Prove that $H$ is a abelian subgroup of odd order

Question is:

Let $G$ be a group of order 2n. Suppose half of the element of G are of order 2 and the other half form a subgroup $H$ of order n . Prove that $H$ is of odd order and is an abelian subgroup of $G$

What could i see is..

if we prove that order of every element of $H$ is odd , then order of $H$ is odd . Also i am unable to use the fact that half of the element of $G$ are of order 2.

I assume no element of $H$ has order $2$. Then every nonidentity element of $H$ can be paired with its own inverse. Since $e\in H$ as well, you get that $H$ is a subgroup of odd order.
Moreover, $[G:H]=2$, so $H$ is normal in $G$. Pick $g\in G\setminus H$. This induces a conjugation automorphism of $H$, $$\varphi_g\colon H\to H:h\mapsto ghg^{-1}=ghg$$ since $g$ has order $2$. In particular, for any $h\in H$, $$h\varphi_g(h)=hghg=(hg)^2=e$$ since $hg\notin H$, lest $g\in H$. So you conclude $\varphi_g(h)=h^{-1}$. It follows that $H$ is abelian, for if $h,k\in H$, $$k^{-1}h^{-1}=(hk)^{-1}=\varphi_g(hk)=\varphi_g(h)\varphi_g(k)=h^{-1}k^{-1}.$$
• @ Ben West : How could you assume that no element of $H$ has order 2 – Struggler Mar 7 '14 at 17:05
• @user112064 Because otherwise the problem is not true. For example $G= (\mathbb Z/2 \mathbb Z)^n$ and $H= (\mathbb Z/2 \mathbb Z)^{n-1}$. So this should be part of the problem. – N. S. Mar 7 '14 at 17:07
• @ N.S : can we prove that no element of $H$ has order 2 by contradiction – Struggler Mar 8 '14 at 3:31