$\mid G:H\mid$=2 Let $G$ a finite group and $H \leq G$ such that $\mid G:H \mid =2$, then $H$ contains all of elements of $G$ of odd order. 
there is a hint: $x^{2k-1}=1$ implies $x=(x^{2})^{k}$.
The first thing I would think about it is, since $\mid G:H \mid =2$, that implies that $H$ is normal in $G$. Hence, $G/H$ is a quotient group. 
would someone please help me out with that. You do not have to use the hint if you have a different way to do that. 

First of all lets show that $H$ is a normal subgroup. Let $h\in G$. If $h\in H$ then $hHh^{-1}=H$. If $h\not\in H$ then, $hHh^{-1} \cap hH = \phi$. Thus, $hHh^{-1}\subset H$. Now look at  the group $G/H$.
Let $x$ be an element of $G$ which is not in $H$. Elements of $G/H$ are $H$ and $Hx$. And its a group. So $Hx .Hx = H $ and hence $Hx^2=H$.So, $x^2 \in$ $H$ $\forall$  x   $\in G$.
Now $x^{2k-1} = e \implies x=x^{2k}=(x^{k})^2 \in H. $
