Suppose $G$ is a finite group, $H\leq G$, and $H$ is a simple group such that $[G:H]=2.$
Prove:
$H$ is the only normal subgroup of $G$ or there is a normal subgroup $K\leq G$, $|K|=2$ such that $G=H \times K.$
My idea:
$[G:H]=2 \implies |H|=\frac{|G|}{2}$
Suppose there is a normal subgroup $K\leq G$.
We know that $H \cap K=\{1\}$ or $H \cap K=H$ since $H$ is a simple subgroup.
If $K=\{e\} \implies H$ is the only normal subgroup (except $\left\{ 1\right\}$ and $G$ itself).
If $K=H \implies H$ is the only normal subgroup (except $\left\{ 1\right\}$ and $G$ itself).
If $K \neq \{1\}$ and $H \cap K=\{1\}$ then $|HK|=|G|$ ,HK is a subgroup since $H,K$ are normal subgroup. Also $|H|=\frac{|G|}{2},|K|=|2|,H \cap K= \{ 1\}, G=HK \implies G =HK \simeq H \times K.$
If $H \cap K=H$ $ |H|=\frac{|G|}{2}\leq |K| \leq |G| \implies |K|=|G| \implies H$ is the normal subgroup (except $\left\{ 1\right\}$ and $G$ itself).
I'd be grateful for your some help!