Contents
MOTIVATION
Permutation
Algebra 1
Sophie, MARQUES
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Contents
MOTIVATION
Permutation
Contents
Set theory
Integers
Group theory
Transformation groups
Permutation Groups
Product Structure in Groups
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MOTIVATION
Permutation
The set of the integers
The Group of the integers (Z, +)
The set of integer
Z = {. . . , −2, −1, 0, 1, 2, 3, . . .}
is embedded with an addition operation that is a map
+:
Z×Z
(n, m)
→
7
→
Z
m+n
that satisfies :
G1 (x + y ) + z = x + (y + z) (addition is associative)
G2 There exists an element in Z which is in fact 0 such that 0 + x = x = x + 0 for all x ∈ Z.
This is the "zero element" of the system.
G3 Every element x ∈ Z has an " additive inverse", denoted −x which has the property that
x + (−x) = 0 = (−x) + x
A set satisfying these three properties is called a group. Thus (Z, +) is a group.
Moreover, we have that + satisfies the commutativity property GC :
x + y = y + x, for any x, y ∈ Z
A set satisfying these four properties is called a commutative group. Thus (Z, +) is a
commutative group.
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MOTIVATION
Permutation
The set of the integers
Exercises
1
2
From only G 2, one can prove that the zero element is unique.
From G 2 and G 3, one can prove that the inverse is unique.
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MOTIVATION
Permutation
The set of the integers
The multiplication on Z
The set of integer Z is embedded with a multiplication operation that
is a map
·: Z×Z →
Z
(n, m) 7→ m · n
Is (Z, ·) a group ? NO. More precisely, G 1 and G 2 hold but not G 3.
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MOTIVATION
Permutation
The set of the integers
Exercise
Is there a maximal subset S of Z such that (S, ·) is a group ?
YES.
One can prove ({±1}, ·) is this maximal subset and thus a group.
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MOTIVATION
Permutation
The set of the integers
Units
The elements such that u ∈ Z and there exist v ∈ Z, such that
uv = vu = 1 are called the units of Z.
They form a group, called the group of units of Z..
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MOTIVATION
Permutation
The Ring of the integers (Z, +, ·)
Definition
R1 (x · y ) · z = x · (y · z) (multiplication is associative) ;
R2 x · (y + z) = x · y + x · z and (x + y ) · z = x · z + y · z (distributive law)
The properties G 1, G 2, G 3, R1 and R2 are characteristic of what we will
call ring. So that, (Z, +, ·) is a ring.
Moreover, the property RC
x · y = y · x, for any x, y ∈ Z
(the multiplication is commutative) together with GC , make Z into what
we will call a commutative ring.
In addition, there exists in Z a "multiplicative identity element", 1
which has the characteristic property that
1 · x = x = x · 1, for all x ∈ Z, for all x ∈ R.
This property RU is characteristic of unitary commutative ring.
Exercise : Prove that the unit element is unique.
Finally, obviously 1 6= 0. This only say that Z is not the trivial ring : the
ring which has 0 as its only element.
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MOTIVATION
Permutation
The Ring of the integers (Z, +, ·)
Definition
R1 (x · y ) · z = x · (y · z) (multiplication is associative) ;
R2 x · (y + z) = x · y + x · z and (x + y ) · z = x · z + y · z (distributive law)
The properties G 1, G 2, G 3, R1 and R2 are characteristic of what we will
call ring. So that, (Z, +, ·) is a ring.
Moreover, the property RC
x · y = y · x, for any x, y ∈ Z
(the multiplication is commutative) together with GC , make Z into what
we will call a commutative ring.
In addition, there exists in Z a "multiplicative identity element", 1
which has the characteristic property that
1 · x = x = x · 1, for all x ∈ Z, for all x ∈ R.
This property RU is characteristic of unitary commutative ring.
Exercise : Prove that the unit element is unique.
Finally, obviously 1 6= 0. This only say that Z is not the trivial ring : the
ring which has 0 as its only element.
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MOTIVATION
Permutation
The Ordered ring of the integers (Z, +, ·, >)
Definition
If we take P = N to be the subset of Z of the positives elements of Z
(x ∈ P if x > 0), then :
O1 Z is a disjoint union Z = −P ∪ {0} ∪ P of the sets P , where
−P = {−x : x ∈ P}, that is for x ∈ Z either x > 0 or x = 0 or x < 0.
O2 P + P = {x + y : x, y ∈ P} ⊂ P, that is x > 0 and
y > 0 ⇒ x + y > 0;
O3 P · P = {x · y : x, y ∈ P} ⊆ P, that is x > 0 and y > 0 ⇒ xy > 0 ;
A ring with a subset P with the properties O1, O2 and O3 is called an
ordered ring.
We refer to P as positive elements of the ring and −P as negative
element of the ring.
As a consequence, (Z, +, ., >) is an ordered ring.
Note that x < 0 ⇔ −x > 0 ( x ∈ −P ⇔ −x ∈ P) and
x < y ⇔ x − y < 0.
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MOTIVATION
Permutation
The Ordered ring of the integers (Z, +, ·, >)
Exercise
Prove that an ordered ring has the following properties :
1
1 > 0;
2
x > y ⇒ x + c > y + c, for all c ∈ Z.
3
x > y and y > z ⇒ x > z. (transitivity of the order relation)
4
x 6= 0 ⇒ x 2 > 0 ;
5
c > 0 and x > y ⇒ xc > yc ; c < 0 and x > y ⇒ xc < yc (reverses inequality).
6
7
x > 0 and y > 0 ⇒ xy > 0 i.e. (+) · (+) = (+) ;
x > 0 and y < 0 ⇒ xy < 0 i.e. (+) · (−) = (−) ;
x < 0 and y < 0 ⇒ xy > 0 i.e. (−) · (−) = (+) ;
x > y ⇔ −x < −y .
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Contents
MOTIVATION
Permutation
The Ordered ring of the integers (Z, +, ·, >)
Proposition
A ordered ring satisfies the following properties :
1 No zero divisors : If xy = 0, then either x = 0 or y = 0 (or both).
2 Cancellation law : If a 6= 0, ax = ay , then x = y .
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MOTIVATION
Permutation
Induction
Definition
1
2
A subset S ⊆ P in the set of the positive elements in a commutative
ordered ring R is an inductive set if
1 The identity element 1 lies in S : 1 ∈ S
2 The successor s + 1 of any element in S is also in S :
s ∈ S ⇒ s + 1 ∈ S.
A subset S ⊆ P in the set of the positive elements in a commutative
ordered ring R is an strong inductive set if
1 The identity element 1 lies in S : 1 ∈ S
2 1, . . . , s ∈ S ⇒ s + 1 ∈ S.
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MOTIVATION
Permutation
The integers as the only inductive system
Definition
We say that an ordered commutative ring R with set of positive integer
P verify the (strong) induction axiom if any (strong) inductive subset
of R is equal to the full set P of positive elements in the ring R.
It can be proved that there exists an only ordered ring with this property.
This system is the system of integers Z.
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MOTIVATION
Permutation
Induction.
Theorem (Proof by induction)
Suppose an assertion P(n) has been assigned to each counting number
n ∈ N, and each statement is either true or false. Suppose we can show
that
1 Statement P(1) is true.
2 If statement P(n) is assumed to be true, we can then prove P(n + 1)
true based on this information. (In symbolic logic shorthand :
P(n) ⇒ P(n + 1) for all n ∈ N.)
Conclusion : the statement P(n) must be true for all n ∈ N.
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MOTIVATION
Permutation
Induction.
Exercise
Show that for any n ∈ N
1 + 2 + · · · + n = n(n + 1)/2, ∀n ∈ N
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MOTIVATION
Permutation
Minimum property.
Theorem (The minimum property)
If S ⊆ N is non-empty, then there exists a unique minimum element in S.
That is an element s0 = min{S} in S such that s0 ≤ s for all s ∈ S.
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Contents
MOTIVATION
Permutation
Minimum property and induction.
Theorem
A commutative unitary ordered ring satisfies the induction property if and
only if it satisfies the minimum property.
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MOTIVATION
Permutation
Divisibility
Definition
If a 6= 0 and b are two integers, we say that b divides a or a is a multiple
of b. , often written as b|a, if there exists a m ∈ Z such that a = mb.
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MOTIVATION
Permutation
Divisibility
Exercises
1
2
3
4
5
a|0 for any a ∈ Z.
0 does not divide any non-zero element in Z.
1|a and −1|a for all a ∈ Z (Thus ±1 are "trivial divisors" of every
a ∈ Z)
Divisibility is transitive : If a, b, c 6= 0 then a|b and b|c ⇒ a|c.
If a, b 6= 0 then a|b and b|a ⇒ b = a or b = −a.
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MOTIVATION
Permutation
The Euclidean ring of the integers
Theorem (Euclidean Division Algorithm)
Let a, b ∈ Z with b 6= 0. Then there exist q, r such that
a = qb + r
with q ∈ Z and 0 ≤ |r | < |b|.
We can always arrange that r ≥ 0, and then the pair (q, r ) is unique.
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MOTIVATION
Permutation
The GCD ring of the integers
Definition (Greatest common divisor)
A greatest common divisor of nonzero elements a, b ∈ Z, denoted by
(a, b) or gcd(a, b) is an element c ∈ Z such that
1 c|a and c|b ;
2 c > 0;
3 If c 0 is any other element of Z satisfying c 0 |a and c 0 |b then c 0 divides
c.
A greatest common divisor is denoted by gcd(a, b).
Lemma
For any pair of nonzero integers there exists a unique greatest common
divisor gcd(a, b), and Za + Zb = gcd(a, b)Z.
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MOTIVATION
Permutation
The GCD ring of the integers
Lemma
For any pair of nonzero integers there exists a unique greatest common
divisor gcd(a, b), and Za + Zb = gcd(a, b)Z.
Main step of the proof
Prove uniqueness. (Suppose there are two and prove they are equal.)
Prove existence.
Prove that (Za + Zb) ∩ N 6= ∅ and deduce it contains a minimal element.
Call it c, we will prove that c = gcd(a, b). Clearly c · Z ⊆ Za + Zb.
Observe that c > 0 by definition.
Prove that c|a and c|b. For this, prove more generally that c|ma + nb,
for any m, n ∈ Z (Do the Euclidean division of ma + nb by c and
deduce that the reminder is in Za + Zb, finally conclude using the
minimality of c). Here, we have proven that Za + Zb ⊆ c · Z and thus
that Za + Zb = gcd(a, b)Z.
Prove that if c 0 is any other element of Z satisfying c 0 |a and c 0 |b then
c 0 divides c.
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Contents
MOTIVATION
Permutation
The GCD ring of the integers
Exercise
Let a, b be nonzero integers. Prove that their greatest common divisor
has the following properties
1 gcd(a, b) = gcd(b, a) ;
2 gcd(a + kb, b) = gcd(a, b), for any k ∈ Z
3 If a, b 6= 0 and a divides b, explain why gcd(a, b) = a.
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MOTIVATION
Permutation
The GCD ring of the integers
Theorem (GCD Algorithm)
Let a and b be positive integers, a > b. Then we apply a series of divisions as follows.
a
b
r1
rn−2
rn−1
=
=
=
.
.
.
=
=
bq0 + r1
r1 q1 + r2
r2 q2 + r3
0 < r1 < b,
0 < r2 < r1 ,
0 < r3 < r2 ,
rn−1 qn−1 + rn
rn qn .
0 < rn < rn−1 ,
The process of division comes to an end when rn+1 = 0. The integer rn is the G. C. D.
of a and b.
Theorem (Extended GCD Algorithm)
Let a and b be integers with gcd(a, b) = c. There exist integers u and v such that
au + bv = c.
Such u, v can be obtained by backward tracing of the Euclidean divisions in finding
the G. C. D.
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MOTIVATION
Permutation
The GCD ring of the integers
Bezout lemma
Let a and b be integers non zero. Then, gcd(a, b) = 1 if and only if there
exist integers u and v such that
au + bv = 1.
In this case, we say that a and b are relatively prime.
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MOTIVATION
Permutation
The GCD ring of the integers
Exercise
1
2
Use the GCD Algorithm to compute the greatest common divisor of
48 and 347 using GCD algorithm.
Find r , s ∈ Z, such that
gcd(48, 347) = r × 347 + s × 48
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MOTIVATION
Permutation
The UFD ring of integers
Definition
A prime is an element p in Z such that
1 p is not a unit, and
2 p does not have a proper factorization p = ab (both factors
non-units).
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MOTIVATION
Permutation
The UFD ring of integers
Gauss lemma
Let p > 1 be a prime
1 If n is a nonzero integer, then either p|n or gcd(p, n) = 1
2 p divides ab if and only if p divides a or p divides b.
More generally, if a prime p > 1 divides a product b1 . . . bk of nonzero
integers, then there exists an index i such that p|bi .
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MOTIVATION
Permutation
The UFD ring of integers
Theorem (Existence and uniqueness of the factorization)
Any integer n > 1 has a unique factorization as a product of nontrivial
positive primes p > 1. That means,
m
Y
p1 , p2 , . . . , pm > 1 are distinct primes
()
n=
piri where
ri > 0 for all 1 ≤ i ≤ m
i=1
The prime pi appearing in () are unique as are their "multiplicities", the
exponents ri .
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MOTIVATION
Permutation
Review on equivalence relation
Definition
1
2
If X is a set, a relation between points in X is defined by specifying
some subset R in the Cartesian product X × X . Once R is given, we
say that a is related to b indicated by writing a ∼R b (or simply
a ∼ b) if the pair (a, b) lies in R.
