Chapter 4- structure of groups

Question 1

define proper subgroup of G

Answer 1

subgroup not equal to e or G

Question 2

G-group, N-proper normal subgroup
On can think of G as being decomposed into 2 groups namely ; N and G/N.
Explain this (2-reasons)

In this context, it is natural to expect that the groups which cannot be decomposed in this way are significant.

Answer 2

For example, the order of G is equal to |N||G/N|. Also, these two groups reflect parts of the subgroup structure of G: N con- tains all subgroups of G contained in N, whereas, by the Correspondence Theorem, the subgroups of G/N are in 1-1 correspondence with subgroups of G containing N.

Question 3

Definition 14.1.
A non-trivial group is said to be simple if…

Answer 3

it has no proper normal subgroups

Question 4

It’s easy to characterise all ABELIAN simple groups.

Thm 14.2
Let G be an abelian group. Then G is simple if and only if…

Answer 4

G is isomorphic to Zp for a prime number p.

Question 5

Thm 14.3
The alternating group An is…

Answer 5

simple for every n≥ 5.

Summarised proof: (The real proof is extremely long. Read through at some point.)

Two facts are established about a normal subgroup N of An:
I) If N contains a non-identity permutation, then N contains a 3-cycle.
II) If N contains one three-cycle, then it contains all 3-cycles.
The above statements are proven in reverse order.
For II), let (i j k) ∈ N, and let (p q r) be an arbitrary three-cycle. It is shown that (p q r) ∈ N by first proving that (i j r) ∈ N and then using this to write out (p q r) in terms of (i j r) and replacing variables to obtain (p q r) ∈ N.
For I), let σ ∈ N be a non-identity permutation. We write σ as a product of disjoint cycles and examine five cases to show that N contains a three-cycle in each case:
(1) γ1 has length at least four;
(2) γ1 and γ2 both have length three;
(3) γ1 has length three, and γ2 has length two;
(4) γ1 has length two and r = 2;
(5) γ1 has length two and r > 2.
Since the set of all 3-cycles generates An, it is concluded that if N is non-trivial, then N = An, completing the proof of the theorem.

Question 6

Another class of simple groups arises from matrix groups. Consider the special liner group SL(m, F), where m ≥ 2 and F is a finite field. The set N=…

The quotient SL(m, F)/N is called…

Answer 6

N = {kI:k∈F,km=1} is easily seen to be a normal subgroup of SL(m, F).

The projective special linear group, and is denoted by PSL(m, F). It turns out that PSL(m, F) is simple.

Question 7

As indicated in the previous section, if a group G is not simple, it can be ‘decomposed’ into two groups N and G/N, where N is a proper normal subgroup.
Then, if they are not simple, one can ‘decompose’ N and G/N. If G is finite, then this process will eventually stop, and one will obtain a collection of simple groups. So, every finite group is in some sense built of simple groups.

Answer 7

This raises 2 questions:
1.given a finite group G, describe different ways of decomposing G into collections of simple groups.
2. The second question is, given two groups H and K, describe different ways in which one can ‘put them together’ to form a new group which ‘decomposes’ into H and K. This is a much harder question

Question 8

Def 5.1
A subnormal series of a group G is…..

The factors of this series are…

The length of the series is…

The series is a composition series if…

Answer 8

a chain of subgroups G=G0 ≥G1 ≥G2 ≥…≥Gn such that Gi+1is a normal subgroup of 􏰌Gi for all i,0≤I<n

the quotients Gi/Gi+1, 0 ≤ i < n.

n.

Gn = {e} and all factors are simple.

Question 9

Thm 15.2
Every finite group G has…

Answer 9

A composition series

Question 10

Def 15.4
2 subnormal series G ≥ G1 ≥ G2 ≥ . . . ≥ Gn and G ≥ H1 ≥ H2 ≥ … ≥ Hm of the same group G are said to be equivalent if…

Answer 10

they have equal lengths and there is a one-one correspondence between the factors such that the corresponding factors are isomorphic.

