Is $G=HK$ if $G/H$ and $G/K$ are simple groups, where $H$ and $K$ are distinct normal subgroups of $G$?
Actually, I want to show that under the above assumptions $G/K\cong H/(H\cap K).$
I know from one of the isomorphism theorems that if $G=HK$ then $G/K\cong H/(H\cap K).$ So if I can prove that $G=HK$ when $G/H$ and $G/K$ are simple groups, I would be done. But I am not sure how to show that $G\subset HK.$
If $G/H$ and $G/K$ are simple then we know that under the projection $\pi_H:G\to G/H$, $\pi_H(K)=G/H$ since $\pi_H(K)$ is a normal subgroup and it cannot be equal to the trivial normal subgroup as $H$ and $K$ are distinct. Similarily we have that $\pi_K(H)=G/K.$ But I am not sure if I can use this to prove that $G=HK$ (if at all that is the case).
If $H$ and $K$ are normal in $G$, then $HK$ is normal in $G$. Therefore, $HK/H$ is normal in $G/H$, and hence $HK/H=H/H$ or $HK/H=G/H$. But if $H\neq K$, then $HK\neq H$, so $HK/H = G/H$, and hence $HK=G$, as desired. Note that you do not need $G/K$ simple.
As to your desired isomorphism, this now follows: $G/K = HK/K\cong H/H\cap K$ by the isomorphism theorems.