MATH 20201: ALGEBRAIC STRUCTURES I (SECTIONS 1-5)
RALPH STÖHR
1. Binary Operations
Definition. A binary operation ∗ on a non-empty set S is a rule that assigns to
each ordered pair of elements of elements of S a uniquely determined element of S.
The element assigned to the ordered pair (a, b) with a, b ∈ S is denoted by a ∗ b.
Remark. In other words, a binary operarion of a set S is a function ∗ : S×S → S
from the Cartesian product S × S to the set S. The only difference is that the value
of the function ∗ at an ordered pair (a, b) is denoted by a ∗ b rather than ∗((a, b)).
Examples. Let S = N = {1, 2, 3, . . .}.
(1) a ⋆ b = max(a, b)
e.g. 2 ⋆ 3 = 3, 3 ⋆ 2 = 3, 3 ⋆ 3 = 3.
(2) a ⋄ b = a
e.g. 2 ⋄ 3 = 2, 3 ⋄ 2 = 3, 3 ⋄ 3 = 3.
(3) azb = ab
e.g. 2z3 = 23 = 8, 3z2 = 32 = 9, 3z3 = 33 = 27.
Definition. A binary operation ∗ on a set S is commutative, if
a∗b=b∗a
∀a, b, ∈ S.
The binary operation ⋆ is commutative, but the binary operations ⋄ and z are
not commutative.
Let ∗ be a binary operation on a set S. and let a, b, c ∈ S. Consider the
expression
a ∗ b ∗ c.
This expression doesn’t have a meaning since ∗ gives only a meaning to ordered
pairs of elements. In fact, there are two ways of making a ∗ b ∗ c respectable, namely
(a ∗ b) ∗ c
and
a ∗ (b ∗ c).
For the operation ⋆ we have
(3 ⋆ 2) ⋆ 4 =
3⋆4=
4
3 ⋆ (2 ⋆ 4) =
3⋆4=
4.
In fact, for all a, b, c ∈ N we have
(a ⋆ b) ⋆ c = a ⋆ (b ⋆ c) = max(a, b, c).
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RALPH STÖHR
For the operation ⋄ we have
(3 ⋄ 2) ⋄ 4
=
3⋄4=
3
3 ⋄ (2 ⋄ 4) =
3⋄2=
3.
In fact, for all a, b, c ∈ N we have
(a ⋄ b) ⋄ c = a ⋄ (b ⋄ c) = a.
Things are different for the operation z. Here we have
(3z3)z3
= (33 )z3 =
3z(3z3) = 3z(33 ) =
(33 )3 = 39
327 .
So, in general,
(azb)zc ̸= az(bzc).
Definition. A binary operation ∗ on a set S is called associative, if
(a ∗ b) ∗ c = a ∗ (b ∗ c).
∀a, b, c ∈ S.
In our examples,
⋆ is both commutative and associative,
⋄ is not commutative, but associative,
z is neither commutative nor associative.
If ∗ is an associative operation on S, then we can write a ∗ b ∗ c for the common
value of (a ∗ b) ∗ c and a ∗ (b ∗ c):
a ∗ b ∗ c = (a ∗ b) ∗ c = a ∗ (b ∗ c).
NB. This works only for associative operations!
Our three examples ⋆, ⋄, z are of course artificially made up operations. But
there are many natural examples of binary operations.
(a) ∗ = +.
Addition of numbers is a binary operation on N, Z, Q, R, C
(b) ∗ = ×. Multiplication of numbers is a binary operation on N, Z, Q, R, C,
also on R+ = {r ∈ R : r > 0} and on {1, −1}.
(c) Addition and multiplication modulo n are binary operations on the set Zn =
{0, 1, . . . , n − 1} of residues modulo n.
(d) Matrix addition and matrix multiplication are binary operations on the set
Mn (R) of all n × n matrices with entries in R, also on Mn (Q), Mn (C), Mn (Zn ).
(e) ∗ = ◦ = composition of functions. This is a binary operation on the set
F(Ω) = {f | f : Ω → Ω}
of all functions from Ω to itself. Recall: If f : Ω → Ω and g : Ω → Ω are functions
from Ω to Ω, then f ◦ g ∈ F(Ω) is the function defined by
(f ◦ g)(x) = f (g(x)) ∀x ∈ Ω.
MATH 20201: ALGEBRAIC STRUCTURES I (SECTIONS 1-5)
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This is also a binary operation on the set Sn , the set of all permutations of the set
Ω = {1, 2, . . . , n}.
(f) Addition of vectors in a vector space is a binary operation. (NB. Scalar
multiplication is not a binary operation.)
ALL the binary operations in Examples a) - f) are associative, and all EXCEPT
matrix multiplication and composition of functions are commutative.
3 important points about binary operations:
(i) The result of the operation must be an element of S. This fails, for example,
for + on the set S = {−1, 0, 1} (as 1 + 1 = 2 ∈
/ S).
(ii) The operation must be defined for all elements of S. This fails, for example
for A ∗ B = A−1 BA on Mn (R) (as the matrix A−1 may not exist).
(iii) The result of the operation must be uniquely determined. This fails, for
example, if we set
a∗b=c
c2 = ab
where
on
C
(as for a = b = 2, c may be 2 or −2).
MULTIPLICATION TABLES
Let S = {a1 , a2 , . . . , an } be a finite set, and let ∗ be a binary operation on S.
The multiplication table of ∗ is the table
∗
a1
···
ai
···
an
a1
a1 ∗ a1
···
···
···
an ∗ a1
a2
a1 ∗ a2
···
···
···
an ∗ a2
···
···
···
···
···
···
aj
···
···
ai ∗ aj
···
···
···
···
···
···
···
···
an
a1 ∗ an
···
···
···
an ∗ an
Example: For multiplication modulo 3, the multiplication table is
×
0
1
2
0
0
0
0
1
0
1
2
2
0
2
1
Remark: Commutativity of a binary operation is instantly recognizable from
the multiplication table: ∗ is commutative if and only if its multiplication table is
symmetric with respect to the main diagonal. There is no easy way of detecting
associativity from the multiplication table.
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RALPH STÖHR
IDENTITY ELEMENTS
Definition. Let ∗ be a binary operation on a set S. An element e ∈ S is an
identity element for ∗ if
e∗a=a∗e=a
∀a ∈ S.
