Suppose is a finite simple group in which every proper subgroup is abelian. If and are distinct maximal subgroups of , prove that .
Let be a simple group in which every proper subgroup is abelian, and suppose and are distinct maximal subgroups of . Consider the normalizer in of the subgroup . Since and is abelian, . Similarly, . Thus . Note that is a subgroup of which properly contains, say, . Thus , and we have that is normal in . Since is simple and is a proper normal subgroup, .
Let and be distinct maximal subgroups. Since and and is simple, we have and . Moreover, because and are abelian, we have and . Finally, note that .
Now let and . We can write for some elements and . Since and both of and are abelian, , so that . In fact, is normal in . Since is a proper normal subgroup and is simple, .