorphism Theorems
59
First Isomorphism Theorem
Let φ ∶ G → G′ be an epimorphism with kernel K. Then G/K ≅ G′ .
Proof. To show that G/K is isomorphic to G′ , we should find an isomorphism
̃ ∶ G/K → G′ that matches
from G/K to G′ . Let us try with the mapping φ
̃
each coset Kx with the element φ(x). That is, φ(Kx)
= φ(x). We first need
̃
to make sure that φ(Kx)
is uniquely defined. To do so, we must make sure
̃
̃
that if Ka is the same coset as Kb, then φ(Ka)
But, if Ka = Kb,
= φ(Kb).
̃
̃
̃ is
then φ(Ka)
= φ(a) = φ(b) = φ(Kb)
by Theorem 3.16, proving that φ
well-defined. Next, for Ka, Kb ∈ G/K we have that
̃
̃
̃
̃
φ(Ka
⋅ Kb) = φ(Kab)
= φ(ab) = φ(a) ⋅ φ(b) = φ(Ka)
⋅ φ(Kb),
̃ is a group homomorphism. It remains to check that φ
̃ is
which shows that φ
an bijective.
̃ is injective: If φ(Ka)
̃
̃
−φ
= φ(Kb),
then φ(a) = φ(b) and Ka = Kb by
Theorem 3.16.
̃ is surjective: every element of G′ is of the form φ(x) = φ(Kx).
̃
−φ
The First Isomorphism Theorem asserts that every homomorphic image
of G is isomorphic to a quotient group of G. If we are given an
homomorphism not surjective φ ∶ G → G′ , we can consider the restriction
φ ∶ G → φ(G) (which is now an epimorphism) and apply the First Isomorphism
Theorem to it. More precisely:
Corollary 3.17. Let φ ∶ G → G′ be a group homomorphism with kernel K.
Then G/K ≅ Imφ.
60
J. Sánchez-Ortega
Example 3.18. Check the following assertions for the set G given below.
⎧
⎫
1 a b ⎞
⎪
⎪
⎪
⎪
⎪⎛
⎪
G ∶= ⎨⎜ 0 1 c ⎟ ∶ a, b, c ∈ R⎬ .
⎪
⎪
⎪
⎪
⎪
⎪
⎩⎝ 0 0 1 ⎠
⎭
(i) G is a group under matrix multiplication;
⎧
⎫
1 0 b ⎞
⎪
⎪
⎪
⎪
⎪⎛
⎪
(ii) Z(G) = ⎨⎜ 0 1 0 ⎟ ∶ b ∈ R⎬ ;
⎪
⎪
⎪
⎪
⎪
⎪
⎩⎝ 0 0 1 ⎠
⎭
(iii) Z(G) ≅ R;
(iv) G/Z(G) ≅ R × R.
Solution:
(i) We first check that G is closed under matrix multiplication:
′
′
′
′
′
⎛ 1 a b ⎞ ⎛ 1 a b ⎞ ⎛ 1 a + a b + ac + b ⎞
1
c + c′
⎟
⎜ 0 1 c ⎟ ⎜ 0 1 c′ ⎟ = ⎜ 0
⎠
⎝ 0 0 1 ⎠⎝ 0 0 1 ⎠ ⎝ 0
0
1
⎛ 1 0 0 ⎞
− The identity matrix ⎜ 0 1 0 ⎟ is in G, and
⎝ 0 0 1 ⎠
−1
⎛ 1 a b ⎞
⎛ 1 −a ac − b ⎞
−c ⎟ .
− G is closed under inverses: ⎜ 0 1 c ⎟ = ⎜ 0 1
⎝ 0 0 1 ⎠
⎝ 0 0
1 ⎠
Since matrix multiplication is an associative operation, we can conclude
that G is a group under matrix multiplication.
′
′
⎛ 1 a b ⎞
(ii) If ⎜ 0 1 c′ ⎟ ∈ Z(G), then
⎝ 0 0 1 ⎠
′
′
′
′
⎛ 1 a b ⎞⎛ 1 a b ⎞ ⎛ 1 a b ⎞⎛ 1 a b ⎞
⎜ 0 1 c′ ⎟ ⎜ 0 1 c ⎟ = ⎜ 0 1 c ⎟ ⎜ 0 1 c′ ⎟ ,
⎝ 0 0 1 ⎠⎝ 0 0 1 ⎠ ⎝ 0 0 1 ⎠⎝ 0 0 1 ⎠
∀ a, b, c ∈ R.
It implies that b′ + ac′ + b = b + a′ c + b′ , for all a, b, c ∈ R. In particular, taking
a = 1 and c = 0 gives c′ = 0. Next, if we take c = 1 we obtain that a′ = 0. Thus:
⎧
⎫
1 0 b ⎞
⎪
⎪
⎪
⎪
⎪⎛
⎪
Z(G) = ⎨⎜ 0 1 0 ⎟ ∶ b ∈ R⎬
⎪
⎪
⎪
⎪
⎪
⎪
⎩⎝ 0 0 1 ⎠
⎭
3. Isomorphism Theorems
61
(iii) The map φ ∶ R → Z(G) given by
⎛ 1 0 b ⎞
b ↦ ⎜ 0 1 0 ⎟,
⎝ 0 0 1 ⎠
is the desired isomorphism (fill in the details yourself!).
(iv) We will apply the First Isomorphism Theorem. We then need to find an
epimorphism from G to R × R with kernel Z(G) ≅ R. Taking into account the
previous calculations, we consider the map ψ ∶ G → R × R given by
⎛ 1 a b ⎞
ψ ⎜ 0 1 c ⎟ = (a, c).
⎝ 0 0 1 ⎠
⎛ 1 x 0 ⎞
Notice that ψ is surjective; if (x, y) ∈ R × R, then (x, y) = ψ ⎜ 0 1 y ⎟ ∈ G.
⎝ 0 0 1 ⎠
Next, we prove that ψ is a group homomorphism:
⎡ 1 a b
⎤
′
′
′
′
′
⎢⎛
⎞ ⎛ 1 a b ⎞⎥⎥
⎛ 1 a + a b + ac + b ⎞
⎢
′
1
c + c′
⎟=
ψ ⎢⎜ 0 1 c ⎟ ⎜ 0 1 c ⎟⎥ = ψ ⎜ 0
⎢⎝
⎥
⎝ 0
⎠
⎢ 0 0 1 ⎠ ⎝ 0 0 1 ⎠⎥
0
1
⎣
⎦
′
′
⎛ 1 a b ⎞
⎛ 1 a b ⎞
= (a + a′ , c + c′ ) = (a, c) + (a′ , c′ ) = ψ ⎜ 0 1 c ⎟ + ψ ⎜ 0 1 c′ ⎟
⎝ 0 0 1 ⎠
⎝ 0 0 1 ⎠
We claim that ker ψ = Z(G); in fact:
⎛ 1 a b ⎞
ψ ⎜ 0 1 c ⎟ = (0, 0) ⇔ (a, c) = (0, 0) ⇔ a = c = 0.
⎝ 0 0 1 ⎠
Thus, an application of the First Isomorphism Theorem gives that G/ ker ψ =
G/Z(G) ≅ R × R.
62
J. Sánchez-Ortega
3.5
The Second Isomorphism Theorem
Let G be a group and H, K subgroups of G. We define their product as
HK = {hk ∣ h ∈ H and k ∈ K}.
Note that H ⊆ HK and K ⊆ HK.
Second Isomorphism Theorem
Let H and K be subgroups of a group G with K G. Then
(i) HK is a subgroup of G.
(ii) K HK
(iii) K ∩ H H and the factor groups HK/K and H/(H ∩ K) are
isomorphic.
We first prove a preliminary lemma.
Lemma 3.19. Let H and K be subgroups of a group G. Then HK is a
subgroup of G if and only if KH ⊆ HK.
Proof. Assume first that HK is a subgroup of G and take kh ∈ KH, where
k ∈ K and h ∈ H. Our goal is to show that kh ∈ HK. Applying that K and H
are both subgroups of G, we have that k −1 ∈ K, h−1 ∈ H, which implies that
h−1 k −1 ∈ HK and so (h−1 k −1 )−1 ∈ HK because we are assuming that HK is a
subgroup of G. From here we obtain that kh = (h−1 k −1 )−1 ∈ HK, as desired.
Conversely, suppose that KH ⊆ HK and let us prove that HK is a subgroup
of G. We will use the Subgroup Test. First notice that 1 = 1 ⋅ 1 ∈ HK. Next
given h1 k1 , h2 k2 ∈ HK with k1 , k2 ∈ K and h1 , h2 ∈ H we have that
(h1 k1 )(h2 k2 ) = h1 (k1 h2 )k2 ,
(3.2)
but k1 h2 ∈ KH ⊆ HK, and so k1 h2 = h′2 k1′ for some k1′ ∈ K and h′2 ∈ H. Thus
from (3.2) we get that
(h1 k1 )(h2 k2 ) = h1 (k1 h2 )k2 = h1 (h′2 k1′ )k2 = (h1 h′2 )(k1′ k2 ) ∈ HK.
To finish notice that (h1 k1 )−1 = k1−1 h−1
1 ∈ KH ⊆ HK. The Subgroup Test allows
us to conclude that HK is a subgroup of G.
3. Isomorphism Theorems
63
Proof of the Second Isomorphism Theorem.
(i) Lemma 5.13 tells us that the result will follow by showing that KH ⊆ HK.
If kh ∈ KH, then kh = h(h−1 kh) ∈ HK because h−1 kh ∈ h−1 Kh = K, since we
are assuming that K is normal in G.
(ii) The subgroup K is normal in HK because K is normal in G (by hypothesis)
and K ⊆ HK.
(iii) We are going to show that K ∩ H is a normal subgroup of H by viewing
it as the kernel of a group homomorphism. Moreover, we are interested in
finding an epimorphism from H to HK/K with kernel H ∩ K because (iii)
will then follow by an application of the First Isomorphism Theorem.
We define φ ∶ H → HK/K by φ(h) = Kh. Notice that Kh ∈ HK/K
because h ∈ HK. It is straightforward to check that φ is an homomorphism.
Moreover, φ is an epimorphism because, if x ∈ HK, say x = hk with k ∈ K and
h ∈ H. Then Kx = K(hk) = (Kh)(Kk) = (Kh)(K) = Kh = φ(h). Let us now
check that ker φ = K ∩ H. If h ∈ ker φ, then φ(h) = K, that is, Kh = K, which
by Cosets Property 7 is equivalent to say that h ∈ K. Thus h ∈ H ∩ K and
ker φ ⊆ H ∩ K. The other containment holds by Cosets Property 7 . Thus
ker φ = H ∩ K, which tells us that H ∩ K H. By the First Isomorphism
Theorem we obtain that
HK
H
≅
,
H ∩K
K
as desired.
We close the section with an application of the Second Isomorphism
Theorem. A group G is called a metacyclic group if there exists a normal
subgroup K of G such that both K and G/K are cyclic. Notice that every
cyclic group is metacyclic (simply take K = {1}). We can use the Second
Isomorphism Theorem to show that every subgroup of a metacyclic group
is again metacyclic.
