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.
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.