Question 11

(Following is given with our proof)
Thm 15.5 (Jordan-holder)

Answer 11

If a group has a compositions series then any 2 composition series are equivalent.

In particular any two composition series of a finite group are equiva- lent. Thus a finite group is composed from simple groups, and the ingre- dients are uniquely determined by the group.

Question 12

Another common idea in group theory is to consider abelian groups as ‘well understood’ (in light of Theorem 13.7, for example), which says…
and then to try to somehow measure how far off a group is from being abelian.
The main tools used in this are..

Answer 12

Every finite abelian group is isomorphic to a direct sum of cyclic groups, each of which has a prime power order.

The centre and derived subgroup.

Question 13

Definition 16.1.
The centre of a group G is the set

Answer 13

Z(G)={x∈G : xa=ax for all a∈G}. i.e. the set of all elements of G that commute with every element of G.

Question 14

Theorem 16.2.
Let G be a group. Then (i) Z(G)􏰌􏰌 is a normal subgroup of…
(ii) G is abelian iff Z(G)=G ; (obvious)

Answer 14

G

Question 15

Definition 16.4.
Let G be a group. The commutator of two elements a, b ∈ G is the element…

The derived subgroup G′ of G is the subgroup of G…..

Answer 15

[a, b] = a−1 b−1 ab

generated by all commutators, i.e.
G′ = ⟨{[a,b] : a,b ∈ G}⟩.

Question 16

Remark 16.5
The product of 2 commutators is not necessarily a commutator and so…

Answer 16

The commutator subgroup doesn’t only contain commutators.

Question 17

Theorem 16.6.
Let G be a group. The derived subgroup of G is

Answer 17

the smallest normal subgroup N of G such that the quotient G/N is abelian.

proving this:
we know its a subgroup.
check normal by conjugation, not that the inverse of a commutator is a commutator, so any element of the group is just a product of commutators.
The prove quotient is abelian
(You should study the last part of this proof and make sure you understand each step)

Question 18

Recall: Every group G acts on itself via conjugation.

We can generalise this to an action an all subsets of G via…

The orbit of S under this action is …

The size or this orbit will be the index of the stabiliser of S given by…

This set (the stabiliser) is called the normaliser of S, and will be denoted by N(S) from now on.

Answer 18

(S, g) 􏰂→ g−1Sg.

CS ={g−1Sg : g∈G}.

GS ={g∈G : g−1Sg = S}.

Question 19

Theorem 17.1.
If H is a subgroup of G then H is a normal subgroup of N(H). Moreover, N(H) is the largest subgroup of G such that…

Answer 19

H is normal in it.
In particular, if H is a normal subgroup of G then N(H) = G.

Easy proof: remember normal=closed under conjugacy, and here we are using the conjugacy action

Question 20

Translating Theorem 11.12 into our context here
(Thm 11.12: (Orbit–Stabiliser Theorem) Let G be a group that acts on a set X. Then for any x ∈ X, the size of the orbit of x is equal to the number of cosets of its stabiliser:
|Cx| = [G : Gx]),
we obtain:
Theorem 17.2. …

Answer 20

Let G be a group, and let S be a non-empty subset of G. The number of distinct conjugates of S in G is equal to [G : N(S)], and divides the order of G by Lagrange’s Theorem.

Question 21

Specialising to singletons, for a ∈ G the normaliser becomes

his is sometimes referred to as the

Answer 21

N(a)={g∈G : g−1ag=a}={g∈G : ag=ga},
the set of all elements that commute with a.

centraliser of a

Question 22

Note that an element a ∈ G has a singleton conjugacy class {a} if and only if

Answer 22

N(a) = G, i.e. if and only if a commutes with all the elements of G.
(This is to do with the orbit-stabilise theorem)

Question 23

Lemma 17.3.
An element x of a group G is the only conjugate of itself if and only if

Answer 23

it belongs to the centre Z(G)

