# Thread: [SOLVED] Every Subgroup of Dn of Odd order is cyclic

1. ## [SOLVED] Every Subgroup of Dn of Odd order is cyclic

Prove that any subgroup of $D_n$ of odd order is cyclic.

So far I have:
Let $H \leq D_n$
Let $r$ denote a simple rotation such that $r^n = e,$ and let $f$ denote a vertical flip
if $|H| = 1$, we're done since $H = {e}$, which is cyclic
otherwise let $|H| = 2k+1$ for some non-negative $k \in \mathbb{N}$

Consider $a \in H$.
If $a = r^m*f$ for some $m \in \mathbb{Z}$, then $ord(a) = 2$. Since the order of any element must divide the order of the group, we have that $a$ cannot be in $H$.

So the only elements of $H$ are of the form $r^m$.

But I can't seem to show that $H$ must be generated by an element of the form $r^m$. Any help would be greatly appreciated.

2. Originally Posted by Haven
Prove that any subgroup of $D_n$ of odd order is cyclic.

So far I have:
Let $H \leq D_n$
Let $r$ denote a simple rotation such that $r^n = e,$ and let $f$ denote a vertical flip
if $|H| = 1$, we're done since $H = {e}$, which is cyclic
otherwise let $|H| = 2k+1$ for some non-negative $k \in \mathbb{N}$

Consider $a \in H$.
If $a = r^m*f$ for some $m \in \mathbb{Z}$, then $ord(a) = 2$. Since the order of any element must divide the order of the group, we have that $a$ cannot be in $H$.

So the only elements of $H$ are of the form $r^m$.

But I can't seem to show that $H$ must be generated by an element of the form $r^m$. Any help would be greatly appreciated.
$H \subseteq \langle r \rangle$ and we know that every subgroup of a cyclic group is cyclic.