Examples. Recall our examples ⋆, ⋄ and z.
• a ⋆ b = max(a, b).
e = 1 is an identity element: 1 ⋆ a = a ⋆ 1 = a ∀a ∈ N.
• a ⋄ b = a.
There is no identity element. Indeed, suppose e ∈ N is an identity element.
Then we must have e ⋄ 1 = 1. But e ⋄ 1 = e. Hence e = 1. But we also
must have e ⋄ 2 = 2. However, for e = 1 we get 1 ⋄ 2 = 1 ̸= 2. Hence there
is no identity element.
• azb = ab .
No identity element. Indeed, if e was an identity element, we would have
2ze = 2, that is 2e = 2. This gives e = 1. At the same time we must have
ez2 = 2, that is e2 = 2. But for e = 1 we get 1z2 = 12 = 1 ̸= 2. Hence
there is no identity element.
Examples of identity elements
(a) ∗ = +: e = 0 (a + 0 = 0 + a = a
(b) ∗ = ×: e = 1 (a1 = 1a = a
∀a)
∀a)
(c) ∗ = + on Zn : e = 0.
∗ = × on Zn : e = 1.
(d) The identity element for addition of n × n matrices is e = On (the zero
matrix).
The identity element for matrix multiplication on Mn (R) is e = In (the identity
matrix).
(e) ∗ = ◦ on F(Ω): e = id (the identity map defined by id(x) = x for all x ∈ Ω).
Indeed, for any function ϕ : Ω → Ω we have ϕ ◦ id = id ◦ϕ = ϕ.
(f) Addition of vectors in a vector space: e = 0, the zero vector.
Fact: If there is an identity element for a binary operation, then this element is
unique.
Proof. Suppose e and f are identity elements for a binary operation ∗ on a set
S. Then e ∗ f = e (since f is an identity element). At the same time, e ∗ f = f
(since e is an identity element). Hence e = e ∗ f = f , and therefore e = f .
MATH 20201: ALGEBRAIC STRUCTURES I (SECTIONS 1-5)
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ASSOCIATIVE POWERS
For an associative binary operation ∗ on a set S, and a natural number n we
define
an = a
| ∗ a ∗{z· · · ∗ a} .
n
Example. Let ∗ = ◦, S = F(R), and let f ∈ F (R) be given by f (x) = 2x + 3 for
all x ∈ R. Then f 3 ∈ F (R) is the function given by
f 3 (x)
= (f ◦ f ◦ f )(x)
= f (f (f (x)))
=
2(2(2x + 3) + 3) + 3
=
8x + 21.
Fact. Let ∗ be an associative binary operation ∗ on a set S. Then, for all a ∈ S
and all natural numbers m and n, we have
(i)
am ∗ an = am+n
(ii)
(am )n = amn .
Proof.
am ∗ an
= |a ∗ a ∗{z· · · ∗ a} ∗ a
| ∗ a ∗{z· · · ∗ a}
m
n
= a
| ∗ a ∗{z· · · ∗ a}
m+n
= a
(am )n
m+n
.
= |a ∗ a ∗{z· · · ∗ a} ∗ · · · ∗ a
| ∗ a ∗{z· · · ∗ a}
m
m
|
{z
}
= |a ∗ a ∗{z· · · ∗ a}
n
mn
= amn .
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RALPH STÖHR
2. Groups
Definition. Let G be a non-empty set with a binary operation ∗ defined on it.
The system (G, ∗) is a group if the following three conditions are satisfied:
(G1) ∗ is associative.
(G2) There exists an element e ∈ G such that
e∗g =g∗e=g
∀g ∈ G
(i.e. G has an identity element. )
(G3) For each g ∈ G there exists an element g ′ ∈ G such that
g ∗ g ′ = g ′ ∗ g = e.
Such g ′ is called an inverse of g.
Remark. We know from §1 that the identity element e in a group is unique.
Lemma 2.1. Let (G, ∗) be a group. Then every element g ∈ G has exactly one
inverse.
Proof. Suppose that g ′ and g ′′ are inverses of g ∈ G. Then
g ′ = e ∗ g ′ = (g ′′ ∗ g) ∗ g ′ = g ′′ ∗ (g ∗ g ′ ) = g ′′ ∗ e = g ′′ .
Hence g ′ = g ′′ , so the inverse of each g ∈ G is unique.
Notation. For any g ∈ G, the unique inverse of g is denoted by g −1 .
Note:
• e−1 = e (since e ∗ e = e by definition)
• For all g ∈ G, (g −1 )−1 = g.
EXAMPLES OF GROUPS
I. Groups of numbers
Additive groups of numbers
(Z, +) is a group.
Indeed, + is a binary operation on Z, addition of numbers is associative (⇒ (G1)
is okay), 0 ∈ Z and 0 + n = n + 0 = n for all n ∈ Z (⇒ (G2) is okay), for any n ∈ Z,
we have −n ∈ Z and n + (−n) = (−n) + n = 0 (⇒ (G3) is okay).
Further examples of groups of numbers under addition:
(Q, +),(R, +),(C, +), (2Z, +), i.e. the even numbers.
Multiplicative groups
Let C∗ = C \ {0}, the set of all non-zero complex numbers.
(C∗ , ×) is a group.
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Indeed, × is a binary operation on C∗ (the product of two non-zero complex
numbers is again a non-zero complex number), multiplication of numbers is associative (⇒ (G1) is okay), 1 ∈ C∗ and 1z = z1 = z for all z ∈ C∗ (⇒ (G2) is okay),
for any z ∈ C∗ , we have 1/z ∈ C∗ and z(1/z) = (1/z)z = 1 (⇒ (G3) is okay). .
Note: The set C of all complex numbers is not a group under multiplication
because O does not have an inverse.
Further examples of groups of numbers under multiplication:
(Q∗ , ×),(R∗ , ×),
but also (R+ , ×) and ({1, −1}, ×).
Note. The set N is not a group under multiplication. There is an identity
element (1 ∈ N), but 2, 3, 4, . . . don’t have inverses.
II. The group of residues (Zn , +)
Addition modulo n is a binary operation on Zn = {0, 1, . . . , n − 1}, it is associative, 0 is the additive identity element and for all k ∈ Zn we have −k + k =
k + (−k) = 0 mod n. Hence Zn is a group under addition modulo n.
Note: Zn is not a group under multiplication modulo n since 0 doesn’t have a
multiplicative inverse modulo n.
Q: Is Zn \ {0} a group under multiplication modulo n?
A: If and only if n is a prime number!
