How is this subgroup abelian?

Question

Let $G$ be a finite group of order $2n$ such that half of the elements of $G$ are of order $2$ and the other half form a subgroup $H$ of order $n$.

Then I know that $H$ is of odd order because for each $x \ne e$ in H, we have $x \ne x^{-1}$; thus after pairing all such elements we are left with the identity.

Also, the subgroup $H$, being of index two, is a normal subgroup of $G$.

How to determine if $H$ is abelian or not?

Answer 1

Try to show that $x \mapsto x^{-1}$ is an automorphism of $H$.

Answer 2

Hint: If $a^2=1$ then $aha^{-1}\cdot h=(ah)^2$.

Answer 3

Try to prove that G is isomorphic to $D_n$ (try to prove if you have an element of order 2,a, and one of order n,b, that generate all the group and $(ab)^2=1$ ) If you prove that, then H is generated by only one element of order n and it's abelian and cyclic