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I've seen this in various proofs, but never justified. I'm guessing it's something very simple, but I haven't been able to do it so far.

Let $G$ be a group of finite order $n$ and $H\triangleleft G$ such that $(G:H)=|G/H|=m$. I want to prove that $(aH)^{m} = H$ for any $a \in G$.

I know that $(aH)^m=a^mH$ and that $a^n=e$ for any element of $G$. These two facts seem to be the key, but I don't see exactly how.

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    are you aware of Lagrange's Theorem? – 2017-01-13

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First consider the following lemma:

Let $G$ be a finite group of order $n$. Then $g^n=e$ for any $g\in G$.

This follows from Lagrange's Theorem and you can find the full proof here:

Does someone know why raising the element of a group to the power of the order of the group yields the identity?

Now back to your case. Since $H$ is normal in $G$ then $G/H$ is a group. Also the order of $G/H$ is equal to $(G:H)$. Apply the lemma above to $G/H$ to get the result.

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    So let me see if I got it right: $aH \in G/H$ and $|G/H|=m$. Thus, using the lemma you stated, $(aH)^m = e$. But the identity $e$ of $G/H$ is $H$. – 2017-01-13
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    @XicoSim Exactly. – 2017-01-13
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G/H is a quotient group of order m. So $\forall aH\in G/H, ord(aH)|m$. Then $(aH)^m=(aH)^{ord (aH)}=H=1_{G/H}$.