$\mid G:H\mid$=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. 

Try the contrapositive, showing that every element in $G \setminus H$ has even order. You've already discovered that $G/H$ is a cyclic group of order 2. So every element $x \in G \setminus H$ gets sent to the nontrivial element $\bar x \in G/H$ under the projection map $\pi: G \rightarrow G/H$. But if $x$ has order $m$, then
$$1_{G/H} = \pi(1_G) = \pi(x^m) =  \bar x^m$$
So $\bar x$ has order dividing $m$. But we know $\bar x$ has order 2 by the structure of $G/H$, so 2 divides $m$ and therefore $m$ is even.