Let G be a metacyclic group and H a subgroup of G. We know that there
exists K G such that K and G/K are both metacyclic. Applying the Second
Isomorphism Theorem to G, K G and H we get that HK is a subgroup of G
having K as a normal subgroup, H ∩ K H and the factor groups H/(H ∩ K),
HK/K are isomorphic.
64
J. Sánchez-Ortega
We know that K is a cyclic group and H ∩ K is a subgroup of it, so H ∩ K
is cyclic. On the other hand, the factor group HK/K is a subgroup of G/K,
which is cyclic and so HK/K is cyclic. The isomorphism H/(H ∩K) ≅ HK/K
implies that H/(H ∩ K) is cyclic. Notice that we are done, H is metacylic
since its subgroup H ∩ K is cyclic and the factor group H/(H ∩ K) is cyclic.
3.6
The Third Isomorphism Theorem
Third Isomorphism Theorem
Let K ⊆ H ⊆ G be groups with K G and H G. Then H/K G/K
and
G/K
≅ G/H
H/K
Proof. We define φ ∶ G/K → G/H by φ(Kg) = Hg for all g ∈ G. The first
thing that we need to do is to check that φ is well defined, that is, it does
not depend on the the representative of the coset. If Kg = Kg1 , we then need
to show that Hg = Hg1 . From Kg = Kg1 we can conclude that gg1−1 ∈ K but
K ⊆ H, so gg1−1 ∈ H, which implies that Hg = Hg1 .
It is straightforward to show that φ is an epimorphism (complete the
details). Next, we claim that ker φ = H/K. In fact, by definition
ker φ = {Kg ∈ G/K ∣ φ(Kg) = H}
But φ(Kg) = Hg = H if and only if g ∈ H, and so ker φ = H/K. By the First
Isomorphism Theorem we get that
G/K
≅ Imφ,
ker φ
that is,
G/K
≅ G/H.
H/K
As an application of the Third Isomorphism Theorem, we will show that
every factor group of a metacyclic groups is also metacyclic. Assume that G
is a metacyclic group and let K G such that K and G/K are both cyclic
groups. Consider G/N with N G. The Second Isomorphism Theorem tells
3. Isomorphism Theorems
65
us that KN ≤ G and KN /N ≅ K/(N ∩ K). In this case, because both K and
N are normal in G, we have that KN is also normal in G. In fact, for g ∈ G
we have that
(KN )g = K(N g) = K(gN ) = (Kg)N = (gK)N = g(KN ),
which shows that KN G. Now, from the Correspondence Theorem we obtain
that KN /N G/N . Since K is cyclic, the factor group K/(N ∩ K) is cyclic
and so is KN /N . On the other hand, by the Third Isomorphism Theorem
G/N
≅ G/KN,
KN /N
which is a cyclic group since
G/KN ≅
G/K
.
KN /K
Notice that we have already showed that G/N is metacylic because its subgroup
G/N
KN /N is cyclic and the factor group KN
/N is also cyclic.
Chapter 4
Cauchy’s Theorem
Let p be a prime and G a group of order pn . A quick application of Lagrange’s
Theorem reveals that the order of any element of G is a power of p. The
converse, known as Cauchy’s Theorem, is also true. If every element of a finite
group G has order a power of p, then ∣G∣ = pn for some n ≥ 0.
To prove Cauchy’s Theorem we need to develop some machinery and learn
a bit more about groups of p-power order.
4.1
Group Action on a Set
We have already seen examples of how groups can act on things like, for
example, the group of symmetries of a square. In this section, we introduce
the general notion of a group acting on a set and present some examples.
Recall that a binary operation ∗ on a set S is a map from S × S into S.
The map ∗ provides us a rule for “multiplying” an element s1 in S and an
element s2 in S to obtain a new element s1 ∗ s2 in S. In a more general way,
for any sets A, B and C we can view a map ∗ ∶ A × B → C as defining a
“multiplication”, where any element a ∈ A times any element b ∈ B gives rise to
some element c ∈ C, usually denoted by a ∗ b = c, or simply ab = c. Here we
are interested in the case of having a group G, a set X, a map ∗ ∶ G × X → X.
To ease the notation, we will write ∗(g, x) as g ∗ x, or gx.
Definition 4.1. Let X be a set and G a group. An action of G on X is a
map ∗ ∶ G × X → X (written as gx for all g ∈ G and x ∈ X) such that
1. (g1 g2 ) ∗ x = g1 ∗ (g2 ∗ x) for all g1 , g2 ∈ G and x ∈ X.
67
68
J. Sánchez-Ortega
2. 1 ∗ x = x for all x ∈ X.
Under these conditions we say that X is a G-set.
Examples 4.2.
1. If X is any set and G is any group, the trivial action is defined by
g ∗ x = x for all x ∈ X and g ∈ G.
2. Let G be a group and let X = G. The operation in G gives a group action
of G on itself, where the action on g2 by g1 is given by left multiplication.
To be more precise, ∗(g1 , g2 ) = g1 g2 (or, g1 + g2 if G is an additive group).
3. Let V a vector space over a field F . The axioms (λµ)v = λ(µv) and
1v = v for scalars λ, µ ∈ F , and a vector v show that V is an F × -set,
where F × denotes the multiplicative group of nonzero scalars.
4. Let X be any set and H a subgroup of the group SX of all permutations
of X. Then X is an H-set, where the action of σ ∈ H on X is its action
as an element of SX , that is, σ ∗ x = σ(x) for all x ∈ X. In particular,
{1, 2, . . . , n} is an Sn -set.
Our first result shows that for every G-set X, the map σg ∶ X → X defined
by σg (x) = g ∗ x is a permutation of X, and that there is a homomorphism
from G to SX .
Theorem 4.3. Let G be a group acting on a set X.
(i) For each g ∈ G, the map σg ∶ X → X defined by σg (x) = g ∗ x for x ∈ X
is a permutation of X.
(ii) The map ϕ ∶ G → SX given by ϕ(g) = σg is a homomorphism.
Proof. (i) Given g ∈ G, to show that σg is a permutation of X, we just need
to prove that σg is bijective. But you can easily check that σg−1 = σg−1 .
(ii) We must show that ϕ(g1 g2 ) = ϕ(g1 ) ○ ϕ(g2 ) for all g1 , g2 . To do so, we need
to prove that boths permutations act in the same way, that is, ϕ(g1 g2 )(x) =
(ϕ(g1 ) ○ ϕ(g2 ))(x). For x ∈ X, apply the two conditions in Definition 4.1:
ϕ(g1 g2 )(x) = σg1 g2 (x) = (g1 g2 ) ∗ x = g1 ∗ (g2 ∗ x) = g1 ∗ σg2 (x) = σg1 (σg2 (x)) =
4. Cauchy’s Theorem
69
= (σg1 ○ σg2 )(x) = (ϕ(g1 ) ○ ϕ(g2 ))(x),
as desired.
Remark 4.4. If X is a G-set, then the elements of G fixing every element
of X is a normal subgroup of G, since it coincides with the kernel of the
homomorphism ϕ ∶ G → SX . In fact,
ker(ϕ) = {g ∈ G ∣ σg = IdX } = {g ∈ G ∣ g ∗ x = σg (x) = x for all x ∈ X}
We call ker(ϕ) the kernel of the action. If ker(ϕ) = {1}, then the identity
element of G is the only element of G that leaves every x ∈ X fixed; in such a
case we say that G acts faithfully on X.
Definition 4.5. Let G be a group acting on a set X. For each x ∈ X the
stabilizer of x in G, denoted S(x), is the set of elements of G that fix the
element x, that is, S(x) = {g ∈ G ∣ g ∗ x = x} . It is easy to check that S(x) is
a subgroup of G.
We next prove that the action of a group G on a set X gives rise to an
equivalence relation on X.
Theorem 4.6. Let G be a group acting on a set X. The relation on X
x1 ∼ x2 if and only if x2 = g ∗ x1 for some g ∈ G
is an equivalence relation on X. For each x ∈ X, the number of elements in
the equivalence class Gx of x is ∣G ∶ S(x)∣, the index of the stabilizer of x.
Proof. Condition 2 in Definition 4.1 shows that ∼ is reflexive. If x1 ∼ x2 , then
x2 = g∗x1 for some g ∈ G. Then g −1 ∗x2 = g −1 ∗(g∗x1 ) = (g −1 g)∗x1 = 1∗x1 = x1 ,
and ∼ is symmetric. Finally if x1 ∼ x2 and x2 ∼ x3 , then x2 = g ∗ x1 and
x3 = g ′ ∗ x2 for some g, g ′ ∈ G. Thus x3 = g ′ ∗ x2 = g ′ ∗ (g ∗ x1 ) = (g ′ g) ∗ x1 , so
x1 ∼ x3 and ∼ is transitive.
Recall that the equivalence class G∗x of x is just the subset of X consisting
of all the elements related to x via ∼, that is,
G ∗ x = {x′ ∈ X ∣ x ∼ x′ } = {g ∗ x ∣ g ∈ G}.
70
J. Sánchez-Ortega
We are going to define a bijection ψ between the left cosets of S(x) in G and
the elements of G ∗ x. Let x1 ∈ G ∗ x, then x1 = g1 ∗ x for some g1 ∈ G. We
define ψ(x1 ) to be the left coset of g1 S(x) of S(x), i.e., ψ(x1 ) = g1 S(x).
To show that ψ is injective, assume that ψ(x1 ) = ψ(x2 ) for some x1 , x2 ∈ [x].
Then there exists g1 , g2 ∈ G such that x1 = g1 ∗ x and x2 = g2 ∗ x. From
ψ(x1 ) = ψ(x2 ) we derive g1 S(x) = g2 S(x). In particular, g2 ∈ g1 S(x), and so
g2 = g1 g for some g ∈ S(x). Thus x2 = g2 ∗x = (g1 g)∗x = g1 ∗(g ∗x) = g1 ∗x = x1 ,
and ψ is injective.
Finally, we show that each left coset of S(x) in G is of the form ψ(x1 )
for some x1 ∈ G ∗ x. Let g1 S(x) be a left coset. Then x1 = g1 ∗ x ∈ G ∗ x and
satisfies that ψ(x1 ) = g1 S(x).
Definition 4.7. Let X be a G-set. The equivalence class G ∗ x = {g ∗ x ∣ g ∈ G}
of x under the equivalence relation described in the previous theorem is called
the orbit of x under G.
The action of G on X is called transitive if there is only one orbit, that
is, for all x1 , x2 ∈ X there exists g ∈ G such that g ∗ x1 = x2 .
Let X be a G-set. Then Theorem 4.6 allows us to characterise the elements
x of X such that its orbit Gx is a singleton. In fact:
G ∗ x = {x} ⇔ ∣G ∶ S(X)∣ = 1 ⇔ S(x) = G ⇔ g ∗ x = x for all g ∈ G.
