1
See my comments above for 1 and 2.
As for 3, I will use the one-step subgroup test.
Fix $m\in\Bbb N$.
Since $e\in H$, ${\rm ord}(e)=1$ and $1\mid m$, we have $e\in Q_m$. Thus $Q_m$ is nonempty.
By definition, $Q_m=\{\color{red}{x\in H}\; :\; {\rm ord}(x)\mid m\},$ so we have $Q_m\subseteq H$.
Let...