III. The symmetric group (Sn , ◦)
Let Ω = {1, 2, . . . , n}, and let Sn denote the set of all permutations of Ω, i.e.
the set of all 1-1 and onto functions from Ω to itself. Then Sn is a group under
composition of functions. Indeed, the composite of two 1-1 and onto functions is
again 1-1 and onto (so ◦ is a binary operation on Sn ), composition of functions is
associative, the identity map,
(
)
1 2 3 ··· n
id =
1 2 3 ··· n
is the identity element, and the inverse of an arbitrary permutation
(
)
1
2
···
n
σ=
σ(1) σ(2) · · · σ(n)
is
(
σ
−1
=
σ(1)
1
σ(2) · · ·
2
···
σ(n)
n
This group is called the symmetric group of degree n.
)
.
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RALPH STÖHR
IV. Groups of Matrices
Let GL(n, R) denote the set of all invertible nxn-matrices with entries in R:
GL(n, R) = {A ∈ Mn (R) | det(A) ̸= 0}.
Then (GL(n, R), ×) is a group under matrix multiplication. Indeed, the product of
two invertible nxn-matrices is again invertible, so we have a proper binary operation
on GL(n, R), matrix multiplication is associative, , the n × n identity matrix I has
the property that IA = AI = A for all (not necessarily invertible) matrices A of
size n × n, so we have an identity element, and, finally, each A ∈ GL(n, R) has an
inverse by definition.
The group (GL(n, R), ×) is called the General Linear Group of degree n over R.
Similarly, we have the general linear groups
GL(n, Q),
GL(n, C) and GL(n, Zp )
where p is a prime.
Note: The set Mn (R) of all n × n-matrices over R is not a group under matrix
multiplication (since it contains matrices which are not invertible). However, Mn (R)
is a group under matrix addition.
ELEMENTARY PROPERTIES OF GROUPS
Lemma 2.2 (The Cancelation Lemma) Let (G, ∗) be a group, and let a, b, c ∈ G.
Then
(i)
(ii)
a∗b=a∗c
b∗a=c∗a
implies
implies
b = c,
b = c.
Proof. (i) Multiplying the equation
a∗b=a∗c
on the left by a−1 gives
a−1 ∗ a ∗ b = a−1 ∗ a ∗ c,
and since a−1 ∗ a = e (e is the identity element) this gives
e ∗ b = e ∗ c,
and hence b = c, as required. The proof of (ii) is similar.
Simplified Notation: From now on we write simply G instead of (G, ∗), and
for g, h ∈ G we write gh instead of g ∗ h. When we say ”Let G be a group”, this
will refer to the system (G, ∗), with the binary operation always understood, never
forgotten.
The order of a group G is the number of elements in the set G, and we say that
G is an infinite group if the set G is infinite.
Notation: |G| denotes the order of G.
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For example,
|Z| = ∞,
|R∗ | = ∞,
|Zn | = n,
|Sn | = n!,
|GL(n, R)| = ∞.
More interestingly, what is the order of GL(n, Zp ), p a prime? In other words,
what is the number of invertible n × n matrices with entries in the Zp ?
An easier question: What is the total number of matrices with entries in Zp ?
2
Answer: pn .
Return to the original question: What is the order of GL(n, Zp ), p a prime? In
other words, what is the number of invertible n × n matrices with entries in the
ZP ?
Recall from Linear Algebra: A matrix A ∈ Mn (Zp ) is invertible if and only if
its column vectors are linearly independent. So we need to count the number of
n × n matrices with linearly independent columns. In an invertible matrix, the first
column can be any column except the zero column. Hence the number of possible
choices for the first column is pn − 1. Once the first column is chosen, the second
column can be any vector that is not in the span of the first column. There are
p vectors in that span, and hence there are pn − p possible choices for the second
column. Once the first and second column are chosen, the third column can be any
vector that is not in the span of the two first column vectors. That span contains p2
vectors. Hence there are pn − p2 possible choices for the third column. Continuing
in this way we see that the number of choices for the (k + 1)-st column, once the
first k have been chosen, is pn − pk . Hence the total number of invertible matrices
in Mn (Zp ) is
|GL(n, Zp )| = (pn − 1)(pn − p)(pn − p2 ) · · · (pn − pn−1 ).
In particular,
|GL(2, Z2 )| = 3 × 2 = 6,
|GL(2, Z3 )| = 8 × 6 = 48,
|GL(3, Z2 )| = 7 × 6 × 4 = 168.
Definition. A group G is called Abelian (or commutative), if the binary operation of G is commutative, i.e. if gh = hg for all g, h ∈ G.
For example,
• all groups of numbers are abelian,
• Zn is abelian,
• Sn is not abelian for n ≥ 3, since
(
)
(
1 2 3
1
= (12)(23) ̸= (23)(12) =
2 3 1
3
2
1
3
2
)
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RALPH STÖHR
• GL(n, R) is not abelian for n ≥ 2, and GL(n, Zp ) is not abelian for all
n ≥ 2. For example,
(
)(
) (
)(
)
0 1
1 1
1 1
0 1
̸
=
.
1 0
0 1
0 1
1 0
Indeed,
(
LHS =
0
1
1
1
)
(
RHS =
1 1
1 0
)
.
POWERS IN A GROUP
Recall that for associative binary operations we have defined the notion of powers
for exponents that are positive integers. In group notation this is an = aa
· · · a} for
| {z
n
n > 0. Now we extend this definition to arbitrary integer exponents by setting