The set of all such elements x is denoted
Xf = {x ∈ X ∣ g ∗ x = x for all g ∈ G}.
and is called the fixed subset of X under the action of G. Notice that Xf is
the union of the singleton orbits.
4. Cauchy’s Theorem
71
Examples 4.8.
1. If a group G acts trivially on a set X, then S(x) = G for all x ∈ X, and
the orbits are precisely the elements of X.
2. The symmetric group Sn acts transitively in its usual action as permutations on Xn = {1, 2, . . . , n}.
Orbit Decomposition Theorem
Let G be a group acting on a finite set X ≠ ∅. Let G ∗ x1 , . . . , G ∗ xn
denote the nonsingleton orbits. Then
n
∣X∣ = ∣Xf ∣ + ∑ ∣G ∶ S(xi )∣
i=1
Proof. The fixed subset is the union of the singleton orbits. Because the
orbits partition X, we have that ∣X∣ = ∣Xf ∣+ ∑ni=1 ∣G∗xi ∣. But from Theorem 4.6
we have that ∣G ∗ xi ∣ = ∣G ∶ S(xi )∣ for each i. Thus ∣X∣ = ∣Xf ∣ + ∑ni=1 ∣G ∶ S(xi )∣,
as required.
4.2
Conjugation
Any group G can act on itself (i.e. X = G) by conjugation:
g ∗ x = gxg −1 for all g, x ∈ G,
where gxg −1 ∈ G is computed in the group G, as usual. This formula satisfies
the two conditions in Definition 4.1. In fact, clearly 1 ∗ g = g for all g ∈ G, and
g1 ∗(g2 ∗x) = g1 ∗(g2 xg2−1 ) = g1 (g2 xg2−1 )g1−1 = (g1 g2 )x(g2−1 g1−1 ) = (g1 g2 )x(g1 g2 )−1 ,
for all g1 , g2 , x ∈ G.
Definition 4.9. Let G be a group and a, b two elements in G. We say that a
and b are conjugate in G if b = gag −1 for some g ∈ G. That is to say that a
and b are in the same orbit of G acting on itself by conjugation.
Elements of the form gag −1 are called conjugates of a. Roughly speaking
a conjugate of a is any product consisting of a sandwiched between any element
and its inverse.
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J. Sánchez-Ortega
The orbits of G acting on itself by conjugation are called the conjugacy
classes of G. The conjugacy class of a ∈ G is denoted by class a. Notice that
it consists on the conjugates of a:
class a = G ∗ a = {g ∗ a ∣ g ∈ G} = {gag −1 ∣ g ∈ G}
The conjugacy classes constitute a partition of G.
Examples 4.10.
1. If G is an abelian group, then the action of G on itself by conjugation is
the trivial action, and class a = {a} for each a ∈ G.
2. class 1 = {1} in any group. More generally, class a = {a} ⇔ a ∈ Z(G).
2. Let G = Sn . Then two permutations in Sn are conjugate if and
only if they have the same cycle structure; that is, when factored
into disjoint cycles they have the same number of cycles of each length.
It follows from the fact that the conjugate of an r-cycle of length is
an r-cycle, as well. More precisely, if (k1 k2 . . . kr ) is an r-cycle and
σ ∈ Sn , then σ(k1 k2 . . . kr )σ −1 = (σk1 σk2 . . . σkr ). If α = γ1 . . . γs is the
decomposition of α into disjoint cycles, then
σασ −1 = (σγ1 σ −1 )(σγ2 σ −1 ) . . . (σγs σ −1 )
It remains to check that (σγi σ −1 ) and (σγj σ −1 ) are disjoint. This is left
as an exercise.
In particular, the conjugacy classes of S3 are
class ε = {ε},
class (1 2) = {(1 2), (1 3), (2 3)},
class (1 2 3) = {(1 2 3), (1 3 2)}.
Theorem 4.11. Let G be a group and g ∈ G. The conjugation mapping
λg ∶ G → G defined by λg (x) = gxg −1 is an automorphism of G.
4. Cauchy’s Theorem
73
Proof. If x, y are in G, then
λg (xy) = g(xy)g −1 = gx(g −1 g)yg −1 = (gxg −1 )(gyg −1 ) = λg (x)λg (y),
which shows that λg is an homomorphism. Moreover, λg is bijective because
λ−1
g = λg −1 .
Such conjugation automorphisms are called inner automorphisms. The
set consisting of all inner automorphisms of G, namely, Inn(G) = {λg ∣ g ∈ G}
is a normal subgroup of the group of automorphisms of G. In terms
of inner automorphisms the conjugacy class of a ∈ G can be written as
class a = {λg (a) ∣ g ∈ G}.
Corollary 4.12. Let G be a group and a, b ∈ G. If a and b are conjugate,
then o(a) = o(b). Moreover, all the elements in a conjungacy class have the
same order.
Proof. If a and b are conjugate, then b = gag −1 = λg (a). The result now
follows from Theorem 4.11 because automorphisms preserve the order of the
elements.
Example 4.13. Find the conjugacy classes of D3 = {1, a, a2 , b, ba, ba2 },
where o(a) = 3, o(b) = 2 and aba = b.
Solution: Recall that ak b = ba3−k for all k ∈ Z. As a and a2 are the only
elements with order 3, we have that class a ⊆ {a, a2 }. But from a2 b = ba we
obtain that a2 = bab−1 ∈ class a and so class a = {a, a2 }. The elements ba and
ba2 have both order 2, so class b ⊆ {b, ba, ba2 }. On the other hand, from aba = b
we obtain that aba2 = ba and since a2 = a−1 , we obtain that ba ∈ class b. To
finish, from a2 b = ba we derive that a2 ba−2 = ba−1 = ba2 and ba2 ∈ class b. Thus:
class b = {b, ba, ba2 }. The conjugacy classes of D3 are
class 1 = {1},
class a = {a, a2 },
class b = {b, ba, ba2 }.
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J. Sánchez-Ortega
We can generalise the conjugacy action by allowing G acts on the set P(G)
of all subsets of G. If T is any subset of G and g ∈ G, we define g ∗ T = gT g −1 ,
where gT g −1 = {gtg −1 ∣ t ∈ T } .
Definition 4.14. We say that two subsets T and U of G are conjugate
in G if U = gT g −1 for some g ∈ G. The subsets of the form gT g −1 are
called conjugates of T . As before, the conjugation relation is an equivalence
relation.
Normal subgroups are self-conjugate: if K G, then gKg −1 = K for
all g ∈ G. In particular, it says that K contains the conjugacy class of each of
its elements. In fact, if a ∈ K then class a = {gag −1 ∣ g ∈ G} ⊆ K. The converse
is also true.
Theorem 4.15. Let H be a subgroup of a group G. Then H G if and only
if H is a union of conjugacy classes.
Proof. If H is normal, then H is the union of the conjugacy classes of its
elements. Conversely, assume that H is a union of conjugacy classes. Suppose
that H = class a1 ∪ . . . ∪ class an . We use the Normality Test: for g ∈ G we
need to show that gHg −1 ⊆ H. If h ∈ H, let h ∈ class ai (remember that the
conjugacy classes are disjoint sets), then h = bai b−1 for some b ∈ G. Thus:
ghg −1 = g(bai b−1 )g −1 = (gb)ai (b−1 g −1 ) = (gb)ai (gb)−1 ∈ class ai ⊆ H,
as desired.
Theorem 4.15 provides us a method to determine the normal subgroups of a group. For example, by Theorem 4.15 any normal subgroup
K of D3 must be an union of conjugacy classes of D3 , which are {1}, {a, a2 },
{b, ba, ba2 }. On the other hand, Lagrange’s Theorem tells us that the order of
K must divide 6 (the order of D3 ). From here we can deduce that the normal
subgroups of D3 are {1}, {1, a, a2 } and D3 .
Let G be a group and T a subset of G. Theorem 4.6 tells us that the
number of conjugates of T equals the index ∣G ∶ S(T )∣ of the stabilizer S(T )
of T . In our case, for action by conjugation the stabilizer
S(T ) = {g ∈ G ∣ gT g −1 = T } = NG (T )
4. Cauchy’s Theorem
75
is called the normalizer of T in G. For the particular case of conjugation
by elements of G, Theorem 4.6 yields that the number of conjugates of an
element a of G is the index of the ∣G ∶ S(a)∣ of the stabilizer S(a) = S({a}) of
a, which is called the centralizer of a in G and denoted CG (a) or C(a).
S(a) = S({a}) = {g ∈ G ∣ gag −1 = a} = NG ({a}) = CG (a) = C(a),
∣G ∶ C(a)∣ = ∣class a∣
Notice that C(a) consists on the elements of G that commutes with a (saying
gag −1 = a is equivalent to say that ga = ag). Moreover we have that
class a = {a} ⇔ a ∈ Z(G) ⇔ C(a) = G
For acting by conjugation the fixed subset of G(= X) coincides with the
center of G. In fact:
Gf = {a ∈ G ∣ gag −1 = a for all g ∈ G} =
= {a ∈ G ∣ ga = ag for all g ∈ G} = Z(G)
Let G be a finite group and class a1 , class a2 , . . . , class an be the nonsingleton
conjugacy classes. The Orbit Decomposition Theorem yields
n
∣G∣ = ∣Z(G)∣ + ∑ ∣G ∶ C(ai )∣
(CE)
i=1
Equation (CE) is called the Class Equation. Note that all the summands
on the right hand side of the Class Equation, namely ∣G ∶ C(ai )∣, are divisors
of ∣G∣ because they are indices of subgroups of G.
Examples 4.16.
1. The Class Equation gives no information in an abelian group since
conjugation is the trivial action and all conjugacy classes have size 1.
2. Let us consider the group D3 . We computed the conjugacy classes in
Example 4.13. Recall that [1] = {1}, [a] = {a, a2 }, and [b] = {b, ba, ba2 }.
The Class Equation tells us that the cardinal of D3 is 1 + 2 + 3 = 6, as
expected. The center is trivial in this case.
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J. Sánchez-Ortega
4.3
Cauchy’s Theorem
We are now in a position to prove Cauchy’s Theorem. In some texts, Cauchy’s
Theorem is first proved for abelian groups and using this abelian version they
then provide a proof for the general case. Here, we will use group actions
to give a proof that works for all the groups abelian or not. This beautiful
proof of Cauchy’s Theorem is due to J. H. McKay. We first need a preliminary
lemma.
Lemma 4.17. Let G be a group of order pn and X be a finite G-set. Then
∣X∣ ≡ ∣Xf ∣ (mod p).
Proof. The Orbit Decomposition Theorem tells us that
m
∣X∣ = ∣Xf ∣ + ∑ ∣G ∶ S(xi )∣,
i=1
which implies that ∣X∣ − ∣Xf ∣ = ∑m
i=1 ∣G ∶ S(xi )∣. The result now follows because
p divides ∣G ∶ S(xi )∣, since they are indices of subgroups of G.
