Suppose $\mathbb{Z}/m\mathbb{Z}$ is a subgroup of $S_n$. Then $D_{2m}$ is too.

Question

I need to prove the statement in the title. If $n \geq m$ then it is easy to take an $m$-cycle and construct a reflection from the action of $D_{2m}$ on $m$ vertices. I think I have a similar way of constructing a reflection if $n < m$, but it is ugly. Is there a clean way to do this?

Answer 1

As $S_{n}$ has a cyclic subgroup of order $m$, we get that $m\le n$. Without lost of generality as $S_{n}\le S_{n+k}$, we may assume that $n=m$.
We know that $D_{2m}= \langle a,b\; |\; a^{m}=b^{2}=1, \; bab=a^{-1} \rangle$. So we look for two permutations that satisfy those conditions in $S_{m}$.
By considering $\langle a, b\rangle \le S_{m}$, where $a=(1\: 2\: 3\: ... \: m)$ and $b=(1\; \; m)(2 \; \; m-1)(3\; \: m-2)...$,($b$ is product of $[m/2]$ transpositsions), we can see that $D_{2m}\cong \: \langle a, b\rangle\: \le \: S_{m}$.