n

for n > 0;
 a ,
n
a =
e,
for n = 0;

 −1 |n|
(a ) , for n < 0.
With this definition the usual power rules remain valid, i.e. we have for any element
of a group G and any integers m, n
am an = am+n
(am )n = amn .
and
For example,
a5 a−3
= aaaaaa−1 a−1 a−1
−1 −1 −1
= aaaa aa
| {z } a a
=e
= aaaaa−1 a−1
= aaa |aa{z−1} a−1
=e
−1
= aa aa
| {z }
=e
= aa
= a2
Finally, for multiplication tables of groups we have an important special property.
Lemma 2.4. Let G = {a1 , a2 , . . . , an } be a finite group. Then in any row and
column of the multiplication table of G, each element of G appears exactly once.
Proof. We assume that the elements a1 , a2 , . . . , an of G are distinct as listed.
The i-th row of the multiplication table is
ai a1 , ai a2 , . . . , ai an .
(∗)
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All these elements are distinct as listed. Indeed, if ai aj = ai ak for j ̸= k, then
the Cancelation Lemma implies that aj = ak contradicting the assumption that
aj ̸= ak for j ̸= k. Hence the n elements in (∗) are distinct, so each element of G
must appear, and it can only appear once.
The proof for columns is similar.
3. Subgroups
Let G be a group and let H be a subset of G. Then, for any two elements
g, h ∈ H, the binary operation of G gives a meaning to the product gh. This will
be an element of G, which may or may not be in H. If gh ∈ H for all g, h ∈ H, we
say that H is closed under the binary operation of G.
For example, consider Z, the additive group of integers ((Z, +) in our old notation), and its subsets
H1
= {. . . , −4, −2, 0, 2, 4, 6, . . .} ⊆ Z
H2
= {1, 2, 3, 4, 5, 6, . . .} ⊆ Z
H3
= {. . . , −3, −1, 1, 3, 5, 7, . . .} ⊆ Z
i.e. the even integers, the positive integers and the odd integers. H1 and H2 are
closed under the operation +, but H3 is not closed, (e.g. 1 + 1 = 2 ∈
/ H3 ). If a
subset H is closed under the binary operation of G, then the group operation of G
gives us a binary operation on the subset H. ”The group operation of G induces a
binary operation of H.”
Q: IF H is closed under the binary operation of G, is H a group with respect to
the induced operation?
For example, the even integers H1 are a group (0 is the identity element and for
each even integer 2k, the element −2k is the (additive) inverse). But the positive
integers H2 do not form a group (e.g. because they do not contain an identity
element).
Definition. A non-empty subset H of a group G is a subgroup, if H is closed
under the binary operation of G and H is itself a group under the induced operation.
Notation: H ≤ G.
In detail: A subset H ⊆ G (H ̸= ∅) is a subgroup if
gh ∈ H. (closure)
(1) ∀g, h ∈ H
(2) ∀g, h, k ∈ H
(gh)k = g(hk).
(3) ∃e′ ∈ H : e′ h = he′ = h ∀h ∈ H.
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RALPH STÖHR
(4) ∀h ∈ H ∃h′ ∈ H :
hh′ = h′ h = e′ .
Condition 1 is the closure condition and Conditions 2-4 are the normal group
axioms.
Comments on conditions 1-4:
(a) Condition 1 is essential. Without it we do not have a binary operation on
H, and then the question of whether or not H is a group doesn’t make sense.
(b) The Condition 2 holds automatically (because it holds for all g, h, k ∈ G,
and hence, in particular, for all g, h, k ∈ H).
(c) If H has an identity element e′ , then e′ = e, i.e. the identity element of H
coincides with the identity element of G.
Proof. ee′ = e′ since e is the identity element of G. But also e′ e′ = e′ (since e′
is the identity element of H and e′ ∈ H). Hence ee′ = e′ e′ in G. Hence e = e′ by
the Cancelation Lemma, applied in G.
Therefore, Condition 3 is equivalent to the condition e ∈ H.
(d) Now Condition 4 reads as follows: ∀h ∈ H ∃h′ ∈ H : hh′ = h′ h = e, and this
is equivalent to the condition ∀h ∈ H : h−1 ∈ H.
After these comments it is not hard to prove the following Theorem which provides an easy criterion for deciding whether of not a given subset of a group is a
subgroup.
Theorem. (The Subgroup Criterion) Let G be a group. A non-empty subset
H of G is a subgroup if and only if the following two conditions hold:
(i) ∀g, h ∈ H :
(ii) ∀h ∈ H :
gh ∈ H.
h ∈ H.
−1
Proof. (⇒) Suppose H is a subgroup of G, i.e. Conditions 1-4 are satisfied.
Then (i) holds as it is the same as Condition 1, and (ii) holds as it is the same as
Condition 4.
(⇐) Suppose that conditions (i) and (ii) hold for a non-empty subset H of a
group G. We need to check that then Conditions 1-4 hold. Now,
Cond. 1 holds as it is the same as (i),
Cond. 2 holds automatically (see Comment (b)),
Cond. 3. Take an element g ∈ H (this can be done since H ̸= ∅). By (ii) we
then have g −1 ∈ H, and then we have by (i) that gg −1 ∈ H. But gg − 1 = e, so
e ∈ H, and by Comment (c) this is equivalent to Cond. 3.
Cond. 4. is equivalent to (ii) by comment (d).
Examples.
In any group G, the identity element on its own forms a subgroup: {e} ≤ G.
This subgroup is called the trivial subgroup.
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Also, in any group G, the group itself is a subgroup: G ≤ G. A subgroup H of
G is called proper if H ̸= G.
We proceed with a list of concrete examples.
(a) Q ≤ R (the sum of two rational numbers is again a rational number, and the
negative of a rational number is also rational).
N is not a subgroup of Z (although the sum of two natural numbers is natural,
N is not closed under taking inverses).
N is not a subgroup of Z (although the sum of two natural numbers is natural,
N is not closed under taking inverses).
(b) H = {z ∈ C : |z| = 1} ≤ C∗ (the product of two non-zero complex numbers
of modulus 1 is again a complex number of modulus 1, and the inverse of a complex
number of modulus 1 is also of modulus 1).
N is not a subgroup of Q∗ (although the product of two natural numbers is
natural, N is not closed under taking inverses: 2 ∈ N but 2−1 = 1/2 ∈
/ N).
(c) {0, 3, 6} ≤ Z9
From the multiplication table
+ 0 3 6
0 0 3 6
3 3 6 0
6 6 0 3
we see, that this set is closed under addition modulo 9, and that each element
has an inverse.
However, H = {0, 3, 6} is not a subgroup of Z8 : 3 + 6 = 1 mod 8, and 1 ∈
/ H.
(d) {e, (123), (132)} ≤ S3
From the multiplication table
×
e
e
e
(123) (123)
(132) (132)
(123) (132)
(123) (132)
(132)
e
e
(123)
we see, that this set is closed under composition of permutations, and that each
element has an inverse.
{(
)
}
1 n
(e) H =
: n ∈ Z ≤ GL(2, R)
0 1
Indeed, we have
so (i) is okay, and
(
1
0
n
1
(
)(
1 n
0 1
1
0
m
1
)
(
=
)−1
(
=
1 n+m
0
1
1 −n
0 1
)
,
)
,
14
RALPH STÖHR
so (ii) is okay.
Now some theory:
Lemma 3.1. Let G be a group, H ≤ G and K ≤ G. Then H ∩ K ≤ G.
Proof. We have that e ∈ H and e ∈ K, so e ∈ H ∩ K, so H ∩ K ̸= ∅, and hence
we can apply the Subgroup Criterion to prove that H ∩ K is a subgroup.
To check Condition (i), suppose that g, h ∈ H ∩ K. Then g ∈ H and h ∈ H,
and hence gh ∈ H (by Condition (i) of the Subgroup Criterion since H ≤ G).
Likewise, g, h ∈ H ∩ K means that g ∈ K and h ∈ K, and hence gh ∈ K (again
by Condition (i) of the Subgroup Criterion since K ≤ G). Consequently, gh ∈ H
and gh ∈ K, and hence gh ∈ H ∩ K. So Condition (i) of the Subgroup Criterion
holds. Now, if g ∈ H ∩ K, then g ∈ H and hence g −1 ∈ H (by Cond. (ii) of the
Subgroup Criterion). Likewise, g ∈ K and hence g −1 ∈ K (again by Cond. (ii)
of the Subgroup Criterion). Hence g −1 ∈ H and g −1 ∈ K, so g −1 is in H ∩ K,
and this means that Cond. (ii) of the Subgroup Criterion is satisfied. The Lemma
follows.
Note: In the proof we have used the Subgroup Criterion in both directions.
Remark: If H, K ≤ G, then H ∪ K is not necessarily a subgroup of G. For
example,
2Z = {2k | k ∈ Z} ≤ Z
and
3Z = {3k | k ∈ Z} ≤ Z,
but
2Z ∪ 3Z Z
since, for example, 2 + 3 = 5 ∈
/ 2Z ∪ 3Z.
CYCLIC SUBGROUPS
For a group G and an element a ∈ G, let
⟨a⟩ = {ak | k ∈ Z} = {. . . , a−2 , a−1 , a0 = e, a, a2 , a3 , . . .}.
denote the set of all powers of a in G.
Examples: (a) In R∗ we have
1 1 1
⟨−2⟩ = {(−2)k | k ∈ Z} = {. . . − , , − , 1, −2, 4, −8, 16, . . .}
8 4 2
and
⟨−1⟩ = {(−1)k | k ∈ Z} = {−1, 1}.
MATH 20201: ALGEBRAIC STRUCTURES I (SECTIONS 1-5)
15
(b) In S3 , consider ⟨(123)⟩. Here we have
(123)1
= (123)
(123)2
= (123)(123) = (132)
(123)
3
= (123)2 (123) = (132)(123) = e.
For an arbitrary integer k, write k = 3m + r with 0 ≤ r ≤ 2. Then
(123)k
= (123)3m+r
= (123)3m (123)r
= ((123)3 )m (123)r
= em (123)r
= (123)r .
Hence