A relation R in a set X is called an RST relation, or equivalence
relation, if it has the following properties
1 x ∼ x for all x ∈ X (the relation is reflexive)
2 x ∼ y ⇒ y ∼ x for all x, y ∈ X (the relation is symmetric)
3 x ∼ y and y ∼ z ⇒ x ∼ z (the relation is transitive)
We say that "x is equivalent to y " if x ∼R y .
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MOTIVATION
Permutation
Review on equivalence relation
Definition
Let R be an equivalence relation on X . Given a point p in X , we define
its equivalence class to be the set
[p]R = {y ∈ X : y ∼R p}
p is called a representative of the equivalence class.
Since p ∈ [p], the equivalence classes fill X . Every RST relation
corresponds to a partition of the underlying set X into disjoint subsets
that fill X .
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MOTIVATION
Permutation
Review on equivalence relation
Lemma
Let R be an equivalence relation on X . Its equivalence classes have the
following properties :
0
0
1 If C = [p] is an equivalence class and p ∈ [p] then [p ] = [p] ;
2 If C = [p ] and C = [p ] are two equivalence classes in X , then
1
1
2
2
either C1 = C2 (the sets are identical) or C1 ∩ C2 = ∅ ( the sets are
disjoint).
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MOTIVATION
Permutation
Review on equivalence relation
Partitions and equivalence relations
The distinct equivalence classes partition X as a union of disjoint sets, in
which two points are equivalent if and only if they lie in the same subset
of the partition.
Conversely, suppose X is a set and P = {Xα : α ∈ I } is a collection of
nonempty subsets indexed by a set of labels I that partition X , so that
[
X =
Xα
α∈I
and Xα ∩ Xβ = ∅ if α 6= β in I .
Then we can define a relation R on X such that the partition PR is the
same as the partition P we started with.
In fact, we define :
x ∼R y ⇔ x and y lie in the same subset Xα of the partition P
Exercise : Prove that this defines an equivalent relation.
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MOTIVATION
Permutation
Review on equivalence relation
Definition
Given a set X and an RST relation R on it, the associated quotient
space X /R is defined to be the set whose elements are the equivalence
classes [x]R in X .
Be careful ! Points in the quotient space X /R are subsets of the original
space X .
Having defined X /R, there is a natural quotient map π : X → X /R
defined by taking
π(x) = [x]R = the equivalence class of x
This map is clearly surjective.
Given an equivalence class C = [p], we refer to p as a representative of
the class. Of course, every other point in C is also a representative.
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MOTIVATION
Permutation
Congruences
Definition
Fix an integer n > 1 and define the following RST relation in X = Z :
a ≡ b (mod n) ⇔
⇔
⇔
⇔
b − a is a multiple of n
b = a + nk for some k ∈ Z
b ∈ a + nZ = {a + nk : k ∈ Z}
b + nZ = a + nZ
In plain English, the relation a ≡ b (mod n) is read as : " a is congruent
to b modulo the integer n".
Exercise : Fix an integer n > 1
1 Prove that the relation mod n is an equivalence relation.
2 What are the equivalence classes for this relations ? They are referred
as congruence classes in Z.
3 Describe the quotient space and the quotient map.
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Permutation
The ring (Z/nZ, +, ·)
Theorem (Algebraic structure in the quotient space Z/nZ)
Fix an integer n > 1. Let Z/nZ be the quotient space of (mod n)
congruence classes and let π : Z → Z/nZ be the quotient map. In Z/nZ
define operations
[a] ⊕ [b] = [a + b] and [a] [b] = [ab]
for a, b ∈ Z. These operations are well-defined despite the fact that class
representatives are used to define them. Furthermore,
1 The element [0] is the zero element with respect to the ⊕ operation
[0] ⊕ x = x for all x ∈ Z/nZ,
2 The element [1] is the multiplicative identity element with respect to
the operation : [1] x = x for all x ∈ Z/nZ.
Exercise
Prove that prove that (Z/nZ, +, ·) is a ring.
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MOTIVATION
Permutation
Units of the ring (Z/nZ, +, ·)
Definition
An element [a] ∈ Z/nZ has a multiplicative inverse if there exists some
[k] ∈ Z/nZ such that [k] [a] = [a] [k] = [1]. If it exists this inverse or
"reciprocal" is denoted by [a]−1 . The invertible elements in Z/nZ are the
units of the system (Z/nZ, +, ·) ; we denote them by Un :
Un = {[k] ∈ Z/nZ| there is [l] ∈ Z/nZ, [k] [l] = [1]}
Exercise
Prove that (Un , ) is a commutative group.
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MOTIVATION
Permutation
Units of the ring (Z/nZ, +, ·)
Exercise
Give an exhaustive list of the elements of U5 , U7 and U12 .
Theorem
If n > 1, the group of units in Z/nZ is
Un = {[k] ∈ Z/nZ : 1 ≤ k ≤ n − 1 and gcd(k, n) = 1}
Theorem
If n > 1 is an integer then all elements [a] 6= [0] in Z/nZ have
multiplicative inverses ⇔ the modulus n is a prime. This means Z/pZ is
a field for each prime p > 1 (A field being a ring such that the units are
exactly all the non zero element of this ring), in which division is allowed
by taking
[a][b]−1
for all pairs such that [b] 6= [0].
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MOTIVATION
Permutation
The rational Q
Exercise
Let F be the set of all "fraction symbols" p/q with p, q ∈ Z and q 6= 0. Into this
system we introduce a relation (∼) by declaring that
p 0 /q 0 ∼ p/q ⇔ p 0 q = pq 0 in Z ()
1
Prove that () is an equivalence relation.
2
Prove that
mp/(mq) ∼ p/q, for all m 6= 0 inZ
3
Given two classes, take any representatives p/q and r /s and define :
[p/q] + [r /s] = [(ps + qr )/(qs)]
and
[p/q] · [r /s] = [(pr )/(qs)]
Prove that the equivalence classes on the right don’t depend on which
representative p/q and r /s of the original classes we chose, that is the operations
on classes are well-defined in spite of the fact that we used class representatives
to determine the outcome.
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Permutation
Groups
Definition
A group is a set G equipped with a binary operation mapping
G ×G
(x, y )
→ G
7
→
xy
such that
G.1 Associativity : (xy )z = x(yz), for all x, y , z ∈ G .
G.2 Unit element : There exists an element e ∈ G such that
ex = x = xe, for all x ∈ G .
G.3 Inverses exist : For each x ∈ G there exists an element y ∈ G such
that xy = e = yx.
The inverse element y = y (x) in G.3 is called the multiplicative
inverse of x, and is generally denoted by x −1 .
The group G is said to be commutative or abelian if the additional
axiom.
G.4 Commutativity : xy = yx, for all x, y ∈ G is satisfied.
We write |G | for the number of elements in G , which could be ∞.
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Permutation
Groups
Lemma
In a group (G , ·) the unite e is unique, and so is x −1 for each x.
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Permutation
Groups
IMPORTANT
When working with a group first of all answer to this question :
WHAT IS THE GROUP LAW (ADDITIVE, MULTIPLICATIVE OR
OTHER) ?
Then, use the appropriate notation.
For a start, it will be good for you to rewrite the definition in additive or
other notation.
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Permutation
Groups
IMPORTANT : BE CAREFUL WITH ADDITIVE AND MULTIPLICATIVE
NOTATION.
Multiplicative notation (G , ·)
Additive notation (G , +)
Identity
Inverse
Product
Powers
e
0
x −1
−x
x ·y
x +y
xk = x . . . x
k · x = x + ··· + x
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Permutation
Groups
Definition
A non-empty subset H in a group G is a subgroup if it has the
properties
1 H is closed under formation of products : H · H ⊆ H, or equivalently
x, y ∈ H ⇒ xy ∈ H ;
2 The identity element e lies in H.
3 H is closed under inverses : h ∈ H ⇒ h −1 ∈ H.
Definition
The trivial groups H = (e) and H = G are subgroups ; all other
subgroups, if any, are referred to as proper subgroups. The suggestive
notation H ≤ G is sometimes used to indicate that a subset H ⊆ G is
actually a subgroup.
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Permutation
Groups
Examples
1
2
3
4
5
6
7
8
9
G = {e}.
G = (Z, +).
G = (Z/nZ, +)
G = (Un , ·).
Let G be any vector space V , equipped with vector addition as the
binary operation.
The set G = (C, +).
The set G = (C× , ·).
The circle group G = (S 1 , ·).
(Ωn = {z ∈ C : z n = 1} = {e 2πik/n : 0 ≤ k ≤ n − 1}, ·)nth roots of
unity).
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Permutation
Matrix Groups
Examples
1
2
3
4
5
(Mn (F), +) with F = Q, R, C, or Z/pZ where p > 1 is a prime.
The general linear group over F,
(GLn (F) = {M ∈ Mn (F) : det(M) 6= 0}, ·)
The special linear group over F,
(SLn (F) = {M ∈ Mn (F) : det(M) = 1}, ·)
The orthogonal group over F,
(On (F) = {M ∈ Mn (F) : MM T = In }, ·)
The special orthogonal group over F,
(SOn (F) = {M ∈ Mn (F) : MM T = In and det(M) = 1}, ·)
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Permutation
Matrix Groups
Examples
1
2
The upper triangular group consists of all n × n matrices of the
form


a11 a12 . . . ann
 0
a22 . . . a2n 


.
A= .



.
0
... 0
ann
such that det(A) = a11 . . . ann 6= 0 and aij = 0 for below diagonal
entries. If all the diagonal entries are equal to 1 we get the group of
strictly upper triangular matrices.
The three-dimensional Heisenberg group of quantum mechanics
consists of all real 3 × 3 matrices of the form
!
1 x z
0 1 y
A=
0 0 1
with x, y , z ∈ R. It plays central role in Quantum Mechanics.
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MOTIVATION
Permutation
Transformation groups
Examples
Many important groups are made up of bijective maps (transformations)
T : X → X of some point set X . For example the n-dimensional group of
rigid motions M(n) on Euclidean space Rn consists of all bijections
T : Rn → Rn that preserve the usual Euclidean distance between points
n
X
||x − y || = [
|xi − yi |2 ]1/2
i=1
Exercise : Prove that (M(n), ◦) is a group.
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MOTIVATION
Permutation
Permutation group
Definition
The permutation group Sn is the collection of all bijective maps
σ : X → X of the set X → X of the set X = {1, 2, . . . , n}, with
composition of maps ◦ as the group operation.
Show that (Sn , ◦) is a group.
It is easily seen that Sn is finite, with |Sn | = n! = n(n − 1) . . . 3 · 2 · 1.
It is non-commutative except when n = 2.
One way to describe elements σ ∈ Sn employs a data array to show
where each k ∈ X ends up :
1 2 ... k ... n
σ=
i1 i2 . . . ik . . . in
where (i1 , i2 , . . . , in ) is some ordered listing of the integers 1 ≤ k ≤ n. In
this notation the identity element is
1 2 ... k ... n
e=
1 2 ... k ... n
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MOTIVATION
Permutation
Permutation group
Definition
For k > 1, a k-cycle is a permutation σ = (i1 , . . . , ik ) that acts on X in
the following way
i1 → i2 → · · · → ik → i1 (a cyclic shift of list entries)
σ maps
j →j
for all j not in the list {i1 , . . . , ik }
The action of σ depends on the particular order of the list entries
i1 , . . . , ik .
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MOTIVATION
Permutation
Permutation group
Exercise
1
2
3
The cycle σ = (1, 2, 3) in S5 ;
The cycle σ = (1, 2) in S5 .
The one-cycles (k) and the identity map idX
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MOTIVATION
Permutation
Permutation group
BE CAREFUL ! ! ! ! ! !
The symbol σ = (i1 , . . . , ik ) denoting a cycle is ambiguous. If we make a
cyclic shift of list entries we get k different symbols that describe the
same mapping of X .
(i1 , . . . , ik ) = (i2 , . . . , ik , i1 ) = (i3 , . . . , ik , i1 , i2 ) = · · · = (ik , i1 , . . . , ik−1 )
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MOTIVATION
Permutation
Permutation group
Example
You must understand how to evaluate the product on an element k ∈ X
can be evaluated by feeding k into the product from the right, as shown
below taking σ = (1, 2) and τ = (1, 2, 3) in S5 .
στ : k
(1,2,3)
/ τ (k)
(1,2)
/ σ(τ (k)) = (σ ◦ τ )(k)
To determine the net effect of στ , start by examining the fate of k = 1,
then look what happens to the image of 1 etc.
Action
1
2
3
4
5
(1,2,3)
(1,2,3)
(1,2,3)
(1,2,3)
(1,2,3)
Net effect
/2
(1,2)
/1
1→1
/3
(1,2)
/3
2→3
/1
(1,2)
/2
3→2
/4
(1,2)
/4
4→4
/5
(1,2)
/5
5→5
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MOTIVATION
Permutation
Characteristic subgroups
Definition
1
The center Z (G ) of a group G is the set of elements that commute
with everyone in G
Z (G )
=
{x ∈ G : gx = xg for all g ∈ G }
= {x ∈ G : gxg −1 = x for all g ∈ G } ≤ G
Obviously G is abelian ⇔ Z (G ) = G .
More generally, given a nonempty subset S ⊆ G we may define
2. The centralizer of S is
ZG (S) = {x ∈ G : xs = sx for all s ∈ S}
3. The normalizer of S is
NG (S) = {x ∈ G : xSx −1 = S }
Both ZG (S) and NG (S) are subgroups of G , with
NG (S) ⊇ ZG (S) ⊇ Z (G ).
