Let G be a group. The commutator subgroup of G, often denoted G' , is the group generated by the set $\{aba^{-1}b^{-1} ~~|~~a, b\in G\}$
Prove that G is abelian if and only if G' equal to the trivial subgroup of G. Note: Elements of the form $aba^{-1}b^{-1}$for a, b in G are referred to as commutators
Please, help

Question

Let G be a group. The commutator subgroup of G, often denoted G' , is the group generated by the set $\{aba^{-1}b^{-1} ~~|~~a, b\in G\}$
Prove that G is abelian if and only if G' equal to the trivial subgroup of G. Note: Elements of the form $aba^{-1}b^{-1}$for a, b in G are referred to as commutators
 Please, help

loser66 · Answer

If G' is commutative, then $aba^{-1}b^{-1} = aa^{-1}bb^{-1}=e$ then, G' ={e} = trivial subgroup of G.

loser66 · Answer

then??

confluxepic · Answer

@Loser66 What course is this?

loser66 · Answer

Abstract algebra

confluxepic · Answer

The set is the collection of all finite products of commutators in G.

ikram002p · Answer

abelians means for any a,b in the group 
ab=ba right ?

ikram002p · Answer

i just need u to post the trivial subgroup definition , if possible :)

loser66 · Answer

:) I am sorry @ikram002p  the net did not let me in.

ikram002p · Answer

wait @nerdguy2535  is here to save the day :P

anonymous · Answer

You have one direction done. The other isn't harder by any means. You want to show if $G'=\{aba^{-1}b^{-1}\mid a,b\in G\}=\{e\}$, then $G$ is abelian. Any ideas?

loser66 · Answer

I think next is xyx^-y^- =e 
                     xy = yx for all x,y in G --> G is abelian. oh, a, b instead of x, y

anonymous · Answer

Yeah. That's it. This problem was just checking you knew your definitions.

loser66 · Answer

and other ways is if G is abelian, then, what??

loser66 · Answer

oh, G' is trivial

anonymous · Answer

Yeah, you have that one in your first post.

ikram002p · Answer

what does trivial mean ?its the only thing that i dint got :O
i thought the same way u did loser but i get confused with the trivial

loser66 · Answer

and how to show G' is normal?? hihihi, to me the most difficult thing is proving the trivial thing

anonymous · Answer

A group or subgroup is trivial when there is only one element in it. The identity.

loser66 · Answer

oh, all is just check in definition??

ikram002p · Answer

ohh :O ok then its solved !

loser66 · Answer

One more question: how many commutator subgroups we can have of a group? let say S_3

loser66 · Answer

Is there any way to find it out? or we have to check one by one? I mean list out all of permutation in S3 then checking?

ikram002p · Answer

for subgroups its 2^n such that n is the order of the union group , but im not sure about commutator ...

loser66 · Answer

My prof never talked about commutator and he gave me those homework. I have to make question here to ask for help

loser66 · Answer

Thanks for the help, my friend. )