TITLE: Generators of the first commutator subgroup
QUESTION [6 upvotes]: 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.

REPLY [10 votes]: 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.