 (123), if k ≡ 1
k
(123) =
(132), if k ≡ 2


e,
if k ≡ 0
mod 3;
mod 3;
mod 3.
Consequently,
⟨(123)⟩ = {e, (123), (132)}.
Lemma 3.2.. Let G be a group, a ∈ G. Then ⟨a⟩ is a
subgroup of G. Moreover, ⟨a⟩ is abelian.
Proof. We use the Subgroup Criterion. Clearly, ⟨a⟩ ̸= ∅ since a ∈ ⟨a⟩. Now, if
x, y ∈ ⟨a⟩, then x = ak and y = am for some integers k, m. But then
xy = ak am = ak+m ∈ ⟨a⟩.
Also
x−1 = (ak )−1 = a−k ∈ ⟨a⟩.
Hence ⟨a⟩ satisfies the two conditions of the Subgroup Criterion, so it is a subgroup.
Finally, ak am = ak+m = am+k = am ak , i.e. any two powers of a commute.
Hence ⟨a⟩ is abelian.
⟨a⟩ is called the cyclic subgroup generated by a in G.
THE ORDER OF AN ELEMENT
Definition. Let G be a group, a ∈ G. The smallest natural number m such
that am = e is called the order of a. If there is no such natural number, we say
that a has infinite order.
Examples:
(a) In R∗ , −2 has infinite order, −1 has order 2.
(b) In S3 , (123) has order 3.
16
RALPH STÖHR
(c) In C∗ , i (the square root of −1) has order 4: i2 = −1, i3 = −i, i4 = 1.
(d) In Z6 ,
2 has order 3: 2, 2 + 2 = 4, 2 + 2 + 2 = 0,
3 has order 2: 3, 3 + 3 = 0,
5 has order 6: 5, 10, 15, 20, 25, 30 = 0 mod 6
(e) In any group G, the identity element e has order 1, and it is the only element
of order 1.
Lemma 3.3. Let a be an element of infinite order in a group G. Then am ̸= an
for all integers m, n with m ̸= n.
Proof. Suppose am = an with m ̸= n. WLOG we may assume that m > n.
Then m − n > 0, and we have
am−n = am a−n = an a−n = an−n = a0 = e,
a contradiction.
Corollary. If a ∈ G has infinite order, then ⟨a⟩ is a subgroup of infinite order.
Proof. ⟨a⟩ consists of all the powers of a, and by the lemma they are all distinct.
Lemma 3.4. Let a be an element of finite order n in a group G. Then
(i) the elements a, a2 , . . . , an−1 , an = e are distinct;
(ii) if s ∈ Z with s = kn + r where 0 ≤ r ≤ n − 1, then as = ar ;
(iii) If s, t ∈ Z, then as = at iff s ≡ t mod n;
(iv) as = e iff n divides s.
Proof. (i) Suppose as = at for some s, t with 0 < s < t ≤ n. Then n > t − s ≥ 1,
and at−s = at a−s = as a−s = a0 = e contradicting the assumption that n is the
smallest natural number such that an = e.
(ii) We have
as = akn+r = (an )k ar = ar
since an = e.
(iii) If s ≡ t mod n, then s = kn + r and t = mn + r for some integers k, m, r
with 0 ≤ r < n. Hence, by (ii), as = at = ar . Conversely, if as = at , write
s = kn + r1 and t = mn + r2 with 0 ≤ r1 , r2 < n. Then as = ar1 and at = ar2 by
(ii), and since the elements a, a2 , . . . , an−1 , an = e are distinct by (i), this implies
r1 = r2 . Hence s ≡ t mod n.
(iv) This is a special case of (iii).
Corollary. Let a be an element of order n in a group G. Then
⟨a⟩ = {a, a2 , . . . , an−1 , an = e}
MATH 20201: ALGEBRAIC STRUCTURES I (SECTIONS 1-5)
with the elements distinct as listed. In particular, the order of ⟨a⟩ is n.
17
THE ORDER OF A PERMUTATION
First we consider a cycle of length d: τ = (i1 , i2 , . . . , id ) ∈ Sn . The order of τ
is the smallest natural number m such that τ m = id = e. By definition, τ fixes all
the elements of Ω = {1, 2, . . . , n} that do not occur in τ , whereas the entries of τ
are moved as follows:
i1 → i2 → i3 → · · · id−1 → id → i1 → i2 → i3 → · · · .
Hence, if we apply τ d times to an arbitrary entry of the cycle, this entry is moved
to itself, i.e. fixed. So τ d = id whereas τ k ̸= id for 1 ≤ k < d because τ k (i1 ) = ik+1 .
Hence:
The order of a cycle of length d is d.
Now let σ be an arbitrary permutation. Then σ can be expressed as a composite
of disjoint cycles:
σ = τ1 τ2 · · · τk
where τ1 , τ2 , . . . , τk are disjoint cycles of length d1 , d2 , . . . , dk , say. Since disjoint
cycles commute, we have
σ m = τ1m τ2m · · · τkm .
Now note that σ m = id if and only if τim = id for i = 1, 2, . . . , k. Indeed, if τim ̸= id
for some i, then the entries of the cycle τi will be moved by τim , and hence by σ m .
Consequently, the smallest natural number m such that σ m = id is the smallest
natural number m such that τim = id for i = 1, 2, . . . , k. Since the order of τi is
di (the length of the cycle τi ), τim = id iff di |m. Hence the number we are looking
for is the smallest natural m such that di |m for all i = 1, 2, . . . , k. This number is
known as the least common multiple of d1 , d2 , . . . , dk .
Result:
Suppose that σ ∈ Sn is a composite
σ = τ1 τ2 · · · τk
where τ1 , τ2 , . . . , τk are disjoint cycles of length d1 , d2 , . . . , dk , respectively. Then
the order of σ is the least common multiple of d1 , d2 , . . . , dk :
|⟨σ⟩| = l. c. m.(d1 , d2 , . . . , dk ).
18
RALPH STÖHR
Examples:
(a) We have
(
σ
=
=
1
3
2 3
4 5
4 5
2 7
6
8
7
9
8
6
9
1
)
(13579)(24)(68).
Here we have one cycle of length 5 and two cycles of length 2. Hence
|⟨σ⟩| = l. c. m.(5, 2, 2) = 10.
(b) Now let
σ = (135)(24)(56).
Here σ is a composite of one cycle of length 3 and two cycles of length 2. However,
it is not true that
|⟨σ⟩| = l. c. m.(3, 2, 2) = 6.
Our rule doesn’t apply here because the cycles in our example are not disjoint.
In order to apply our rule, we first need to rewrite σ as a composite of disjoint
cycles:
σ = (1356)(24).
Then we get
|⟨σ⟩| = l. c. m.(4, 2) = 4.
SOME SUBGROUPS OF GL(n, R)
The general linear groups have lots of subgroups. One prominent example is the
Special Linear Group:
SL(n, R) = {A ∈ GL(n, R) | det A = 1}
This is clearly a subgroup: The subset SL(n, R) is not empty (it contains, e.g., the
identity matrix). Moreover, if det A = det B = 1, then det(AB) = det A det B = 1
and detA−1 = 1/ det A = 1. So the two conditions of the subgroup criterion are
satisfied.
Some more examples of subgroups of GL(n, R):
(a) The subgroup of all scalar matrices:

λ 0 ···



 0 λ · · ·

Scal(n, R) = 

 · · ···



0 · ···
0
0
·
λ








 : λ ∈ R, λ ̸= 0





MATH 20201: ALGEBRAIC STRUCTURES I (SECTIONS 1-5)
(b) The subgroup of diagonal matrices:

λ1 0 · · · 0




 0 λ2 · · · 0
D(n, R) = 

 ·
· ···
·



0
· · · · λn
19








 : λi ∈ R, λ1 · · · λn ̸= 0





(c) The subgroup of upper triangular matrices:



