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I found this statement in some lecture notes, and I am having trouble proving it, so I just want to make sure that I understand the statement:

"Let $G$ be a group generated by a subset $S$. Then the first commutator subgroup of $G$ is generated by conjugates of commutators of elements in $S$."

I think it means that $$G'= \langle g[a,b]g^{-1} : g\in G, a,b \in S \rangle,$$ am I right?

In case I am right: I am trying to show that every commutator $[x,y]$ ($x,y \in G$) can be represented in this form. But I can't find a way of doing it. Any hint will be appreciated!

Thank you.

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I guess reason comes from the idendity $[x,zy]=[x,y][x,z]^y$ then $[a,b]=[s_1s_2...s_k,r_1...r_j]$ where $r_i s_j \in S$ then try to apply the idendity. you can find more idendity in en.wikipedia.org/wiki/Commutator –  mesel Mar 9 at 13:16
    
@mesel Thanks, I tried following this direction a while ago and it became too messy, so I left it. Maybe I'll give it another try... –  Ludolila Mar 9 at 14:47

1 Answer 1

up vote 2 down vote accepted

The statement says that $G'$ equals the normal closure N of the set of commutators of the elements of $S$, which is the smallest normal subgroup containing the set of such commutators. To prove this, divide G by N. The quotient group is abelian since it's generators commute. Thus, $N$ contains $G'$. It is also clearly contained in $G'$. Hence, they are equal.

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Did you mean that $N=<[S,S]>^G$ ? –  mesel Mar 9 at 13:40
    
@mesel: I do not know what your notation means; what I mean is $N=<<[S,S]>>$, which is the usual notation for the normal closure of a subset, see en.wikipedia.org/wiki/Normal_closure. –  studiosus Mar 9 at 13:56
    
I don't see the notation $\langle\langle S \rangle\rangle$ on the wikipedia page, and I have never seen it before. The standard notation for the normal closure of $S$ is $\langle S^G \rangle$ or $\langle S \rangle^G$, which is used on the linked page to conjugate closure. I like your answer to the question though! –  Derek Holt Mar 9 at 14:37
    
I know what normal closure is, but didn't see it this way, so thank you for rephrasing! Just one more question: how do you see that the generators of the quotient group commute? –  Ludolila Mar 9 at 14:44
    
@Ludolila: Because you divide by subgroup containing elements of the form $[s_i,s_j]$ where $s_i, s_j$ are elements of $S$. –  studiosus Mar 9 at 15:03

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