Suppose that half of the elements of G have order 2 and the other half form a subgroup H of order n. Prove that H is an abelian subgroup of G.

Question

Let $n>1$ be a positive integer. Let $G$ be a group of order $2n$. Suppose that half of the elements of G have order 2 and the other half forma a subgroup $H$ of order $n$. Prove that H is an abelian subgroup of G.

I can only deduce from the index of $H$ that $H$ is normal. I am not sure how to move on.

Answer 1

Suppose $x\notin H$. Then the coset $xH$ consists of all of the elements of order $2$. Thus if $h\in H$, then $$xhxh=1$$ so $$xhx=h^{-1}$$ Suppose $a,b\in H$. Then $$xabx=b^{-1}a^{-1}$$ and this is also equal to $$(xax)(xbx)=a^{-1}b^{-1}$$ Thus $$b^{-1}a^{-1}=a^{-1}b^{-1}$$ for all $a,b\in H$, which is equivalent to saying that $H$ is abelian.