λ1 ∗ · · · ∗






 0 λ · · · ∗ 



2
Tu (n, R) = 
 : λi ∈ R, λ1 · · · λn ̸= 0


 ·
· ··· ∗ 






0
· · · · λn
(b) The subgroup of upper unitriangular

1



 0

U Tu (n, R) = 

 ·



0
matrices:
∗ ···
1 ···
· ···
· ···
∗
∗
∗
1











Similarly, we have the subgroups of lower triangular and lower unitriangular matrices Tl (n, R) and U Tl (n, R).
Note:
U Tu (n, R)
≤ SL(n, R)
U Tu (n, R)
≤ Tu (n, R)
D(n, R)
≤ Tu (n, R)
Scal(n, R) and D(n, R) are abelian for all n ≥ 1. Tu (n, R) is not abelian for all
n ≥ 2, and U Tu (n, R) is not abelian for all n ≥ 3.
All these subgroups may also be defined for the general linear groups over Q,C
and Zp . Of course, over Zp these subgroups will be finite, and it is a nice exercise
to work out their orders. For example,
|D(n, Zp )| = (p − 1)n ,
|U Tu (n, Zp )| = p
n(n−1)
2
.
CENTRALIZERS
Let G be an arbitrary group, and a ∈ G. Consider the set
C(a) = {g ∈ G : ga = ag},
i.e. the set of all elements that commute with the given (fixed) element a.
Example: Let G = S4 , a = (1234). Then (13)(24) ∈ C(a), since
(1234)(13)(24) = (13)(24)(1234) = (1432),
but (123) ∈
/ C(a) since (1234)(123) ̸= (123)(1234). Indeed, we have (1234)(123) =
(1324), but (123)(1234) = (1342).
20
RALPH STÖHR
The set C(a) is called the centralizer of a (in G).
Fact: The centralizer C(a) is a subgroup of G.
Proof. Clearly, e ∈ C(a), so C(a) ̸= ∅, and we can apply the Subgroup Criterion.
Now, if g, h ∈ C(a), i.e. ga = ag and ha = ah, then gh ∈ C(a) since
(gh)a = g(ha) = g(ah) = (ga)h = (ag)h = a(gh),
i.e. gh ∈ C(a). Also, if g ∈ C(a), then g −1 ∈ C(a) since
g −1 a = g −1 ae = g −1 agg −1 = g −1 gag −1 = eag −1 = ag −1 .
Hence C(a) satisfies the two conditions of the Subgroup Criterion, so C(a) ≤ G. Remarks:
• In any group G, C(e) = G.
• In any abelian group G, C(a) = G for all a ∈ G. In fact, a group G is
abelian iff C(a) = G for all a ∈ G.
• In any group G, ⟨a⟩ ≤ C(a) for all a ∈ G (since a commutes with all its
powers).
Examples: Since groups of numbers and the groups Zn are abelian, centralizers
are not of interest in them. However, they are very interesting in symmetric groups
and matrix groups.
(a) Consider S3 = {e, (12), (13), (23), (123), (132)}. What is C((123))? This can
easily be answered be checking the six elements of this group one-by-one. Of course,
e ∈ C((123))
and
(123) ∈ C((123)).
Also, (123)(132) = (132)(123) = e, so (132) ∈ C((123)). But the three transpositions do not commute with (123): (123)(12) = (13) but (12)(123) = (23),
(123)(13) = (23) but (13)(123) = (12) and (123)(23) = (12) but (23)(123) = (13).
Hence
C((123)) = {e, (123), (132)} = ⟨(123)⟩.
(b) More generally, in Sn we have for the full cycle of length n
C((12 . . . n)) = ⟨(12 . . . n)⟩.
Proof. Let τ = (12 . . . n). Of course, ⟨τ ⟩ ⊆ C(τ ), so we need to show the inverse
inclusion C(τ ) ⊆ ⟨τ ⟩. Let σ ∈ C(τ ) with σ(1) = k (1 ≤ k ≤ n). We will show that
σ = τ k−1 . For that we need to verify that σ(m) = τ k−1 (m) for all m ∈ {1, 2, . . . , n}.
In the proof, we will use the fact that
τ k−1 (1) = k
and
τ m−1 (1) = m.
MATH 20201: ALGEBRAIC STRUCTURES I (SECTIONS 1-5)
21
Now
σ(m) =
σ(τ m−1 (1))
= (σ ◦ τ m−1 )(1)
= (τ m−1 ◦ σ)(1)
(since σ ∈ C(τ ))
= τ m−1 (σ(1))
= τ m−1 (k)
= τ m−1 (τ k−1 (1))
= (τ m−1 ◦ τ k−1 )(1)
= (τ k−1 ◦ τ m−1 )(1)
= τ k−1 (τ m−1 (1))
= τ k−1 (m).
as required. Hence σ ∈ ⟨τ ⟩, so C(τ ) ⊆ ⟨τ ⟩ and hence C(τ ) = ⟨τ ⟩.
(
)
1 1
(c) Let A =
∈ GL(2, R). What is C(A)?
1 0
(
)
a b
A 2x2 matrix B =
is in C(A) if and only if AB = BA and B is
c d
invertible.
Now: AB = BA is the same as
(
)(
) (
)(
)
1 1
a b
a b
1 1
=
1 0
c d
c d
1 0
Here
(
LHS =
Again
(
LHS =
a+c
a
b+d
b
a+c
a
b+d
b
)
(
RHS =
)
(
RHS =
Now, LHS=RHS iff
a+c = a+b
a
= c+d
b+d
= a
b
= c.
This holds iff
b=c
So AB = BA iff
and
(
B=
a = b + d.
b+d b
b
d
)
a+b
c+d
a
c
a+b
c+d
a
c
)
.
)
.
22
RALPH STÖHR
Hence
((
C
1 1
1 0
(
(d) Let D =
))
{(
=
b+d b
b
d
)
}
: b, d ∈ R, (b + d)d − b ̸= 0 .
2
)
1 0
∈ GL(2, R). What is C(D)? Again, equating
0 2
(
)(
) (
)(
)
1 0
a b
a b
1 0
=
0 2
c d
c d
0 2
we get
(
LHS =
a
2c
b
2d
)
(
RHS =
a
c
2b
2d
)
.
Here we have LHS = RHS iff b = 2b and c = 2c, i.e. iff b = c = 0. Hence
((
)) {(
)
}
1 0
a 0
C
=
: a, d ∈ R, ad ̸= 0 = D(2, R).
0 2
0 d
THE CENTRE
Let G be an arbitrary group. The centre of G is the set
Z(G) = {g ∈ G : gx = xg ∀x ∈ G}.
In other words, the centre of G is the set all those elements that commute with all
other elements in G.
Fact: The centre Z(G) is a subgroup of G.
Proof. Clearly, e ∈ Z(G), so Z(G) ̸= ∅, and we can apply the Subgroup
Criterion. Now, if g, h ∈ Z(G), i.e. gx = xg and hx = xh for all x ∈ G, then
gh ∈ Z(G) since for all x ∈ G we have
(gh)x = g(hx) = g(xh) = (gx)h = (xg)h = x(gh),
i.e. gh ∈ Z(G).
Also, if g ∈ Z(G), then g −1 ∈ Z(G) since for all x ∈ G we have
g −1 x = g −1 xe = g −1 xgg −1 = g −1 gxg −1 = exg −1 = xg −1 .
Hence Z(G) satisfies the two conditions of the Subgroup Criterion, so Z(G) ≤
G.
Some obvious facts about the centre:
• In any abelian group G, Z(G) = G.
In fact, a group G is abelian if and only if Z(G) = G.
• For all a ∈ G, Z(G) ≤ C(a).
In fact,
∩
Z(G) =
C(a).
a∈G
MATH 20201: ALGEBRAIC STRUCTURES I (SECTIONS 1-5)
23
Examples: Again, since groups of numbers and the groups Zn are abelian, they
just coincide with their centres. However, things are different for symmetric groups
and matrix groups.
(a) Consider S3 = {e, (12), (13), (23), (123), (132)}. Here we have
(12)(13) ̸=
(13)(12)
⇒
(12), (13) ∈
/ Z(S3 )
(23)(123) ̸=
(123)(23)
⇒
(23), (123) ∈
/ Z(S3 )
(23)(132) ̸=
(132)(23)
⇒
(132) ∈
/ Z(S3 ).
Hence,
Z(S3 ) = {e}.
(b) More generally,
Z(Sn ) = {e}
∀n ≥ 3.
Proof. We need to show that no element of Sn except e is in the centre. In other
words, we need to verify that that for any σ ∈ Sn with σ ̸= e there exists a
permutation τ ∈ Sn such that στ ̸= τ σ. Let σ ̸= e, then σ moves at least one
element of Ω = {1, 2, . . . , n}, so ∃a, b ∈ Ω such that a ̸= b and σ(a) = b. Since
n ≥ 3, ∃c ∈ Ω such that c ̸= a and c ̸= b. Now put τ = (ac). Then σ ◦ τ ̸= τ ◦ σ.
Indeed,
(σ ◦ τ )(a) = σ(τ (a)) = σ(c),
(τ ◦ σ)(a) = τ (σ(a)) = τ (b) = b.
