math42001 - group theory

Group Theory

Michael [email protected]

January 18, 2013

About

These are my own notes from lectures, they may contain errors.

1

CONTENTS 2

Contents

1 Prerequisite Knowledge 4

2 The Symmetric Group (Sn) 52.1 The Alternating Group (An) . . . . . . . . . . . . . . . . . . . 7

3 Subgroups 83.1 Centralizers, Normalizers, Generators . . . . . . . . . . . . . . 83.2 Subgroup Products . . . . . . . . . . . . . . . . . . . . . . . . 11

4 Conjugacy 134.1 Some Properties of Conjugacy . . . . . . . . . . . . . . . . . . 134.2 Counting Conjugates . . . . . . . . . . . . . . . . . . . . . . . 15

4.2.1 Worked Example . . . . . . . . . . . . . . . . . . . . . 164.3 The Class Equation . . . . . . . . . . . . . . . . . . . . . . . . 174.4 Conjugacy in Sn . . . . . . . . . . . . . . . . . . . . . . . . . . 18

5 Group Actions 205.1 G-orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225.2 Cauchy’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . 255.3 Burnside’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . 26

6 Finitely Generated Abelian Groups 286.1 Finitely Generated Abelian Groups . . . . . . . . . . . . . . . 29

6.1.1 Classification of Finitely Generated Abelian Groups . . 29

7 Normal Subgroups 317.1 Quotient Groups . . . . . . . . . . . . . . . . . . . . . . . . . 32

7.1.1 Subgroups of Quotient Groups . . . . . . . . . . . . . . 357.2 Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . 36

7.2.1 Isomorphism Theorems . . . . . . . . . . . . . . . . . . 38

8 Simple Groups 408.1 Composition Series . . . . . . . . . . . . . . . . . . . . . . . . 42

8.1.1 The Jordan-Holder Theorem . . . . . . . . . . . . . . . 43

9 Sylow’s Theorems 459.1 Simple Groups and Sylow’s Theorems . . . . . . . . . . . . . . 47

CONTENTS 3

10 Extension Material 4910.1 Commutator Subgroups . . . . . . . . . . . . . . . . . . . . . 4910.2 Finitely Generated Abelian Groups . . . . . . . . . . . . . . . 5210.3 An is simple for n ≥ 5 . . . . . . . . . . . . . . . . . . . . . . 5410.4 The unique simple group of order 60 . . . . . . . . . . . . . . 55

1 PREREQUISITE KNOWLEDGE 4

1 Prerequisite Knowledge

We should be familar with the definition of a group and subgroup.

Definition 1.1. If G is a group and S ⊆ G, the right cosets of S in G arethe collection of sets

Sg = xg | x ∈ S

for each g ∈ G.

Remark If S ≤ G is a subgroup the following cosets properties hold:

(i) ∀g ∈ G : g ∈ Hg

(ii) ∀a, b ∈ G : Ha = Hb ⇐⇒ ab−1 ∈ H

(iii) ∀a, b ∈ G : Ha = Hb or Ha ∩Hb = ∅

(iv) ∀g ∈ G : |H| = |Hg|

(v) The right cosets are a partition of G

Definition 1.2. If H ≤ G, then the number of right cosets of H in G iswritten [G : H] and is called the index of H in G.

Theorem 1.3 (Lagrange’s Theorem). If G is a finite group and H ≤ G asubgroup, then |G| = |H|[G : H].

Corollary 1.4. If K ≤ H ≤ G are finite, then [G : K] = [G : H][H : K].

2 THE SYMMETRIC GROUP (SN) 5

2 The Symmetric Group (Sn)

Definition 2.1. In Sn a cycle of length 2 is called a transposition.

Lemma 2.2. Let n ≥ 2, then every permutation in Sn can be written as aproduct of transpositions.

Proof. Let σ ∈ Sn and write σ = σ1σ2 . . . σt as a product of disjoint cycles.For each cycle σi we have

σi = (α1, α2, . . . , αr) = (α1α2)(α1α3) · · · (α1αr)

as required.

Definition 2.3. Let n ≥ 2 and σ ∈ Sn, then σ is an even (respectively odd)permutation if it can be written as a product of an even (repsectively odd)number of transpositions.

Theorem 2.4. Any permutation is either even or odd, and cannot be both.

Proof. Let σ ∈ Sn and define C(σ) to be the number of cycles in σ whenwritten as a disjoint union of cycles (including those of length one), anddefine S(σ) = (−1)n−C(σ). The following two lemmas are needed:

Lemma 2.5. If τ ∈ Sn is a transposition, then C(στ) = C(σ)± 1.

Proof. Write σ = (α1, α2, . . . , αa)(β1, β2, . . . , βb)(γ1, γ2, . . . , γc) · · · as a prod-uct of disjoint cycles and suppose τ = (α, β). There are two possibilities:

(i) Suppose α and β both lie in the same cycle of σ, without loss of gener-ality suppose this is

σ = (α1 = α, α2, . . . , αi−1, αi = β, αi+1, . . . , αa) · · ·

then

στ = (α, α2, . . . , αi−1, β, αi+1, . . . , αa)(β1, . . . , βb) · · · (α, β)

= (α, α2, . . . , αi−1, β, αi+1, . . . , αa)(α, β)(β1, . . . , βb) · · ·= (α, α2, . . . , αi−1)(β, αi+1, . . . , αa)(β1, . . . , βb) · · ·

and so C(στ) = C(σ) + 1.


(ii) Suppose α and β lie in seperate cylces of σ, without loss of generalitysuppose we have

σ = (α1 = α, α2, . . . , αa)(β1 = β, β2, . . . , βb) · · ·

then

στ = (α, α2, . . . , αa)(β, β2, . . . , βb)(γ1, . . . , γc) · · · (α, β)

= (α, α2, . . . , αa)(β, β2, . . . , βb)(α, β)(γ1, . . . , γc) · · ·= (α, α2, . . . , αa, β, β2, . . . , βb)(γ1, . . . , γc) · · ·

and so C(στ) = C(σ)− 1.

Lemma 2.6. If σ can be written as a product of r transpositions, then S(σ) =(−1)r.

Proof. Suppose σ = τ1τ2 · · · τr. Now, since τ1 has the form

τ1 = (αβ) (γ1)(γ2) · · · (γn−1)︸︷︷︸(n-1) 1-cycles

we have C(τ1) = n − 1, hence n − C(τ1) = 1. Making repeated use of thefirst lemma, we have

n− C(τ1τ2) = n− (C(τ)± 1) ∈ 0, 2n− C(τ1τ2τ3) = n− (C(τ1τ2)± 1) ∈ 1, 3

n− C(τ1τ2τ3τ4) = n− (C(τ1τ2τ3)± 1) ∈ 0, 2, 4

As so on, it is clear that n−C(σ) = n−C(τ1τ2 · · · τr) is even if r is even andodd if r is odd, therefore

S(σ) = (−1)n−C(σ) = (−1)r

as claimed.

Suppose σ can be written as the product of r1 transpositions and also asa product of r2 transpositions, then by the second lemma

(−1)r1 = S(σ) = (−1)r2

So either both r1 and r2 are even, or both r1 and r2 are odd. Therefore σcannot be both an even and odd permutaiton.


2.1 The Alternating Group (An)

Definition 2.7. For n ≥ 2, define An = σ ∈ Sn | σ is even, the alternatinggroup of degree n

Proposition 2.8. An is a subgroup of Sn.

Proof. Firstly, (1) = (1, 2)(1, 2) is even so (1) ∈ An and An 6= ∅.Suppose σ1, σ2 ∈ An, then we can write

σ1 = (α1, β1)(α2, β2) · · · (αr, βr)σ2 = (α1, β1)(α2, β2) · · · (αs, βs)

then

σ−12 = ((α1, β1)(α2, β2) · · · (αs, βs))−1

= (αs, βs)−1 · · · (α2, β2)−1(α1, β1)−1

= (αs, βs) · · · (α2, β2)(α1, β1)

where r and s are both even, so

σ1σ−12 = (α1, β1) · · · (αr, βr)(αs, βs) · · · (α1, β1)

can be written as a product of r + s transpositions, and r + s is even soσ1σ

−12 ∈ An.Therefore, by the subgroup criterion An ≤ Sn.

3 SUBGROUPS 8

3 Subgroups

3.1 Centralizers, Normalizers, Generators

When studying a group, the existence and properties of its subgroups givesus information about the group as a whole. In this section we exhibit varioussubgroups that exist for arbitrary groups.

Definition 3.1. Let G be any group, and A,B ⊆ G be any two subsets, forg ∈ G define:

(i) AB = ab | a ∈ A, b ∈ B

(ii) Ag = ag | a ∈ A, where ag = g−1ag is conjugation by g.

(iii) A− = a−1 | a ∈ A

We will use these to construct subgroups.

Definition 3.2. Suppose G is a group and S ⊆ G is a nonempty subset,define

(i) CG(S) = g ∈ G | ∀x ∈ S : gx = xg

(ii) NG(S) = g ∈ G | Sg = S

(iii) 〈S〉 = x1x2 · · ·xm | m ∈ N, xi ∈ S ∪ S−

we call CG(S) the centralizer of S in G, NG(S) the normalizer of S in G,and 〈S〉 the subgroup of G generated by S.

Proposition 3.3. Let G be a group and S ⊆ G a nonempty subset, thenCG(S), NG(S), 〈S〉 are all subgroups of G.

Proof. Let S ⊆ G be given.

(i) We show CG(S) ≤ G.

Firstly, 1x = x = x1 for all x ∈ S, so 1 ∈ CG(S) and CG(S) 6= ∅.Let g, h ∈ CG(S), then

(gh)x = g(hx) = g(xh) = (gx)h = (xg)h = x(gh)

so gh ∈ CG(S).

Since gx = xg for all x ∈ S, we have xg−1 = g−1x for all x ∈ S, and sog−1 ∈ CG(S).

Therefore, CG(S) ≤ G by the subgroup criterion.

3 SUBGROUPS 9

(ii) We show NG(S) ≤ G.

Firstly,S1 = 1−1x1 | x ∈ S = x | x ∈ S = S

so 1 ∈ NG(S) and NG(S) 6= ∅.Suppose g, h ∈ NG(S), then

(Sg)h = g−1xg | x ∈ Sh

= h−1yh | y ∈ g−1xg | x ∈ S= h−1(g−1xg)h | x ∈ S= (gh)−1x(gh) | x ∈ S= S(gh)

Now both g, h ∈ NG(S) so we know Sg = S = Sh, hence

S(gh−1) = (Sg)h−1

= (Sh)h−1

= S(hh−1) = S1 = S

Therefore, gh−1 ∈ NG(S) and NG(S) ≤ G by the subgroup criterion.

