= Dihedral subgroups of every divisor order
{title2=$k\mid2n\Longrightarrow D_{2n}\text{ has a subgroup of order }k$}
Write the <dihedral group> as $D_{2n}=\langle r,s:r^n=s^2=e,\ srs=r^{-1}\rangle$. If $k\mid n$, the <cyclic subgroup> $\langle r^{n/k}\rangle$ has order $k$. For an even divisor $k=2m$ of $2n$, $m\mid n$ and $\langle r^{n/m},s\rangle$ has $m$ rotations and $m$ reflections, hence order $k$. An odd divisor of $2n$ always divides $n$. This proves that the necessary divisibility condition from <Lagrange's theorem> is also sufficient in a <dihedral group>, although it is not sufficient in arbitrary <finite groups>.
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