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MOTIVATION
Permutation
Exponent laws
Theorem
Let (G , ·) be a group. For any element a ∈ G and any k ∈ N define
ak = a · · · · · a (k times) ;
a0 = e (the identity element) ;
a−k = (a−1 ) · · · · · (a−1 ) (k times).
Then the following exponent laws are valid for all m, n ∈ Z.
m
1 a
· an = am+n ;
2 (am )−1 = (a−1 )m ;
3 (am )n = amn .
4 If G is ABELIAN we also have (ab)n = an · b n .
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MOTIVATION
Permutation
Characterization of finite groups
Theorem
Let H be a nonempty finite subset of a group G , such that
H · H = {h1 h2 : h1 , h2 ∈ H} = H
Then the identity element e automatically lies in H and H is a subgroup
of G .
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MOTIVATION
Permutation
Homomorphisms
Definition
1
A homomorphism between two groups (G , · ) and (G 0 , ∗) is any map
φ : G → G 0 that intertwines the group operations, in the sense that
φ(x · y ) = φ(x) ∗ φ(y )
2
for all x, y ∈ G
(1)
The map is an isomorphism if it is a homomorphism and is also a
bijection.
Corollary
Suppose φ : G → G 0 is an isomorphism. Then φ−1 : G 0 → G is an
isomorphism.
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MOTIVATION
Permutation
Homomorphisms
Definition
Certain terminology is standard in discussing homomorphisms
φ : G → G 0 of groups.
1 The kernel denoted Ker (φ) of φ is the set of elements that get
“killed” by φ :
Ker (φ) = {x ∈ G : φ(x) = e 0 } ,
0
where e is the identity element in G 0 . The kernel is a subgroup of
the initial group G .
2 The range denoted by Range(φ) is the forward image of the initial
group
Range(φ) = φ(G ) = {φ(x) : x ∈ G }
The range is always a subgroup of the target group G 0 , but it may be
a proper subgroup.
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MOTIVATION
Permutation
Homomorphisms
Corollary
If φ : G → G 0 is a homomorphism and e ∈ G , e 0 ∈ G 0 are the respective
identity elements, then φ(e) = e 0 .
Corollary
If φ : G → G 0 is a homomorphism and x ∈ G ,
φ(x −1 ) = (φ(x))−1
.
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MOTIVATION
Permutation
Homomorphisms
Lemma
A homomorphism φ : G → G 0 is one-to-one ⇔ ker(φ) = (e)
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MOTIVATION
Permutation
Homomorphisms
Example
1
2
3
4
5
6
The trivial homomorphism φ0 : G → G 0 sending g in G to e 0 in G 0
The identity map Id : G → G sending g in G to itself.
The inversion map J : G → G , for G ABELIAN sending g ∈ G to
g −1 . For instance, if G = Z, J(k) = −k for k ∈ Z or if G = Z/nZ,
J([k]n ) = [−k]n for [k] ∈ Z/nZ. (The inversion map is not a
homomorphism if G is NOT COMMUTATIVE).
The quotient map π : Z → Z/nZ sending k in Z to [k]n in Z/nZ.
The exponential map e : R → R× sending x in R to e x .
For G ABELIAN, the power map p : G → G sending g in G to g k
where k is an integer. For instance, for G = Z, p(n) = k · n for n
integer and for G = Z/nZ, p([s]n ) = k · [s]n for any [s]n ∈ Z/nZ.
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MOTIVATION
Permutation
Homomorphisms
Example
1
The important linear algebra map given by
L : Mn (F) → Hom(Fn )
A
7→ LA
with LA (x) = A · x for all x ∈ Fn . Inducing a morphism
L : GLn (F) →
A
7→
GL(Fn )
LA
with LA (x) = A · x for all x ∈ Fn . and
L : SO(k) → RO(k)
A
7→ LA
with LA (x) = A · x for all x ∈ Fn , with k = 2, 3
2
The map ψ : Z → Ωn sending k in Z to e
2πik
n
.
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MOTIVATION
Permutation
Generated subgroups
Fact
Given any family {Hα : α ∈ I } of subgroups in a group G , their
intersection
H = ∩α∈I Hα = {x ∈ G : x ∈ Hα for all α ∈ I }
is also a subgroup, even if there are infinitely many Hα .
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MOTIVATION
Permutation
Generated subgroups
Definition
Let S be a nonempty subset of a group G . The intersection
< S >= ∩{H : H is a subgroup and S ⊆ H}
is a subgroup. It is called the subgroup generated by S, and the
elements of S are referred to as "generators" of this group.
Note that
< S >= {a1 · · · as : with ai ∈ S ∪ S −1 }
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MOTIVATION
Permutation
Generated subgroups
Definition
Subgroups generated by a single element are called cyclic subgroups.
Remark
A cyclic subgroup can have various generators, so that
H =< a >=< b >
with a 6= b. The case when a = e is of no interest since < e > is the
trivial subgroup.
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MOTIVATION
Permutation
Generated subgroups
Definition
Let (G , · ) be a group. The order o(a) of a group element a ∈ G is the
smallest positive exponent k > 0 such that ak = e. If no such exponent
exists the group element is said to have infinite order, which we indicate
by writing o(a) = ∞.
Example
1
2
3
Let a 6= 0 in (Z, +) o(a) = ∞ order ;
If G is a finite group, o(a) ≤ |G |.
By definition o(a) ≥ 1, and we have o(a) = 1 ⇔ a = e.
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MOTIVATION
Permutation
Generated subgroups
Theorem (Structure of Cyclic Subgroups)
Let (G , · ) be a group. A cyclic subgroup has the form
H = hai = {ak : k ∈ Z}
for some a ∈ G . There are two possibilities, which depend on the order
o(a) of the generator.
k
1 o(a) = ∞. Then all powers a , k ∈ Z, are distinct and
H ' (Z, +)
2
embedded in the abstract group G .
o(a) = k < ∞ . Then
H = {e, a, a2 , . . . , ak−1 },
with ak = e. In this case
H ' (Z/kZ, +)
embedded in the abstract group G .
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MOTIVATION
Permutation
Generated subgroups
Idea of the proof of the Theorem (Structure of Cyclic Subgroups)
We separate the study in two case either the set S = {ak , k ∈ N} has
repeat or it has not.
1
Case 1 : S has no repeat.
1
2
3
2
Prove that that {ak , k ∈ Z} has also no repeat.
Deduce that o(a) = ∞.
Prove that the map φ : Z →< a > sending k to ak is an homomorphism and
an bijection thus an isomorphism.
Case 2 : S has a repeat. (That is there are l > k such that ak = al )
1
2
3
Deduce that o(a) = k < ∞.
Prove that < a >= {e, · · · , ak−1 }. That is prove that H = {e, · · · , ak−1 } is
a subgroup of G and that all the elements of H are distinct. Why is this
enough to prove that it is the smallest subgroup containing a ?
Prove that the map φ : Z/kZ →< a > sending [k]n to ak is well defined that
0
is if k, k 0 ∈ Z are such that [k]n = [k 0 ]n then ak = ak , that it is an
homomorphism and an bijection thus an isomorphism.
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Contents
MOTIVATION
Permutation
Generated subgroups
Proposition
Every subgroup of a cyclic group is also cyclic.
Idea of the proof
Let G =< a > be a cyclic group generated by a and H ≤ G .
Consider the smallest k ∈ N such that ak ∈ H and Prove that
H =< ak >. (Hint : use the Euclidean division to conclude).
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Contents
MOTIVATION
Permutation
Generated subgroups
Particular case : subgroup of Z
The subgroup of (Z, +) are cyclic, and have the form Hm = Z · m for
some integer m ≥ 0, with H0 = {0} and H1 = Z, H2 = 2·Z, etc.
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Contents
MOTIVATION
Permutation
Generated subgroups
Group of order 2
Let (G , ·) be a group of order |G | = 2. G must be cyclic, hence
commutative, and that G is isomorphic to (Z/2Z, +).
Group of order 3
Let (G , ·) be a group of order |G | = 3. G must be cyclic, hence
commutative, and that G is isomorphic to (Z/3Z, +).
Idea for the proof
Determine what are the possible order for the non trivial elements in
those groups.
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Contents
MOTIVATION
Permutation
Generated subgroups
Theorem
For n > 1, a nonzero element x = [k] in Z/nZ is a cyclic generator under
the (+) operation if and only if gcd(k, n) = 1 (i.e. if and only if [k] lies in
the set Un of multiplicative units in Z/nZ).
Idea for the proof
Bezout theorem should help you. Remember that as soon as [1] is in the
group generated by < [k] > then this [k] generate the all Z/nZ.
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MOTIVATION
Permutation
Cosets
Definition
Given any subgroup H ⊆ G , its left cosets are the subsets of the form
xH = {xh : h ∈ H} with x ∈ G . These are of interest because the whole
group splits into a disjoint union of its distinct cosets xH.
One can also define right cosets as sets of the form Hx, right translates
of H by elements x ∈ G .
There is no difference between left- and right cosets if the group G is
abelian, but we will encounter many noncommutative groups where the
distinction must be recognized.
The group element x is a coset representative for xH. A coset can have
various representatives, (below we will determine when x, y ∈ G yield the
same coset, xH = yH).
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MOTIVATION
Permutation
Cosets
Lemma
Let H be a subgroup in G and let x, y be points in G . Then
1 We have xH = yH ⇔ there is some h ∈ H such that y = xh. In
particular, xH = H ⇔ x ∈ H.
2 Two cosets xH and yH are either identical sets in G or are disjoint.
The cosets form a partition of G .
3 The relation x ∼ y ⇔ xH = yH is reflexive, symmetric, and
R
transitive, and the equivalence classes for this relation are : for any x
the class [x] = {g ∈ G : g ∼
x} is equal to xH.
R
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MOTIVATION
Permutation
Cosets
Proposition
If φ : G → G 0 is a homomorphism of groups and K = Ker (φ) is its kernel,
then
0
1 All points in a coset xK map to a single point in G under φ. Thus a
homomorphism is constant on each coset of its kernel.
2 Distinct cosets xK 6= yK in G are disjoint, with xK ∩ yK = ∅, and
they map to distinct points in G 0 .
Furthermore φ is one-to-one, and hence an isomorphism from G to the
subgroup Range(φ) ⊆ G 0 , if and only if its kernel is trivial : Ker (φ) = {e}.
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MOTIVATION
Permutation
Cosets
Definition
Let H be a subgroup in a group (G , · ). The left cosets are the subsets
having the form xH = {xh : h ∈ H} for some x ∈ G , and the collection of
all such cosets is denoted by G /H. Similarly we could define the space
H\G of right cosets, which have the form Hx. We will mostly deal with
G /H.
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Permutation
Quotient space
Definition
The space of left cosets denoted by G /H is just the set of the left
coset. It is the quotient space of equivalence classes under the rst relation
x∼
y ⇔ xH = yH. Note carefully :
R
Points in the quotient space G /H are subsets in the original
group G .
The quotient map π : G → G /H for this relation is given by
π(x) = xH
(since xH is the equivalence class for x)
General properties of this surjective map follow directly from this
definition.
1 Under π, each coset xH ⊆ G collapses to a single point in the
quotient space G /H.
2 Distinct (disjoint) cosets xH 6= yH in G are mapped by π to distinct
points in the quotient space G /H.
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MOTIVATION
Permutation
Quotient space
Exercise
1
2
The coordinate plane R2 becomes an abelian group G = (R2 , +)
when equipped with the usual vector addition operation (+), and the
x-axis H = {(x, 0) : x ∈ R} is easily seen to be a subgroup.
1 Describe right, left cosets. When two cosets are equal ?
2 Describe G /H.
If G = (Z, +) and n ≥ 2 the set H = nZ = {nk : k ∈ Z} is a
subgroup in Z.
1 Describe left cosets. When two cosets are equal ?
2 Describe G /H and the quotient map. Note that we have defined
a group operation in G /H from the operation on G = Z.
WARMING
For more general groups G and subgroups H it is NOT ALWAYS
POSSIBLE to impose a GROUP STRUCTURE IN THE QUOTIENT
SPACE G /H ; it worked in the last example largely because the group
was abelian.
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MOTIVATION
Permutation
Quotient space
Exercise
Let G = (R, +) and H = Z.
1
Describe right, left cosets. When two cosets are equal ? How many left coset are
there ? How can we choose the representative ?
2
Describe G /H and the quotient map. Can we define a group structure on G /H
3
But what is the mysterious quotient group (R/Z, ⊕) ? ? ? ? We can prove it is
isomorphic to something quite concrete and familiar, namely the “circle group”
S 1 = {z ∈ C : |z| = 1}
equipped with complex multiplication as the group law. (IMPORTANT
PROOF ! ! ! !)
Define the exponential map φ(θ) = e 2πiθ = cos(2πθ) + i sin(2πθ) Prove that it is
an homomorphism from (R, +) to the circle group (S 1 , · ). Surjective ? Kernel ?
One-to-one ?
Important idea : φ induces a bijection φ̃ between the quotient space R/Z and
S 1 that turns out to be a group isomorphism.
Define φ̃ using coset representatives, letting
φ̃(x + Z) = φ(x) = e 2πix
for all x ∈ R
(2)
Well defined ? Homomorphism ? One-to-one ? surjective ?
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MOTIVATION
Permutation
Normal subgroup
Why do we define normal subgroup ?
Let G be any group and H be a subgroup. How could we naturally
define a group operation
: G /H × G /H → G /H?
Is it well define ?
How could we make it well define ?
That give a simple condition on H meaning what we will call normality
insures that this construction does work, even if G is nonabelian.
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MOTIVATION
Permutation
Normal subgroup
Definition
A subgroup N in G is a normal subgroup if it has the property
xN = Nx
for all x ∈ G ,
(3)
which means there is no difference between left- and right-cosets of N.
