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Let $G$ be a group, $N \triangleleft G$ and $\bar{G} = G/N$.

Prove that for every two elements $\bar{a},\bar{b} \in \bar{G}$ the following is true:

$\bar{a}\bar{b} = \bar{b}\bar{a} \Leftrightarrow a^{-1}b^{-1}ab \in N$

So this looks a lot like the Correspondence Theorem, so I have an understanding as to how these are connected, but I'm at a loss when it comes to the proof of this. I am supposed to prove this without having that G/N is Abelian.

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    This is actually not a duplicate of the linked question. In that question, both the question and the answer use that $G/N$ is abelian. (I'm not quite sure why, but it's used) – 2017-02-26
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    Doesn't look like the correspondence theorem to me. – 2017-02-26
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    @Nameless It may not be, it just appeared to be related to it. We have not covered this in class and my professor's personally written book doesn't really cover anything else like this. – 2017-02-26
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    @mixedmath The questions asks to show that $G/N$ is abelian if and only if $a^{-1}b^{-1}ab\in N$. This is clearly a duplicate (OK, the above is really not complete, I admit). It is most popular at the web (see e.g. [here](http://www.chegg.com/homework-help/questions-and-answers/prove-g-n-abelian-ab-1-b-1-exist-n-b-exist-g-q3261302)). – 2017-02-26

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If $\overline{a}\overline{b} = \overline{b}\overline{a}$, then we necessarily have that $\overline{a^{-1}}\overline{b^{-1}}\overline{a}\overline{b} = 1$ in $\overline{G}$.

What does this mean on the level of cosets? This means that the coset $a^{-1}b^{-1}ab N$ is the same as the coset $1N = N$. So in particular, every element of $a^{-1}b^{-1}abN$ is contained within $N$, and thus $a^{-1}b^{-1}ab \in N$.