(iii) We show 〈S〉 ≤ G.

By assumption S 6= ∅, so there exists x ∈ S, then x ∈ S ∪ S−1 sox ∈ 〈S〉 and 〈S〉 6= ∅.Suppose g, h ∈ 〈S〉, then

g = x1x2 · · · xm, h = y1y2 · · · yn

for some m,n ∈ N, xi, yi ∈ S ∪ S−, then

gh−1 = (x1x2 · · ·xm)(y1y2 · · · yn)−1 = x1x2 · · ·xmy−1n y−1

n−1 · · · y−11

Each yi ∈ S ∪ S− so also each y−1i ∈ S ∪ S−.

So gh−1 ∈ 〈S〉 and therefore 〈S〉 ≤ G by the subgroup criterion.

3 SUBGROUPS 10

Remark We can say more about S, CG(S), NG(S).

(i) Observe that

g ∈ CG(S)⇒ ∀x ∈ S : gx = xg

⇒ ∀x ∈ S : xg = g−1xg = x

⇒ Sg = xg | x ∈ S = x | x ∈ S = S

⇒ g ∈ NG(S)

Therefore, CG(S) ⊆ NG(S) and moreover CG(S) ≤ NG(S).

(ii) Suppose S ≤ G is actually a subgroup of G, furthermore that S is anabelian group, then for g ∈ S we have gx = xg, ∀x ∈ S, since S isabelian, hence g ∈ CG(S).

Therefore, S ⊆ CG(S) and moreover S ≤ CG(S).

(iii) Suppose S ≤ G is actually a subgroup of G, let g, x, y ∈ S

xg = yg ⇐⇒ g−1xg = g−1yg ⇐⇒ gg−1x = ygg−1 ⇐⇒ x = y

So conjugation by a fixed element within a group is one-to-one, hence

x ∈ S ⇒ Sx = yx | y ∈ S = S ⇒ x ∈ NG(S)

Therefore, S ⊆ NG(S) and moreover S ≤ NG(S).

There is one particular centralizer that is given its own name:

Definition 3.4. Consider when S = G itself, then

CG(G) = g ∈ G | ∀h ∈ G : gh = hg

is called the centre of G and usually denoted Z(G).

3 SUBGROUPS 11

3.2 Subgroup Products

Suppose that G is a group and H,K ≤ G are two arbitrary subgroups, whenis HK = hk | h ∈ H, k ∈ K a group?

Lemma 3.5. Let G be a group and H,K ≤ G be subgroups, then HK ≤ Gif and only if HK = KH.

Proof. (i) Suppose HK ≤ G, observe that

H = H1 = h1 | h ∈ H, 1 ∈ G ⊆ HK

and similarly K ⊆ HK, hence

KH = kh | k ∈ K ⊆ HK,h ∈ H ⊆ HK ⊆ HK

since HK is a group closed under multiplication.

Let g ∈ HK, since HK is a group g−1 ∈ HK, so g−1 = hk for someh ∈ H, k ∈ K, hence

g = (hk)−1 = k−1h−1 ∈ KH

as h−1 ∈ H, k−1 ∈ K.

Thus, HK ⊆ KH and therefore HK = KH.

(ii) Conversely, suppose that HK = KH, with H,K ≤ G.

Firstly, 1 ∈ H and 1 ∈ K, so 1 = 1 · 1 ∈ HK and HK 6= ∅.Suppose x, y ∈ HK = KH, then x = h1k1 for some h1 ∈ H, k2 ∈ Kand y = k2h2 for some h2 ∈ H, k2 ∈ K. Hence

xy−1 = (h1k1)(k2h2)−1 = h1k1h−12 k−1

2

As k1h−12 ∈ KH = HK, then k1h

−12 = h′k′ for some h′ ∈ H, k′ ∈ K.

Thusxy−1 = h1h

′k′k−12 ∈ HK

since h1h′ ∈ H and k′k−1

2 ∈ K.

Therefore, HK ≤ G by the subgroup criterion.

3 SUBGROUPS 12

Lemma 3.6. Let G be a finite group and H,K ≤ G, then

|HK| = |H||K||H ∩K|

Proof. Recall that H ∩K ≤ K.Let k1, . . . , kn ∈ K be a complete set of right coset representatives of

H ∩K in K (n = [K : H ∩K] = |K|/|H ∩K|).First observe that HK =

⋃k∈K Hk is the union of right cosets of H in G

by elements of K.Let k ∈ K, so k ∈ (H ∩ K)kj for some kj, and we can write k = xkj

where x ∈ H ∩K, now

kk−1j = xkjk

−1j = x ∈ H ∩K ⊆ H

so Hk = Hkj.Therefore every right coset of H in G by an element of K can be repre-

sented by some ki, hence HK =⋃k∈K Hk =

⋃ni=1 Hki — we now show that

this union is disjoint.Observe that (H ∩K)ki (for i = . . . , n) are pairwise disjoint (by choice

of ki). Suppose Hki = Hkj, then kik−1j ∈ H. But ki, kj ∈ K by definition,

so kik−1j ∈ K and kik

−1j ∈ H ∩K, thus ki and kj lie in the same right coset

of H ∩K in K, and therefore ki = kj by choice of these respresentatives.Therefore

HK =⋃k∈K

Hk =n⋃i=1

Hki

which is a disjoint union and so

|HK| =

∣∣∣∣∣n⋃i=1

Hki

∣∣∣∣∣ =n∑i=1

|Hki| = n|H| = |H||K||H ∩K|

since all right cosets of H in G are equinumerous and H itself is such a coset.

4 CONJUGACY 13

4 Conjugacy

Definition 4.1. Recall, given a group G and subset S ⊆ G, we defined

Sg = g−1xg | x ∈ S = xg | x ∈ S

we say that Sg is a conjugate (in G) of S.In particular, when S = x is a singleton set, Sg = xg. So, for any

two elements x, y ∈ G we say x and y are conjugate if there exists g ∈ Gsuch that xg = y.

It is easily seen that conjugacy of elements in a group is an equivalencerelation, we call the equivalence class [x] = xg | g ∈ G (all elementsconjugate to x in G) the conjugacy class of x ∈ G, and use notation [x] = xG.

4.1 Some Properties of Conjugacy

Given a group G, a subset S ⊆ G, and element g ∈ G, some basic propertiesfollow:

(i) |S| = |Sg|.

Proof. The map f : S → G defined by x 7→ xg is bijective

xg = yg ⇐⇒ g−1xg = g−1yg ⇐⇒ x = y

Hence, |Sg| = |f(S)| = |S|.

(ii) If S ≤ G, then Sg ≤ G.

Proof. Firstly, 1 ∈ S as S ≤ G, so 1 = 1g ∈ Sg, so Sg 6= ∅.Let x, y ∈ Sg, so x = ag, y = bg for some a, b ∈ S, then

xy−1 = ag(bg)−1 = g−1ag(g−1bg)−1 = g−1xgg−1b−1g = (ab−1)g

where ab−1 ∈ S since S is a group, therefore xy−1 ∈ Sg and Sg ≤ G bythe subgroup criterion.

(iii) If x ∈ Z(G), then xG = x, in particular 1G = 1.

Proof. Let x ∈ Z(G), then xg = g−1xg = xg−1g = x, since x commuteswith all elements of G. Hence, xG = xg | g ∈ G = x.

(iv) Conjugate elements have the same order.

4 CONJUGACY 14

Proof. Observe that

(xg)2 = (g−1xg)2 = (g−1xg)(g−1xg) = g−1x2g = (x2)g

so by induction (xg)n = (xn)g, hence if y = xg, then

yn = 1 ⇐⇒ (xg)n = 1 ⇐⇒ (xn)g = 1 ⇐⇒ xn = 1

where the last step follows since 1G = 1, so 1 can be the only elementconjugate to 1.

4 CONJUGACY 15

4.2 Counting Conjugates

Lemma 4.2. Suppose G is a group, then G is the disjoint union of its con-jugacy classes.

Proof. As remarked earlier, it is easily shown that conjugacy of elements isan equivalence relation. Any equivalence relation on a set partitions that setinto disjoint equivalence classes.

Lemma 4.3. Suppose G is a group and S ⊆ G (S 6= ∅). Let N = NG(S)and choose gi | i ∈ I to be a complete set of right coset representatives forN in G, then Sgi | i ∈ I is a distinct collection of all possible conjugatesof S in G.

Proof. Let g ∈ G, then g = hgi for some h ∈ N and i ∈ I (since g lies in aright coset Ngi of N), thus

Sg = S(hgi) = (Sh)gi = Sgi

(recall, by definition h ∈ N ⇐⇒ Sh = S). So every conjugate of S in G isof this form.

Now suppose that Sgi = Sgj , then

Sgig−1j = Sgjg

−1j = S1 = S

so gig−1j ∈ N , that is, Ngi = Ngj is same right coset of N , a contradiction

to the choice of these representatives — therefore, each Sgi is a distinctconjugate.

Corollary 4.4. When G is finite, the number of conjugates of S in G is[G : NG(S)].

Remark In particular, when considering a singleton set S = x, we haveNG(x) = CG(x). Therefore, |xG| = [G : CG(x)] and moreover |xG| divides|G|— that is, the size of each conjugacy class divides the order of the group.

4 CONJUGACY 16

4.2.1 Worked Example

Example 4.5. Let G = S4 and S = 〈(1, 2), (1, 3, 2, 4)〉. We will find allconjugates of S in G.

First, let’s find all the elements of S.We know 〈(1, 3, 2, 4)〉 ≤ S so |〈(1, 3, 2, 4)〉| = 4 divides |S| (and as (1, 2) ∈

S, we know |S| > 4). Similarly, S ≤ S4 so |S| divides |S4| = 24, hence |S| =8 or 12.

We can find

S = (1), (1, 3, 2, 4), (1, 2)(3, 4), (1, 4, 2, 3),

(1, 2), (1, 3)(2, 4), (3, 4), (1, 4)(2, 3)

by calculating each (1, 3, 2, 4)n(1, 2) for n = 0, 1, 2, 3. And observing that nomore elements can be formed from finite products in S ∪ S−.

