Chapter 1 - Introduction

Question 1

What is a normal group? What are simple groups? How can you classify finite simple groups?

Answer 1

A normal subgroup of G is N<G with gN=Ng for all g in G
A simple group is one with no normal subgroups.

We classify simple groups into abelian and non-abelian groups.

SIMPLE ABELIAN GROUPS:
These are just cyclic groups of prime order.

NON-ABELIAN SIMPLE GROUPS:

1) An the alternating group of Sn
2) Certain matrix groups over finite fields
3) 26 others

Question 2

State the 3 ISOMORPHISM theorems. 
Hint:
1) is about kernels
2) is about subgroups of G/K
3) is about H^K

Answer 2

1) φ : G—>H a HOM. and K=kerφ then K◀G (is normal) and G/KΞImφ

2) Let K◀G then every subgroup of G/K is of the form H/K for some K<G.
Also, H is normal in G IFF H/K is normal in G/K.
If this is true, then GmodH is ISOM to (G/K) mod (H/K)

3) H a sbgp of G, K normal in G. Then H^K is normal in H and HK is a sbgp of G with the property:
H/(HΛK) Ξ HK/K

Question 3

What is the automorphism group of G?
What is the inner automorphism group of G?
What are the outer automorphisms?
What is the centre of G?
Prove:

G/Z(G) Ξ InnG ◀ AutG

Answer 3

Automorphism group of G is AutG, the set of ISOMS of G
The inner automorphism group is a subgroup of AutG called InnG. It is generated by all elements of the form:
θg : G –> G
θg : x |–> g^(-1)•x•g

α in AutG is an outer automorphism if α € AutG \ InnG

The centre of G is:
Z(G)={z in G : zg=gz for all g in G}

Let Θ : G –> AutG be defined by
Θ : g |–> θg

Use the ISOM theorem for the first part and then show InnG◀AutG is easy (do it!)

Question 4

What is the Outer automorphism group?

Answer 4

OutG = AutG/InnG

NOTE: it’s elements are NOT outer automorphisms

Question 5

What do we mean if we say a subgroup H<G is characteristic?Suppose KcharH then show:

i) if H◀G then K◀G
ii) if HcharG then KcharG

Answer 5

HcharG if α(H)=H for all α in AutG. So if H is preserved under all automorphisms of G.

Let A=InnG for i) and A=AutG for ii)

Then for all α in A we have α(H)=H so then α restricted to H is an automorphism of H, and so α(K)=α|H(K)=K ◼

Question 6

What is the commutator of two elements [g,h]?
What is the commutator of two subgroups [H,K]?
What is the commutator subgroup or derived subgroup of G?

Answer 6

[g,h] =(g^-1)•(h^-1)•g•h
[H,K]=the subgroup of G generated by all [h,k] with h in H, k in K

The commutator subgroup is G’=[G,G], which is the group generated by all commutators of elements in G

Question 7

Prove:

G’charG AND given N◀G the quotient G/N is abelian IFF G’<N

Answer 7

Proof:
Given α in AutG, for all h,g in G we have α[g,h]=(α(g)^-1)•(α(h)^-1)•α(g)•α(h)
=[α(g),α(h)],
So α(G’)=G’ so G’charG

Given N◀G, g,h in G with:
Ng•Nh=Nh•Ng IFF
Ngh=Nhg         IFF
N[g,h]=N          IFF
[g,h] in N

Thus G/N abelian IFF G’<N