Show that $H=\{x^2\mid x\in D_n\}$ is a subgroup of $D_n$.

Question

This question was asked in two parts:

Part I: If $G$ is abelian, must $H=\left\{x^n\mid x\in G\right\}$ be a subgroup of $G$?

I said yes:
Let $x^n, y^n\in H$.
Then $(xy)^n=(xy)(xy)\cdots(xy)$ and
$(xy)^n=x^ny^n$ shows $(xy)^n\in H$.
$(x^{-1})^n=x^{-n}=(x^n)^{-1}$ implies $(x^{-1})^n\in H$.
Hence, $H$ is a subgroup of $G$.

Part II: Show that $H=\left\{x^2\mid x\in D_n\right\}$ is a subgroup of $D_n$.

Thinking about it, all elements with a reflection return to the identity element $e$. If $n$ is odd, then $H$ is a cyclic subgroup generated by $R_\frac{360}{n}$. If $n$ is even, then $H$ is also a cyclic subgroup. I know what this subgroup should look like, i.e.: in $D_6$, we have the subgroup $H=\{e, R_{120}, R_{240}\}$. How do I show that this applies for any even $n$?

Answer 1

Note that $D_n$ is a semidirect product $C_n \rtimes C_2$, taking the inverse function as an automorphism. So we have $(a, b)(a,b) = (a^2, b^ab)= (a^2,1)$, so $H$ is a subgroup of $C_n$. Now taking the square is an autormorphism iff $2$ is not a divisor of $n$, in the other case we have $H = C_{\frac{n}{2}}$.

Answer 2

From the proof for the abelian case one can deduce it for the dihedral groups too.

First look at all the squares of elements in the cyclic subgroup of index 2 in $D_n$. By the proof for the abelian case this is a subgroup (irrespective of $n$ being odd or even). Now the squares of other elements of $D_n$ give the identity element and so it is the same subgroup.