Let’s find NG(S).Firstly S ≤ NG(S) ≤ G since S ≤ G, so |S| divides |NG(S)| which divides

|G|, this implies NG(S) = 8 or 24, meaning NG(S) = S or NG(S) = G.If NG(S) = G, then every element of G conjugates S to S, so let’s try

(1, 2, 3) ∈ G in particular:

(1, 2)(1,2,3) = (1, 2, 3)−1(1, 2)(1, 2, 3) = (1, 3, 2)(1, 2)(1, 2, 3) = (2, 3) 6∈ S

so S(1,2,3) 6= S and (1, 2, 3) 6∈ NG(S), thus NG(S) 6= G — hence NG(S) = S.Therefore, S has just [G : NG(S)] = |G|/|NG(S)| = 24/8 = 3 conjugates.

It is easily found that (1), (1, 2, 3), (1, 3, 2) is a complete set of right cosetrepresentatives of NG(S) in G, therefore the conjugates of S in G are

S = S(1), S(1,2,3), S(1,3,2)

Finally, as S ≤ G, we know S(1,2,3) and S(1,3,2) are subgroups of G also.

4 CONJUGACY 17

4.3 The Class Equation

Theorem 4.6. Let G be a finite group and x1, . . . , xk ∈ G be representativesof the conjugacy classes of G. Let ni = |xGi | and without loss of generalityassume n1 = n2 = · · · = nm = 1 and ni > 1 for i > m, then

|G| =k∑i=1

ni

=k∑i=1

[G : CG(xi)]

= |Z(G)|+k∑

i=m+1

[G : CG(xi)]

Proof. Recall, G is the disjoint union of its conjugacy classes, hence

|G| =

∣∣∣∣∣n⋃i=1

xGi

∣∣∣∣∣ =n∑i=1

∣∣xGi ∣∣ =n∑i=1

ni

by choice of xi.We have shown that [G : CG(xi)] = |xGi | = ni, now splitting the sum gives

|G| =m∑i=1

ni +n∑

i=m+1

ni = |Z(G)|+n∑

i=m+1

[G : CG(xi)]

as ni = 1 ⇐⇒ xGi = xi ⇐⇒ xi ∈ Z(G).

The following is an example of the usefulness of the class equation.

Definition 4.7. Let p be a prime number, a group G is called a p-group ifand only if |G| = pn for some n ∈ N∪0 (that is, G has prime power order).

Lemma 4.8. If G is a p-group and G 6= 1, then Z(G) 6= 1 (a non-trivialp-group has a non-trivial centre).

Proof. We know each ni divides |G| = pn, so if ni 6= 1, then p divides ni.By the class equation

|Z(G)| = |G| −k∑

i=m+1

ni

where both terms of the right-hand side are divisble by p, hence p dividesZ(G) and so |Z(G)| > 1, therefore Z(G) 6= 1.

4 CONJUGACY 18

4.4 Conjugacy in Sn

Finding conjugacy classes in symmetric groups Sn is particularly simple.

Lemma 4.9. Let σ, τ ∈ Sn, if ασ = β, then (ατ)στ = βτ , in other words:

if ασ7→ β, then ατ

στ7→ βτ .

Proof. By direct calculation

(ατ)στ = (ατ)(τ−1στ) = α(ττ−1στ) = (ασ)τ = βτ

as claimed.

Suppose σ ∈ Sn is written as a product of pairwise disjoint cycles

σ = (α1, α2, . . . , αa)(β1, β2, . . . , βb)(γ1, γ2, . . . , γc) · · ·

then, for instance, we have the cycle (α1, α2, . . . , αa), that is

α1σ7→ α2

σ7→ α3σ7→ · · · σ7→ αa

σ7→ α1

so in στ = τ−1στ we have the cycle

α1τστ7→ α2τ

στ7→ α3τστ7→ · · · σ

τ

7→ αaτστ7→ α1τ

similarly for the other cycles in σ, hence

στ = (α1τ, α2τ . . . , αaτ)(β1τ, β2τ, . . . , βbτ)(γ1τ, γ2τ, . . . , γcτ)

Allowing us to easily calculate conjugates in Sn.

Example 4.10. In S8, let σ = (1, 5, 7, 6, 2)(3, 8)(4) and τ = (1, 2)(5, 8)(3, 4, 6, 7),then

στ = (1τ, 5τ, 7τ, 6τ, 2τ)(3τ, 8τ)(4τ) = (2, 8, 3, 7, 1)(4, 5)(6)

Moreover, notice how the number and lengths of cycles remains the same.

Definition 4.11. A permutation in Sn that has Cr cycles of length r (forr ∈ N) is said to have cycle type 1C1 , 2C2 , 3C3 , . . . (terms where Cr = 0 areusually omitted).

Example 4.12. Let σ = (1, 5, 7, 6, 2)(3, 8)(4) as in the previous example,then σ has cycle type 11, 21, 51.

4 CONJUGACY 19

Lemma 4.13. Permutations σ1, σ2 ∈ Sn are conjugate in Sn if and only ifσ1 and σ2 have the same cycle type.

Proof. (i) Supose σ1, σ2 are conjugate, then σ2 = στ1 for some τ ∈ Sn, andas seen above σ1, σ

τ1 have exaclty the same cycle type.

(ii) Conversely, suppose σ1 and σ2 have the same cycle type, we can thenwrite

σ1 = (α1, . . . , αa)(β1, . . . , βb)(γ1, . . . , γc) · · ·σ2 = (α′1, . . . , α

′a)(β

′1, . . . , β

′b)(γ

′1, . . . , γ

′c) · · ·

now define τ ∈ Sn by

αiτ7→ α′i, βj,

τ7→ β′j, γk,τ7→ γ′k, . . .

then σ2 = στ1 , and σ1 and σ2 are conjugate.

Example 4.14. In S26, the permutations:

σ1 = (1, 2, 3, 4)(10, 11, 12)(20, 21, 22, 25, 26)

σ2 = (17, 18, 19)(1, 7, 8, 25)(2, 4, 6, 9, 10)

both have cycle type 114, 31, 41, 51 and are therefore conjugate.

Warning! Whilst two elements may be conjugates in Sn, they may not beconjugate when considered as elements of a subgroup (e.g.: An), the followingexample illustrates this.

Example 4.15. In S4 clearly (1, 4)(2, 3) and (1, 3)(2, 4) are conjugate (asthey have the same cyclic type), by the method in the lemma we find explic-itly:

(1, 4)(2, 3)(3,4) = (1, 3)(2, 4)

But (3, 4) 6∈ A4 so we cannot conclude that (1, 4)(2, 3) and (1, 3)(2, 4) areconjugate in A4. However, we also have

(1, 4)(2, 3)(1,2,3) = (1, 3)(2, 4)

where here (1, 2, 3) = (1, 2)(1, 3) ∈ A4, so in fact they are conjugate in A4.On the other hand, (1, 2, 3) and (1, 3, 2) are conjugate in S4, but in this

case no τ ∈ A4 exists such that (1, 2, 3)τ = (1, 3, 2), so they are not conjugatein A4.

5 GROUP ACTIONS 20

5 Group Actions

Definition 5.1. Let G be a group and Ω a nonempty set, we say that Gacts on Ω if for each g ∈ G and α ∈ Ω there is a corresponding α · g ∈ Ω (seefootnote1) such that

(A1) ∀α ∈ Ω : ∀g1, g2 ∈ G : α · (g1g2) = (α · g1) · g2.

(A2) ∀α ∈ Ω : α · 1 = α.

we also say Ω is a G-set to mean that G acts on Ω.

Remark More formally, we have defined a map Ω × G → Ω given by(g, α) 7→ α · g, which satisfies the given properties.

Example 5.2. Naturally permutations act on the set they are permuting.Let G ≤ Sn, then G acts on Ω = 1, . . . , n by action α · g = β where

αg7→ β (i.e.: the permutation g maps α to β), for g ∈ G and α, β ∈ Ω. The

properties are easily checked.

Example 5.3. Let G be a group, then G acts on itself Ω = G by the actionα · g = αg = g−1αg, i.e.: by conjugation.

Let α ∈ Ω, g1, g2 ∈ G, then (recalling properties of conjugation)

α · (g1g2) = αg1g2 = (αg1)g2 = (α · g1)g2 = (α · g1) · g2

andα · 1 = α1 = α

so (A1) and (A2) are satisfied.

Example 5.4. Let G be a group and H ≤ G a subgroup. Let Ω = Hx |x ∈ G be the right cosets of H in G, then G acts on Ω by (Hx) · g = H(xg)for all Hx ∈ Ω, g ∈ G.

Let g1, g2 ∈ G,Hx ∈ Ω, then

(Hx · g1) · g2 = H(xg1) · g2 = H((xg1)g2) = H(x(g1g2)) = Hx · (g1g2)

where we have used associativity in H, so (A1) holds, and

Hx · 1 = H(x1) = Hx

so (A2) holds, and this is indeed a group action.In particular, if H = 1 is the trivial subgroup, then Ω ∼= G and this

group action of G on Ω is just the group product.

1I am using · to distinguish between the action and the group product (which willremain juxtaposition of symbols) — usually it is left to context to make clear which isintended.

5 GROUP ACTIONS 21

Example 5.5. Let G be a group, p a prime and

Ω = (x1, x2, . . . , xp) | xi ∈ G and x1x2 · · ·xp = 1

Let σ = (1, 2, . . . , p) ∈ Sp.For α = (x1, x2, . . . , xp) ∈ Ω, define

(x1, . . . , xp−1, xp) · α = (x1α, . . . , x(p−1)α, xpα) = (x2, . . . , xp, x1)

in effect σ rotates the p-tuple to the left.We can define

α · σn =

σ acts n-times︷︸︸︷((α · σ) · σ) · · ·σ

then Ω is a 〈σ〉-set.

Example 5.6. Let G be a finite group, k ∈ N, and Ω = X ⊆ G | |X| = kbe all k-element subsets of G.

Given g ∈ G,X = g1, g2, . . . , gk ∈ Ω define actionX·g = g1g, g2g, . . . , gkg,then Ω is a G-set.

5 GROUP ACTIONS 22

5.1 G-orbits

Proposition 5.7. Suppose G is a group that acts of Ω. Let α, β ∈ Ω andwrite α ∼ β if and only if ∃g ∈ G : α · g = β, this is an equivalence relationon Ω.

Proof. (i) Let α ∈ Ω, then α · 1 = α by (A2) so α ∼ α.

(ii) Let α, β ∈ Ω, suppose α ∼ β, then α · g = β for some g ∈ G, hence,

β · g−1 = (α · g) · g−1 = α · (gg−1) = α · 1 = α

using (A1) then (A2), as g−1 ∈ G we conclude β ∼ α.