Cauchy’s Theorem
Let G be a finite group. If a prime p divides the order of G, then G
has an element of order p and, consequently, a subgroup of order p.
Proof. Let X be the set of all p-tuples (g1 , . . . , gp ) of elements of G such
that the product of the coordinates in G equals 1 (the identity element of G).
That is, X = {(g1 , . . . , gp ) ∣ gi ∈ G and g1 g2 . . . gp = 1}. We claim that p divides
∣X∣. In fact, ∣X∣ = ∣G∣p−1 because g1 , . . . , gp−1 can be arbitrary elements of G
but once those are fixed, gp has necessarily to be (g1 g2 . . . gp−1 )−1 . Now since
p divides ∣G∣, then p divides ∣G∣p−1 = ∣X∣.
Let σ = (1 2 . . . p) ∈ Sp and let σ acts on X by
σ(g1 , . . . , gp ) = (gσ(1) , . . . , gσ(p) ) = (g2 , g3 , . . . , gp , g1 )
The element (g2 , g3 , . . . , gp , g1 ) is in X. In fact, from g1 g2 . . . gp = 1 we get that
g1−1 = g2 . . . gp , and so g2 g3 . . . gp g1 = 1. Then we can define the action of the
4. Cauchy’s Theorem
77
subgroup ⟨σ⟩ of Sp on X just by iterating the action of σ on X in the natural
way. Let us now consider the fixed subset Xf of X under the action of ⟨σ⟩.
By Lemma 4.17 we have that ∣X∣ ≡ ∣Xf ∣ (mod p). Now since p divides ∣X∣
we can then conclude that p also divides ∣Xf ∣. On the other hand a p-tuple
(g1 , . . . , gp ) belongs to Xf if and only if it is fixed by σ, and this happens
if and only if g1 = g2 = . . . = gp . To finish, since p divides ∣Xf ∣, the set Xf
must contain at least p elements. So, there exists a ∈ G with a ≠ 1 such that
(a, . . . , a) ∈ Xf . Thus ap = 1, so a has order p, and the subgroup ⟨a⟩ of G has
order p.
Cauchy’s Theorem can be used to characterise groups having order the
power of a prime. This might be one of the most important of its applications.
Definition 4.18. Let p be a prime. A group G is called a p-group if every
element of G has order a power of the prime p. A subgroup of a group G is a
p-subgroup of G if the subgroup is itself a p-group.
Cauchy’s Theorem provides this characterisation of finite p-groups.
Corollary 4.19. Let G be a finite group and p a prime. Then G is a p-group
if and only if the order of G is a power of p.
Proof. Assume that G is a p-group. Then o(g) is a power of p for all g ∈ G.
We claim that ∣G∣ is a power of p. Suppose on the contrary that ∣G∣ is not
a power of p. Then we can find a prime q different from p that divides ∣G∣.
Applying Cauchy’s Theorem we can guarantee the existence of an element of
G with order q, but this contradicts our hypothesis. Hence, ∣G∣ is a power of
p. Conversely, if ∣G∣ = pn , then an application of Lagrange’s Theorem reveals
that every element of G has order a power of p.
We finish this chapter showing that p-groups have nontrivial centers.
Theorem 4.20. If p is a prime and G ≠ {1} is a finite p-group. Then
Z(G) ≠ {1}.
Proof. By Corollary 4.19 we have that G has order a power of p. In
particular, p divides ∣G∣. On the other hand, ∣G∣ ≡ ∣Z(G)∣ (mod p) by Lemma
4.17. Thus p divides ∣Z(G)∣, and so Z(G) ≠ {1}.
Chapter 5
Sylow Theorems
Lagrange’s Theorem asserts that the order of each subgroup of a finite group
G is a divisor of ∣G∣, the order of G. We have seen that the converse of
Lagrange’s Theorem does not hold in general. The Sylow Theorems provide a
“weak” converse. They show that if a divisor d of ∣G∣ is a power of a prime, then
G contains a subgroup of order d. In addition, the Sylow Theorems provide
some information about the number of such subgroups and their relationship
to each other.
5.1
The Sylow Theorems
The Norwegian mathematician Peter Ludwig Sylow (1832-1918) discovered in
1872 that a “converse result” to Lagrange’s Theorem is true when d is a prime
power. More precisely, if p is a prime number such that pk divides the order
of a group G, then G must contain a subgroup of order pk . Sylow stated the
theorems in terms of permutation groups (since the abstract definition of a
group had not been given yet). Later on, in 1887, Georg Frobenius re-proved
the Sylow Theorems for abstract groups. His proofs made use of the Class
Equation. The proof presented here is a more modern proof, which uses the
theory of group actions.
Let G be a group and S(G) the collection of all subgroups of G.
Recall from the previous chapter that we can let G act on S by conjugation,
that is, if H ∈ S(G) and g ∈ G, then g acting on H yields the conjugate
subgroup gHg −1 of H. To avoid confusion with the notation of left cosets, we
79
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J. Sánchez-Ortega
will not write this action as gH. Along this chapter we will use left cosets
because they are more natural when dealing with group actions.
We called the stabilizer of H ∈ S(G) the normalizer of H in G.
N (H) = {g ∈ G ∣ gHg −1 = H}
Recall that N (H) is a subgroup of G and notice that H N (H). Actually,
the normalizer receives its name because it is indeed the largest subgroup
of G having H as a normal subgroup. Notice that N (H) = G if and only
if H G.
Lemma 5.1. Let H be a subgroup of a group G. Then g ∈ N (H) if and only
if ghg −1 ∈ H for all h ∈ H.
Proof. Let g ∈ G such that ghg −1 ∈ H for all h ∈ H. Then the inner automorphism λg (x) = gxg −1 of G restricted to H gives rise to an automorphism of H.
Thus H = λg (H) = gHg −1 and g ∈ N (H). The converse trivially holds.
Lemma 5.2. Let H be a p-subgroup of a finite group G. Then
∣N (H) ∶ H∣ ≡ ∣G ∶ H∣ (mod p)
Proof. Let L be the set of all left cosets of H in G, so that ∣L∣ = ∣G ∶ H∣.
Let H act on L by left translation, so that h ∗ (xH) = (hx)H. You can easily
check that L becomes an H-set under this action. Let us determine the fixed
subset Lf of L under the action on H. Notice that Lf consists on the left
cosets of H that are fixed under the action by all the elements of H, namely,
Lf = {xH ∈ L ∣ h ∗ (xH) = (hx)H = xH for all h ∈ H}.
Now (hx)H = xH for all h ∈ H if and only if x−1 hx = x−1 h(x−1 )−1 ∈ H for all
h ∈ H if and only if x−1 ∈ N (H) by Lemma 5.1. It shows that the left cosets
in Lf are those contained in N (H). The number of such cosets is ∣N (H) ∶ H∣,
and so ∣N (H) ∶ H∣ = ∣Lf ∣ ≡ ∣L∣ = ∣G ∶ H∣ (mod p) by Lemma 2.18.
Corollary 5.3. Let H be a p-subgroup of a finite group G. If p divides
∣G ∶ H∣, then N (H) ≠ H.
Proof. From Lemma 5.2 we have that p divides ∣N (H) ∶ H∣, which yields
that ∣N (H) ∶ H∣ ≠ 1, and so N (H) ≠ H.
5. Sylow Theorems
81
We are now ready to prove the first of the Sylow theorems, which asserts
the existence of p-subgroups of a group G for any prime power dividing ∣G∣.
First Sylow Theorem
Let G be a finite group with order ∣G∣ = pn m, where n ≥ 1 and the
prime p does not divide m. Then
(i) G contains a subgroup of order pi for each i where 1 ≤ i ≤ n.
(ii) Every subgroup H of G of order pi is a normal subgroup of a
subgroup of order pi+1 for 1 ≤ i < n.
Proof. (i) We proceed by induction on n. We know that G contains a
subgroup of order p by Cauchy’s Theorem. Assume that G contains a subgroup
H of order pi , and let us prove the existence of a subgroup of order pi+1 . From
i < n we derive that p divides ∣G ∶ H∣, and by Lemma 5.2 we have that p also
divides ∣N (H) ∶ H∣. On the other hand, we can consider the factor group
N (H)/H because H is normal in N (H). By Cauchy’s Theorem, the factor
group N (H)/H has a subgroup K of order p. By the Correspondence Theorem,
the set K = {g ∈ N (H) ∣ gH ∈ K} is a subgroup of N (H), and hence of G,
that contains H and such that K = K/H. To finish note that the order of K
is pi+1 , since p = ∣K∣ = ∣K∣/pi . The result now follows by induction.
(ii) Repeat the construction in (i) and note that H ≤ K ≤ N (H), where
∣K∣ = pi+1 . Then H is normal in K, since it is normal in N (H).
Definition 5.4. Let G be a group. A Sylow p-subgroup P of G is a maximal
p-subgroup of G, that is, a p-subgroup contained in no larger p-subgroup.
If G is a finite group with order ∣G∣ = pn m, where n ≥ 1 and the prime
p not dividing m. Then a Sylow p-subgroup of G is nothing more than a
subgroup of G of order pn . The First Sylow Theorem guarantees the existence
of Sylow p-subgroups.
For instance, if G is a group of order 100 = 22 ⋅ 52 , a Sylow 2-subgroup will
have order 4, while a Sylow 5-subgroup will have order 25.
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J. Sánchez-Ortega
Examples 5.5.
1. Consider Z12 . Since 12 = 22 ⋅ 3, Sylow 2-subgroups will have order 4, and
Sylow 3-subgroups will have order 3. More precisely, the only Sylow
2-subgroup of Z12 is ⟨3̄⟩ = {0̄, 3̄, 6̄, 9̄}, and the only Sylow 3-subgroup of
Z12 is ⟨4̄⟩ = ⟨0̄, 4̄, 8̄}.
2. Let D3 = {1, a, a2 , b, ba, ba2 }, where o(a) = 3, o(b) = 2, and aba = b. The
order of D3 is 6 = 2⋅3. The only Sylow 3-subgroup of D3 is {1, a, a2 }. But
there are three Sylow 2-subgroups, namely, {1, b}, {1, ba} and {1, ba2 }.
If P is a Sylow p-subgroup, then every conjugate gP g −1 of P is also a
Sylow p-subgroup (the order of P and any of its conjugates is the same). The
second Sylow Theorem states that every Sylow p-subgroup arises as this form,
that is, any two Sylow p-subgroups are conjugate.
Second Sylow Theorem
Let P1 and P2 be Sylow p-subgroups of a finite group G. Then P1 and
P2 are conjugate subgroups of G.
Proof. Let L be the collection of left cosets of P1 , and let P2 act on L
by y ∗ (xP1 ) = (yx)P1 for y ∈ P2 . It is straightforward to check that L is a
P2 -set. By Lemma 2.18, ∣Lf ∣ ≡ ∣L∣ = ∣G ∶ P1 ∣ (mod p). Now p does not divide
∣G ∶ P1 ∣ because P1 is a Sylow p-subgroup, and so ∣Lf ∣ ≠ 0. If xP1 ∈ Lf , then
yxP1 = xP1 for all y ∈ P2 , so x−1 yx ∈ P1 for all y ∈ P2 . Thus x−1 P2 x ⊆ P1 . Since
∣x−1 P2 x∣ = ∣P2 ∣ = ∣P1 ∣, we have that P1 = x−1 P2 x, which says that P1 and P2 are
conjugate subgroups.
