# 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 $\displaystyle D_n$ of odd order is cyclic.

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

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

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

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

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

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

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

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

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