(iii) Let α, β, γ ∈ Ω, suppose α ∼ β and β ∼ γ, then α ·g1 = β and β ·g2 = γfor some g1, g2 ∈ G, hence

α · (g1g2) = (α · g1) · g2 = β · g2 = γ

so α ∼ γ.

Definition 5.8. The ∼ equivalence classes of Ω are called the G-orbits. Letα ∈ Ω, the G-orbit of Ω containing α is denoted by αG.

Definition 5.9. Let G be a group and Ω a G-set, for α ∈ Ω, define

Gα = g ∈ G | α · g = α

this is called the stabalizer in G of α.

Lemma 5.10. Let G be a group and Ω a G-set, let α ∈ Ω, then Gα ≤ G.

Proof. Firstly, α · 1 = α by (A1), so 1 ∈ Gα and Gα 6= ∅.Suppose g, h ∈ Gα, then

α · (gh−1) = (α · g) · h−1 = (α · h) · h−1 = α · (hh−1) = α · 1 = α

using (A1), (A2) and the fact that α · g = α = α · h, hence gh−1 ∈ Gα

Therefore Gα ≤ G by the subgroup criterion.

Lemma 5.11. Let G be a finite group, Ω a finite G-set, and ∆ a G-orbit ofΩ, then

(i) ∀α ∈ ∆ : |∆| = |αG| = [G : Gα]

(ii) ∀α, β ∈ ∆ : if α · g = β, then (Gα)g = Gβ.

5 GROUP ACTIONS 23

Proof. (i) Let g1, g2, . . . , gn be a complete set of right coset representativesfor Gα in G (where n = [G : Gα]). By definition ∆ = α · g | g ∈ G,given g ∈ G we know g = hgi for some h ∈ Gα and i ∈ 1, . . . , n, so

α · g = α · (hgi) = (α · h) · gi = α · gi

therefore the action of G on Ω is completely determined by the actionsof g1, . . . , gn, thus ∆ = α · g1, α · g2, . . . , α · gn.Suppose α · gi = α · gj, then

(α · gi) · g−1j = (α · gj) · g−1

j

andα · (gig−1

j ) = α · (gjg−1j ) = α · 1 = α

hence gig−1j ∈ Gα and so gi and gj represent the same right coset of Gα

in G, so gi = gj by choice of these elements.

Therefore α · g1, α · g2, . . . , α · gn are distinct and |∆| = n = [G : Gα].

(ii) Let α, β ∈ ∆, so α · g = β for some g ∈ G.

Let x ∈ (Gα)g, then x = g−1hg for some h ∈ Gα, then

β · x = (α · g) · (g−1hg)

= α · (gg−1hg)

= α · (hg)

= (α · h) · g= α · g= β

using action properties, and α · h = α since h ∈ Gα. Thus, x ∈ Gβ andso (Gα)g ⊆ Gβ.

Let x ∈ Gβ, then β · x = β and α · (gx) = (α · g) · x = β · x = β, so

α · (gxg−1) = (α · (gx)) · g−1 = β · g−1 = (α · g) · g−1 = α

hence h = gxg−1 ∈ Gα, so y = g−1hg ∈ (Gα)g — so, Gβ ⊆ (Gα)g andGβ = (Gα)g as claimed.

5 GROUP ACTIONS 24

We summarize the properties of orbits we’ve seen above in the followingtheorem.

Theorem 5.12. Let G be a finite group and Ω a finite G-set.

(i) Ω is a disjoint union of G-orbits Ω =⋃mi=1 ∆i.

(ii) ∀α ∈ Ω : Gα ≤ G.

(iii) If αi ∈ ∆i (for i = 1, . . . ,m), then |Ω| =∑m

i=1 |∆i| =∑m

i=1[G : GαI ].

(iv) |∆i| divides |G|, for each i = 1, . . . ,m.

Remark Recall, that we can define a group to act on itself by conjugacy(i.e.: ∀α ∈ Ω, g ∈ G : α · g = αg), then the G-orbit αG is just the conjugacyclass of α (also denoted αG) and the stabalizer Gα is just the centralizer ofα in G (previously denoted CG(α)).

Thus, the class equation for conjugacy classes is a special case of theabove theorem for group actions.

5 GROUP ACTIONS 25

5.2 Cauchy’s Theorem

The following theorem is an application of group actions.

Theorem 5.13. Let p be a prime number and G a finite group, if p divides|G|, then G has at least one element of order p.

Proof. Let Ω = (x1, x2, . . . , xp) | xi ∈ G and x1x2 · · ·xp = 1 and σ =(1, 2, . . . , p) ∈ Sp. Recall, from a previous example that 〈σ〉 acts on Ω.

Write Ω =⋃ni=1 ∆i where ∆i are the 〈σ〉-orbits of Ω. We know |∆i|

divides |〈σ〉| for each i = 1, . . . , n and |〈σ〉| = p. Therefore |∆i| = 1 or p.Suppose that |∆i| = 1 for some 1 ≤ i ≤ n, so that ∆i = α for some

α = (x1, x2, . . . , xp). As α is the only element in its orbit, it follows thatα · σm = α for all m ∈ Z — that is:

(x1, x2, . . . , xp) = (x2, x3, . . . , x1) · σ = (x3, x4, . . . , x2) · σ2 = · · ·

and comparing elements of the p-tuple, we see that

x1 = x2 = · · · = xp−1 = xp

and so x1x2 · · · xp = xp1 = 1, hence x1 ∈ G is an element of order 1 or p, ifx1 6= 1, then we have our required element of order p.

Note ∆1 = (1, 1, · · · , 1) is such a singleton 〈σ〉-orbit and if there are noother singleton orbits then |Ω| =

∑ni=1 |∆i| = 1 + (n− 1)p.

But |Ω| = |G|p−1 because to satisfy x1x2 · · ·xp = 1 we may choosex1, x2, . . . , xp−1 arbitarily (having |G| choices for each), so xp = x−1

p−1 · · ·x−11

is uniquely determined. Thus, since p divides |G|, we know p divides |Ω|.Contradicting |Ω| = 1 + (n− 1)p, therefore there must exist another single-ton orbit ∆j = (x, x, . . . , x) for which x 6= 1, and x is then the requiredelement of order p.

5 GROUP ACTIONS 26

5.3 Burnside’s Lemma

Definition 5.14. Let G be a group and Ω a G-set, if Ω is a G-orbit itself,then we say G acts transitively on Ω, i.e.: for any α ∈ Ω

αG = α · g | g ∈ G = Ω

since the G-orbits partiation Ω (in this case, trivially, into just the set Ωitself).

Definition 5.15. Let G be a group and Ω a G-set, for g ∈ G define

fixΩ(g) = α ∈ Ω | α · g = α

the set of elements fixed by g under its action on Ω.

Notice that fixΩ(1) = Ω for any action.

Theorem 5.16. Let G be a finite group and Ω a finite G-set. If G hast orbits on Ω, then t = 1

|G|∑

g∈G | fixΩ(g)|. In other words, the number oforbits is equal to the average number of points fixed by elements of G underits action.

Proof. For the case t = 1, we have a single orbit so G acts transitively on Ω,define

S = (α, g) ∈ Ω×G | α · g = α

we will count |S| in two ways.

(i) Consider g ∈ G fixed and let α ∈ Ω vary, then

(α, g) ∈ S ⇐⇒ α · g = α ⇐⇒ α ∈ fixΩ(g)

therefore, |S| =∑

g∈G | fixΩ(g)|.

(ii) Consider α ∈ Ω fixed and let g ∈ G vary, then

(α, g) ∈ S ⇐⇒ α · g = α ⇐⇒ g ∈ Gα

therefore

|S| =∑α∈Ω

|Gα| =∑α∈Ω

|G|[G : Gα]

=∑α∈Ω

|G||αG|

=∑α∈Ω

|G||Ω|

= |Ω| |G||Ω|

= |G|

5 GROUP ACTIONS 27

Hence|G| = |S| =

∑g∈G

| fixG(g)|

and1

|G|∑g∈G

| fixΩ(g)| = 1 = t

as required.Now in the general case (t ≥ 1) let ∆1, . . . ,∆t be the G-orbits and observe

that if G acts on Ω, then G also acts (transitively) on each orbit ∆i consideredas its own set. So using the t = 1 case on each orbit and our knowledge thatΩ is partitioned by these orbits, we have

1

|G|∑g∈G

| fixΩ(g)| =t∑i=1

1

|G|∑g∈∆i

| fix∆i(g)| =

t∑i=1

1 = t

as claimed.

6 FINITELY GENERATED ABELIAN GROUPS 28

6 Finitely Generated Abelian Groups

Definition 6.1. A group G is finitely generated (f.g.) if there exists a finitesubset S ⊆ G such that 〈S〉 = G.

Lemma 6.2. Zn × Zm ∼= Znm if and only if gcd(n,m) = 1.

Proof. (⇒) Suppose gcd(n,m) = 1, then if for some k ∈ N

(1, 1)k = (1 + 1 + · · ·+ 1︸︷︷︸k-times in Zn

, 1 + 1 + · · ·+ 1︸︷︷︸k-times in Zm

) = (0, 0)

then n|k and m|k, so k ≥ nm and as gcd(n,m) = 1, then nm|k. Also(1, 1)nm = (0, 0).

Hence, the order of (1, 1) ∈ Zn × Zm is nm and as Zn × Zm containsjust nm elements it must be cyclic: Zn × Zm = 〈(1, 1)〉, thereforeZn × Zm ∼= Znm (as cyclic groups of the same order are isomorphic).

(⇐) Suppose Zn × Zm ∼= Znm and for contradiction d = gcd(n,m) > 1.

Write n = n′d and m = m′d (possible as d|n and d|m).

Let (x, y) ∈ Zn × Zm be an arbitrary group element and observe

(x, y)n′dm′ = ( x+ · · ·+ x︸︷︷︸

nm′-times in Zn

, y + · · ·+ y︸︷︷︸n′m-times in Zm

) = (0, 0)

as n′dm′ = nm′ and n′dm′ = n′m.

Hence, the order of (x, y) is at most n′dm′ but n′dm′ < n′ddm′ = nm,this is a contradiction since Zn × Zm ∼= Znm must contain at least oneelement of order nm (as it is a cyclic group).


6.1 Finitely Generated Abelian Groups

Example 6.3. Some examples of f.g. abelian groups:

(i) Any finite abelian group is a f.g. abelian group.

(ii) (Z,+) is a f.g. abelian group (generted by 1).

(iii) Z×Z is a finitely generated abelian group (generated by (0, 1), (1, 0)).

