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Let $G$ be a group, $g, h\in G$, and $N\trianglelefteq G$. Prove that $gN = hN$ if and only if $g^{-1}h\in N$.

The first part is easy:

Suppose $gN = hN$, then $\exists n_1, n_2\in N$ such that $gn_1 = hn_2 \iff n_1 = g^{-1}hn_2\iff n_1 n_2^{-1} = g^{-1}h \implies g^{-1}h\in N$.

Now, I can also prove a half of the second part, but I don't get how to go about the rest of it.

Suppose $g^{-1}h\in N$, then $\exists n_1, n2$ such that $g^{-1}h = n_1 \iff h=gn_1\implies h\in gN\implies hn_2\in gNn_2=gN$. Thus $hN\subseteq gN$. But I don't see how to prove that $gN\subseteq hN$. I can show that $g^{-1}\in h^{-1}N$, but this doesn't help much.

Would appreciate some help.

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    Hint: $g^{-1}h \in N \implies (g^{-1}h)^{-1} = h^{-1}g \in N$ – 2017-02-01
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    Got it, thanks :) – 2017-02-01
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    Pleasure was all mine! – 2017-02-01
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    Cosets partition groups, and in particular distinct cosets must be disjoint. Anytime you see $aH\subseteq bH$ that automatically implies $aH=bH$. – 2017-02-02

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