= Solution
<Lagrange's theorem> says that for a <finite group> $G$ and a <subgroup> $H$,
$$
|G|=[G:H]|H|.
$$
In particular, $|H|$ divides $|G|$. To prove it, consider the left <cosets> $gH$. Every element lies in one, and if $gH$ and $g'H$ intersect, then $gh=g'h'$ for some $h,h'\in H$. Hence $g=g'h'h^{-1}$ and $gH=g'H$. Distinct <cosets> are therefore disjoint and partition $G$. Multiplication by $g$ is a <bijection> $H\to gH$, so every <coset> has $|H|$ elements. Counting this partition proves the formula.
Write the <dihedral group> as
$$
D_{2n}=\langle r,s:r^n=s^2=e,\ srs=r^{-1}\rangle,
$$
where $r$ is a rotation and $s$ a reflection. If $k\mid n$, the <cyclic subgroup> $\langle r^{n/k}\rangle$ has order $k$. If $k\mid2n$ but $k\nmid n$, then $k$ is even: an odd divisor of $2n$ would divide $n$. Put $m=k/2$, so $m\mid n$. The <subgroup>
$$
\boxed{\langle r^{n/m},s\rangle}
$$
contains exactly the $m$ rotations $r^{jn/m}$ and the $m$ reflections $r^{jn/m}s$, for $0\leq j<m$. The inversion relation makes this set closed under multiplication and inverses. These $2m=k$ elements are distinct, so it has the required order. This proves the <dihedral subgroups of every divisor order> property. The construction works for every even divisor, including those already covered by the cyclic construction.
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