# Subgroups-commutator, normal, Abelian

1. Oct 6, 2008

### fk378

Subgroups--commutator, normal, Abelian

1. The problem statement, all variables and given/known data
Let G be a group and g,h in G. Define the commutator of g and h as [g,h]= gh(hg)^-1. Then define the commutator subgroup, denoted [G,G], of G as the subgroup generated by all the commutators of elements of G, i.e. [G,G]=<{[g,h]: g,h in G}>.

(a) What does an element of [G,G] look like?
Hint: it is not enough to only consider elements of the form [g,h].

(b) Prove that [G,G] is a normal subgroup of G.

(c) Prove that G/[G,G] is an Abelian group.

3. The attempt at a solution

(a) Since [G,G] is cyclic, I thought an element would be of some form of [gh(hg)^-1]^n for some n in N. But the hint kind of threw me off...

(b) Initially I wanted to show that gh(hg)^-1 is mm^-1 but I have to show commutativity there and I don't know how to...so there must be another way? I can't seem to get started on this one...

(c) Is this just a consequence of part (b)? Since to show a subgroup is normal you need to use their cosets as an example.

2. Oct 6, 2008

### morphism

Re: Subgroups--commutator, normal, Abelian

(a) [G,G] is not necessarily cyclic - it's generated by a set. I suggest you look up what this means.

(b) I don't understand what you're saying here. What's the definition of "normal subgroup"?

(c) It's a consequence of (b) that G/[G,G] is a group. Why is it an abelian one?

3. Oct 6, 2008

### fk378

Re: Subgroups--commutator, normal, Abelian

A normal subgroup is some H such that xH=Hx for x in G.
Equivalently, xhx^-1 is in H.

4. Oct 7, 2008

### morphism

Re: Subgroups--commutator, normal, Abelian

Right. Now use the definition to show that [G,G] is normal in G.

5. Oct 7, 2008

### fk378

Re: Subgroups--commutator, normal, Abelian

Well...that's my question here.

This is what I attempted:
Let H be a subgroup of G,
g,g-inverse in G
Then gh is in gH, and (g-inverse)(h-inverse) is in (g-inverse)H.
So then [g,h]=gh(g-inverse)(h-inverse)=gH(g-inverse)H=g(g-inverse)H=eH=H
(since the product of left cosets equals a left coset).

Is that correct? If not...any hints?

If it is right...then I have shown that an element of [G,G] looks like the subgroup H, and then all I need to do now is show xH=Hx?

Subgroup of Index $n$ for every $n \in \Bbb{N}$. Feb 25, 2018