6.1.1 Classification of Finitely Generated Abelian Groups

For abelian groups that are finitely generated, we have a complete classifica-tion as follows.

Theorem 6.4. Any finitely generated abelian group G is isomorphic to adirect product of cyclic groups and can be written:

G ∼= Zm1 × Zm2 × · · · × Zmk × Zs

for some mi, s ∈ N such that mi|mi+1 for i = 1, . . . , k − 1. (We call s therank of G and m1, . . . ,mk the torsion coefficients of G).

Proof. Extension material.

Corollary 6.5. Any finite abelian group is isomorphic to Zm1 × · · · × Zmkfor some mi ∈ N such that mi|mi+1. (In this case, |G| = m1m2 · · ·mk).

Corollary 6.6. A finitely generated abelian group which has no elements offinite order (except the identity) is isomorphic to Zs for some s ∈ N.

The following theorem tells us for any f.g. abelian group, its rank andtorsion coefficients are uniquely determined.

Theorem 6.7. Suppose

G1 = Zm1 × Zm2 × · · · × Zmk × Zr where mi|mi+1

andG2 = Zn1 × Zn2 × · · · × Znl × Zs where ni|ni+1

then G1∼= G2 if and only if k = l and mi = ni for i = 1, . . . , k and r = s.

Proof. Extension material.


Example 6.8. By the classification of f.g. abelian groups, there are onlytwo possibly abelian groups of order 12:

(i) G ∼= Z12, or

(ii) G ∼= Z2 × Z6.

Note that Z3 × Z4 doesn’t satisfy the conditions of the theorem (3 doesn’tdivide 4) so doesn’t give a third group (as gcd(3, 4) = 1, we see that in factZ3 × Z4

∼= Z12).

Example 6.9. The group Z6 × Z10 doesn’t appear on our classiciation, but

Z6 × Z10∼= (Z2 × Z5)× Z10

∼= Z2 × (Z3 × Z10) ∼= Z2 × Z30

which is of the required form.

Example 6.10. The abelian groups of order 24 are Z24, Z2×Z12, and Z2×Z2 × Z6. No two of which isomorphic.

7 NORMAL SUBGROUPS 31

7 Normal Subgroups

Definition 7.1. Suppose G is a group and N ≤ G is a subgroup, then N isa normal subgroup if ∀g ∈ G : N g = N .

We write N EG to denote that N is a normal subgroup of G.

There are equivalent definitions given in the following lemma.

Lemma 7.2. Let G be a group and N ≤ G, then the following are equivalent:

(i) N EG,

(ii) G = NG(N),

(iii) The conjugates of N in G are N,

(iv) ∀g ∈ G, n ∈ N : g−1ng ∈ N ,

(v) ∀g ∈ G : Ng = gN .

Remark Part (iv) says that N E G is equivalent to N being the union ofconjugacy classes in G.

Example 7.3. Some normal subgroups include:

(i) For any group G, we have 1EG and GEG.

(ii) If G is an abelian group, then all subgroups are normal.

(iii) More generally, if H ≤ G and H ≤ Z(G), then H is normal in G.

Lemma 7.4. If G is a group and H ≤ G and [G : H] = 2, then H EG.

Proof. We verify property (v) of the above lemma. Let g ∈ G, then either

(i) g ∈ H, then Hg = H = gH simply because H is a group and closedunder its product, otherwise

(ii) g 6∈ H, so g ∈ G\H. Now by hypothesis [G : H] = 2, meaning Hhas precisely two right cosets in G, namely H and G\H. Since g 6∈ H,we know Hg = G\H. But by the same argument, G has precisely thesame two left cosets and so G\H = gH, therefore Hg = gH.

Either way, Hg = gH for all g ∈ G. Therefore H EG.

Corollary 7.5. An E Sn for all n ≥ 2.


7.1 Quotient Groups

Given a group G and a subgroup H ≤ G we can form a set of right cosetsHg | g ∈ G. Furthermore, when H E G is a normal subgroup, we candefine a product of two cosets:

Let g, h ∈ G, then

(Ng)(Nh) = n1gn2h | n1, n2 ∈ N= N(gN)h= N(Ng)h= N(gh)

making use of the fact that Ng = gN , since N EG.

Definition 7.6. Let G be a group and N E G be a normal subgroup, thenG/N = Ng | g ∈ G is a group whose product is defined to be

(Ng1)(Ng2) = N(g1g2)

then, G/N is called the qoutient group of G by N .

Notation If G is a group and N EG, we may sometimes write Ng = g asa shorthand.

Lemma 7.7. G/N as defined above is actually a group.

Proof. Suppose G is a group and N EG.

(G1) We must show that the product is a well-defined operation (since it isdefined in terms of a choice of coset representative, we must show thatit is actually independent of this choice).

Let g1, g2, h1, h2 ∈ G such that Ng1 = Ng2 and Nh1 = Nh2, then

N(g1h1) = (Ng1)(Nh1) = (Ng2)(Nh2) = N(g2h2)

as required.

(G2) Let x, y, z ∈ G, then

[(Nx)(Ny)](Nz) = N(xy)N(z)

= N((xy)z)

= N(x(yz))

= N(x)N(yz)

= N(x)[N(y)N(z)]

using the associativity of G.


(G3) The element N = N1 is the identity element.

(N1)(Ng) = N(1g) = Ng = N(g1) = (Ng)(N1)

for all g ∈ G.

(G4) Given Ng its inverse is (Ng)−1 = N(g−1) since

(Ng)(Ng−1) = N(gg−1) = N1 = N(g−1g) = (Ng−1)(Ng)

for each g ∈ G.

We have verified all the group axioms, therefore G/N is a group.

Remarks

(i) The elements of the group G/N are right cosets of N in G.

(ii) 1 = N is the identity of the group.

(iii) For all right cosets g ∈ G/N we have (g)−1 = (g−1).

(iv) If G is finite, then |G/N | = [G : N ] = |G|/|N |.

(v) let g ∈ G/N , then (g)n = (gn) so (g)n = 1 if and only if gn ∈ N , hencethe order of g is the smallest n such that gn ∈ N .


The following two lemmas use quotient groups to establish informationabout the group.

Lemma 7.8. Let G be a group, if G/Z(G) is cyclic, then G is abelian.

Proof. Suppose G/Z(G) is cyclic, then there exists x ∈ G/Z(G) such thatG/Z(G) = 〈x〉 = xn | n ∈ Z.

Since xn = xn, this means every right coset of Z(G) is of the form Z(G)xn

for some n ∈ Z.Let g, h ∈ G, then g = z1x

i and h = z2xj for some z1, z2 ∈ Z(G) and

i, j ∈ Z, hence

gh = z1xiz2x

h = z1z2xixj = z1z2x

i+j = z2xjz1x

i = hg

using the fact that elements in Z(G) commute with those in G, therefore Gis an abelian group.

Lemma 7.9. If G is a group and |G| = p2 (where p is prime), then G isabelian (and hence G ∼= Zp × Zp or G ∼= Zp2).

Proof. Since G is a p-group and G 6= 1, by lemma 4.8 we know thatZ(G) 6= 1, hence |Z(G)| = p or p2 by Lagrange’s theorem.

Suppose |Z(G)| = p, then |G/Z(G)| = [G : Z(G)] = p2/p = p. HenceG/Z(G) ∼= Zp and in particular is cyclic. So by lemma 7.8 above, G isabelian. This implies Z(G) = G and |Z(G)| = |G| = p2, a contradiction.

Therefore, |Z(G)| = p2 and Z(G) = G (this time without a contradic-tion!) and hence G is abelian.


7.1.1 Subgroups of Quotient Groups

Lemma 7.10. Let G be a group and N EG be a normal subgroup. All sub-groups of G/N are of the form H/N where N EH ≤ G and vice versa. Thatis, there is a one-to-one correspondence between subgroups of the quotientgroup G/N and subgroups of G containing N .

Proof. (i) Assume K ≤ G/N and set

H =⋃x∈K

x ⊆ G

Note 1 ∈ K and so 1 = N ⊆ H and H 6= ∅.We show H ≤ G, let x, y ∈ H, so that x, y ∈ K. Since K ≤ G/N ,xy−1 = x(y)−1 ∈ K and hence xy−1 ∈ H. Therefore, H ≤ G by thesubgroup criterion. Moreover, N EH ≤ G as N ⊆ H and N EG.

(ii) Conversely, assume H ≤ G with N ≤ H, it follows easily that N EH.

We know that since H ⊆ G, that H/N ⊆ G/N , we show this is actuallya subgroup.

Firstly, 1 ∈ H/N , so H/N 6= ∅.Let x, y ∈ H/N , then x(y)−1 = xy−1 ∈ H/N if and only if xy−1 ∈ H,this holds since H is a group and x, y ∈ H. Therefore, H/N ≤ G/N bythe subgroup criterion.

Remark We can extend this theorem so show that H/N is normal in G/Nif and only if H is normal in G.

Knowing the subgroups of G/N gives us all the subgroups of G containingN and vice versa.


7.2 Homomorphisms

Definition 7.11. Let G and K be groups, a map θ : G→ K is a homomor-phism from G to K provided

∀g1, g2 ∈ G : (g1g2)θ = (g1θ)(g2θ)

we will writeGθ = im(θ) = gθ | g ∈ G ⊆ K

for the image of θ, and

ker(θ) = g ∈ G | gθ = 1K ∈ K ⊆ G

for the kernal of θ.

Lemma 7.12. Let θ : G→ K be a group homomorphism, then

(i) ∀g ∈ G : (g−1)θ = (gθ)−1,

(ii) 1Gθ = 1K,

(iii) ker(θ) EG,

(iv) Gθ ≤ K.

Proof. All four properties are easily checked.

Example 7.13. These are numerous examples of group homomorphims.

(i) Let G be any group and NEG a normal subgroup, define θ : G→ G/Nby gθ 7→ g, then θ is a homomorphism (the natural homomorphism fromG to G/N). Note, ker(θ) = N and Gθ = G/N .

(ii) Let G = GL(n,F) where n ∈ N and F is any field, let K = (F\0,×F)be the group of units in F under field multiplication, define θ : G→ Kby gθ = det(g), then θ is a homomorphism. Note, ker(θ) = SL(n,F) EGL(n,F) and Gθ = K.

(iii) Let G be a group and H ≤ G a subgroup. Suppose [G : H] = n, letΩ = Hx | x ∈ G be the set of right cosets of H in G (note Ω isn’tnecessarily a group as H isn’t assumed to be a normal subgroup of G).