All subgroups are normal if G is abelian. Normality of a subgroup is
indicated by writing N C G .
REMEMBER
Normal subgroups were defined in order to put a group structure on the
quotient space.
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MOTIVATION
Permutation
Normal subgroup
Lemma
If N is a subgroup of G , each condition below implies the others.
(a) The subgroup N is normal : xN = Nx for all x ∈ G .
(b) xNx −1 = N for all x ∈ G .
(c) xNx −1 ⊆ N for all x ∈ G .
(d) xnx −1 ∈ N for all x ∈ G , n ∈ N.
In partice
In practice, I advice you to first try to use (d). For this pick x ∈ G and
n ∈ N, and try to prove that xnx −1 ∈ N.
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MOTIVATION
Permutation
Quotient group
Theorem (Quotient groups)
Let N be a normal subgroup in a group G .
Then the operation is well defined : the outcome does not depend on
the particular coset representatives x and y .
This product satisfies all the group axioms, making the coset space G /N
into a group in its own right.
Finally, the quotient map π : G → G /N becomes a surjective
homomorphism of groups with ker(π) = N.
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MOTIVATION
Permutation
Quotient group
Lemma
A subgroup N in a group G is normal if and only if N is the kernel
ker φ = {x ∈ G : φ(x) = e 0 } for some homomorphism φ : G → G 0 .
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MOTIVATION
Permutation
Isomorphism theorem
Exercise
Prove that the surjective homomorphism ψ : Z → Ωn
ψ(k) = ω k = e 2πik/n
where ω = e 2πi/n (the primitive nth root of unity)
induce a isomorphism ψ̃ : Z/nZ → Ωn .
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MOTIVATION
Permutation
Isomorphism theorem
Exercise
Let G be the set C× = {z ∈ C : z 6= 0} of nonzero complex numbers,
equipped with multiplication as the group operation.
Within this abelian group we have the two-element normal subgroup
N = {+1, −1}.
What are the left coset of N ?
What is G /N ?
Prove that G /N is isomorphic to the original group C× .
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MOTIVATION
Permutation
Isomorphism theorem
(First Isomorphism Theorem)
Let φ : G → G 0 be a homomorphism, let K = ker (φ), and let
π : G → G /K be the quotient homomorphism. There is a unique map
φ̃ : G /K → R = range(φ) that makes the diagram
G
π↓
G /K
φ−→
%
φ̃
R ⊆ G0
commute : φ̃ ◦ π = φ.
This map is a group homomorphism and is bijective, so it is an
isomorphism from the quotient group G /K to R = range(φ).
In particular, when φ is surjective we have G 0 ∼
= G /K .
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MOTIVATION
Permutation
Isomorphism theorem
Idea for the proof
1
Explain why since φ̃ ◦ π = φ, we have no choice to define φ̃ it has to
send xK 7→ φ(x).
2
We need to prove that φ̃ is well defined, that is for two x, y ∈ G such
that xK = yK , then φ(x) = φ(y ).
3
Prove that φ̃ is an homomorphism.
4
Prove that φ̃ is one-to-one.
Deduce the rest of the theorem.
5
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MOTIVATION
Permutation
Isomorphism theorem
Exercise
Let G be the matrix group GL(n, C) of all n × n matrices A with complex
entries and det (A) 6= 0.
This is a group under matrix multiplication, and so is the subgroup
N = SL(n, C) of matrices with determinant +1.
Prove that N is normal in G , and that the quotient group G /N is
isomorphic to the group (C× , · ) of nonzero complex numbers under
multiplication.
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MOTIVATION
Permutation
Isomorphism theorem
Theorem (Second Isomorphism Theorem)
Let A be any subgroup in G and let N be a normal subgroup. Then
(a) The product set AN is a subgroup in G , with N C AN.
(b) A ∩ N is a normal subgroup in A.
(c) AN/N ∼
= A/(A ∩ N)
Proof left as Exercise
In (c) consider the map ψ : A/(A ∩ N) → AN/N given by
ψ(a(A ∩ N)) = aN for a ∈ A. Start by showing this map is well-defined :
if a(A ∩ N) = a0 (A ∩ N) then aN = a0 N Algebra 1
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MOTIVATION
Permutation
Isomorphism theorem
Theorem (Third Isomorphism Theorem)
Let G ⊇ A ⊇ B be groups such that A and B are both normal subgroups
in G . Prove that (G /B)/(A/B) ∼
= G /A.
Note : This is the group-theory analog of the arithmetic relation
(a/c)/(b/c) = a/b
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MOTIVATION
Permutation
Lagrange theorem
Theorem (Lagrange)
If G is a group of finite order |G | = n and H is a subgroup, then |H|
must divide |G |. In fact, we have
|G | = |G /H| · |H|
(4)
so the number of left cosets in G /H also divides |G |.
Corollary
If G is a finite group and a ∈ G then the order o(a) of this element must
divide |G |.
Corollary
If a group G has finite order |G | = n then an = e for all elements a ∈ G .
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MOTIVATION
Permutation
Lagrange theorem
Corollary
If G is a finite group whose order is a prime |G | = p > 1, then G = hai
for every element a 6= e and G ∼
= (Z/pZ, +). In particular, every finite
group of prime order is cyclic, abelian.
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MOTIVATION
Permutation
Subgroup of Z/nZ
Theorem
In (Z/nZ, +), for every divisor d of n, (1 ≤ d ≤ n) there is a unique
(cyclic) subgroup Hd such that |Hd | = d
Idea for the proof
1
2
Existence : Just prove that Hd = h[n/d]i is a subgroup of order d.
Uniqueness : Suppose there were two subgroups A, B of order d.
Consider
A + B = {x + y : x ∈ A, y ∈ B}
Prove that A = B = A + B.
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MOTIVATION
Permutation
Counting principle
Theorem
Let G be a group and A, B subgroups. Then
1 The product set AB is a subgroup ⇔ AB = BA.
2 Whether or not AB is a subgroup, we always have
|AB| =
|A| · |B|
|A ∩ B|
Idea for the proof
1
2
Good exercise.
Look at the map ρ : A × B → AB ⊆ G defined by setting
ρ(a, b) = ab, and ask :
Question : For how many pairs (a, b) in the Cartesian
product set A × B do the group elements ρ(a, b) = ab take
on the same value ?
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MOTIVATION
Permutation
Automorphisms and Inner Automorphisms
Definition
An automorphism of a group is an isomorphism from G to itself.
We denote
Aut(G ) = {φ : G → G : φ isomorphism}
the set of all automorphism.
Prove that (Aut(G ), ◦) is a group.
An inner automorphism of a group is a automorphism sending g to
aga−1 , for some a ∈ G .
We denote
In(G ) = {αa : G → G with αa (g ) = aga−1 for some a ∈ G }
Prove that (In(G ), ◦) is a normal subgroup of (Aut(G ), ◦).
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MOTIVATION
Permutation
Computation of some automorphisms groups
Automorphisms for some classic group
1
2
Aut (Z, +) = {id G , −id G } ;
(Aut (Z/nZ, +), ◦) ' (Un , · ).
Idea for the proof (Important to remember)
Let G =< a > be a cyclic group. Prove that there is a unique
homomorphism φ : G → G 0 such that φ(a) = b where b ∈ G 0 . That
means that a homomorphism whose domain is a cyclic group is fully
determined by knowing where the generator goes.
Then think of how you need to choose b in order to have an isomorphism.
Algebra 1
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MOTIVATION
Permutation
Inner automorphisms
Theorem
For any group G we have
Int (G ) ∼
= G /Z (G )
Idea for proof
Use first isomorphism theorem applied to the map :
Φ : G → Int (G ) ⊆ Aut (G )
sending a ∈ G to φa ∈ Aut(G ) sending g ∈ G to aga−1 .
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MOTIVATION
Permutation
Left group actions
Definition
A (left) group action on X by G denoted by (G , X ) is a map
τ : G ×X
(g , x)
→
X
7
→
τg (x) = g · x
We get a mapping
τg : X
x
→
X
7
→
g ·x
for each g ∈ G . We require the action to have the following properties.
1 For each g1 , g2 ∈ G we have τg g (x) = τg (τg (x)) – i.e.
1 2
1
2
g1 g2 · x = g1 · (g2 · x). discussion.
2 τ (x) = x for all x ∈ X , so that τ = id
and e · x = x.
e
e
X
Note
From 1 and 2 we have that for any g ∈ G , τg is invertible and its inverse
is τg −1 .
Algebra 1
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MOTIVATION
Permutation
Left group actions
Fact
The existence of a map
τ : G ×X
(g , x)
→
X
7
→
τg (x) = g · x
defining a group action on X by G is equivalent to the existence of a
group homorphism
Φ: G
g
→ (Per (X ), ◦)
7→
τg
Algebra 1
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MOTIVATION
Permutation
Left group actions
Definition
The normal subgroup of G
K = ker(Φ) = {g ∈ G |τg = id X },
is often referred to as the kernel of the action G × X → X , or simply the
action kernel.
Fact
A group action on X by G induces a group action on X by G /Ker (Φ)
and this action has trivial kernel.
Algebra 1
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MOTIVATION
Permutation
Left group actions
Definition
Given a group action G × X → X , each point x0 ∈ X has a G-orbit
G · x0 = {g · x0 : g ∈ G }.
The action G × X → X is transitive if there is just one orbit :
For all x, y ∈ X , there exists some g ∈ G such that g · x = y
(5)
or equivalently, G · x = X for any x.
Algebra 1
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MOTIVATION
Permutation
Left group actions
Fact
In X there is a natural an RST relation R
x∼
y ⇐⇒ ∃g ∈ G such that y = g · x
R
(6)
Furthermore, the equivalence class of a point x0 under R is precisely its
G -orbit.
Thus X splits into disjoint G -orbits which fill X .
Algebra 1
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MOTIVATION
Permutation
Left group actions
Example
1
2
Take X = G and τg (x) = gx (left translation by the element g ).
Prove that τ define an action and compute the action kernel. Is this
action transitive ?
Take G = Sn (permutations on n objects) and X = {1, 2, . . . , n},
with the obvious action of σ ∈ Sn on the integers 1 ≤ k ≤ n meaning
τσ (i) = σ(i) for σ ∈ Sn and i ∈ {1, 2, . . . , n}. Prove that τ define an
action and compute the action kernel. Is this action transitive ?
Algebra 1
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MOTIVATION
Permutation
Left group actions
Example (Permutation Action of G on a Coset Space G /H)
Here G is any group, H any subgroup, and X = the space of cosets
G /H. We define a left action G × G /H → G /H letting
τg (xH) = gxH
1
2
3
4
5
τg : G /H → G /H
Prove that τg is well defined.
Prove it is a group action. This action is called the permutation
action of G on G /H.
What can you say about the action kernel ?
Is this action transitive ?
What happens if H = {e} ? H = G ?
Algebra 1
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MOTIVATION
Permutation
Left group actions
Example (The adjoint Action- Action by conjugation)
Here we take X = G and let G act on itself by conjugation :
αg (x) = gxg −1
1
2
3
for all x ∈ X = G
Prove it is a group action.
Compute the action kernel.
Compute the orbits. They are called conjugation classes.
Algebra 1
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MOTIVATION
Permutation
Left group actions
Example (The adjoint Action- Action by conjugation)
1
2
Under the adjoint action G × X → X , with X = G and action
αg (x) = gxg −1 , prove that the action kernel which is the kernel of
the homomorphism Φ : G → Int (G ) is precisely the center Z (G ).
Deduce using the first isomorphism theorem that for any group G we
have Int (G ) ∼
= G /Z (G ).
Orbits under the adjoint action are the conjugacy classes in G ,
Cx = G · x = {gxg −1 : g ∈ G } = {αg (x) : g ∈ G }
3
4
Prove that the action by conjugation is not transitive. (Consider
e ∈ G .)
Prove that if x ∈ G then Cx = {x} is a one-point class if and only if
x is in the center Z (G ).
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MOTIVATION
Permutation
Left group actions
Theorem (The class equation)
If G is finite and if C1 = Ce , C2 , . . . , Cr are the distinct conjugacy classes,
then
|G | = |C1 | + . . . + |Cr |
=
#(one-point classes) +
X
|Ck |
(7)
nontrivial Ck
= |Z (G )| +
X
|Ck |
nontrivial Ck
Furthermore, if G is finite and Cx is any conjugacy class, the number of
points |Cx | in the class always divides |G |.
Algebra 1
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MOTIVATION
Permutation
Left group actions
Example
The group of matrices
G = GL n (F) = {n × n matrices A : det A 6= 0}
acts as F-linear operators on the vector space X = Fn of n-tuples
x = (x1 , . . . , xn ) if we define τA = LA , the left multiplication operator
LA (x) = Ax (matrix product (n × n) · (n × 1) = (n × 1))
1
2
3
(8)
Prove it is an action.
Compute the action kernel.
Is this action transitive ?
Algebra 1
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MOTIVATION
Permutation
Left group actions
Example
Let G = R and X = R2 . We get a group action G × X → X if we let
θ ∈ R act as the rotation operator Rθ , a linear operator that rotates each
vector x counterclockwise about the origin by θ radians.
1 Prove that it is an action.
2 Is it transitive ?
3 Compute the action kernel.
Algebra 1
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MOTIVATION
Permutation
Left group actions
Example (Similarity Transforms of a Matrix)
Let X = Mn (F) be the set of all n × n matrices with entries in a number
field F and let G = GL n (F) = {S : det(S) 6= 0}. Elements S ∈ GL n (F)
act on arbitrary matrices A by similarity transformations
τS (A) = SAS −1 .
