This is Exercise 2.1.8 of "Combinatorial Group Theory: Presentations of Groups in Terms of Generators and Relations," by W. Magnus et al.
The Details:
Theorem 1: Let $G=\langle a, b, c, \ldots \mid P, Q, R, \dots\rangle$ under the mapping $$a\mapsto g, b\mapsto h, c\mapsto k, \dots$$ and let $N$ be the normal subgroup of $G$ generated by $S(g, h, k, \dots), T(g, h, k, \dots), \dots$ Then the factor group $G/N$ has the presentation $$\langle a, b, c, \ldots \mid P, Q, R, \dots , S, T, \dots\rangle\tag{1}$$ under the mapping $$a\mapsto gN, b\mapsto hN, c\mapsto kN, \dots.\tag{2}$$
The Question:
The commutator subgroup $G'$ of a group $$G=\langle a, b, c, \dots \mid P, Q, R, \dots\rangle$$ is the normal subgroup of $G$ generated by the commutators $$aba^{-1}b^{-1}, aca^{-1}c^{-1}, \dots , bcb^{-1}c^{-1}, \dots.$$ Show that, for any normal subgroup $N$ of $G$, $G/N$ is Abelian if and only if $G'\subseteq N$.
[Hint: Use Theorem 1.]
My Attempt:
($\Leftarrow$) Suppose $G'\subseteq N$. Then the generators of $G/N$ commute by Theorem 1. Hence $G/N$ is Abelian.
($\Rightarrow$) Since $G/N$ is Abelian, $$aba^{-1}b^{-1}, aca^{-1}c^{-1}, \dots , bcb^{-1}c^{-1}, \dots$$ can be added to the relators of the presentation of $(1)$ of $G/N$, so that, via the mapping $(2)$, we have that $G'\subseteq N$.
Is that correct? I'm not sure about $(\Rightarrow)$.
Please help :)
($\Leftarrow$) What are the generators of $G/N$? What happens is that, since $G$ is generated by $a,b,c,\ldots$, then $G/N$ is generated by $aN,bN,cN,\ldots$ But, for instance $(aN)(bN)=(bN)(aN)$, since this is equivalent to $aba^{-1}b^{-1}\in N$, which is true, since $aba^{-1}b^{-1}\in G'$ and $G'\subset N$.
($\Rightarrow$) Since $G/N$ is abelian, it is easy to prove that, say, $aba^{-1}b^{-1}\in N$. Since this happens for any two generators of $G$ (I mean $a,b,c,\ldots$), $G'\subset N$.