For each g ∈ Ω, define a map ψg : Ω → Ω by (Hx)ψg = H(xg). Notethat ψg is a permutation of Ω, as follows:


Suppose (Hx)ψg = (Hy)ψg, then H(xg) = H(yg) and (xg)(yg)−1 =xy−1 ∈ H, hence Hx = Hy, i.e.: Ω is injective. Since |Ω| = n is finite,ψg is therefore a bijectiion Ω→ Ω, i.e.: a permutation of Ω.

We now define Ψ : G→ Sym(Ω) by gΨ→ ψg, then:

Let g1, g2 ∈ G, then (g1g2)Ψ = ψg1g2 whereas (g1Ψ)(g2Ψ) = ψg1ψg2 ,these maps are equal as given x = Hx ∈ Ω

xψg1g2 = x(g1g2) = (xg1)g2 = xg1ψg2 = (xψg1)ψg2 = xψg1ψg2

therefore Ψ : G→ Sym(Ω) is a homomorphism.

We have

ker(Ψ) = g ∈ G | gΨ = ψg = 1Sym(Ω)= g ∈ G | ∀x ∈ G : H(xg) = Hx= g ∈ G | ∀x ∈ G : xgx−1 ∈ H= g ∈ G | ∀x ∈ G : g ∈ Hx

=⋂x∈G

Hx

= coreG(H)

we define coreG(H) to be the core of H in G — it is the intersection ofall the conjugates of H in G.

All we can say in general about GΨ is that GΨ ≤ Sym(Ω).


7.2.1 Isomorphism Theorems

Theorem 7.14 (1st Isomorphism Theorem).Let θ : G→ K be a group homomorphism, then G/ ker(θ) ∼= Gθ.

Theorem 7.15 (2nd Isomorphism Theorem).Let G be a group and N,M EG be normal subgroups such that N ≤M ≤ G,then

(G/N)/(M/N) ∼= G/M

Proof. Firstly, N EG and N ≤M , so N EM and M/N is defined.For g ∈ G, let g = Ng and g = Mg be shorthand notation of cosets of N

and M respectively.Define θ : G/N → G/M by gθ = g.

Suppose g = h, then gh−1 ∈ N so gh−1 ∈ M since N ⊆ M , hencegθ = g = h = hθ and so θ is well-defined.

Moreover, let g1, g2 ∈ G, then

(g1g2)θ = (g1g2)θ = g1g2 = g1g2 = (g1θ)(g2θ)

and so θ is a homomorphism.We see that

ker(θ) = g ∈ G/N | g = M= Ng |Mg = M= Ng | g ∈M= M/N

Finally, θ is surjective, given g ∈ G/M it is easily verified that gθ = g forg ∈ G/N .

Therefore, by the first isomorphism theorem

(G/N)/(M/N) = (G/N)/ ker(θ) ∼= (G/N)θ = G/M

as claimed.


Theorem 7.16 (3rd Isomorphism Theorem).Let G be a group, H ≤ G and N EG, then HN/N ∼= H/(H ∩N).

Proof. Define θ : H → G/N by hθ = x, then θ is a homomorphism, for leth1, h2 ∈ H, then

(h1h2)θ = h1h2 = h1 h2 = (h1θ)(h2θ)

as required.Now

ker(θ) = h ∈ H | h = N = h ∈ H | h ∈ N = H ∩N

andHθ = h | h ∈ H = Nh | h ∈ H = NH/H

therefore, by the first isomorphism theorem

H/(H ∩N) = H/ ker(θ) ∼= Hθ = NH/H

as claimed.

8 SIMPLE GROUPS 40

8 Simple Groups

Definition 8.1. A group G is a simple group if G 6= 1 and the only normalsubgroups of G are 1 and G itself.

Example 8.2. Some simple groups include:

(i) Let p be prime, then Zp is simple.

(ii) Let n ≥ 5, then An is simple (we wil prove the n = 5 case below).

(iii) Let G = SL(n, p), for n ∈ N, p prime. Then G/Z(G) is simple provided(n, p) 6= (2, 2), (2, 3).

We will now prove that A5 is a simple group. The folllowing lemmas willbe useful.

Lemma 8.3. Let G be a group and N E G a normal subgroup, for g ∈ Gand n ∈ N , then [n, g] = n−1g−1ng ∈ N .

Proof. N E G so g−1ng ∈ N , now as n−1 ∈ N , then [n, g] = n−1g−1ng ∈ Nalso.

Lemma 8.4. Let n ≥ 3, then every element of An can be written as a productof 3-cycles (i.e.: 3-cycles generate An).

Proof. Consider two transpositions (α, β), (γ, δ) ∈ An, then either

(i) (α, β)(γ, δ) = (1), if α, β = γ, δ, or

(ii) (α, β)(γ, δ) = (α, δ, β), if β = γ and α 6= δ, else

(iii) (α, β)(γ, δ) = (α, β)(β, γ)(β, γ)(γ, δ) = (α, γ, β)(β, δ, γ)

exactly one of the preceeding statements hold, but in any case we see thatthe product of two transpositions is either the identity (1) or a product ofthree cycles.

Let σ ∈ An, then σ = τ1τ2 · · · τk for k transpositions τi, where k is even.Hence, σ = (τ1τ2)(τ2τ4) · · · (τk−1τk) and by the above consideration, we canwrite σ as a product of 3-cycles.

8 SIMPLE GROUPS 41

Lemma 8.5. Let n ≥ 5, then the 3-cycles are all conjugate in An.

Proof. Firstly, 3-cycles are all conjugate in Sn – since they have the samecycle type.

Let σ = (α1, α2, α3) and µ = (β1, β2, β3) be arbitrary 3-cycles in An, thenthere exists g ∈ Sn such that σg = µ.

Since n ≥ 5, we can choose α4, α5 ∈ Ω\α1, α2, α3, set τ = (α4, α5), noteτ commutes with σ, so

στg = (τg)−1στg = g−1τ−1στg = g−1σg = σg

Since τ is a transposition, either g or τg is even, therefore σ and µ areconjugate in An.

Proposition 8.6. A5 is simple.

Proof. Suppose N EA5 and N 6= 1, let σ ∈ N and σ 6= (1), then the cycletype of σ is one of the following: (i) 11, 22, (ii) 12, 31, or (iii) 51 (since noother cycle type will given an even permutation).

Write Ω = α, β, γ, δ, ε, we consider the three cases seperately.

(i) Write σ = (α, β)(γ, δ), let τ = (α, ε, β), then [σ, τ ] = (α, β, ε) ∈ N .

(ii) Write σ = (α, β, γ).

(iii) Write σ = (α, β, γ, δ, ε), let ρ = (α, β, γ), then [σ, ρ] = (α, β, δ) ∈ N .

In any case, N must contain at least one three cycle, say λ. Any other 3-cycle, say µ ∈ A5, is conjugate to λ by lemma 8.5. Since N is a normalsubgroup of A5, then µ ∈ N .

Therefore, N contains all 3-cycles and therefore N = A5 by lemma 8.4.

8 SIMPLE GROUPS 42

8.1 Composition Series

Definition 8.7. Let G be a group, if we have subgroups G1, G2, . . . , Gn ≤ Gsuch that

(i) GDG1 DG2 D · · ·DGn = 1, and

(ii) G/G1, G1/G2, . . . , Gn−1/Gn are all simple groups,

then we call G,G1, G2, . . . , Gn a composition series for G and Gi/Gi+1 thecomposition factors.

Definition 8.8. Let G be a group, then K is a maximal normal subgroupof G if K E G (K 6= G) and if N E G with K ≤ N , then either N = K orN = G.

Four useful results

(A) K is a maximal normal subgroup of G if and only if G/K is simple.

(B) If N1, N2 EG, then N1N2 EG.

(C) If N1, N2 EG, then N1 ∩N2 EG.

(D) If N EG, H ≤ G, then HN/N ∼= H/(H ∩N).

We will make use of these in the following theorems.

Lemma 8.9. Any finite group, except 1, has at least one compositionseries.

Proof. Let G be a finite group, we will prove by induction on |G|.When G = Z2 (the only group of order 2, up to isomorphism), then G is

itself simple so GD 1 is a composition series for G.In the general case, letK be a maximal normal subgroup ofG (which must

exist as G is finite), by (A) we know G/K is simple and GDK. If K = 1,then we have a composition series GD 1, otherwise by induction K has acomposition series K DK1 D · · ·DKn = 1, hence GDK DK1 D · · ·DKn,is a composition series of G.

8 SIMPLE GROUPS 43

8.1.1 The Jordan-Holder Theorem

Theorem 8.10. Let G be a finite group (G 6= 1), suppose

GDH1 DH2 D · · ·DHr = 1

andGDK1 DK2 D · · ·DKs = 1

are two composition series for G, then

(i) r = s, and

(ii) the two series of r factor groups G/H1, . . . , Hr−1/Hr and G/K1, . . . , Kr−1/Kr

are equal up to isomorphism and multiplicities.

Proof. We will say two such composition series are “equivalent” if they satisfyproperties (i) and (ii) of the theorem (note this is an equivalence relation).

If H1 = 1, then G ∼= G/H1 and so G is simple and has exactly onecomposition series (namely G D 1) and any two given composition seriesmust therefore be equivalent. For induction on |G| this covers the base caseG = Z2.

Otherwise, we may suppose H1 6= 1 and K1 6= 1. There are twopossibilities:

(i) If H1 = K1, thenH1 DH2 D · · ·DHr = 1

andH1 DK2 D · · ·DKs = 1

are two composition series for H1, so by induction they are equivalent.Hence, the original composition series for G are equivalent also.

(ii) If H1 6= K1, then H1K1 E G by (B), we cannot have both H1 = H1K1

and K1 = H1K1, so without loss of generality suppose H1 6= H1K1,then H1 ≤ H1K1 is a proper subgroup. Since G/H1 is simple, H1 is amaximal normal subgroup of G by (A), therefore it must be the casethat H1K1 = G.

Consider G/K1 = H1K1/K1∼= H1/(H1 ∩K1) by (D), let L = H1 ∩K1.

If L = 1, do nothing, else choose a composition series for L (existingby lemma 8.9), say LD L1 D · · ·D Lt = 1.

8 SIMPLE GROUPS 44

We now have four composition series for G, as follows

(1) : GDH1 DH2 D · · ·DHr = 1(2) : GDK1 DK2 D · · ·DKs = 1(3) : GDH1 D LD L1 D · · ·D Lt = 1(4) : GDK1 D LD L2 D · · ·D Lt = 1

For compositions series (1) and (3) we may apply case (i) and concludethey are equivalent. Similarly, (2) and (4) are equivalent.

