Prove that $K=\{e\}$

Question

If a simple group $G$ has a subgroup $K$ that is a normal subgroup of two distinct maximal subgroups ,prove that $K=\{e\}$.

Attempt:

Let $H_1,H_2$ be two maximal subgroups of $G$ in which $K$ is normal.

Now considering $H_1H_2$ in which $K$ is normal but $H_1H_2$ may not be a subgroup of $G$ as $G$ is simple.

Then how should I proceed here.

Any hints

Answer 1

Hint: consider the subgroup generated by $H_1$ and $H_2$: $\langle H_1,H_2 \rangle$. Since $H_1$ and $H_2$ are distinct we must have $H_1 \subsetneq \langle H_1,H_2 \rangle$. Observe that $K \unlhd \langle H_1,H_2 \rangle$. Can you take it from here?

Answer 2

$N_G(K)$ is the largest normal subgroup containing $K$. Therefore, $H_1, H_2 \leq N_G(K)$. But, $H_1, H_2$ are maximal subgroups. This implies that $N_G(K) = G$. Thus, $K$ is normal in $G$. So $K$ is either $G$ or $\{e\}$ (because $G$ is simple). Has to be the latter due to the distinctness.