Question 24

Now, consider an arbitrary finite group, and denote by C1 , C2 ,. . . , Cn all the conjugacy classes of G. Further, assume that Cr+1,. . . , Cn are those of them which contain only one element. Then, by Lemma 17.3, we have
Z(G)=
Also, if for each i, 1 ≤ i ≤ n, we choose a representative xi ∈ Ci, then by
orbit-stabiliser , we have…
Finally, we note that
|G| = |C1| + . . . + |Cr| + |Cr+1| + . . . + |Cn| = the sum of (􏰏[G : N(xi)] )(up to r,i.e. the non-singletons)+ |Z(G)|

Answer 24

Z(G)=Cr+1 ∪…∪Cn.

|Ci| = [G : N(xi)].

Question 25

Theorem 17.4 (The Class Equation)
Let G be a group, let Ci, 1 ≤ i ≤ n, be its conjugacy classes, and let xi ∈ Ci, 1 ≤ i ≤ n, be arbitrary. Furthermore, let us assume that each of the classes Ci, 1 ≤ i ≤ r, contains at least two elements, while the remaining classes Ci, r + 1 ≤ i ≤ n, are one-element. Then…

Answer 25

|G| = |Z(G)| + 􏰏(The sum of: [G : N(xi)]…for each of the classes up to r)

Question 26

Definition 18.1.
A group of order pn, where p is a prime, is called…

Answer 26

a p-group.

Question 27

Lemma 18.2.
If G is a group of order p^n, where p is a prime, then…

Answer 27

G has a no trivial centre
Z(G)={x∈G : xa=ax for all a∈G}.

SEE proof: uses class equation

Question 28

Lemma 18.3
Let G be a group. If G/Z(G) is a cyclic group, then

Answer 28

G is abelian and G = Z(G).

Question 29

Theorem 18.4
Every group of order p^2 where p is a prime is either isomorphic to

Answer 29

Zp^2 or to Zp ⊕Zp.

proof uses previous 2 lemma and FTFAG

Question 30

Remark 18.5.
It is not true that every p-group is abelian. Indeed, Q8 and D4 are non-abelian 2-groups.

Question 31

Theorem19.2 (The first Sylow theorem) Let G be a finite group, and let|G|= (p^n)*m, where p is a prime and gcd(p,m) = 1. Then…

Answer 31

Then for each i, 1 ≤ i ≤ n, G con- tains a subgroup of order p^i.

Proof is long but understandable but I still need to run through sylow proofs.

Question 32

Definition 19.3.
If G is a finite group of order pnm, where p is a prime and gcd(p, m) = , then a sylow p-subgroup is …

Answer 32

any subgroup of G of order p^n

Question 33

The first Sylow theorem states that a finite group G contains a Sylow p-subgroup for every prime p dividing |G|. The Second Sylow Theorem asserts that all Sylow p-subgroups are related via conjugation.

Question 34

Theorem 19.4 (The second Sylow theorem) Let G be a finite group and let p be a prime dividing the order of G. Every conjugate of a Sylow p-subgroup of G is…
Conversely…

Answer 34

again a Sylow p-subgroup of G.

Conversely, every two Sylow p-subgroups are conjugate in G.

Question 35

Lemma 19.5.
Let P be a Sylow p-subgroup of a finite group G, and let g∈G be an element whose order is a power of P. Then…

Answer 35

If g−1Pg=P then g∈P x.

used in proof of second sylow thm

Question 36

Lemma 19.6.
Let G be a group, and let H, T ≤ G. The number of distinct conjugates of H of the form t−1Ht with t ∈ T is

Answer 36

[T : T ∩ N(H)].

Used in proof of second sylow thm

Question 37

The third sylow theorem
Let G be a finite group, and let p be a prime dividing the order of G. The number of Sylow p-subgroups of G

Answer 37

Let G be a finite group, and let p be a prime dividing the order of G. The number of Sylow p-subgroups of G divides |G| and has the form ps + 1 for s ≥ 0.