Since G/H1 = H1K1/H1∼= K1/(H1 ∩ K1) = K1/L and similarly

G/K1∼= H1/L, we may conclude composition series (3) and (4) are

equivalent.

Therefore, composition series (1) and (2) are equivalent, as claimed.

Example 8.11. Some composition series.

(i) If G is a simple group, then G D G1 = 1 is a composition series forG. We have a single composition factor G/G1

∼= G.

(ii) For G = Z6 we know G = 〈g〉 for some g ∈ G with order 6 (as G iscyclic). We can give two composition series:

GDG1 DG2 where G1 = 〈g2〉, G2 = 1GDH1 DH2 where G1 = 〈g3〉, G2 = 1

We have composition factors G/G1∼= Z2 and G1/G2

∼= Z3 (as |G| = 6and |G1| = 3), similarly G/H1

∼= Z3 and H1/H2∼= Z2 (as H1 = 2).

Note that the composition factors are equal (but in different order) inaccordance with the Jordan-Holder theorem.

(iii) For G = S4 we have composition series GDG1 DG2 DG3 DG4 = 1,where G1 = A4, G2 = (1), (12)(34), (13)(24), (14)(23), and G3 =(1), (12)(34). and composition factors G/G2

∼= Z2, G1/G2∼= Z3,

G2/G3∼= Z2, and G3/G4

∼= Z2.

9 SYLOW’S THEOREMS 45

9 Sylow’s Theorems

Theorem 9.1. If G is a finite group and p prime, write |G| = prm wherep - m and r ∈ N ∪ 0, then

(i) G has at least one subgroup P of order pr,

(ii) The subgroups of G of order pr form a conjugacy class in G (these Sylowp-subgroups are all conjugate to each other),

(iii) If X is a p-subgroup of G, then ∃g ∈ G : X ≤ P g,

(iv) If n is a number of subgroups of G with order pr, then n | m andn ≡ 1 (mod p) (i.e.: p | (n− 1)).

Proof. By induction on |G| we prove (i) as follows.

Let p be a prime and write |G| = prm where p - m.

If p - |G|, then r = 0 and pr = 1. In which case P = 1 is the requiredsubgroup — in particular, this deals with the case |G| = 1.

If p | |G|, then suppose there exists a subgroup H G such thatpr | |H|, then |H| = prm′ where p - m′ so by induction H contains atleast one subgroup P of order pr, in which case P ≤ G is the requiredsubgroup.

Otherwise, for all H G we have pr - |H| and by Lagrange’s theorem|G| = |H|[G : H] so p | [G : H].

If Z(G) = G, then p | |Z(G)| simply because p | |G|. Otherwise, chooseg ∈ G\Z(G), then CG(g) G so p | [G : CG(g)] = |gG|. Hence, by theclass equation |Z(G)| = |G| −

∑ki=l ni, we see p | |Z(G)|.

Therefore, p | |Z(G)| and by Cauchy’s theorem Z(G) must have anelement x ∈ Z(G) of order p (see theorem 5.13). Let X = 〈x〉, then|X| = p and X ≤ Z(G) ≤ G. Moreover X E G so consider G/X. Wehave |G/X| = |G|/|X| = prm/p = pr−1m, thus by induction |G/X|has at least one subgroup K of order pr−1. Hence K = P/X for someX ≤ P ≤ G (see lemma 7.10) where |P | = [P : X]|X| = |P/X||X| =pr−1p = pr, therefore P ≤ G is the required subgroup.


Remarks

(i) For a group G and p prime Sylp(G) denotes the set of Sylow p-subgroupsof G (i.e.: P ∈ Sylp(G) means P ≤ G and |P | = pr where |G| = prmsuch that p - m).

(ii) The number of Sylow p-subgroups will be denoted np = | Sylp(G)|.

(iii) For any P ∈ Sylp(G) it follows from (ii) in Sylow’s theorem (9.1) thatSylp(G) = P g | g ∈ G. Recall, the number of distinct conjugates ofP in G is np = | Sylp(G)| = [G : NG(P )].

Useful results The following things to remember will be used in succeed-ing lemmas:

(A) If P ∈ Sylp(G), then P EG if and only if np = 1.

Proof. As np = [G : NG(P )], then np = 1 if and only if NG(P ) = G ifand only if P EG.

(B) If |G| = prm with p - m, then np | m and np ≡ 1 (mod p).

(C) If P ∈ Sylp(G) and Q ∈ Sylq(G) where p, q are distinct primes, thenP ∩Q = 1.

Proof. As P ∩Q ≤ g and P ∩Q ≤ P , then |P ∩Q| | |P | by Lagrange’stheorem (1.3), similarly |P ∩Q| | |Q|. As |P | = pr and |Q| = qs for somer, s ∈ N ∪ 0 and p 6= q, it can only be the case that |P ∩ Q| = 1 andtherefore P ∩Q = 1.

Theorem 9.2. If |G| = pq where p, q are distinct primes and p - (q − 1),then G has a normal Sylow p-subgroup.

Proof. Using (B) we know np | q so np = 1 or np = q (as q is prime). Supposenp = q, we also know np ≡ 1 (mod p), so np = 1 ≡ 0 (mod p), i.e. p | (np− 1),but np − 1 = q − 1 contradicting our hypothesis. Therefore np = 1 and thetheorem follows by (A).


Corollary 9.3. If |G| = pq where p, q are distinct primes such that p < qand p - (q − 1), then G is cyclic.

Proof. Let P ∈ Sylp(G) and Q ∈ Sylq(G), by the above theorem P E G.Since p ≤ q we also know q - (p− 1), so similarly QEG. Since |P | = p and|Q| = q are both prime, then P ∼= Zp and Q ∼= Zq are cyclic.

Let P = 〈x〉 and Q = 〈y〉 (where x has order p and y has order q). Bylemma 8.3, [x, y] ∈ P ∩ Q = 1, hence xy = yx and it follows that xy hasorder pq and therefore G = 〈xy〉 is cyclic.

Example 9.4. A group of order 15 is cyclic.

9.1 Simple Groups and Sylow’s Theorems

Lemma 9.5. No simple groups exist of orders (i) 200, or (ii) 30.

Proof. (i) Suppose |G| = 200 = 23 × 52.

For p = 5, by Sylow’s theorems 9.1, we know n5 | 23 so n5 ∈ 1, 2, 4, 8but also n5 = 1 (mod 5) so n5 ∈ 1, 6, 11, . . .. Therefore, it can only bethe case that n5 = 1. So there exists P ∈ Syl5(G) such that |P | = 52

and P EG, therefore G is not simple.

(ii) Suppose |G| = 30 = 2× 3× 5.

Assume G is simple, then n3 > 1 and n5 > 1.

By Sylow’s theorems 9.1, we know n3 | 10 so n3 ∈ 1, 2, 5, 10 andn3 ≡ 1 (mod 3) so n3 ∈ 1, 4, 7, 10, . . ., therefore n3 = 10. Similarlyn5 | 6 so n3 ∈ 1, 6 and n5 ≡ 1 (mod 5) so n3 ∈ 1, 6, 11, . . ., thereforen5 = 5.

Let P1, P2 ∈ Syl3(G) and P1 6= P2, so |P1| = |P2| = 3. By Lagrange’stheorem 1.3, |P1 ∩ P2| | |P1|, so necessarily P1 ∩ P2 = 1. So thenumber of elements (excluding 1) contained in the Sylow 3-subgroupsof G is 2× n2 = 20.

Similarly, the number of elements (excluding 1) contained in the Sylow5-subgroups of G is 4× n5 = 24.

By (C) we are assured we have counted 44 distinct elements in G, acontradiction. Therefore G cannot be simple.


Theorem 9.6. If |G| = pqr where p, q, and r are distinct primes, then G isnot simple.

Proof. Suppose G is simple, then np > 1, nq > 1, and nr > 1. As in theprevious lemma 9.5, we count the number of elements contained in Sylow p-subgroups (respectively q-subgroups, r-subgroups) of G, there are np(p− 1)(respectively nq(q − 1), nr(r − 1)) — excluding 1G in each case (and we areassured we have not double counted), hence

|G| ≥ 1 + np(p− 1) + nq(q − 1) + nr(r − 1) (1)

Without loss of generality assume p > q > r, then

(1) np | qr so np ∈ q, r, qr; np ≡ 1 (mod p) so np ≥ 1 + p > q, r, so np = qr.

(2) nq | pr so nq ∈ p, r, pr; nq ≡ 1 (mod q) so nq ≥ 1 + q > r, so nq ≥ p.

(3) nr | pq so nr ∈ p, q, pq, so nr ≥ q.

hence, substituting into equation (1), we have

pqr ≥ 1 + qr(p− 1) + p(q − 1) + q(r − 1)

rearranging implies

0 ≥ 1 + pq − p− q = (p− 1)(q − 1) ≥ (2− 1)(2− 1) = 1

a contradiction — therefore G cannot be simple.

10 EXTENSION MATERIAL 49

10 Extension Material

10.1 Commutator Subgroups

Definition 10.1. Let G be a group and x, y ∈ G, define

[x, y] = x−1y−1xy

to be the commutator of x and y.Let H,K ≤ G be two subgroups, then the commutator subgroup is defined

to be[H,K] = 〈[h, k] | h ∈ H, k ∈ K〉

For a group G its derived group is G′ = [G,G].

Properties of Conjugation

(i) (xy)g = xgyg,

(ii) xgh = (xh)g,

(iii) (x−1)g = (xg)−1.

Properties of Commutators

(i) [x, y]z = [xz, yz],

(ii) [xy, z] = [x, z]y[y, z],

(iii) [x, yz] = [x, z][x, y]z,

(iv) [h, k]−1 = [k, h].

Lemma 10.2. For subgroups H,K ≤ G, then [H,K] = [K,H].

Proof. Let [h, k] ∈ [H,K], then [h, k] = [k, h]−1. Since [k, h] ∈ [K,H], wehave [h, k] = [k, h]−1 ∈ [K,H]. Hence, the generators of [H,K] all lie in[K,H], it follows that [H,K] ≤ [K,H]. A similar argument shows that[K,H] ≤ [H,K], therefore [H,K] = [K,H].


Lemma 10.3. For subgroups H,K ≤ G, then [H,K] E 〈H,K〉.

Proof. Let x ∈ [H,K], then x can be written x = [h1, k1][h2, k2] · · · [hn, kn]for some [hi, ki] ∈ [H,K], because we can write any [hj, kj]

−1 = [kj, hj] inthe form [h′j, k

′j], as [H,K] = [K,H].

