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**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.