1 Prove it is an action.
2 Compute the action kernel.
3 Compute the Orbits. Those are called the similarity classes
(remember from linear algebra those correspond to change of basis).
Try to think about diagonalization in term of similarity classes. In linear algebra the
similarity invariants of a matrix A are properties that hold for all matrices in a similarity
class. These include such properties as
1
2
3
The determinant det
P:n Mn (F) → F is constant on each similarity class.
The trace Tr (A) =
i=1 aii (sum of diagonal entries) is also constant on similarity
classes.
Another invariant is the spectrum, the distinct eigenvalues of A in F
sp F (A) = {λ ∈ F : det(A − λI ) = 0}
Matrices that are not similar might by accident have the same spectrum, but the spectra
of A, B always agree if they are similar.
Algebra 1
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MOTIVATION
Permutation
Cayley’s Theorem
Action kernel
The action kernel of the action G × G /H → G /H on cosets is the largest
subgroup N in G such that : (i ) N is normal in G , (ii ) N ⊆ H. In
particular, ker Φ is trivial and Φ is a faithful (one-to-one) embedding of G
in Per (G /H) if and only if there are no nontrivial normal subgroups
N 0 C G that lie within H.
Hint for proof
Consider the action by permutation, Φ : G → Per (G /H) and prove that
\
ker (Φ) =
xHx −1
x∈G
and prove that ker (Φ) is the largest group with this properties (i) and
(ii).
Algebra 1
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MOTIVATION
Permutation
Cayley’s Theorem
Theorem (Cayley’s Theorem)
If G is any finite group then G is isomorphic to some subgroup of the
permutation group Sn , n = |G |.
Hint for proof
Consider the action by permutation Φ : G → Per (G ) ' Sn Prove that G
is isomorphic to the Range(Φ) the latest can be seen as a subgroup of Sn .
Algebra 1
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MOTIVATION
Permutation
Transitive actions
Given a group action G × X → X , its orbits Ox0 = G · x0 are invariant
under the action of G . The restricted action G × Ox0 → Ox0 has the
properties required of a left action and this action is transitive ;
Since the orbits form a partition of X , we can split the study of the action
of G on a set X into the actions of G on the different orbits involved.
Algebra 1
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MOTIVATION
Permutation
Stabilizer
Definition
Let G × X → X be any group action, and let x0 be any point in X . The
stabilizer of x0 is the subgroup of G of points that leave x0 fixed. That is,
Stab G (x0 ) = {g ∈ G : g · x0 = x0 }
Exercise
Show that Stab G (x0 ) is a subgroup in G . If x, y are points in the same
G -orbit under a group action, say with y = τg (x), prove that
1
2
In G , Stab G (y ) is a conjugate g ·Stab G (x)·g −1 of the stabilizer of x.
Stabilizers of points in the same G -orbit have the same size.
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MOTIVATION
Permutation
Orbits and stabilizer
Theorem
If G is finite and G × X → X is a transitive group action, then the space
X must be finite. Furthermore, if we fix a base point x0 ∈ X and let
H = Stab G (x0 ), we have
|X | = |G /H|
and
|G | = |G /H| · |H| = |X | · |Stab G (x0 )|
(9)
In particular, for transitive actions |X | must always divide |G |. For a
general group action the size |Ox0 | of any individual orbit must divide |G |.
Hint for proof
Let H = Stab G (x0 ) and
ψ : G /H → X
where
ψ(gH) = g ·x0
Prove that
1
ψ is well defined. That is for any g , g 0 ∈ G such that gH = g 0 H then
g ·x0 = g 0 ·x0 ;
2
ψ is a bijection.
3
Conclude with Lagrange theorem.
Algebra 1
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MOTIVATION
Permutation
Orbits and stabilizer
Corollary
If Cx is a conjugacy class in a finite group G (an orbit under the adjoint
action), then the cardinality |Cx | of the class must divide |G |.
Algebra 1
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MOTIVATION
Permutation
Orbits and stabilizer
Theorem
Let G × X → X be a transitive group action, fix a base point
x0 ∈ X , let H = Stab G (x0 ), and define the bijection ψ : G /H →
X
where
ψ(gH) = g · x0 . Then ψ intertwines the two group
actions and the diagram commutes :
ψ
G /H
λg ↓
−→
G /H
−→
That is
ψ ◦ λg = τg ◦ ψ
ψ
X = G · x0
↓ τg
X
(or equivalently, that τg = ψ ◦ λg ◦ ψ −1 , ∀g ∈ G ).
This property is often described by saying that the map ψ : G /H → X intertwines the
actions of G on the two spaces, or that ψ is an equivariant map.
What all this means in practice is that there is a way to identify points in G /H with points in
X so that the action of λg on G /H becomes the given action of G on X . Intuitively, the
actions λ : G × G /H → G /H and τ : G × X → X are different models of the same action–
they are “isomorphic” group actions, in the same sense that certain groups are isomorphic.
(The notion of “isomorphism” applies to all sorts of algebraic structures.) The preceding
discussion also shows that permutation actions on coset spaces G /H are “universal models” for
all transitive groups actions.
Algebra 1
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MOTIVATION
Permutation
Orbits and stabilizer
Exercise
The matrix group
SO(3) := {A ∈ GL(3, R) : AA
T
= Id and det(A) = 1}
Given A ∈ SO(3) we get a linear operator
3
3
LA : R → R
via
LA (v) = Av
(a matrix product (3 × 3) · (3 × 1)).
It is well known from linear algebra (Euler’s Theorem) that the resulting group G of linear
operators on R3 consists of all
R`,θ =
(
Rotation counterclockwise by θ radians
about an oriented axis ` through the origin.
)
Next consider the unit sphere in R3 , X = {x ∈ R3 : kxk = 1}, where kxk2 = x12 + x22 + x32 . We
get a group action G × X → X and this action is transitive (Exercise). Take x0 = (0, 0, 1) in
X . The stabilizer H = Stab G (x0 ) consists of all rotations (by any angle θ) about the z-axis ; the
corresponding group of matrices in SO(3) is




 cos θ − sin θ 0


sin θ
cos θ
0
:θ∈R
H =


0
0
1
By our discussion, there is a bijection between G /H (a purely algebraic construct) and the
sphere X , given by ψ(AH) = Ax0 . This bijection transfers the purely algebraic action
λA : G /H → G /H (with λA (BH) = AB ·H) to the geometric action LA : X → X . Thus the
geometric action of SO(3) on the sphere can be studied by algebraic methods by examining the
permutation action G × G /H → G /H. Algebra 1
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MOTIVATION
Permutation
Class equation
Definition
The centralizer ZG (x) := {g ∈ G : gxg −1 = x} is the stabilizer of x
when considering the action by conjugation.
Corollary (Class equation)
If G is a nontrivial finite group then
|G | = |Z (G )| +
X
x∈S 0
X
|G |
= |Z (G )| +
|Cx |
|ZG (x)|
0
(10)
x∈S
where S 0 is a set of representatives of the distinct conjugacy classes such
that |Cx | > 1.
Algebra 1
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MOTIVATION
Permutation
The center of a p-group is non trivial.
Definition
Let p be a prime. A finite p-group G is a group with order |G | = p n for
some integer n.
Theorem (Cauchy)
Let G be a nontrivial finite “p-group". Then the center is nontrivial :
Z (G ) 6= {e}.
Idea of the proof
Use the class equation.
Algebra 1
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MOTIVATION
Groups of order
Permutation
p2
Theorem
If |G | = p 2 for some prime p > 1, then G is abelian.
Idea of the proof
1
2
3
Deduce from the last result that |Z (G )| = p or p 2 .
What does it means if |Z (G )| = p 2 ?
Suppose |Z (G )| = p. Take a ∈ G \Z (G ). Prove that ZG (a) = G and
deduce a contradiction.
Algebra 1
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MOTIVATION
Permutation
Cauchy theorem
Theorem (Cauchy’s theorem for abelian finite groups)
Let G be a nontrivial finite abelian group. If p > 1 is a prime that divides
n = |G |, then there is an element a ∈ G of order exactly equal to p – i.e.
G contains an isomorphic copy of (Z/pZ, +).
Idea of the proof
We argue by induction on the of order of G n. The result is trivially true when n = 1
even n = 2, why ? We procede to the induction step : for n ≥ 3 we assume the result
true for all groups G 0 of order less than n and all primes p > 1 that divide |G 0 | ; we
must prove the result holds for any group G of order n and prime divisor p|n. We
distinguish several possibilities.
1
G has no proper subgroups H. Prove that if G is non cyclic it has a proper
subgroup, prove that if G is cyclic with order n which is not a prime it has a
proper subgroup. Deduce that G ' Z/pZ with p prime and deduce the theorem.
2
There is a proper subgroup N.
1
2
Suppose p||N| and deduce the theorem from the induction hypothesis.
Suppose p - |N|. Prove that p||G /N|. Apply the induction hypothesis to G /N (why is it
important to know that G is abelian in order to apply the induction hypothesis ?) and get
a inG /N such that o(a) = p with a preimage of a via the quotient map π : G → G /N. Prove
that ap ∈ ker (π) = N but a ∈
/ ker (π) = N. Prove that (ap )|N| = e and than b = a|N| is an
element of order p in G . That finishes the proof.
Algebra 1
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MOTIVATION
Permutation
Cauchy theorem
Theorem (Cauchy’s theorem for finite groups)
Let G be any nontrivial finite group. If p > 1 is a prime that divides
n = |G |, then there is an element a ∈ G of order exactly p.
Idea of the proof
We argue by induction on the of order of G n. The result is trivially true when n = 1
even n = 2. We procede to the induction step : for n ≥ 3 we assume the result true for
all groups G 0 of order less than n and all primes p > 1 that divide |G 0 | ; we must prove
the result holds for any group G of order n and prime divisor p|n. We distinguish
several possibilities.
1
there is a subgroup H 6= G such that p divides |H|. Deduce the theorem by
induction.
2
p does not divide the order of any subgroup H 6= G . Deduce that p||Cx | for
any |Cx | > 1. Deduce that p||Z (G )| (Hint : class equation) and deduce a
contradiction and thus that this case cannot happens unless G is abelian and that
this finishes the proof of this theorem.
Algebra 1
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Contents
MOTIVATION
Permutation
What you need to know at least about this chapter.
You need to know
1
the definition of a group action, orbits, stabilizer, how to define the homomorphism
associated to an action and the action kernel, what is a transitive action and how to
prove all this with the correct notations and operations.
2
what is an action by permutation on a quotient space, by conjugation and they
properties, what is a conjugacy class, centralizer, the two version of the class equations
(and how to obtain them.
3
Caley theorem.
4
that G /Stab(x) is in bijection with O(x) for each x ∈ X and so |O(x)|||G |
5
what is a p-group, that the center of a p-group is non-trivial, that group of order p 2 are
abelian.
6
Cauchy theorem.
Algebra 1
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MOTIVATION
Permutation
The Structure of a Permutation
Definition
The permutation group Sn is the collection of all bijective maps
σ : X → X of the set X = {1, 2, . . . , n}, with composition of maps ◦ as
the group operation.
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MOTIVATION
Permutation
The Structure of a Permutation
Definition
For k > 1, a k-cycle is a permutation σ = (i1 , . . . , ik ) that acts on X in
the following way
i1 → i2 → . . . → ik → i1
(a cyclic shift of list entries)
σ maps
j →j
for all j not in the list {i1 , . . . , ik }
One-cycles (k) are redundant ; every one-cycle reduces to the identity
map id X , so we seldom write them explicitly, though it is permissible and
sometimes useful to do so.
The support of a k-cycle is the set of entries supp(σ) = {i1 , . . . , ik }, in
no particular order. The support of a 1-cycle (k) is the one-point set {k}.
Warming
Recall that the order of the entries in a cycle (i1 , . . . , ik ) matters, but
cycle notation is somewhat ambiguous : k different symbols obtained by
cyclic shifts of the entries in σ all describe the same operation on X .
(i1 , . . . , ik ) = (i2 , . . . , ik , i1 ) = (i3 , . . . , ik , i1 , i2 ) = . . . =
(ik , i1 , . . . , ik−1 )
Algebra 1
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MOTIVATION
Permutation
The Structure of a Permutation
Fact
If σ = (m1 , . . . , mk ) and τ = (n1 , . . . , nr ) are disjoint cycles, so that
supp(σ) ∩ supp(τ ) = {m1 , . . . , mk } ∩ {n1 , . . . , nr } = ∅
then these operations commute στ = τ σ. If supports overlap, the cycles
may or may not commute.
Algebra 1
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MOTIVATION
Permutation
The Structure of a Permutation
Theorem
Every σ ∈ Sn has a factorization σ =
X =
r
[
supp(σi )
Qr
i=1
σi into cycles whose supports are disjoint and fill X
and
supp(σi ) ∩ supp(σj ) = ∅
for i 6= j
(11)
i=1
Some factors may be trivial 1-cycles, which must be written down to get the support condition (12).
The factors σi are uniquely determined, and they commute.
Idea for the proof
If σ = e the result is clear (why ?)
So, assume σ 6= e and consider the cyclic group it generates H = hσi ⊆ Sn .
1
Prove that τ : H × X → X sending (σ i , x) to σ i (x) is a group action that determines various
disjoint H-orbits that partition X = (H · x1 ) ∪ . . . ∪ (H · xr ).
Let’s label the orbits Oi = H · xi in order of increasing size, so that 1 ≤ |O1 | ≤ . . . ≤ |Or |. For
each orbit Oi we are going to define a cycle τi such that supp(τi ) = Oi .