So if σ ◦ τ = τ ◦ σ, then σ(c) = b. But this is impossible since σ(a) = b, a ̸= c, and
σ is 1-1. Hence σ ◦ τ ̸= τ ◦ σ and so σ ∈
/ Z(Sn ).
(c) For the general linear group we have
Z(Gl(2, R)) = Scal(2, R).
Clearly, the scalar matrices are contained in the centre, since multiplying a matrix
by a scalar matrix (no matter if on the right or on the left) amount to multiplying
all entries of the matrix by the scalar in question:
(
)(
) (
) (
)(
)
λ 0
a b
λa λb
a b
λ 0
=
=
0 λ
c d
λc λd
c d
0 λ
(
)
a b
Now suppose A =
∈ Z(Gl(2, R)). Then
c d
(
)(
) (
)(
)
a b
1 1
1 1
a b
=
.
c d
0 1
0 1
c d
In this matrix equation we have
(
)
a a+b
LHS =
c c+d
(
RHS =
a+c b+d
c
d
)
Hence equality holds if and only if
a=a+c & a+b=b+d & c+d=d
⇒
c = 0 & a = d.
24
RALPH STÖHR
)
a b
. But to be a
So if a matrix is in the centre, it must be of the form
0 a
central matrix, it must commute with all matrices in Gl(2, R). In particular, we
must have
)(
) (
)(
)
(
1 0
1 0
a b
a b
=
.
0 a
1 1
1 1
0 a
(
(
Here
LHS =
a+b b
a
a
)
(
RHS =
a
b
a a+b
)
.
So we have equality if and only if
a+b=a
(
⇒
)
b = 0.
a 0
, i.e. it is a scalar matrix. Of
0 a
course we must have a ̸= 0 for the matrix to be invertible. So we have proved that
Z(Gl(2, R)) = Scal(2, R).
Alternatively, we could have used our earlier examples of centralizers to show
that Z(Gl(2, R)) = Scal(2, R). Recall that
((
)) {(
)
}
1 0
a 0
C
=
: a, d ∈ R, ad ̸= 0 = D(2, R).
0 2
0 d
Hence a central matrix is of the form
and that
C
((
1 1
1 0
))
{(
=
b+d b
b
d
)
}
: b, d ∈ R, (b + d)d − b ̸= 0 .
2
Any matrix in the centre must belong to both of these centralizers, i.e. it must
be in the intersection of the two centralizers, which is the group of scalar matrices
Scal(2, R).
Our first proof, however, can nicely be modified to establish a more general
result, namely that Z(Gl(n, R)) = Scal(n, R).
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4. Cyclic Groups
Definition. A group G is called cyclic, if there exists an element a ∈ G such
that G = ⟨a⟩.
In other words, G is cyclic if all its elements are powers of a single element. If
G = ⟨a⟩, then a is called a generator for G.
Examples: (a) Z is an infinite cyclic group:
Z = ⟨1⟩ = ⟨−1⟩
(b) For any natural n,
Zn = ⟨1⟩
is a cyclic group of order n. In particular, for any natural an there exists a finite
group of order n.
Theorem 4.1. Any subgroup of a cyclic group is itself cyclic.
Proof. Let G = ⟨a⟩ be a cyclic group with generator a, and let H be a subgroup
of G. If H = {e}, then H = ⟨e⟩ is cyclic with generator e. Now suppose H ̸= {e}.
Then H contains some non-zero power of a, say ak . Being a subgroup, H also
contains the inverse of ak , that is a−k . Since one of k and −k is positive, we may
conclude that H contains a positive power of a.
Now let m be the smallest positive integer such that am ∈ H. We will show that
H = ⟨am ⟩.
Clearly, ⟨am ⟩ ⊆ H, so we need to show that H ⊆ ⟨am ⟩. Let h be an arbitrary
element of H. Then h = as for some s ∈ Z. Write s = mt + r with 0 ≤ r < m.
Then we have
h = as = atm+r = (am )t ar
It follows that
ar = (am )−t h.
Consider the RHS: Here am ∈ H, and hence (am )−t ∈ H, and also h ∈ H. Since
H is a subgroup, it is closed under products, and so (am )−t h ∈ H. But this means
ar ∈ H. Now recall that r < m, and m was the smallest positive integer such
that am ∈ H. Hence r = 0. But then h = (am )t , i.e. h ∈ ⟨am ⟩. Consequently,
H ⊆ ⟨am ⟩, as required.
.
Suppose the G = ⟨a⟩ is a finite cyclic group of order n, so the order of the
generator a is n. By our Theorem, any subgroup H of G is itself cyclic, so it is of
the form H = ⟨am ⟩ for some integer m. In fact, we can be more precise.
Lemma 4.2. Let G = ⟨a⟩ be a finite cyclic group of order n. Then, for any
integer m, ⟨am ⟩ = ⟨ad ⟩ where d = (m, n) is the greatest common divisor of n and
m.
26
RALPH STÖHR
Proof. Since d divides m, am is a power of ad , and we clearly have ⟨am ⟩ ⊆ ⟨ad ⟩.
It remains to prove the inverse inclusion. By the Euclidean algorithm there exist
integers s and t such that d = sm + tn. But then
ad = asm+tn = (am )s (an )t = (am )s (e)t = (am )s ,
so ad is a power of am . Consequently, ⟨ad ⟩ ⊆ ⟨am ⟩, and the result follows.
An immediate consequence is the following
Corollary 1. The order of am is equal to the order of ad , namely n/d.
Proof. Since ⟨am ⟩ = ⟨ad ⟩, the respective generators must be of the same order.
Since d is a divisor of n, the order of ad is obviously n/d.
Corollary 2. The element am ∈ G is a generator for G iff (m, n) = 1.
Proof. The subgroup ⟨am ⟩ has order n/d, so it coincides with G iff d = 1.
Another consequence of the Lemma is the following
Theorem 4.3. The subgroups of a finite cyclic group of order n are in one-to-one
correspondence with the divisors of n.
Proof. Let a be a generator for G. Then, by Theorem 4.1 and Lemma 4.2, every
subgroup of G is of the form ⟨ad ⟩ where d is a divisor of n. Also, if d1 and d2 are
distint divisors of n, then ⟨ad1 ⟩ ̸= ⟨ad2 ⟩ (as they have different orders, namely n/d1
and n/d2 , respectively).
Remark: In view of Corollary 1, the order of a subgroup of a finite cyclic group
G divides the order of G. This is a special case of a very important Theorem,
Lagrange’s Theorem, which states that if G is any finite group and H is a subgroup
of G, then the order of H divides the order of G.
5. Cosets and Lagrange’s Theorem
Let G be a group and let H be a subgroup of G. For a (fixed) element g ∈ G,
set
gH = {gh : h ∈ H},
i.e. gH is the set of all products gh where g is a fixed element of G and h runs over
all elements of the subgroup H. So if H = {h1 , h2 , h3 , . . .}, then
gH = {gh1 , gh2 , gh3 , . . .}.
Definition. Let G be a group, H ≤ G. A set of the form gH with g ∈ G is
called a left coset of H in G.
MATH 20201: ALGEBRAIC STRUCTURES I (SECTIONS 1-5)
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Examples:
(a)
Z ≥ ⟨3⟩ = {. . . , −6, −3, 0, 3, 6, 9, . . .}.
For a ∈ Z, the left coset a + ⟨3⟩ is
a + ⟨3⟩ = {. . . , a − 6, a − 3, a, a + 3, a + 6, a + 9, . . .}.
So we have 3 distinct cosets of ⟨3⟩ in Z:


