For every integer n, show Dn has a subgroup of order 4.
We know Dn is a cyclic group of order n. I claim H=. We can rotate by 1/n th the way around. |H|=|R|=n. H is a cyclic subgroup of order n. And this is a contradiction. So, |Dn|=2n
How does this prove Dn is a subgroup of order 4?
There's a few bugs with the claim to begin with. (a) $D_n$ only makes sense for positive integers $n$. (b) $D_n$, for odd $n$, cannot have a subgroup of order $4$ since it would violate Lagrange's Theorem.
So the claim could only possibly be true for even $n \geq 2$.
If $n=2m$ for odd $m$, then "rotate by $\tfrac{n}{4}$" (not $\tfrac{1}{n}$-th) is not actually possible (since $m$ is not divisible by $2$). It will, however, have a subgroup isomorphic to $D_2$ which has order $4$ (combine "rotate by $\tfrac{n}{2}$" with "flip").
If $n=4m$ for $m \geq 1$, then the element whose action defines "rotate by $\tfrac{n}{4}$" exists, and consequently it will indeed generate a cyclic subgroup of order $4$.