2
Let x ∈ X . Prove that there exist a k such that O = H ·x = {x, σ(x), . . . , σ k−1 (x)} with
σ k (x) = x and {x, σ(x), . . . , σ k−1 (x)} all distinct.
3
Consider the k-cycle τx = (x, σ(x), . . . , σ k−1 (x)) with support supp(τ ) = O, show that for any
choice y ∈ O, we have τx = τy , (the resulting cycle does not depend of the base point x ∈ O)
4
Define
S the cycles τ1 , . . . , τr , one for each orbit as before. Why are they disjoint ? Why
X = ri=1 supp(τi ) ?
Qr
5 Prove that σ =
i=1 τi .
6
Why uniqueness of the cycles τi is built into the above construction ?
Algebra 1
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MOTIVATION
Permutation
Cycle Types of Permutations.
Definition
If σ ∈ Sn and H = hσi the cycle type of σ is the list of integers
1 ≤ n1 ≤ n2 ≤ . . . ≤ nr
such that
n1 + . . . + nr = n
(12)
determined by listing the H-orbits in X in order of increasing size and
taking ni = |Oi |. The ni are just the lengths of the cycles in the unique
disjoint cycle decomposition of σ.
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MOTIVATION
Permutation
Cycle Types of Permutations
Exercise
Complete the following table.
Cycle Type
Example
?
?
#(Elements in S5 )
?
Algebra 1
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MOTIVATION
Permutation
Parity of a Permutation
Lemma
For n ≥ 2, every σ ∈ Sn can be written as a product of finitely many
2-cycles.
Idea for the proof
Prove that
(1, 2, . . . , k) = (1, k)(1, k − 1) · · · (1, 3)(1, 2)
and deduce the lemma.
Warming
This decomposition is far from being unique but the parity of the number
of 2-cycles on this decomposition is unique and that is an important
invariant as we will see.
Algebra 1
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MOTIVATION
Permutation
Parity of a Permutation
Theorem (The Parity sgn(σ) of a Permutation)
If σ ∈ Sn is decomposed as a product σ = σ1 · · · σr of 2-cycles, then the
Parity :
sgn(σ) = (−1)
r
(r = number of factors)
(13)
is uniquely determined. Furthermore, sgn(e) = 1 and parity is multiplicative
sgn(στ ) = sgn(σ) · sgn(τ )
for all
σ, τ ∈ Sn
(14)
Thus the parity map sgn : Sn → {±1} is a homomorphism from the group of operators (Sn , ◦)
to the 2-element multiplicative group ({±1}, · ).
Idea for the proof
Each σ ∈ Sn can be thought of as a permutation of vectors in the standard orthonormal basis
X = {e1 , . . . , en } in Rn , with σ̃ : ek 7→ eσ(k) . That action induces a linear operator σ̃ : Rn → Rn given
by


n
n
X
X
σ̃ 
ak ek  =
ak eσ(k)
where ak ∈ R, 1 ≤ k ≤ n
k=1
k=1
The matrix [σ̃] = [σ̃]X of σ̃ with respect to the standard basis is called a permutation matrix. These
matrices are characterized by the following (i) Each entry is 0 or 1, (ii) Each column contains exactly
one “1” and (iii) Each row contains exactly one “1”.
1
Prove that The correspondence Φ : σ → σ̃ is a homomorphism mapping Sn into the group
(GL(n), ◦) of invertible linear operators on Rn .
2
Prove that det (Φ(σ)) = sgn(σ) and deduce the Theorem.
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MOTIVATION
Permutation
Parity of a Permutation
Definition
We say a permutation is even if it can be written as a product of an even
number of 2-cycles, so that sgn(σ) = +1, and otherwise it is odd.
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MOTIVATION
Permutation
Parity of a Permutation
Proposition
In Sn all k-cycles have the same parity, namely
sgn(σ) = (−1)k−1
for k = 1, 2, . . .
(15)
If σ is a product
σ = σ1 . . . σr of cycles of various lengths then
Qr
sgn(σ) = j=1 sgn(σj ), regardless of whether or not the cycles are
disjoint.
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MOTIVATION
Permutation
The alternating group
Definition
The alternating group An is the set of all even permutations,
An = ker (sgn) = {σ ∈ Sn : sgn(σ) = +1}
(16)
An is a normal subgroup in Sn , with index |Sn /An | = 2
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MOTIVATION
Permutation
Conjugacy Classes in Sn
Theorem
Let σ = (m1 , . . . , mk ) be any cycle. Conjugation by τ ∈ Sn yields another cycle of the same
length
−1
τ (m1 , . . . , mk )τ
= (τ (m1 ), . . . , τ (mk ) )
In other words, the conjugate of σ is a new k-cycle whose entries are the τ -images of the
entries in σ, in the same cyclic order.
Corollary
For n ≥ 2, let σ ∈ Sn and let τ στ −1 be any conjugate. Then
1
The support of the conjugate is the τ -image of supp(σ), so that
−1
supp(τ στ ) = τ supp(σ)
2
If σ = σ1 · · · σr is the disjoint cycle decomposition of σ then the decomposition of the
conjugate is τ στ −1 = τ σ1 τ −1 · . . . · τ σr τ −1 .
3
If σi = (m1 , . . . , mk ) then τ σi τ −1 = (τ (m1 ), . . . , τ (mk )).
4
Conjugate elements in Sn have the same cycle types.
Corollary
Let n ≥ 2. Two elements σ, σ 0 are conjugate in Sn if and only if they have the same cycle type.
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MOTIVATION
Permutation
Simple group
Definition
A group G is simple if it has no proper normal subgroups.
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MOTIVATION
Permutation
An simple for n ≥ 5.
Lemma
For n ≥ 3 the alternating group An is generated by the set of all 3-cycles
in Sn .
Idea for the proof
1
Explain why it is enough to prove that every product (i, j)(k, `) is a product of 3-cycles.
2
Write e as a product of two 3-cycle, compute (i, j)(j, `) when j = k and prove that
(i, j)(k, `) = (i, j, k)(j, k, `) if the 2-cycles have no entries in common. Deduce the
theorem.
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MOTIVATION
Permutation
An simple for n ≥ 5.
Lemma
If n ≥ 5 all 3-cycles are conjugate in An – i.e. if σ, σ 0 are 3-cycles, then
there exists some τ ∈ An such that σ 0 = τ στ −1 .
Idea for the proof
1
Why all 3-cycles σ and σ 0 are conjugate in Sn , that is here exists some τ ∈ An such that
σ 0 = τ στ −1 ?
2
Since n ≥ 5, how can you find (r , s) two cycles such that τ 0 = τ · (r , s) still conjugates σ
and σ 0 . Use this to deduce the theorem.
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MOTIVATION
Permutation
An simple for n ≥ 5.
Theorem
If n ≥ 5 the alternating group An is simple.
Idea for the proof
We need to prove that An has no proper normal subgroup. For this, let N be a non trivial
normal subgroup of An , we need to prove that N = An .
Why even if An is generated by the 3-cycles we do not know that N contains a 3-cycle ? Why
the theorem is proven if we prove that N contains a 3-cycle ?
Let σ ∈ N,
1
Prove that σ is of one of this 3 forms σ = (1, · · · , r )τ , r ≥ 4 with τ disjoint from
(1, · · · , r ), σ = (1, 2, 3)τ with τ disjoint from (1, 2, 3) or (1, 2)(3, 4)τ , with τ disjoint
from (1, 2)(3, 4).
2
Case 1 : σ = (1, · · · , r )τ , r ≥ 4 take δ = (1, 2, 3) and prove that
σ
−1 −1
δ
σδ = (r , . . . , 1)(1, 3, 2)(1, . . . , r )(1, 2, 3) = (2, 3, r )
Deduce the theorem in this case.
3
Case 2 : σ = (1, 2, 3)τ , take δ = (1, 2, 3) prove that σ −1 δ −1 σδ = (1, 4)(2, 3) and take
µ = (1, 5, 2), prove that µ(1, 4)(2, 3)µ−1 = (1, 3)(4, 5) and
(1, 4)(2, 3)(1, 3)(4, 5) = (1, 2, 3, 4, 5). Deduce the theorem in this case from case 1.
4
Case 3 : (1, 2)(3, 4)τ . Deduce the theorem in this case from case 2.
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MOTIVATION
Permutation
Subgroups of S3 .
Subgroups of S3
1
2
3
What is the cardinality of S3 ? Deduce the possible order for the
cardinality of the subgroups of S3 ? Describe the different subgroup
of S3 up to isomorphism in terms of well known groups ?
What is the cardinality of A3 ? Prove that there is only one subgroup
of cardinality 3 in S3 , use counting principle ? Is it a normal subgroup
of S3 ? Is A3 simple ?
How many subgroup of order 2 is there in S3 ? Describe them. Are
they normals ?
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MOTIVATION
Permutation
Useful criterion for normal subgroup
Lemma
Let C0 = (e), C1 , . . . , Cr be the distinct conjugacy classes in a group G ,
and let H be any subgroup. Then H is normal in G if and only if it is a
union H = Ci1 ∪ . . . ∪ Cis of whole conjugacy classes from G .
Consequence
If H is a normal subgroup of G , we have the following purely numerical
constraints
(i ) |H| = |Ci1 | + . . . + |Cis |
(ii ) |H| must divide |G |
that restrict the combinations of classes whose union can be a subgroup
H.
Remember the conjugacy class of e which is {e} is always here.
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MOTIVATION
Permutation
Normal subgroup of S4
Normal subgroup of S4
1
2
3
4
What is the cardinality of S4 ? Deduce the possible order for the
cardinality of the subgroups of S4 ? Using the previous lemma and
consequence, what cardinality cannot be considered for normal
subgroups of S4 ?
What is the cardinality of A4 ? Is it a normal of S4 Prove that A4 is
the only normal subgroup of cardinality |A4 |.
Prove that there is a unique normal subgroup of S4 of order 4 using
the previous lemma and consequence. Is this group abelian ? Write
the table for this group.
How many normal subgroup are there in S4 ? Is A4 simple ?
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Permutation
What do you need at least to know about this chapter.
You should know
1 the definition of the permutation group, cardinality of the
permutation group, parity of a permutation, k-cycle, support, cycle
type of a permutation, how is defined signature homomorphism, the
definition of the alternative group (that it is simple, normal in Sn and
its properties), how to decompose a permutation in disjoint cycles
(and that this decomposition is unique) and how to decompose a
permutation in in 2-cycle, compute the parity, order of a permutation.
2 How to count the number of permutation of a certain cycle type,
how to obtain the subgroup of S3 and the normal subgroup of S4
(using cycle types) and that A3 is simple but not A4 .
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MOTIVATION
Permutation
Direct products of groups
Definition (External Direct Product)
Given groups A1 , . . . , An their external direct product is the Cartesian product set G = A1 × . . . × An
equipped with component-by-component multiplication of n-tuples. If
a = (a1 , . . . , an ), b = (b1 , . . . , bn ) in the Cartesian product set G , their product is
a · b = (a1 , . . . , an ) · (b1 , . . . , bn ) = (a1 b1 , . . . , an bn )
for all ai , bi ∈ Ai
The identity element is e = (e1 , . . . , en ) where ei is the identity element in Ai ; the inverse of an element
−1
a is a−1 = (a1
, . . . , an−1 ).
There is a natural isomorphism between Ai and the subgroup
Ai = (e1 ) × . . . × Ai × . . . (en ) ,
the n-tuples whose entries are trivial except for ai . It follows easily that
1 Each Ai is a subgroup in G .
2
The bijective map Ji (ai ) = (e1 , . . . , ai , . . . , en ) defines an isomorphism from Ai to Ai .
3
The Ai commute with each other in the sense that xy = yx if x ∈ Ai , y ∈ Aj and i 6= j. Note
carefully what 3. does not say : the subgroup Ai need not commute with itself (the case when
i = j) unless the group Ai happens to be abelian.
4
Each Ai is a normal subgroup in G .
5
The product set
A1 · . . . · An = {x1 . . . xn : xi ∈ Ai , 1 ≤ i ≤ n}
is all of G .
6
Every x ∈ G has a unique factorization
x = x 1 ·. . .·x n
with xi ∈ ai
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Permutation
Direct products of groups
Example
1
2
G = (Rn , +) is precisely the direct product group (R × . . . × R, +)
made up of n copies of the real line (R, +).
G = (Zn , +) is precisely the direct product group (Z × . . . × Z, +)
made up of n copies of the group of the integers (Z, +).
Let
Λ = {a1 u1 + . . . + an un : ai ∈ Z} = Zu1 + . . . + Zun
Prove that (Λ, +) is an abelian group isomorphic to (Zn , +).
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MOTIVATION
Permutation
Direct products of groups
Theorem (Internal Direct Product)
Let G be a group and A1 , . . . , An subgroups such that
(i ) Each Ai is a normal subgroup in G
(ii ) The product set A1 · · · An is equal to G .
(iii ) Each g ∈ G decomposes uniquely as g = a1 · · · an with
ai ∈ Ai .
Then G is isomorphic to the direct product group A1 × . . . × An . In
particular, elements of Ai and Aj automatically commute if i 6= j.
We say that G is the internal direct product of A1 , · · · , and An
(Note that (ii) insures the existence of such a factorization while (iii)
insures its uniqueness.)
Idea for the proof
1
Prove that if i 6= j we have pairwise disjointness Ai ∩ Aj = (e)
2
Deduce that Ai and Aj commute when i 6= j.
3
Prove that the map p : A1 × . . . × An → G is an isomorphism.
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Permutation
Direct products of groups
Corollary
Let G be a group and A, B two subgroups such that
(i ) A and B are normal subgroups in G .
