Order of the subgroup <g>H of G

Question

This is an old exam question I'm trying to solve:

Having a group $G$ and $H$ a normal subgroup of $G$ with order $n$ and taking $g$ in $G$ to be such that $gH$ has order $m$ in $G/H$, I wish to prove that $\langle g \rangle H$ is a subgroup of $G$ of order $mn$.

My attempt:

If it's a subgroup, I can totally prove it has order $mn$. But I can't make it to prove that it is a subgroup for some reason. I guess I'll need to use the fact that h is closed under conjugacy but even with that I haven't made it... Can you help me? Any hint would be awazing! Thank you!

Answer 1

First of all $\langle g\rangle H$ is nonempty, since $e=ee$ and $e$ is an element of both $\langle g\rangle$ and $H$. Now let $g^i h$ and $g^j h'$ be elements of $\langle g\rangle H$. Their product is $$ g^i h g^j h' = g^i g^j g^{-j} h g^j h' = \underbrace{g^{i+j}}_{\in\langle g\rangle}\ \underbrace{(g^{-1} h g)^j}_{\in H}\ h' \in \langle g\rangle H, $$ where we used that $H$ is normal, so $g^{-1}hg\in H$.

I leave it to you to check the same trick works for showing that $(g^i h)^{-1} \in \langle g\rangle H$, which concludes the proof that $H$ is in fact a subgroup of $G$.

Answer 2

More generally, if $H, K$ are subgroups of $G$, with $H$ normal in $G$, then $$ H K = \{ h k : h \in H, k \in K \} $$ is a subgroup of $G$.

This is because, if $h_{i} \in H$ and $k_{i} \in K$, then $$ (h_{1} k_{1}) (h_{2} k_{2}) = (h_{1} k_{1} h_{2} k_{1}^{-1}) (k_{1} k_{2}) \in HK, $$ as $k_{1} h_{2} k_{1}^{-1} \in H$, and then $$ (h k)^{-1} = k^{-1} h^{-1} = (k^{-1} h^{-1} k) k^{-1} \in H K, $$ or use the argument in the comment by Christoph.