 {. . . , −6, −3, 0, 3, 6, 9, . . .}, if a ≡ 0 mod 3;
a + ⟨3⟩ =
{. . . , −5, −2, 1, 4, 7, 10, . . .}, if a ≡ 1 mod 3;


{. . . , −4, −1, 2, 5, 8, 11, . . .}, if a ≡ 2 mod 3.
(2)
Z15 ≥ ⟨5⟩ = {0, 5, 10}.
For k ∈ Z15 , the left coset k + ⟨5⟩ is
k + ⟨5⟩ = {k, k + 5, k + 10}.
So we have 5 distinct cosets of ⟨5⟩ in Z15 :

{0, 5, 10},



 {1, 6, 11},
k + ⟨5⟩ =

···



{4, 9, 14},
(3)
if k ≡ 0 mod 5;
if k ≡ 1 mod 5;
···
if k ≡ 4 mod 5.
S3 ≥ ⟨(12)⟩ = {e, (12)}
Here we get
e⟨(12)⟩
= {ee, e(12)} = {e, (12)}
(12)⟨(12)⟩
= {(12)e, (12)(12)} = {(12), e}
(13)⟨(12)⟩
= {(13)e, (13)(12)} = {(13), (123)}
(23)⟨(12)⟩
= {(23)e, (23)(12)} = {(23), (132)}
(123)⟨(12)⟩
= {(123)e, (123)(12)} = {(123), (13)}
(132)⟨(12)⟩
= {(132)e, (132)(12)} = {(132), (23)}
Hence we have three distinct cosets of ⟨(12)⟩ in S3 :
e⟨(12)⟩ = (12)⟨(12)⟩ = {e, (12)}
(13)⟨(12)⟩ =
(123)⟨(12)⟩ = {(13), (123)}
(23)⟨(12)⟩ =
(132)⟨(12)⟩ = {(23), (132)}
Ex.: Work out the cosets of ⟨(123)⟩ in S3 .
Remarks: (a)The subgroup H itself is always one of the cosets of H in G (since
eH = H, obviously). In fact, for all h ∈ H, we have hH = H. Indeed, of course we
have hH ⊆ H, but we have also the inverse inclusion H ⊆ hH because, if k ∈ H,
then k = h(h−1 k) ∈ hH.
28
RALPH STÖHR
(b) For all g ∈ G, g ∈ gH (because g = ge ∈ gH).
Now we are going to deduce Lagrange’s Theorem, a fundamental result at the
very heart of Group Theory and arguably the most important Theorem in this
course. We need two Lemmas, and then the theorem will follow easily.
Lemma 5.1. Let H ≤ G, and x, y ∈ G. Then
either
xH = yH
xH ∩ yH = ∅.
or
Proof. Assume that xH ∩ yH ̸= ∅. We need to show that then xH = yH. Since
xH ∩ yH ̸= ∅, there exists an element z ∈ xH ∩ yH. Then
z = xh1 for some h1 ∈ H (as z ∈ xH) and
z = yh2 for some h2 ∈ H (as z ∈ yH).
Therefore,
xh1 = yh2
x = yh2 h−1
1 .
and hence
Now let a ∈ xH. Then a = xh3 for some h3 ∈ H. But then
a = xh3 = yh2 h−1
1 h3 ∈ yH
(since h2 h−1
1 h3 ∈ H.).
Consequently, xH ⊆ yH. Likewise, yH ⊆ xH, so xH = yH.
Corollary. If H ≤ G, then G is the disjoint union of the distinct left cosets of
H in G.
Proof. Any element of G belongs to some left coset of H in G (for example,
g ∈ gH), so G is the union of those left cosets. Then the Lemma tells us that
distinct cosets are disjoint.
Definition. The number of (distinct) left cosets of H in G is called the index
of H in G, and is denoted by [G : H].
In our examples we have seen that
[Z : ⟨3⟩] = 3,
[Z15 : ⟨5⟩] = 5,
[S3 : ⟨(12)⟩] = 3.
The examples also illustrate what we know from the Corollary, namely, that any
group G is the disjoint union of the left cosets of H in G.
Lemma 5.2. Let H be a finite subgroup of a group G. Then |gH| = |H| for all
g ∈ G.
Proof. Let
H = {h1 , h2 , . . . , hn }
and assume that the hi are distinct as listed. Then
gH = {gh1 , gh2 , . . . , ghn }.
MATH 20201: ALGEBRAIC STRUCTURES I (SECTIONS 1-5)
29
But these elements too are distinct as listed. Indeed, if ghi = ghj , then hi = hj by
the Cancelation Lemma. Hence gH has exactly n elements, so |gH| = |H|.
LAGRANGE’S THEOREM. Let G be a finite group and H ≤ G. Then
|G| = [G : H]|H|
where [G : H] is the index of H in G.
Proof. Let [G : H] = r and let
g1 H, g2 H, . . . , gr H
be the distinct left cosets of H in G. By the Corollary to Lemma 5.1, G is the
disjoint union of those cosets:
G = g1 H ∪ g2 H ∪ · · · ∪ gr H
Since the union is disjoint we have
|G| = |g1 H| + |g2 H| + · · · + |gr H|.
By Lemma 5.2, |gi H| = |H| for i = 1, 2, . . . , r. Hence
|G| = |H| + |H| + · · · + |H| = r|H| = [G : H]|H|.
|
{z
}
r
So |G| = [G : H]|H| as required.
Corollary 1. If H ≤ G, then the order of the subgroup H divides the order of
the group G.
Corollary 2. If g ∈ G, then the order of the element g divides |G|, the order of
the group G.
Proof. This is because the order of the element g is equal to the order of the
cyclic subgroup ⟨g⟩.
Corollary 3. If |G| = p, where p is a prime, then G is cyclic.
Proof. Let g ∈ G with g ̸= e. By Lagrange’s Theorem, |⟨g⟩| divides |G|. Since
|G| = p, a prime, this gives that |⟨g⟩| = 1 or |⟨g⟩| = p. But the former is impossible
since ⟨g⟩ contains at least two elements, namely g and e. Hence |⟨g⟩| = p = |G| and
therefore ⟨g⟩ = G.