Subgroup of $D_n$ with order $m$ as $m \mid 2n$

Question

I have to prove that as $m \mid 2n$ then there is a subgroup of $D_n$ with order $m$. How can i do that ?

Answer 1

Hint: $D_n$ is generated by a reflection $s$ of order 2 and a rotation $r$ of order $n$. See what kind of subgroup orders you can get from forming subgroups generated by powers of $r$, either including or excluding $s$.

Answer 2

Thanks. As $m \mid n$ then $\langle r^k \rangle$ is such a group as $mk = n$. Further we have the case $m = 2m'$ and $m' \mid n$. If $m'k = n$ then $\langle r^k, s \rangle$ is a group with order $m$ because it consists of the elements $r^{lk}$ with $l \in \{0,\cdots,m'-1\}$ and $sr^{lk}$ for those $l$. Thus there are $2 m' = m$ elements in this subgroup.