Exercise 4.5.52 (If $G$ is a finite simple group in which every proper subgroup is abelian, and if $M$ and $N$ are distinct maximal subgroups of $G$ then $M \cap N = 1$)

Proof. Put $H=M\cap N$, and consider its normalizer $N_G(H)$.

The maximal subgroups $M$ and $N$ are proper subgroups by definition, so $M$ and $N$ are abelian by hypothesis. Therefore $M$ and $N$ normalize $H=M\cap N$:

Since $M$ and $N$ are maximal subgroups

If $N_G(H)\ne G$, then $M=N_G(H)=N$: this is impossible, because $M,N$ are distinct subgroups. Hence

This proves $H⊴G.$ Moreover, $G$ is a simple group, therefore $H=\{1\}$ or $H=G$. Since $H\le M$, where $M$ is a proper subgroup, then $H\ne G$. Therefore $H=\{1\}$:

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