Hence, for any h ∈ H, we have

[hi, ki]h = h−1(h−1

i k−1i hiki)h

= h−1h−1i k−1

i hi(hki(hki)−1)kih

= [hih, ki][ki, h]

= [hih, ki][h, ki]−1 ∈ [H,K]

Therefore

xh = ([h1, k1] · · · [hn, kn])h = [h1, k1]h · · · [hn, kn]h ∈ [H,K]

So [H,K] is closed under conjugation by H.By a similar argument, we can show [H,K] = [K,H] is closed under

conjugation by K.Therefore, [H,K] E 〈H,K〉 as claimed.

Theorem 10.4. If G = AB where A,B ≤ G are abelian subgroups, thenG′′ = 1 (meaning G′ is an abelian subgroup of G).

Proof. Observe that [A,B] E 〈A,B〉, now G = AB so G = 〈A,B〉, hence[A,B] EG.

We show G′ = [A,B]. Certainly [A,B] ≤ [G,G] = G′, so suppose [g, h] ∈[G,G], as g, h ∈ G = AB we can write g = a1b1, h = a2b2 for some a1, a2 ∈ A,b1, b2 ∈ B, thus using the properties of conjugation and commutators

[g, h] = [a1b1, a2b2]

= [a1, a2b2]b1 [b1, a2b2]

= ([a1, b2][a1, a2]b2)b1 [b1, b2][b1, a2]b2

= [a1, b2]b1([a2, b1]−1)b2

∈ [A,B]

using the fact that A,B are abelian so [a1, a2] = [b1, b2] = 1, and that [A,B]E〈A,B〉. Therefore [G,G] ≤ [A,B]. We have shown G′ = [A,B] and so G′EG.

Let a1, a2 ∈ A and b1, b2 ∈ B. As AB = G certainly AB ≤ G soAB = BA, thus we can write ba11 = a3b3 and ab21 = b4a4 for some a3, a4 ∈ A


and b3, b4 ∈ B, thus

[a1, b1]a2b2 = [aa21 , ba21 ]b2

= [a1, a3b3]b2

= ([a1, b3][a1, a3]b3)b2

= [a1, b3]b2

= [ab21 , bb23 ]

= [b4a4, b3]

= [b4, b3]a4 [a4, b3]

= [a4, b3]

Similarly, [a1, b1]b2a2 = [a4, b3], thus [a1, b1]a2b2 = [a1, b1]b2a2 , therefore

[a1, b1]a2b2a−12 b−1

2 = 1

Meaning [a2, b2] = a2b2a−12 b−1

2 commutes with [a1, b1].Given [α1, β1], [α2, β2], by taking a1 = α1, b1 = β1, a2 = α−1

2 and b2 =β−1

2 gives that [α1, β1] and [α2, β2] commutes. Since all such commutatorsgenerate [A,B], it follows that G′ = [A,B] is abelian. Therefore G′′ = 1.


10.2 Finitely Generated Abelian Groups

We will prove the classification theorem for finitely generated abelian groups.

Lemma 10.5. Let q ∈ N, then the number of solutions to xq = 1 in Zm isgcd(m, q).

Proof. In the context of Zm we have xq = 1 if and only if qx ≡ 0 (modm) ifand only if m | qx.

Hence, we find how many integers r satisfy 0 ≤ r < m and m | qr.Let d = gcd(m, q) and write m = m′d and q = q′d, then gcd(m′, q′) = 1.If m | qr, then m′d | q′dr, hence m′ | q′r, so m′ | r as gcd(m′, q′) = 1.

Therefore r is one of 0,m′, 2m′, . . . , (d− 1)m′.Conversely, each such r satisfies 0 ≤ r < m and m | qr.Thus, there are d choices for r, and therefore d = gcd(m, q) solutions.

Corollary 10.6. Let q ∈ N, then the number of solutions to xq = 1 inH = Zm1 × Zm2 × · · · × Zmk is gcd(m1, q) gcd(m2, q) · · · gcd(mk, q).

Proof. Let x = (a1, a2, . . . , ak), then xq = 1 if and only if aqi = 1 for alli = 1, . . . , k.

By the previous lemma, there are gcd(mi, q) solutions to aqi = 1 in Zmifor each i = 1, . . . , k.

Therefore, there are gcd(m1, q) gcd(m2, q) · · · gcd(mk, q) possibilities forx, all solutions.

Theorem 10.7. Let G1 = Zm1×· · ·×Zmr with mi | mi+1 (for i = 1, . . . , r−1)and G2 = Zn1 × · · · × Zns with ni | ni+1 (for i = 1, . . . , s− 1), then G1

∼= G2

if and only if r = s and mi = ni (for i = 1, . . . , r).

Proof. Suppose G1∼= G2 (the converse is trivial), then the number of solu-

tions to xq = 1 must the the same in G1 and G2 for all q ∈ N Without lossof generality suppose r ≥ s.

Set q = m1, then we have

gcd(m1,m1) · · · gcd(m1,mr) = gcd(m1, n1) · · · gcd(m1, ns)

The left-hand side is mr1 as gcd(m1,m1) = m1 and m1 | mi for i = 1, . . . , r,

whereas the right-hand side is at most ms1. Hence, mr

1 ≤ ms1, but r ≥ s so

mr1 ≥ ms

1, therefore mr = ms and r = s.Furthermore, as the right-hand side must then attain its maximum value

we must have gcd(m1, n1) = m1, thus m1 | n1.Similarly, setting q = n1 tells us n1 | m1, hence m1 = n1.


For induction assume mi = ni for i = 1, . . . , k−1 (k < r). Setting q = mk

gives

gcd(m1,mk) · · · gcd(mk−1,mk) gcd(mk,mk) gcd(mk+1,mk) · · · gcd(mr,mk)

is equal to

gcd(n1,mk) · · · gcd(nk−1,mk) gcd(nk,mk) gcd(nk+1,mk) · · · gcd(nr,mk)

We may cancel the first k − 1 factors as ni = mi for i = 1, . . . , k − 1, giving

gcd(mk,mk) · · · gcd(mr,mk) = gcd(nk,mk) · · · gcd(nr,mk)

Now the left-hand side is exactly mr−(k−1)k and the right-hand side is at most

mr−(k−1)k , attaining this maximum only when nk = mk. As we know equality

holds, we can conclude nk = mk.


10.3 An is simple for n ≥ 5

Theorem 10.8. For n ≥ 5, the alternating group An is simple.

Proof. We have seen that A5 is simple, this is a base for our inductive proof.Suppose n ≥ 6 and An−1 is simple, and let G = An and 1 6= N E G be anormal subgroup.

Since N 6= 1, there exists a permutation σ ∈ N with σ 6= (1). Writeσ = (a, b, . . . , c)(. . .) · · · as a product of disjoint cycles. Consider the cycleσ1 = (a, b, . . . , c), the first cycle of σ of length at least 2 (we may have b = c).If σ1 has length is greater than 3 choose d in this cycle with d 6∈ a, b, c,otherwise choose d in one of the other cycles (such at d must exist as n ≥ 6),then dσ = e for some e and in particular e 6= a.

Let ρ = (a, b, d) ∈ G, then [σ, ρ] = σ−1σρ ∈ N as N EG, now

[σ, ρ] = (ρ−1)σρ = (aσ, dσ, bσ)(a, d, b) = (b, e, ?)(a, d, b)

First [σ, ρ] 6= (1) as e 6= a, and secondly [σ, ρ] is a product of 3 cycles whichshare a common point b, therefore [σ, ρ] can move at most 5 points. As n ≥ 6,then [σ, ρ] must have a fixed point α. Hence [σ, ρ] ∈ Gα.

Therefore [σ, ρ] ∈ Gα∩N , hence Gα∩N 6= 1. As N EG it follows thatGα ∩N E Gα, but Gα

∼= An−1 which is simple by our induction hypothesis.Thus Gα = Gα ∩N ≤ N .

It follows that Gα contains a 3-cycle, as N is normal in An it must there-fore contain all 3-cycles (they are conjugate in An for n ≥ 5), thereforeN = G (as 3-cycles generate An for n ≥ 3) and G = An must be simple.


10.4 The unique simple group of order 60

Theorem 10.9. If G is a simple group of order 60, then G ∼= A5.

Proof. Let G be a simple group of order 60, we first show that G must containa subgroup of index 5.

If H ≤ G and [G : H] = n > 1, then we know |G| | n!, as |G| = 60 it isnecessary that n ≥ 5 (as 60 - 2!, 60 - 3! and 60 - 4!).

Suppose for contradiction that G does not contain a subgroup of index 5.As |G| = 60 = 22 × 3× 5, we have Sylow-2, Sylow-3 and Sylow-5 subgroups.

Now n2 ∈ 1, 3, 5, 15 but by our assumptions only np = 15 is allowed,similarly n3 ∈ 1, 2, 4, 10, 20 but only n3 = 10 is allowed.

Let P1, P2 ∈ Syl2(G) and P1 6= P2. If P1∩P2 6= 1, let x ∈ P1∩P2 wherex 6= 1. As |P1| = |P2| = 22 has prime-squared order, P1, P2 are abelian.Hence P1, P2 ≤ CG(x), we cannot have CG(x) = P1 as then P2 = CG(x) =P1, so |Cg(x)| > 4. CG(g) 6= G, otherwise 〈x〉 E G is a normal subgroup,contradicting the fact that G is simple. So by Lagrange |CG(x)| = 12 or|CG(x)| = 20, giving index 5 or index 3 respectively, both contrary to ourassumptions. Therefore P1 ∩ P2 = 1.

More easily, if Q1, Q2 ∈ Syl3(G), then Q1 ∩Q2 = 1, as |Q1| = |Q2| = 3is prime and there can only exist the trivial proper subgroup.

So by counting distinct elements, as Sylow subgroups for different primesonly share the identity element, we have

60 = |G| ≥ 15(4− 1) + 10(3− 1) + 1 = 66

a contradiction.Therefore, ∃K ≤ G such that [G : K] = 5. Hence, G ∼= L for some

subgroup L ≤ S5. Now L ∩ A5 E L since A5 E S5 and

120 = 5! ≥ |LA5| =|L||A5||L ∩ A5|

=602

|L ∩ A5|

so we must have |L ∩ A5| > 1, so L ∩ A5 6= 1, but L ∼= G is simple, soL ∩ A5 = L. Since L = L ∩ A5 ≤ A5 and |L| = 60 = |A5|, we must haveL = A5, that is G ∼= A5.

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