Corollary 5.6. A Sylow p-subgroup P of a finite group G is normal in G
if and only if P is the unique Sylow p-subgroup of G.
Proof. Assume that P is the only Sylow p-subgroup of G, then gP g −1 = P
because gP g −1 is also a Sylow p-subgroup of G, which shows that P is normal.
Conversely, assume that P G. By the Second Sylow Theorem, any other
Sylow p-subgroup P1 of G must be a conjugate of P , that is, P1 = g1 P g1−1 = P
because P is normal. Thus P is the only Sylow p-subgroup of G.
5. Sylow Theorems
83
Example 5.7. Let D3 = {1, a, a2 , b, ba, ba2 }, where o(a) = 3, o(b) = 2, and
aba = b. The Sylow 2-subgroups of D3 are
P1 = {1, b},
P2 = {1, ba} and P3 = {1, ba2 }.
By the Second Sylow Theorem we know that P1 , P2 and P3 are conjugates.
Let us illustrate this here. The conjugacy class of b is class b = {b, ba, ba2 } and
aba−1 = ba and a2 ba−2 = ba2 . Therefore P2 = aP1 a−1 and P3 = a2 P1 a−2 .
The last of the Sylow Theorems provides us quite a lot of information
about the number of Sylow p-subgroups.
Third Sylow Theorem
Let G be a finite group and p a prime dividing ∣G∣. Then the number np
of Sylow p-subgroups of G is congruent to 1 modulo p and divides ∣G∣.
Proof. Let P be a Sylow p-subgroup of G (we know that such a P exits by
the First Sylow Theorem). Let S be the set of all Sylow p-subgroups and let
P act on S by conjugation, so that x ∈ P carries T ∈ S into xT x−1 . By Lemma
2.18, np = ∣S∣ ≡ ∣Sf ∣ (mod p). Let us find Sf . If T ∈ Sf , then xT x−1 = T for all
x ∈ P . This says that P is a subgroup of N (T ), but also T is a subgroup of
N (T ). On the other hand, P and T are both Sylow p-subgroups of N (T ),
since they are Sylow p-subgroups of G. Now by the Second Sylow Theorem, P
and T are conjugate in N (T ). But T is normal in N (T ), and so it is its only
conjugate in N (T ). Thus P = T , and we have proved that Sf = {P }. From
np = ∣S∣ ≡ ∣Sf ∣ = 1 (mod p), we see that the number of Sylow p-subgroups of
G is congruent to 1 modulo p.
To finish, let G act on S by conjugation. The Second Sylow Theorem
yields that there is only one orbit in S under G. If P ∈ S, then by the Orbit
Decomposition Theorem np = ∣S∣ = ∣class P ∣ = ∣G ∶ N (P )∣, where N (P ) is the
normalizer of P . Since ∣G ∶ N (P )∣ divides ∣G∣, we can conclude that the
number of Sylow p-subgroups of G divides ∣G∣.
Corollary 5.8. Let G be a finite group of order pn m, where the prime p
does not divide m. Then the number np of Sylow p-subgroups of G divides m.
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J. Sánchez-Ortega
Proof. From Second Sylow Theorem we know that np ≡ 1 (mod p) and
np ∣∣G∣. Since p does not divide np , we have that np ∣m.
5.2
Applications of the Sylow Theorems
In this section we derive some applications of the Sylow Theorems. We will
use the Sylow Theorems to show that a group of a particular order can not
be prime. More generally, we will learn different techniques to prove that
all groups of certain orders are abelian, or that they have nontrivial proper
subgroups, that is, they are not simple.
Application 5.9. If p and q are primes, show that no group of order
pq is simple.
Solution: Let G be a group of order pq. If p = q, then ∣G∣ = p2 and we
proved in the previous chapter that in such a case G is abelian, and either
G ≅ Zp2 or G ≅ Zp × Zp . In any case, this shows that G is not simple.
Assume now that p ≠ q. Let us say that p > q. By the First Sylow Theorem,
G has at least one Sylow p-subgroup. On the other hand, if np denotes the
number of Sylow p-subgroups of G, then by the Third Sylow Theorem we have
that np ≡ 1 (mod p) and np ∣q. It implies that np = 1 because p > q, which tells
us that there is only one Sylow p-subgroup. This unique Sylow p-subgroup is
normal by Corollary 5.6. Thus G is not simple.
Application 5.10. No group of order pr for r > 1 is simple.
Solution: Let G be a group of order pr , for r > 1. By the First Sylow
Theorem G contains a subgroup of order pr−1 , which is normal in a subgroup
of order pr , that is, the group G. We have then showed that G contains a
nontrivial normal subgroup, so G can not be simple. For example, a group of
order 16 is not simple because it has a normal subgroup of order 8.
Application 5.11. No group of order 20 is simple.
Solution: Let G be a group of order 20 = 22 ⋅ 5. By the First Sylow
Theorem G contains Sylow 5-subgroups. Moreover if n5 denotes the number
of Sylow 5-subgroups of G, then by the Third Sylow Theorem n5 ≡ 1 (mod 5)
5. Sylow Theorems
85
and n5 ∣4. Thus n5 = 1 and the unique Sylow 5-subgroup is normal in G, and
so G is not simple.
Notice that the technique used in all the previous examples has been to show
that our group G has only one Sylow p-subgroup for some prime p dividing
the order of G. This unique Sylow p-subgroups yields the nonsimplicity of G.
Let us use once more time to show that there are no simple groups of order
30.
Application 5.12. No group of order 30 is simple.
Solution: Let G be a group of order 30. Let n5 and n3 denote the number
of Sylow 5-subgroups and 3-subgroups of G, respectively. The third Sylow
Theorem yields that either n5 = 1 or 6, and n3 = 1 or 10. If G has 6 Sylow
5-subgroups, let us say, P1 , . . . , P6 . Then Pi ∩ Pj (for i ≠ j) is a subgroup of
Pi and of Pj of order dividing 5, and hence just {1}. Thus each Pi contains
4 elements of order 5 that are not in Pj for i ≠ j. This gives 24 elements of
order 5 in G. If G has 10 Sylow 3-subgroups, then proceeding as before we
obtain 10 elements of order 3 in G. Altogether we count 44 elements in G. So
necessarily either n5 = 1 or n3 = 1, which shows that G is not simple.
For the last two applications, we need another lemma.
Lemma 5.13. Let H and K be finite subgroups of a group G. Then
∣HK∣ =
∣H∣ ⋅ ∣K∣
∣H ∩ K∣
Proof. Recall that HK = {hk ∣ h ∈ H, k ∈ K}. Let ∣H∣ = r, ∣K∣ = s, and
∣H ∩ K∣ = t. Notice that HK has at most rs elements because we might
have h1 , h2 ∈ H and k1 , k2 ∈ K such that h1 k1 = h2 k2 . But if that happens,
that is, if h1 k1 = h2 k2 , then x = (h2 )−1 h1 = k2 (k1 )−1 . Now (h2 )−1 h1 ∈ H and
k2 (k1 )−1 ∈ K, so x ∈ H ∩ K, and h2 = h1 x−1 and k2 = xk1 .
On the other hand, if for y ∈ H ∩ K we let h3 = h1 y −1 ∈ H and k3 = yk1 ∈ K,
then clearly h3 k3 = h1 k1 . Thus each element u ∈ HK can be written in the form
hk, for h ∈ H and k ∈ K, as many times as there are elements of H ∩ K, i.e., t
times. It shows that the number of elements in HK is rs/t, as desired.
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J. Sánchez-Ortega
We will use Lemma 5.13 in the following way: a finite group G can not
have subgroups H and K of large order having a very small intersection. The
reason for that is that the order of HK would exceed the order of G, which
clearly makes no sense. For instance, a group of order 24 can not have two
subgroups of order 12 and 8 with an intersection of order 2.
Remark 5.14. A subgroup H of index 2 in a finite group G is always
normal. The left cosets are H itself, and the coset consisting of all elements in
G that they are not in H (remember that the set of left cosets are a partition
of G). The right cosets are exactly these two, so left and right cosets coincide
and H is normal in G.
Application 5.15. No group of order 48 is simple.
Solution: Let G be a group of order 48 = 24 ⋅ 3. By the Third Sylow
Theorem G has either one or three Sylow 2-subgroups of order 16. If there is
only one Sylow 2-subgroup, proceeding as before, we can conclude that G is
not simple.
Suppose that there are three Sylow 2-subgroups, let us say P1 , P2 and P3 .
If ∣Pi ∩ Pj ∣ ≤ 4, then by Lemma 5.13 Pi Pj would have at least 64 elements,
which is impossible because ∣G∣ = 48. Thus Pi ∩ Pj has order 8. But then
∣Pi ∶ Pi ∩ Pj ∣ = ∣Pj ∶ Pi ∩ Pj ∣ = 2, and so Pi ∩ Pj is normal in both Pi and Pj by
the Sylow First Theorem. Hence the normalizer N (Pi ∩ Pj ) of Pi ∩ Pj contains
Pi and Pj , and must have order a multiple > 16 of 16 and a divisor of 48, so
48. Thus N (Pi ∩ Pj ) = G and Pi ∩ Pj G.
Chapter 6
Finite Abelian groups
In this chapter, we provide a classification of all finite abelian groups. To ease
the notation, in what follows, we will use additive notation + for all of the
groups. Notice that even though we will use the same symbol +, the additive
operations might not be the same.
6.1
Direct Sums and Products
Let G1 , G2 , . . . , Gn be abelian groups (not necessarily, finite). Their direct
product or external direct sum G1 × G2 × . . . × Gn is the abelian group
consisting on all the n-tuples (g1 , g2 , . . . , gn ), where gi ∈ Gi for all i, with
component wise addition:
(g1 , g2 , . . . , gn ) + (g1′ , g2′ , . . . , gn′ ) = (g1 + g1′ , g2 + g2′ , . . . , gn + gn′ ).
A very easy example is the plane R2 . We introduce now a new definition
(which turns out to be equivalent) using subgroups of a given group G.
Definition 6.1. Let H and K be subgroups of an abelian group G. We say
that G is the (internal) direct sum of H and K, written as G = H1 ⊕ H2
if the two following conditions are satisfied.
(D1) H1 + H2 = G; this means that any element of g of G can be written as
g = h1 + h2 , for some h1 ∈ H1 and h2 ∈ H2 ;
(D2) H1 ∩ H2 = {0}.
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J. Sánchez-Ortega
Proposition 6.2. The following are equivalent for an abelian group G and
subgroups H1 and H2 of G.
(i) G = H1 ⊕ H2 .
