$ G/N$ has an element of order $m$ so, $G$ also has an element of order $m$

Question

I swear to God, I've done about a trillion of these sorts of problems and I still don't know how to prove them. Is it as easy as saying that $G/N$ is a subgroup of $G$, and therefore $G$ contains this element of $m$? Of course it's not that simple.

Answer 1

Suppose $G/N$ has an element of order $m$. This means there is an equivalence class $gN$ with the property that if you multiply it by itself $m$ times, you get the equivalence class $N$. So $g^mN=N$, which implies $g^m\in N$ and $g^k \notin N$ for $k < m$. In particular, for such $k$, $g^k$ is not the identity, so the order of $g$ is at least $m$. In fact, it must be a multiple of $m$, since otherwise dividing the order by $m$ and looking at the remainder would contradict $g^k \neq e$ (the identity) for $k < m$.

Write the order of $g$ in $G$ as $mn$ for some $n\ge1$. Then $G$ has a cyclic subgroup of order $mn$. Knowing the structure of subgroups of cyclic groups, we know that $G$ also has to have a subgroup of order $m$, and we're done.