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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.

Shaun
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Ludolila
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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 http://en.wikipedia.org/wiki/Commutator – mesel Mar 09 '14 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 09 '14 at 14:47

1 Answers1

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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.

Jendrik Stelzner
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Moishe Kohan
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  • Did you mean that $N=<[S,S]>^G$ ? – mesel Mar 09 '14 at 13:40
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    @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 http://en.wikipedia.org/wiki/Normal_closure. – Moishe Kohan Mar 09 '14 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 09 '14 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 09 '14 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 $S$. – Moishe Kohan Mar 09 '14 at 15:03