(ii) For every g ∈ G, there are unique elements h1 ∈ H1 and h2 ∈ H2 such
that g = h1 + h2 .
(iii) There exist group homomorphisms πi ∶ G → Hi , ιi ∶ Hi → G, for i = 1, 2,
such that πi ιi = 1Hi , for i = 1, 2, πi ιj = 0 for i ≠ j and ι1 π1 + ι2 π2 = 1G .
Moreover, π1 and π2 are epimorphisms, called projections, and ι1 , ι2
are monomorphisms, called injections.
Proof. (i) ⇒ (ii) Suppose first that G is the direct sum of H1 and H2 . For
g ∈ G, using (D1), we know that g = h1 + h2 , for some h1 ∈ H1 and h2 ∈ H2 .
It remains to show that such an expression is unique. To do so, assume
that g can also be written as g = h′1 + h′2 , for h′1 ∈ H1 and h′2 ∈ H2 . Thus,
h1 + h2 = g = h′1 + h′2 and h1 − h′1 = h′2 − h2 is an element in H1 ∩ H2 = {0} by
(D2). From here, we obtain that h1 = h′1 and h2 = h′2 , proving (ii).
(ii) ⇒ (iii) Given g ∈ G, we know that there exists unique h1 ∈ H1 and h2 ∈ H2
such that g = h1 + h2 . We define π1 ∶ G → H1 and π2 ∶ G → H2 by πi (g) = hi ,
for i = 1, 2. Notice that these maps are well defined because of the uniqueness
in (ii). We define ι1 and ι2 in the usual way, that is, ιi (hi ) = hi , for all hi ∈ Hi ,
i = 1, 2. It is straightforward to check that ιi is a group homomorphism. We
check here that π1 , π2 are group homomorphisms. For g, g ′ ∈ G, applying (ii)
we can find unique elements h1 , h′1 ∈ H1 and h2 , h′2 ∈ H2 such that g = h1 + h2
and g ′ = h′1 +h′2 . Then g +g ′ = (h1 +h′1 )+(h2 +h′2 ) and h1 +h′1 ∈ H1 , h2 +h′2 ∈ H2
are the unique elements satisfying that condition. From here, we obtain that
π1 (g + g ′ ) = h1 + h′1 = π1 (g) + π1 (g ′ ),
π2 (g + g ′ ) = h2 + h′2 = π2 (g) + π2 (g ′ ).
I leave as an exercise for you to check that πi ιi = 1Hi , for i = 1, 2, πi ιj = 0
for i ≠ j and ι1 π1 + ι2 π2 = 1G . Lastly, notice that the surjectivity of πi and
injectivity of ιi follow from πi ιi = 1Hi , for i = 1, 2.
(iii) ⇒ (i) For g ∈ G, using the equation ι1 π1 + ι2 π2 = 1G , we obtain that
g = ι1 (π1 (g)) + ι2 (π2 (g)) ∈ H1 + H2 , since Im ιi = Hi , for i = 1, 2. It remains to
6. Finite Abelian groups
89
show that H1 ∩ H2 = {0}. Suppose that h ∈ H1 ∩ H2 . Then h = ι1 (h) and from
π1 ι1 = 1H1 we get that π1 (h) = π1 (ι1 (h)) = h. But since h ∈ H2 we get that
h = ι2 (h) and h = π1 (h) = π1 (ι2 (h)) = (π1 ι2 )(h) = 0, finishing the proof.
A consequence of this result, we can prove that internal and external sums
are equivalent notions. Therefore, there is no need to refer to internal of
external.
Corollary 6.3. Let H and K be subgroups of an abelian group G. If
G = H ⊕ K, then H ⊕ K ≅ H × K. Conversely, suppose that H and K are two
abelian groups. Then H ′ = {(h, 0) ∣ h ∈ H}, K ′ = {(0, k) ∣ k ∈ K} are subgroups
of H × K isomorphic to H and K, respectively, such that H × K = H ′ ⊕ K ′ .
Proof. Suppose first that G = H ⊕K. Then for g ∈ G = H ⊕K an application
of Proposition 6.2 (ii) allows us to write g = h + k, for unique elements h ∈ H
and k ∈ K. We consider the map φ ∶ H ⊕ K → H × K given by φ(g) = (h, k).
It is straightforward to check that φ is the required isomorphism.
Conversely, assume that H and K are abelian groups. Using the Subgroup
Test you can easily check that H ′ and K ′ are subgroups of H × K. These are
trivially isomorphic to H and K, respectively. Notice that H ′ ∩ K ′ = {(0, 0)}.
Moreover, any element x ∈ H × K is in H ′ + K ′ ; in fact: x = (h, k) = (h, 0) +
(0, k) ∈ H ′ + K ′ . Thus, H × K = H ′ ⊕ K ′ .
We can define the direct sum of a finite number of subgroups inductively:
Definition 6.4. Let n ≥ 2 and H1 , H2 , . . . , Hn be subgroups of an abelian group
G. We define the finite direct sum H1 ⊕ H2 ⊕ . . . ⊕ Hn using induction on
n. We have already defined the direct sum of two subgroups, assuming that
H1 ⊕ H2 ⊕ . . . ⊕ Hn−1 has been defined, we let
H1 ⊕ H2 ⊕ . . . ⊕ Hn ∶= (H1 ⊕ H2 ⊕ . . . ⊕ Hn−1 ) ⊕ Hn .
Proposition 6.5. Let H1 , . . . , Hn be subgroups of an abelian group G such
that G = H1 + . . . + Hn . Then the following are equivalent:
(i) G = H1 ⊕ H2 ⊕ . . . ⊕ Hn .
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J. Sánchez-Ortega
(ii) Every g ∈ G can be written in a unique way as g = h1 + h2 + . . . + hn ,
where hi ∈ Hi for all i.
(iii) Hi ∩ (H1 + . . . + Hi−1 + Hi+1 + . . . Hn ) = {0}, for all i.
Proof. (i) ⇒ (ii) We proceed by induction on n ≥ 2. For n = 2 the result
follows from Proposition 6.2. Suppose that (ii) holds for n − 1 and we prove
it for n. Consider H ∶= H1 + H2 + . . . + Hn−1 ; then G = H ⊕ Hn . Given g ∈ G,
there are unique elements h ∈ H and hn ∈ Hn such that g = h + hn . Now, using
the hypothesis induction we can find unique hi ∈ Hi , for i = 1, . . . , n − 1, such
that h = h1 + . . . + hn−1 . Thus, g = h1 + . . . + hn−1 + hn , and (ii) follows.
(ii) ⇒ (iii) Suppose that x ∈ Hi ∩ (H1 + . . . + Hi−1 + Hi+1 + . . . Hn ). Thus
x = hi = h1 + . . . + hi−1 + hi+1 + . . . + hn . But then, we have that
0 = h1 + . . . + hi−1 + (−hi ) + hi+1 + . . . + hn , and also 0 = 0 + . . . + 0.
This shows that the element 0 admits two different expressions as sum of
elements of Hi . Therefore, (i) implies that hi = 0 for all i and so x = 0.
(iii) ⇒ (i) We proceed by induction on n. For n = 2, this is just the definition of
direct sum of subgroups. Suppose that the result follows for n − 1 and we prove
it for n. From Hn ∩ (H1 + . . . + Hn−1 ) we get that G = Hn ⊕ (H1 + . . . + Hn−1 ).
Moreover, from
Hi ∩(H1 +. . .+Hi−1 +Hi+1 +. . . Hn−1 ) ⊆ Hi ∩(H1 +. . .+Hi−1 +Hi+1 +. . . Hn ) = {0},
an application of the inductive hypothesis tells us that H1 + . . . + Hn−1 =
H1 ⊕ . . . ⊕ Hn−1 , finishing the proof.
We close this section with an example using vector spaces:
Example 6.6. Let V denote an n-dimensional vector space over a field K.
Suppose that {v1 , . . . , vn } is a basis of V . Then V = ⟨v1 ⟩ ⊕ . . . ⊕ ⟨vn ⟩, where
⟨vi ⟩ = {λvi ∣ λ ∈ K} consists on the scalar multiples of vi , for i = 1, . . . , n. In
particular, we have that R3 = ⟨(1, 0, 0)⟩ ⊕ ⟨(0, 1, 0) ⊕ ⟨(0, 0, 1)⟩.
6. Finite Abelian groups
6.2
91
Fundamental Theorem of Finite Abelian
Groups
The Fundamental Theorem of Finite Abelian Groups provides a precise description of all finite abelian groups.
Fundamental Theorem of Finite Abelian Groups
Every finite abelian group G can be written as a direct product of cyclic
groups of prime-power order. Moreover the number of cyclic groups in such
decomposition and their orders are uniquely determined by G.
The first proof of the Fundamental Theorem of Finite Abelian Groups
dates back to 1858 and it is due to Leopold Kronecker. Although the proof
we have today has been refined, it is still very long and technical. Due to this
reason, we start with very powerful and useful application.
6.2.1
Constructing all abelian finite groups
Here, we will use the Fundamental Theorem of Finite Abelian Groups to
construct all finite abelian groups of a given. We begin by computing all
abelian groups of a prime power order pk , for p a prime and k ≥ 1.
Step 1 To find all the partitions of k, that is, all the different ways of
writing k as a sum of positive integers k = n1 + n2 + . . . + nk , where ni ≥ 1.
Step 2 Each of the partitions determined in Step 1 gives rise to an abelian
group Zpn1 1 × Zpn2 2 × . . . × Zpnk of order pk .
k
Given a prime p, in the table below, you can find all the possible abelian
groups of order pk , for k = 1, 2, 3. Notice that the uniqueness part in the
Fundamental Theorem of Finite Abelian Groups guarantee that different
partitions of k give rise to non-isomorphic groups. For instance, the groups
Zp2 and Zp × Zp are not isomorphic. This should not surprise us, since we
know that Zp2 is cyclic while Zp × Zp is not.
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J. Sánchez-Ortega
Value of k
k=1
k=2
k=3
k=4
Abelian groups of order pk
Order of G
Partitions of k
Possible options for G
p
1
Zp
p2
2
Zp2
1+1
Zp × Zp
3
p
3
Zp3
2+1
Zp2 × Zp
1+1+1
Zp × Zp × Zp
4
p
4
Zp4
3+1
Zp3 × Zp
2+2
Zp2 × Zp2
2+1+1
Zp2 × Zp × Zp
1+1+1+1
Zp × Zp × Zp × Zp
Next, we construct all abelian groups of order n, where n is a
positive integer, not necessarily a prime power.
Step 1 To compute the prime factorisation of n: n = pn1 1 pn2 2 . . . pnk k .
Step 2 Proceed, as before, to determine all abelian groups of orders pni i ,
for i = 1, . . . , k.
Step 3 To form all the possible direct products of the abelian groups found
in Step 2 . Notice that we will need to grab one factor of each order pni i .
Example 6.7. Any abelian group of order 16 = 24 is isomorphic to one of
the groups given below.
Z16 ,
Z8 × Z2 ,
Z4 × Z4 ,
Z4 × Z2 × Z2 ,
Z2 × Z2 × Z2 × Z2 .
