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Lagrange's theorem says that for a finite group $G$ and a subgroup $H$,In particular, $|H|$ divides $|G|$. To prove it, consider the left cosets $gH$. Every element lies in one, and if $gH$ and $g^{\prime }H$ intersect, then $gh=g^{\prime }{h}^{\prime }$ for some $h,h^{\prime }\in H$. Hence $g=g^{\prime }{h}^{\prime }{h}^{-1}$ and $gH=g^{\prime }H$. 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 aswhere $r$ is a rotation and $s$ a reflection. If $k\mid n$, the cyclic subgroup $⟨r^{n/k}⟩$ has order $k$. If $k\mid 2n$ 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 subgroupcontains exactly the $m$ rotations $r^{jn/m}$ and the $m$ reflections $r^{jn/m}s$, for $0\le j<m$. 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.