Prove that every subgroup of D_n of odd order is cyclic
Remember that $\displaystyle D_n = \{ a^ib^j| 0\leq i\leq n-1, 0\leq j\leq 1, ba=a^{n-1}b\}$
Notice that $\displaystyle (a^kb)^2 = a^k ba^k b = a^k a^{-k} bb = b^2 = 1$
This means that each $\displaystyle b,ab,...,a^{n-1}b$ have order $\displaystyle 2$.
Thus, this subgroup cannot contain these elements for then Lagrange's theorem would imply $\displaystyle 2$ two divides its order.
This means if $\displaystyle H$ is a subgroup of $\displaystyle D_n$ of odd order it contains the elements $\displaystyle e,a,...,a^{n-1}$. Thus, it is a subgroup of $\displaystyle \left< a\right>$. But every subgroup of a cyclic group is cyclic. Thus, $\displaystyle H$ is cyclic itself.