6. Finite Abelian groups
93
Example 6.8. Let us determine all the abelian groups of order 1176. We
begin by factorising 1176 into primes: 1176 = 23 ⋅ 3 ⋅ 72 . Next, we find using
Steps 1, 2 all groups of orders 23 , 3 and 72 , respectively. These are:
Z23 , Z22 × Z2 , Z2 × Z2 × Z2 ,
Z3 ,
Z72 , Z7 × Z7
Thus, an abelian group of order 1176 is isomorphic to one of the following:
Z8 × Z3 × Z49 ,
Z8 × Z3 × Z7 × Z7 ,
Z4 × Z2 × Z3 × Z49 ,
Z4 × Z2 × Z3 × Z7 × Z7 ,
Z2 × Z2 × Z2 × Z3 × Z49 ,
Z2 × Z2 × Z2 × Z3 × Z7 × Z7 .
A natural question arises: Given an abelian group G of order 16, how
can we determine to which of the five groups from Example 6.7 is G
isomorphic to? The key here will be to look at the orders of the element of
G and compare it with the orders of the elements of those five groups. This
will work since two abelian groups of order n are isomorphic if and
only if they contain the same number of elements of each order.
Recall that the order of a ∈ G is the least positive integer such that na = 0
(in additive notation) or an = 1 (in multiplicative notation). For instance,
suppose that G has an element of order 8, then G is either isomorphic to Z16
or Z8 × Z2 . If, on the contrary, all the elements of G have order 2, then G is
isomorphic to Z2 × Z2 × Z2 × Z2 . Let us see a more concrete example.
Example 6.9. Let G be the (multiplicative) group consisting on all the
units, that is, the invertible elements of Z65 . Recall that an element ā ∈ Z65
has a multiplicative inverse in Z65 if and only if gcd(a, 65) = 1. With a bit of
patience, one can show that
G = {1̄, 8̄, 12, 14, 18, 21, 27, 31, 34, 38, 44, 47, 51, 53, 57, 64},
with operation multiplication modulo 65. Since we are going to be looking at
the orders of the elements of G, it makes sense to compute them all. You can
find them all in the table below.
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J. Sánchez-Ortega
Element
Order
1̄
1
8̄
4
12
4
14
2
18
4
21
4
27
4
31
4
34
4
38
4
44
4
47
4
51
2
53
4
57
4
A quick look to the table helps us to discard Z16 and Z8 × Z4 , since G
doesn’t contain an element of order 16 and 8, respectively, and to eliminate
Z2 × Z2 × Z2 × Z2 , since not every element in G has 2. Thus, we are left with
two possible groups, namely, Z4 × Z4 and Z4 × Z2 × Z2 . Notice that Z4 × Z2 × Z2
contains a subgroup isomorphic to Z2 × Z2 × Z2 , so Z4 × Z2 × Z2 has (at least)
five elements of order 2 and therefore Z4 × Z2 × Z2 can not be isomorphic to
G. These considerations allow us to conclude that G ≅ Z4 × Z4 .
To finish, let us find two subgroups H and K of G such that G = H ⊕ K ≅
Z4 × Z4 . Pick the first element from the table of order 4, namely, 8̄ and take
H ∶= ⟨8̄⟩ = {1̄, 8̄, 64, (8̄)3 = 57}. It remains to find a subgroup K of G such that
G = HK and H ∩K = {1̄}. Notice that G is a group under multiplication, so in
the definition of direct sum we have HK instead of H + K. An application of
the Second Isomorphism Theorem tells us that G/H = HK/H ≅ K/H ∩K ≅ K,
which implies that ∣K∣ = ∣G∣/∣H∣ = 16/4 = 4. Thus, we need to find an element
a ∈ G of order 4 such that ⟨a⟩ ∩ H = {1̄}. Let us try with 12, which has order
4. We have that ⟨12⟩ = {1̄, 12, 14, 38}. You can check that G = ⟨8̄⟩ ⊕ ⟨12⟩.
Remark 6.10. If n, m ≥ 2 are relatively prime, then Zn × Zm ≅ Znm . In
fact, define the map φ ∶ Znm → Zn × Zm by φ([a]nm ) = ([a]n , [a]m ). It is
straightforward to check that φ is a group homomorphism. Since both groups
are finite and of the same order, it is enough to check that φ is injective.
Suppose that φ([a]nm ) = φ([b]nm ). Then [a]n = [b]n and [a]m = [b]m , which
implies that n∣(a − b) and m∣(a − b). From here, Theorem 1.8 (1) yields that
nm∣(a − b), and so [a]nm = [b]nm .
Example 6.11. Consider the group G consisting on all the units (i.e. invertible elements) of Z135 , that is,
G = {1̄, 8̄, 17, 19, 26, 28, 37, 44, 46, 53, 62, 64, 71, 73, 82, 89, 91, 98, 107,
109, 116, 118, 127, 134}.
64
2
6. Finite Abelian groups
95
Recall that G is a group under multiplication modulo 135. Notice that
∣G∣ = 24 = 23 ⋅ 3. Using the algorithm described above we obtain that G must
be isomorphic to one of the following three groups:
Z8 × Z3 ≅ Z24 ,
Z4 × Z2 × Z3 ≅ Z12 × Z2 ,
Z2 × Z2 × Z2 × Z3 ≅ Z6 × Z2 × Z2 .
Consider 8̄ ∈ Z135 ; its order is 12, since
(8̄)2 = 64, (8̄)3 = 512 = 107, (8̄)4 = 8̄ ⋅ 107 = 856 = 46, (8̄)5 = 8̄ ⋅ 46 = 368 = 98,
(8̄)6 = 8̄ ⋅ 98 = 784 = 109, (8̄)12 = (109)2 = 11881 = 1̄.
Thus, G contains an element of order 12, which allows us to discard the group
Z6 × Z2 × Z2 , since it does not contain any elements of order 12. Moreover,
109 and 134 are elements of G of order 2, which implies that G can not
be isomorphic to Z24 , which is a cyclic group. Recall that the order of an
element k̄ coincides with the order of the subgroup ⟨k̄⟩, which divides 24. And,
conversely, each positive divisor d of 24 gives rise to a unique subgroup ⟨24/d⟩
of Z24 of order d. Thus, ⟨24/2⟩ = ⟨12⟩ is the unique subgroup of Z24 of order
2, and so Z24 has only one element of order 2, namely 12. Therefore, G is
isomorphic to Z12 × Z2 .
We know that the converse of Lagrange’s Theorem does not hold in general.
Using the Fundamental Theorem of Finite Abelian Groups we can prove that
any finite abelian group satisfies the converse of Lagrange’s Theorem.
Corollary 6.12. Let G be a finite abelian group of order n. If d is a positive
divisor of n, then G contains a subgroup of order d.
Proof. An application of the Fundamental Theorem of Finite Abelian Groups
yields that G ≅ Zpn1 1 × Zpn2 2 × . . . × Zpnk , where n = pn1 1 pn2 2 . . . pnk k . Now, any
k
positive divisor d of n is of the form d = pd11 pd22 . . . pdkk , where di ∣ni for all i.
For each i, we have Zpni is a cyclic group of order pni i and since pdi i ∣pni i we
i
can find a (cyclic) subgroup Hi of order pdi i , by the Fundamental Theorem of
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J. Sánchez-Ortega
Finite Cyclic Groups. It is straightforward to check that H1 × H2 . . . × Hk is a
subgroup of Zpn1 1 × Zpn2 2 × . . . × Zpnk of order d. This implies that G contains a
k
subgroup of order d, as desired.
Example 6.13. Suppose you are given an abelian group G of order 72 and
asked to construct a subgroup of order 12. Using the Fundamental Theorem of
Finite Abelian Groups, we know that G is isomorphic to one of the following:
Z8 × Z9 ≅ Z72 ,
Z8 × Z3 × Z3 ≅ Z24 × Z3 ,
Z4 × Z2 × Z9 ≅ Z36 × Z2 ,
Z4 × Z2 × Z3 × Z3 ≅ Z12 × Z6 ,
Z2 × Z2 × Z2 × Z9 ≅ Z18 × Z2 × Z2 ,
Z2 × Z2 × Z2 × Z3 × Z3 ≅ Z6 × Z6 × Z2 .
Notice that Z12 × {0̄} is a subgroup of Z12 × Z6 of order 12; Z6 × {0̄} × Z2 and
{0̄} × Z6 × Z2 are subgroups of Z6 × Z6 × Z2 of order 12 and ⟨6̄⟩ is a subgroup of
Z72 of order 12. It remains to find subgroups of order 12 in Z24 × Z3 , Z36 × Z2
and in Z18 × Z2 × Z2 . Notice that both Z24 and Z36 are cyclic and contain
subgroups of order 12, namely, ⟨2̄⟩, ⟨3̄⟩, respectively. Thus ⟨2̄⟩ × {0̄} and
⟨3̄⟩ × {0̄} are subgroups of order 12 of Z24 and Z36 , respectively. Lastly, to
contsruct a subgroup of order 12 of Z18 × Z2 × Z2 , we find a subgroup of order
6 of Z18 , namely, ⟨3̄⟩ and consider, for instance, ⟨3̄⟩ × Z2 × {0̄}.
6.3
Proof of the Fundamental Theorem of Finite
Abelian Groups
In this section, to make the notation in the proofs easier, we will use multiplicative notation to refer to the operation of the group. Due to the difficulty
and length of the proof of the Fundamental Theorem of Finite Abelian Groups,
we split it up in a few lemmas.
Lemma 6.14. Let G be a finite abelian group of order pn m, where p is a prime
that does not divide m. The following assertions hold:
6. Finite Abelian groups
97
(i) H = {g ∈ G ∣ g p = 1} and K = {g ∈ G ∣ g m = 1} are subgroups of G;
n
(ii) G = H ⊕ K;
(iii) ∣H∣ = pn .
Proof. (i) Let k be a positive integer and consider the map φk ∶ G → G given
by φk (g) = g k . Using that G is abelian one can show that φk is a group
homomorphism. Notice that H = ker φpn and K = ker φm , which yields that
both H and K are subgroups of G.
(ii) Since we are using multiplicative notation, we need to show that G = HK
and H ∩ K = {1}. From gcd(m, pn ) = 1, applying Bézout’s Identity we find
integers s and t such that 1 = sm + tpn . For any g ∈ G we have that
g = g 1 = g sm+tp = g sm ⋅ g tp .
n
n
On the other hand, an application of Lagrange’s Theorem tells us that the
order of the element g divides the order of G, that is, pn m. Thus:
(g sm )p = (g p m )s = 1 ⇒ g sm ∈ H,
n
n
(g tp )m = (g p m )t = 1 ⇒ g tp ∈ K,
n
n
n
which shows that g ∈ HK, and so G = HK. It remains to show that H∩K = {1}.
Suppose that a ∈ H ∩ K, then ap = 1 = am . From here, we obtain that the
n
order of a divides both m and pn , which yields a = 1, since gcd(pn , m) = 1.
