Commutator Subgroup with Abelian Factor Group

I'm trying to show that if $Y$ is the commutator subgroup of a group $G$ (assumed to be normal in $G$ with $G/Y$ Abelian), and $N$ is a a subgroup with those same properties, then $Y$ is a subgroup of $N$.

My approach has been to try to show that $\forall y\in Y$, $yN = N \Leftrightarrow xwx^{-1}w^{-1}N=N$, where $y=xwx^{-1}w^{-1}$ for some $x,w\in G$. Then I would have $y\in N$.

Is this the correct approach? I'm getting stuck.

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Suppose that $Y$ were not contained in $N$; then there is a commutator $aba^{-1}b^{-1} \in Y$ not in $N$. Now consider the commutator of $a + N$ and $b + N$ in the quotient group $G/N$:

\begin{align*} (a+N)(b+N)(a+N)^{-1}(b+N)^{-1} &= aba^{-1}b^{-1}+N \ne N \end{align*}

since $aba^{-1}b^{-1} \notin N$. It follows that $G/N$ isn't abelian, since there's a non-identity commutator.

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Could you explain the statement "It follows that $G/N$ isn't abelian, since there's a non-identity commutator"? – Paul Malinowski Mar 5 '14 at 2:31
The elements $a +N$ and $b+N$ don't commute. (Note that elements commute $\iff$ their commutator is the identity.) – user61527 Mar 5 '14 at 2:31