Let $H \le G$ and let $g \in G$. Prove that if the right coset $Hg$ equals some left coset of $H$ in $G$ then it equals the left coset of $gH$ and $g$ must be in $N_G(H)$.
I see that, if $hg \in Hg$, then $(hg)^{-1} = g^{-1} h^{-1} \in g^{-1}H$. And, since the number of left cosets equals the number right cosets, it seems plausible that there must be a bijection between $gH$ and $Hg$ (presumably of the form $gH \mapsto g^{-1}H$); but getting to $gH = Hg$ is escaping me.
Since I've yet to use the fact that the right coset $Hg$ equals some left coset of $H$ in $G$, I assume that that's the key. So, suppose $Hg = aH$ for some $a \in G$. Then, $hg = ah'$, for some $h' \in H$. Well, this doesn't tell me much… I've got that $g \in h^{-1}aH$, $a\in Hg(h')^{-1}$, but I'm not seeing much else.
Any help here would be appreciated.
Best Answer
Hint: If two left cosets share an element (have nonempty intersection), then they are equal.