(ii ) The product set AB is equal to G .
(iii ) A ∩ B = (e).
Then G is isomorphic to the direct product A × B under the map
p(a, b) = a · b.
Remark
If you prove that in a group G , you have two subgroup A and B such
that satisfying (i), (ii) and (iii). Then you will know that G is isomorphic
to the direct product A × B.
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MOTIVATION
Permutation
Direct products of groups
Groups of order 4
Up to isomorphism, describe all groups of order |G | = 4.
Solution : We first consider the largest possible order o(b) for an element
of G , what are the possibilities for this largest possible order ? Justify.
1 Case 1 : o(b) = 4. Describe G .
2 Case 2 : o(b) = 2. Prove that there exist two distinct elements x
and y in G of order 2. Let A =< x > and B =< y >, prove that G
is the internal direct product of A and B. Describe G .
3 Are the groups found in case 1 and 2 isomorphic ?
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MOTIVATION
Permutation
Chinese Remainder Theorem
Exercise
Here are two systems of congruences
x ≡ 5 (mod 3)
(a)
x ≡ 1 (mod 12)
(b)
x ≡5
x ≡1
(mod 3)
(mod 5)
Taking a “bare hands” approach we shall verify that the system (a) has
no solutions and that the solutions of (b) are of the form x0 + k · 15
where x0 = 11 and k ∈ Z.
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MOTIVATION
Permutation
Chinese Remainder Theorem
Theorem (Chinese Remainder Theorem)
If m, n > 1 are relatively prime, so that gcd(m, n) = 1, then
Z/mZ × Z/nZ ∼
= Z/mnZ as additive groups. Furthermore, the system of
congruences
x ≡ a1
x ≡ a2
(mod m)
(mod n)
has a solution for every choice of a1 , a2 ∈ Z, and if x0 ∈ Z is one solution
the full set of solutions in Z is the congruence class [x0 ]mn = x0 + Zmn.
Idea for the proof
Define
ψ:
Z/mnZ
[k]mn
→
7
→
Z/mZ × Z/nZ
([k]m , [k]n )
1
Prove that ψ is well defined. That is, if [k]mn = [k 0 ]mn , then [k]m = [k 0 ]m and [k]n = [k 0 ]n .
2
Prove that ψ is an isomorphism.
3
Prove that the unique preimage of ([a1 ]m , [a2 ]n ) via ψ is the solution of the congruence
equation of the theorem.
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MOTIVATION
Permutation
Chinese Remainder Theorem
Algorithm for Solving the Remainder Problem.
If gcd(m, n) = 1 we know that solutions exist and are unique up to an added multiple of mn ;
Prove that
[x0 ]m = [a1 ]m
x0 = a1 + Km
⇔
there exist K , L ∈ Z such that
[x0 ]n = [a2 ]n
x0 = a2 + Ln
Thus x0 = a1 + Km = a2 + Sn which implies a2 − a1 = Km − Ln.
Since gcd(m, n) = 1 we can find r , s ∈ Z such that 1 = rm + sn = rm − (−s)n. Multiplying
both sides by (a2 − a1 ) yields
(a2 − a1 ) = [(a2 − a1 )r ] m − [(a2 − a1 )(−s)] n
from which we can read off suitable values for K , L.
K = (a2 − a1 )r
L = −(a2 − a1 )s
A particular solution is then x0 = a1 + Km = a2 + Ln (both values being equal) and the full
solution set will be the set x0 + Z/mnZ.
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MOTIVATION
Permutation
Chinese Remainder Theorem
Corollary
Z/n1 n2 . . . nr Z ∼
= Z/n1 Z × . . . × Z/nr Z
if the ni are pairwise relatively prime (so gcd(ni , nj ) = 1 if i 6= j)
Idea for the proof
Use induction.
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MOTIVATION
Permutation
Chinese Remainder Theorem
Corollary
If n1 , . . . , nr are pairwise relatively prime, with gcd(ni , nj ) = 1 for i 6= j,
use induction on r to prove that
Un1 n2 ...nr ∼
= Un1 × . . . × Unr
and
|Un1 n2 ...nr | = |Un1 | · . . . · |Unr | Idea for the proof
Prove that a unit is sent to a unit via ψ and thus ψ
induces(Umn , · ) ∼
= (Um , · ) × (Un , · ) and use induction.
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MOTIVATION
Permutation
Semi-direct product)
Proposition (Internal semi direct product)
Let G be a group and N, H two subgroups such that
(i ) N is normal in G
(iii ) N ∩ H = (e)
(ii ) NH = G
Then, each g ∈ G has a unique factorization as g = nh, so there is a natural bijection between
G and the Cartesian product set N × H.
The group action H × N → N via conjugation operators φh (n) = hnh−1 in Aut (N) determines
the following group operation in the Cartesian product set N × H.
0
0
0
0
(n, h) ∗ (n , h ) = (nφh (n ) , hh )
0
0
for n, n ∈ N and h, h ∈ H
(17)
This operation makes the Cartesian product set into a group,
and the product map p(n, h) = nh becomes an isomorphism from (N × H, ∗ ) to the original
group G .
If the condition of the previous theorem are satisfied, we say that G is the internal semidirect
product of the subgroups N and H, which we indicate by writing G = N×|H or N×|φ H to
distinguish it from the direct product N × H.
Of course if the action H × N → N is trivial, as when the subgroups H and N commute (so
that φh = id N for all h ∈ H), then the semidirect product reduces to the ordinary direct
Algebra 1
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Permutation
Semi-direct product)
Idea for the proof
1
2
Prove first that (N × H, ∗ ) is a group. Closure is clear, associativity
is annoying to prove but not hard. Prove that the identity element is
(e, e) and the inverse of an element (n, h) ∈ N × H is
(φh−1 (n−1 ), h−1 ).
Prove then that p is an isomorphism.
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MOTIVATION
Permutation
Semi-direct product
Theorem (External semi direct product)
Given abstract groups N, H and a homomorphism Φ : H → Aut (N),
define the binary operation on the Cartesian product set G = N × H :
(n, h)∗(n0 , h0 ) = (nφh (n0 ) , hh0 )
for all n, n0 ∈ N and h, h0 ∈ H (18)
where φh = Φ(h) ∈ Aut (N).
Then (G , ∗) is a group, the external semidirect product, which we
denote by N×|φ H. The identity element is (eN , eH ) and the inverse of an
element (n, h) ∈ N × H is
(n, h)−1 = (φh−1 (n−1 ), h−1 )
The subsets N = {(n, eN ) : n ∈ N} and H = {(eH , h) : h ∈ H} are
subgroups in G that are isomorphic to N and H.
They satisfy the conditions N ·H = G , N ∩ H = {(eN , eH )}, and N is
normal in G . Thus G is the internal semidirect product of these
subgroups.
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MOTIVATION
Permutation
Internal semi-direct product
Example (The Dihedral Groups Dn )
These groups of order |Dn | = 2n, defined for n ≥ 2, are the full symmetry
groups of regular n-gons. Let θ = 2π/n radians, and define the basic
symmetry operations
ρθ
σ
= (counterclockwise rotation about the origin by θ radians)
= (reflection across the x-axis)
The dihedral group Dn is the subgroup Dn = hρθ , σi generated by ρθ
and σ in the group O(2) of linear isometries (distance-preserving
transformations) of the plane. Obviously N = hρθ i is a cyclic subgroup
isomorphic to Z/nZ and H = hσi is a copy of Z/2Z embedded in Dn .
Show that N is normal and Dn is the semidirect product N×|H.
Note that Dn is fully define as follow :
Dn =< σ, ρ > with o(σ) = 2, o(ρ) = n and σρσ = σρσ −1 = ρ−1
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Permutation
Semi-direct product
Example (The Group of Affine Mappings Aff(V))
The affine mappings on a finite dimensional vector space V are the
bijective maps of the form
T (v ) = A(v ) + a
where A is a bijective linear map and a ∈ V
The set of all those affine mapping is denoted as Aff (V ) and it is a
group under composition.
Within G = Aff (V ) we find two natural subgroups
Translations : N = {ta : a ∈ V }. This subgroup is abelian,
normal and the map j : a 7→ ta is an isomorphism between
(V , +).
Invertible Linear Operators :
H = GL(V ) = {A|A : V → V is a bijective linear map}.
Prove that Aff (V ) is the semi direct product N×|φ H.
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MOTIVATION
Permutation
Semidirect products Z/nZ o Z/mZ
Proposition
If Φ : (Z/mZ, +) → (Un , · ) is the homomorphism that sends the
generator a = [1]m of Z/mZ to an element b = [r ]n in Un that satisfies
the compatibility condition b m = [1]n , then the group operation in the
semidirect product Z/nZ×|Z/mZ takes the form
([i]n , [j]m ) · ([k]n , [`]m ) = ([i + r j k]n , [j + `]m )
(19)
Idea of the proof :
In order to determine all the possible external semi-direct product of Z/nZ o Z/mZ it is
enough to determine all the homomorphism Φ : Z/mZ → Aut (Z/nZ).
1
Why finding a homomorphism Φ : Z/mZ → Aut (Z/nZ) is equivalent to finding a
homomorphism Φ̃ : (Z/mZ, +) → (Un , · ) ?
2
Explain why such a homomorphism is defined sending the generator a = [1]m of Z/mZ to
an element b = [r ]n in Un that satisfies the compatibility condition b m = [1]n .
3
Describe the homomorphism Φ : Z/mZ → Aut (Z/nZ) corresponding to Φ̃.
4
Describe the group operation define by [r ]n .
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MOTIVATION
Permutation
Group of order 6
Group of order 6
Let G be an arbitrary group of order |G | = 6.
1 What are the possible order for subgroup of G ?
2 Why subgroups of G of order 2 and 3 exist ? Denote H
3 a subgroup
of G of order 3 and H2 a subgroup of G of order 2.
3 To which well known groups are isomorphic H
3 and H2 ?
4 Is H
3 necessary normal in G ? What about H2 ?
5 Prove that H ∩ H = {e} and G = H H .
2
3
2 3
6 Prove that G is a semi direct product of H
2 and H3 .
7 Describe all the possible external semi direct product Z/3Z o Z/2Z.
8 Deduce all the possible group of order 6 up to isomorphism ? How
many are there ? Which ones are commutative ? Find Z/6Z, D3 and
S3 .
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MOTIVATION
Permutation
Semi direct product
Z/5Z×|Z/3Z
1
2
3
Determine the group of units (U5 , · ) and its isomorphism type.
Find all possible homomorphisms Φ : (Z/3Z, +) → (U5 , · ).
Describe the possible semidirect products G = Z/5Z×|Z/3Z.
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MOTIVATION
Permutation
Semi direct product
Z/9Z×|Z/3Z
1
2
3
Determine the group of units (U9 , · ) and its isomorphism type.
Find all possible homomorphisms Φ : (Z/3Z, +) → (U9 , · ).
Describe the possible semidirect products G = Z/9Z×|Z/3Z.
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MOTIVATION
Permutation
Group extensions
Exact sequence
Suppose N C G is a normal subgroup. Then there is a natural exact
sequence of homomorphisms
φ1 =id
φ2 =π
e −→ N −−−−−−→ G −−−−−−→ H = G /N −→ e
(20)
where π : G → G /N is the quotient map. Here “exact” means
range(φi−1 ) = ker (φi ) at every step in the sequence.
The middle group G in this exact sequence is called an extension of the
group G /N by the group N.
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Permutation
Group extensions
Definition
Let G be a group and N a normal subgroup, as in
φ1 =id
φ2 =π
e −→ N −−−−−−→ G −−−−−−→ H = G /N −→ e
A subgroup H is a cross-section for G /N if each coset in G /N meets
the set H in a single point ; in particular, H ∩ N = (e).
If such a subgroup exists we say that the exact sequence splits.
Such subgroups are generally not unique.
The idea of the cross section is that
Every coset C in G /N has a unique representative x (i.e.
xN = C ) such that x lies in the cross-section subgroup H.
In particular H ∼
= G /N for any cross-section.
In effect, cross-sections H are copies of the quotient group G /N
embedded back inside G .
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Permutation
Group extensions
Lemma
Let G be a group, H a subgroup and N a normal subgroup. Then the
following statements are equivalent
(a) The product set NH is equal to G and N ∩ H = (e)
(b)
Each g ∈ G has a unique factorization g = nh with n ∈ N, h ∈ H.
(c) H is a cross-section for G /N cosets
These conditions are satisfied precisely when the previous exact sequence
splits, and then we have H ∼
= G /N.
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Permutation
Group extensions
Example
If we take G = R and N = Z, the extension e → Z → R → G /N → e
does not split, so G is not a semidirect product of N and another
subgroup H. To see why, prove that G /N is isomorphic to the circle
group S 1 = {z ∈ C : |z| = 1}.
Arguing by contradiction, we now show that no subgroup H in R can
cross-section the cosets x + Z in R/Z. Suppose such an H actually
exists. Consider any rational value 0 < θ < 1.
1 Prove that there is x ∈ H such that x − θ ∈ Z,
2 Explain why there is a m such that mθ ∈ Z.
3 Deduce that φ(mx) = φ(mθ) = 1.
4 Explain why, the restricted homomorphism φ
: H → S 1 is a
H
bijection.
5 Deduce that x = 0 and the contradiction.
Algebra 1
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Contents
MOTIVATION
Permutation
Sylow theorem
Definition
Qr
Let G be a nontrivial finite group whose order has n = i=1 pini as its
prime factorization.
A subgroup H ⊆ G is a pi -group if its order is some power of pi .
It is a Sylow pi -subgroup if its order is as large as possible, namely
|H| = pini .
