Show $H$ is the only subgroup in $G$ of index 2, when $|G| \not = 4$ and $[G:H]=2$ 
Let $G$ be a group where $|G| \not = 4$. Now let $H \leq G$ where $[G:H]=2$. Suppose $H$ is simple. Show that $H$ is the only subgroup in $G$ of index 2 en then deduce $H \underset{\text{char}}{\leq} G$. What happens when $|G|=4$?

Can anyone give some pointers on how to prove this? 
I don't see how I can use $|G|\not = 4$. I guess the theorem is not true when $|G|=4$, Klein's group has 3 simple subgroups of index 2. So the the fact of $|G|\not = 4$ should play a decisive role in the proof.
The usual way of proving would be, let $M$ be another subgroup of index 2, and then showing how $M$ must equal $H$...
On the other hand, I also know how $H \trianglelefteq G$.
 A: Let $G$ be a group such that $\mid G\mid \ne 4.$ $H$ be a subgroup of $G$ such that $H$ is simple and $\mid G:H\mid=2.$
Let $K$ be a subgroup of $G$ such that $\mid G:K\mid=2.$ Then both of $H$ and $K$ are normal in $G.\Rightarrow H\cap K\trianglelefteq H\Rightarrow H\cap K=H$ or $\{e_G\}$, as $H$ is simple.
If possible let $H\cap K=\{e_G\}.$ Now $x^2\in H, \forall x\in G,$as $\mid G:H\mid=2.$ Similarly $x^2\in K, \forall x\in G$ $\Rightarrow x^2=e_G,\forall x\in G$ i.e. $G$ is  commutative. $\Rightarrow H$ is commutative. 
Then $H\cong \mathbb Z_p$ for some prime.$\Rightarrow \mid G\mid =2p.$
As $G$ is commutative, $G\cong Z_{2p}.$ But $Z_{2p}$ has only one subgroup (the subgroup $\langle 2\rangle$) of index $2.$(this is possible as $p\ne 2$). This contradicts the fact that $H\cap K=\{e_G\}.$
So we get, $H\cap K=H\Rightarrow H\subseteq K.$ Now $\mid G:H\mid =\mid G:K\mid \mid K:H\mid \Rightarrow \mid K:H\mid =1\Rightarrow K=H.\Rightarrow H$ is unique subgroup of index $2.$
Let $\phi \in Aut(G).$
Since, $\mid G:H \mid=2$, $\mid G:\phi(H) \mid=2.$
From the previous result, $\phi(H)=H$ i.e. $H \underset{\text{char}}{\leq} G.$
However, if $\mid H\mid =4$ then , the results are not true when $G\cong \mathbb Z _2×\mathbb Z_2$ , we get three subgroups of $G$ having index $2.$ 
If $G\cong \mathbb Z _4$ then we get a unique $H$, as $Z_4$ has only one subgroup of index $2.$ Also, $H \underset{\text{char}}{\leq} G.$