(iii) Applying the Second Isomorphism Theorem we get that HK/H ≅ K/H∩K,
which implies that pn m = ∣G∣ = ∣HK∣ = ∣H∣ ⋅ ∣K∣. Thus, either p∣ ∣H∣ or p∣ ∣K∣.
If p∣ ∣K∣, then Cauchy’s Theorem tells us that K contains an element a of
order p. But since a ∈ K we have that am = 1 and so p∣m, a contradiction.
Therefore, p∣ ∣H∣ and so ∣H∣ = pn .
Corollary 6.15. Let G be a group of order ∣G∣ ∶= pn1 1 pn2 2 . . . pnk k , where the
p1 , p2 , . . . , pk are distinct primes. Then G = Hp1 ⊕ Hp2 ⊕ . . . ⊕ Hpk , where
ni
Hpi = {g ∈ G ∣ g pi = 1} is a subgroup of G of order pni i .
Proof. Applying Lemma 6.14 we can write G as G = Hp1 ⊕ K, where Hp1
is a subgroup of G of order pn1 1 and K is a subgroup of G of order pn2 2 . . . pnk k .
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J. Sánchez-Ortega
Repeat this argument with K to obtain that G = Hp1 ⊕ Hp2 ⊕ K ′ , where Hpi
is a subgroup of G of order pni i and K ′ is a subgroup of G of order pn3 3 . . . pnk k .
The result follows by repeating this process until we run out of factors.
Lemma 6.16. Let G be an abelian group of order pn , where p is a prime.
Suppose that a is an element of G of maximum order. Then G = ⟨a⟩ ⊕ K.
Proof. We proceed by (Strong) Induction on n. If n = 1, then G is a cyclic
group of order p. Taking a a generator of G, we can always write G as
G = ⟨a⟩ ⊕ {1}. Suppose that the result is true for all abelian groups of order pk ,
for k < n. Notice that any element of G has order a power of p by Corollary
4.19. We choose a ∈ G of maximum order pm . In particular, g p = 1 for all
m
g ∈ G. If G = ⟨a⟩, then we are done. Suppose that G ≠ ⟨a⟩. Then there exists
g ∈ G such that g ∉ ⟨a⟩. Consider the order of such elements (which is a
nonempty subset of N) and apply the Well-Ordering Axiom to choose b ∈ G
with minimal order such that b ∉ ⟨a⟩. We claim that ⟨a⟩ ∩ ⟨b⟩ = {1}. In fact,
let us begin by noticing that the order of bp is o(b)/p, which is, in particular,
smaller that o(b). This implies that bp ∈ ⟨a⟩, and so bp = ai for some i. From
here we obtain that
1 = bp = (bp )p
m
m−1
= (ai )p
m−1
,
which yields that o(ai ) ≤ pm−1 < pm and so ai is not a generator of ⟨a⟩. Thus
gcd(pm , i) = 1 (notice that ⟨a⟩ is a cyclic group of order pm ), which yields
that p∣i. From here we obtain that i = pj, for some j and so bp = ai = apj .
Consider the element c ∶= (a−1 )j b = a−j b ∉ ⟨a⟩ (otherwise, b ∈ ⟨a⟩, which is a
contradiction). Notice that cp = a−jp bp = b−p bp = 1, which implies that o(c) = p
and so o(b) = p, since b had minimal order. Lastly, if g ∈ ⟨a⟩ ∩ ⟨b⟩ with g ≠ 1,
then g must be a generator of ⟨b⟩ (since o(b) = p prime). In particular, we
could write b as b = g k ∈ ⟨a⟩, a contradiction. This shows that ⟨a⟩ ∩ ⟨b⟩ = {1}.
Next, let K ∶= ⟨b⟩, consider the factor group G ∶= G/K and the coset
of a in G, namely, ā = Ka. Notice that (ā)p
m
= (Ka)p
m
= Kap
m
= K. If
o(ā) < pm , then Kas = (ā)s = K for some s < pm . Then as ∈ K = ⟨b⟩, which
implies as ∈ ⟨a⟩ ∩ ⟨b⟩ = {1}. But then as = 1, which is impossible since we are
assuming that o(a) = pm . This proves that o(ā) = pm and so ā is an element of
maximum order in G. Applying the inductive hypothesis to G, we can write
6. Finite Abelian groups
99
G = ⟨ā⟩ ⊕ H, for some subgroup H of G. The Correspondence Theorem yields
that H = {x ∈ G ∣ Kx ∈ H} is a subgroup of G containing K. We will prove
now that this H is the desired subgroup of G. We first show that ⟨a⟩∩H = {1}.
To do so, take g ∈ ⟨a⟩ ∩ H. Then g is of the form g = ar for some r ≥ 1, and
so ḡ = Kg = Kar = (Ka)r ∈ ⟨Ka⟩ = ⟨ā⟩. But then ḡ ∈ H ∩ ⟨ā⟩ = K, by the
inductive hypothesis. This implies that g ∈ ⟨a⟩ ∩ ⟨b⟩ = {1}, and so g = 1. Lastly,
notice that G = ⟨a⟩H; in fact, given x ∈ G consider x̄ ∈ G = ⟨ā⟩H, and so
x̄ = (ā)t y = at y, for some t ≥ 1 and ȳ ∈ H. Notice that y ∈ H by the definition
of H. Moreover, using Coset Property 6 we can write x as x = (at y)z = at (yz)
for some z ∈ H, finishing the proof.
Corollary 6.17. Any finite abelian p-group is a direct sum of cyclic p-groups.
Proof. Let G be a finite abelian p-group, which means that the order of G
is a power of p, say ∣G∣ = pn , for some n ≥ 1. We proceed by (strong) induction
on n. If n = 1, then G is isomorphic to Zp , which is a cyclic group and the
result follows. Suppose the result true for all p-groups of order pr for r < n.
Applying Lemma 6.16 we can write G as G = ⟨a⟩ ⊕ K, where a ∈ G and K is a
subgroup of G. Notice that ⟨a⟩ is a cyclic group (by definition) and K is a
p-group of order pm , for m < n. The result follows by applying the inductive
hypothesis to K.
We are ready to prove the existence of the decomposition in the
Fundamental Theorem of Finite Abelian Groups. Suppose that G is an abelian
group of order n ∶= pn1 1 . . . pnk k , where the p1 , . . . , pk are distinct primes. From
ni
Corollary 6.15 we get that G = Hp1 ⊕ . . . ⊕ Hpk , where Hpi = {g ∈ G ∣ g pi = 1}
is a subgroup of G of order pni i . In particular, each Hpi is a pi -group and so
Corollary 6.17 tells us that each Hpi is a direct sum of cyclic pi -groups. These
considerations show that G is a direct sum of cyclic groups of prime-power
order. It remains to show the uniqueness of such decomposition. Notice
that it is enough to prove it for abelian groups of prime-power order since
the subgroups Hpi are uniquely determined by G. Our goal is to prove that
there is only one way (up to reordering of the factors) to write an abelian
group G of prime-power order pn as a direct sum of cyclic groups. We proceed
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J. Sánchez-Ortega
by (Strong) Induction on n. If n = 1, then G ≅ Zp and the uniqueness of the
decomposition trivially holds. Suppose that the result is true for all p-groups
of order pk with k < n. If G is cyclic, then G ≅ Zpn and the uniqueness of
the decomposition follows. Otherwise, applying Lemma 6.16 we get that
G = ⟨a⟩ ⊕ K = K ⊕ ⟨a⟩, where a ∈ G is an element of maximum order pi for
some i < n. Then G ≅ Zpi ⊕ K = K ⊕ Zpi , and the result follows by applying
the inductive hypothesis to K.
Example 6.18. Suppose that G is an abelian group of order 120 and that
G has exactly three elements of order 2. Determine G up to isomorphisms.
Let us begin by noticing that 120 = 23 ⋅ 3 ⋅ 5. Thus, the Fundamental
Theorem of Finite Abelian Groups tells us that G must be isomorphic to one
of the following groups:
Z120 ≅ Z8 × Z15 ,
Z4 × Z2 × Z15 ,
Z2 × Z2 × Z2 × Z15 .
The Fundamental Theorem of Finite Cyclic Groups tells us that Z15 does not
contain any elements of order 2, so the three elements of order 2 must belong
to the other factors in the decompositions. Next, Z120 is a cyclic group so they
contain only one element of order 2 by the Fundamental Theorem of Finite
Cyclic Groups. Thus, we are left with two possible groups, namely, Z4 ×Z2 ×Z15
and Z2 × Z2 × Z2 × Z15 . But there are 7 elements of order 2 in Z2 × Z2 × Z2 :
(1̄, 1̄, 1̄), (0̄, 1̄, 1̄), (1̄, 0̄, 1̄), (1̄, 1̄, 0̄), (1̄, 0̄, 0̄), (0̄, 1̄, 0̄), (0̄, 0̄, 1̄). Lastly, there are
exactly three elements of order 2 in Z4 × Z2 : (2̄, 1̄), (2̄, 0̄), (0̄, 1̄). Thus, we
can conclude that G ≅ Z4 × Z2 × Z15 .
Example 6.19. Prove that any abelian group of order 45 has an element of
order 15. Does every abelian group or order 45 have an element of order 9?
Let G be an abelian group of order 45 = 32 ⋅5. The Fundamental Theorem of
Finite Abelian Groups yields that there are (up to isomorphisms) two possible
abelian groups of order 45, namely, Z9 × Z5 ≅ Z45 and Z3 × Z3 × Z5 . The
Fundamental Theorem of Finite Cyclic Groups tells us that Z45 has exactly
one subgroup of order 15, and so it has an element of order 15. To be more
precise, 3 ∈ Z45 has order 15. An element of order 15 in Z3 × Z3 × Z5 is, for
example, (1̄, 1̄, 1̄).
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101
We know, by the Fundamental Theorem of Finite Cyclic Groups that
Z45 has exactly one elements of order 9. But the group Z3 × Z3 × Z5 does
not contain any elements of order 9. The possible orders of the elements in
Z3 × Z3 × Z5 are 3, 5 and 15.
Example 6.20. Up to isomorphisms, how many (additive) abelian groups
G of order 16 satisfy that g + g + g + g = 0 for all g ∈ G?
Let us begin by noticing that G must be isomorphic to one of the five
groups given below, by an application of the Fundamental Theorem of Finite
Abelian Groups.
Z16 ,
Z8 × Z2 ,
Z4 × Z4 ,
Z4 × Z2 × Z2 ,
Z2 × Z2 × Z2 × Z2 .
On the other hand, notice that the property 4g = g + g + g + g = 0, for all g ∈ G,
can be rephrased as “G does not contain any elements of order 8”. From the
groups above, we can see that there are three of them which do not contain
any elements of order 8, namely, Z4 × Z4 , Z4 × Z2 × Z2 , Z2 × Z2 × Z2 × Z2 .
Thus, up to isomorphisms, there are three abelian groups satisfying the given
property.