For any prime p > 1 we write Syl p (G ) to indicate the collection of all
Sylow p-subgroups in G .
Unless p is a divisor of |G | this collection will be empty ; otherwise
Syl p (G ) might contain several distinct Sylow p-subgroups for each divisor
p.
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Contents
MOTIVATION
Permutation
Sylow theorem
Remark
In analyzing the structure of groups, special interest attaches to the
p-groups, for which |G | = p k , so it is worth noting that
A finite group G is a p-group ⇔ the order of each element is a power of
p.
Algebra 1
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Contents
MOTIVATION
Permutation
Sylow theorem
lemma
Let X be a finite set acted on by p-group G . Let
X G = Fix G (X ) = {x ∈ X : g · x = x, all g ∈ G }
be the set of G -fixed points in X . Then |X G | ≡ |X | (mod p). In
particular fixed points must exist if |X | 6≡ 0 (mod p).
Idea of the proof
|X | = |X ∼ X G | + |X G |
Prove that |X ∼ X G | is the disjoint union of non trivial orbits and
conclude.
Algebra 1
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Contents
MOTIVATION
Permutation
Sylow theorem
The Sylow theorems
Qr
Let G be a nontrivial finite group of order n = i=1 pini . Then for each
prime factor pi
n
1 G contains a subgroup Sp that has exactly p i elements.
i
i
2 If S
pi is a fixed Sylow pi -subgroup, any subgroup H whose order is a
power of pi can be conjugated to lie within Spi – i.e. there is some
g ∈ G such that gHg −1 ⊆ Sp . In particular all Sylow pi -subgroups
are conjugates of one another.
3 The number of distinct Sylow p -subgroups is equal to 1 + mp for
i
i
some m, so their number is congruent to 1 (mod pi ). The number of
pi -Sylow subgroups must also be a divisor of |G |.
For each prime divisor p > 1, all the Sylow p-subgroups are isomorphic
since they are conjugates.
If p k is the largest power dividing n, one can generalize the Sylow
theorems to prove that there exist subgroups of order p r for every
1 ≤ r ≤ k.
Algebra 1
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Contents
MOTIVATION
Permutation
Sylow theorem
The Sylow theorems
Qr
Let G be a nontrivial finite group of order n = i=1 pini . Then for each
prime factor pi , G contains a subgroup Spi that has exactly pini elements.
Idea of the proof
We start with the abelian case, working by induction on n = |G | with n > 1. Prove the
theorem for n = 2. So, we may assume n > 2 and p > 1 is one of its prime divisors, say with
p k the largest power dividing n and suppose that the theorem is true for all the groups of order
strictly less than n.
1
Prove that G contains a subgroup H of order p and that if k = 1, we are done.
2
Suppose k > 1. Prove that p k−1 ||G /H| and construct a Sylow group using the induction
assumption
For general groups G , we argue again by induction on n = |G |, and again the case n = 2 is
trivial (not to mention abelian).So, assume n > 2 and that p > 1 is a prime divisor whose
largest power in n is p k and suppose that the theorem is true for all the groups of order strictly
less than n.
1
If there is a nontrivial class Cx such that |ZG (x)| is divisible by p k , conclude for the
theorem by induction.
2
If there is no such a class, deduce that p||Z (G )|. Take a Sylow p-group Hp for the center.
Consider G /Hp and deduce the theorem by induction.
Algebra 1
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Contents
MOTIVATION
Permutation
Sylow theorem
The Sylow theorems
Qr
Let G be a nontrivial finite group of order n = i=1 pini . Then if Spi is a
fixed Sylow pi -subgroup, any subgroup H whose order is a power of pi
can be conjugated to lie within Spi – i.e. there is some g ∈ G such that
gHg −1 ⊆ Sp . In particular all Sylow pi -subgroups are conjugates of one
another.
Idea of the proof
What if |G | = p m ?
Suppose now that |G | = n with n = p k m with gcd(m, p k ).
Fix a Sylow p-group Sp and let H be a subgroup whose order is a power
of p.
Consider the permutation action H × X → X on the coset space
X = G /Sp .
Deduce from the previous lemma on fix points that this action has a fix
point and conclude.
Algebra 1
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Contents
MOTIVATION
Permutation
Sylow theorem
The Sylow theorems
Qr
Let G be a nontrivial finite group of order n = i=1 pini . Then, the
number of distinct Sylow pi -subgroups is equal to 1 + mpi for some m, so
their number is congruent to 1 (mod pi ). The number of pi -Sylow
subgroups must also be a divisor of |G |.
Idea of the proof
1
We fix a Sylow p-subgroup Sp and look at the group action Sp × X → X of Sp by
conjugation on the set X = Syl p of all Sylow p-groups in G .
Is the action transitive ?
Give a fix point for this action and prove it is unique. (Hint : For uniqueness :
Prove that Sp ⊆ N := NG (H). Deduce that there is n ∈ N such that
Sp = nHn−1 , and conclude)
Deduce that |X | ≡ 1 mod p.
2
Let P be any base point in Syl p (G ) and consider the action G × Syl p → Syl p .
Prove it is transitive. and deduced that |G |||Sylp |, using StabG (P).
Algebra 1
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Contents
MOTIVATION
Permutation
Sylow theorem
Remark
By the Sylow theorem, a Sylow p-subgroup is normal in G if and only if
there is just one such subgroup.
Sometimes we can determine when this happens, and a lot can be
learned about the pattern of Sylow subgroups by looking at intersections
of conjugates Sp ∩ gSp g −1 with Lagrange and the Sylow theorems (part
3) in mind. Indeed, the number of pi -Sylow N has to divide |G | and be of
the form 1 + mpi . This gives constraint on N which together with
consider the union of all p-Sylow and it possible cardinality might lead to
contradiction and permit us to conclude the uniqueness of the p-Sylow.
Algebra 1
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Contents
MOTIVATION
Permutation
Sylow theorem
Theorem
A finite abelian
Qr group is isomorphic to a direct product Sp1 × . . . × Spr
where n = i=1 pini is the prime decomposition of the order n = |G | and
Spi is the unique Sylow pi -subgroup in G of order pini . This direct product
decomposition is canonical : the subgroups Spi are uniquely determined,
as are the primes pi and their exponents ni . Warning : The
components Spi need not be cyclic groups. (Think about groups of
order 4, for instance).
Idea of the proof
1
Why all the Sylow subgroup have to be normal ?
2
Use induction in r and counting principle, to prove that G = Sp1 · · · Spr .
3
Note to check the uniqueness of the factorization of g into a product of elements
on the Sylow subgroups, one can show that e = a1 · · · ar with ai ∈ Spi implies
ai = e for all i ∈ {1, · · · , r }. Why ? Prove it.
4
Prove that G is the internal direct product of it Sylow subgroups.
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Contents
MOTIVATION
Permutation
Sylow theorem
Theorem
Let G be a nontrivial finite group and let S be the subgroup generated by
all the Sylow subgroups in G :
[
S= h
{H : H ∈ Syl pi (G ), 1 ≤ i ≤ r } i
where the pi > 1 are the prime divisors of |G |. Then S is all of G .
Idea of the proof
When G is abelian the result has already been proven. We argue by induction on n = |G |. The
result is trivial when n = 2 why ? So we may assume that n > 2 and that the theorem is true
for all groups of order at most n − 1.
Suppose by contradiction, S 6= G , there exist elements a ∈
/ S. Consider the cyclic sugroup of G
M =< a > and denote by m its cardinality.
1
Which prime appear in the factorization of m into prime ?
2
Let p be one of those prime suppose that p n |||G |. What can you say about the order of a
p-Sylow in M.
3
Deduce that such a subgroup for M has to be contained on one of the Sylow p-subgroups
for G .
4
Deduce that the product of all the Sylow subgroup of M is in S and deduce a
contradiction using the abelian case.
Algebra 1
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Contents
MOTIVATION
Permutation
Sylow theorem
Study of group of order 12
Let G be a group of order 12.
1 Describe the Sylow subgroup of G .
2 Why 2-Sylow subgroup and 3 Sylow subgroups exist ? Denote H
2 a
2-sylow subgroup for G and H3 a 3-Sylow subgroup for G .
3 Describe H
2 and H3 up to isomorphism.
4 Why H ∩ H = {e} and G = H H ?
2
3
2 3
5 How many 2-Sylow subgroups can we have ? How many 3-Sylow
subgroups can we have ?
6 Using a counting argument and the union of 3-Sylow subgroups,
prove either H3 or H2 has to be a normal subgroup.
7 Deduce that G is an internal semi direct product either H o H
2
3 or
H3 o H2 .
Algebra 1
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Contents
MOTIVATION
Permutation
Sylow theorem
Study of group of order 12
Let G be a group of order 12.
1 First, suppose H3 is normal and H4 ' Z/4Z. Determine all the
semidirect products of the form Z/3Z o Z/4Z.
2 Now, suppose H
3 is normal and H4 ' Z/2Z × Z/2Z. Determine all
the semidirect products of the form Z/3Z o (Z/2Z × Z/2Z). For this
determine all possible homomorphism
Φ : Z/2Z × Z/2Z → Aut (Z/3Z).
3 Suppose H ' Z/4Z is normal. Determine all the semidirect products
4
of the form Z/4Z o Z/3Z.
4 Suppose H ' (Z/2Z × Z/2Z) is normal. Determine all the
4
semidirect products of the form (Z/2Z × Z/2Z) o Z/3Z.
Algebra 1
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Contents
MOTIVATION
Permutation
Sylow theorem
Study of group of order 12 : Study of the case H3 is normal and
H4 ' Z/2Z × Z/2Z.
Determine all the semidirect products of the form
Z/3Z o (Z/2Z × Z/2Z). For this determine all possible homomorphism
Φ : Z/2Z × Z/2Z → Aut (Z/3Z). We use multiplicative notation and
write
H2 = {u i v j : i, j ∈ Z/2Z} = {e, u, v , uv }
H3 = {e, a, a2 }
1
2
3
with u 2 = v 2 = (uv )2 = e
with a3 = e
Prove that we have either the trivial morphism or Φ(e) = Φ(u) = I
and φ(v ) = J (Inversion map J : ai → a−i on H3 )
Deduce that we have either a direct product or the group law is given
by
j
(ak , u i v j ) ? (a` , u r v s ) = (ak+(−1) ` , u i+r v j+s )
Prove that the later case is D6 . (Hint : Prove that N = H3 · < u > is
a cyclic normal subgroup of G and that G is a semi direct product
No < v > and if ρ = au and σ = v then σρσ −1 = ρ−1 .)
Algebra 1
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Contents
MOTIVATION
Permutation
Sylow theorem
Study of group of order 12 : Study of the case H2 ' Z/2Z × Z/2Z is normal.
Determine all the semidirect products of the form (Z/2Z × Z/2Z)Z o Z/3Z. For this determine all
possible homomorphism Φ : Z/3Z → Aut (Z/2Z × Z/2Z). Important insight is achieved if we relabel
elements of H2 = {u i v j : i, j ∈ Z/2Z} as
x0 = e,
x1 = u,
x2 = v ,
x3 = uv
Observe that xi2 = e and xi xi+1 = xi+2 = xi−1 for 1 ≤ i ≤ 3 when subscripts are reckoned (mod 3).
(Thus for instance, x1 x2 = uv = x3 , x2 x3 = v · uv = u = x1 etc.)
1
Prove that Aut (Z/2Z × Z/2Z) = S3 .
2
Prove that Φ([1]) = σ with σ 3 = Id . Thus we have two cases.
3
Suppose Φ(a) = (1, 2, 3),
Φ(a)(e) = e
4
Φ(a)(u) = v ,
Φ(a)(v ) = uv ,
k
Φ(a)(uv ) = u
Φ(a) xi = xi+k
The group law is given by :
i j
k
r s
`
i j
k
r s
(u v , a ) ? (u v , a ) = (u v · Φ(a) (u v ) , a
k+`
(21)
)
and
k
`
=
(e, a
`
=
(e · Φ(a) (xi ) , a
=
(xi+k , a
`
=
(xi · Φ(a) (e) , a
`
=
(xi · Φ(a) (xj ) , a
(e, a ) ? (e, a )
k
(e, a ) ? (xi , a )
k
(xi , a ) ? (e, a )
k
(xi , a ) ? (xj , a )
5
Prove that this group is isomorphic to A4 .
k+`
)
k
k+`
k
k
k+`
)
k+`
) = (xi , a
)
k+`
k+`
)
) = (xi xj+k , a
k+`
)
Algebra 1
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Contents
MOTIVATION
Permutation
Sylow theorem
Study of group of order 8
Let G be a group of order 8. We argue about the possible of the center
Z = Z (G ).
1
If |Z | = 8 then G is abelian. Prove that G is isomorphic to Z/8Z, Z/4Z × Z/2Z, or
Z/2Z × Z/2Z × Z/2Z. (Hint : argue in term of maximal order).
2
If |Z | = 4, prove that G /Z is cyclic and deduce that G is abelian and deduce a
contradiction.
3
if |Z | = 2. Prove that we must have an element x of order 4. Let N =< x >. Why N is
normal ? So we have the exact sequence :
e −→ N ∼
= Z/4Z −→ G −→ Q = G /N ∼
= Z/2Z −→ e
If it split prove that G ' D4 if not prove that G is the group of unit quaternions
Q8 = {±1, ±i, ±j, ±k} in which the elements ±1 are central,
−i = (−1)i, . . . , −k = (−1)k and
2
(−1) = 1
2
2
2
i = j = k = ijk = −1
from which we get other familiar relations such as ij = k = −ji, jk = i = −kj, and
ki = j = −ik.
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