Prove that if is any nonabelian group in which every proper subgroup is abelian then is not simple.
We begin with a lemma.
Lemma: If is maximal, then for all , is maximal. Proof: Suppose to the contrary that is maximal and that is not; let . Then , with all containments proper, a contradiction.
Let be a finite nonabelian group in which every proper subgroup is abelian. Suppose that is simple.
Let be a maximal subgroup. By this previous exercise, ; let such that is not in a conjugate of . Now for some maximal subgroup . Note that is not conjugate to , so that the conjugates of and are pairwise distinct. Moreover, by this previous exercise, the conjugates of and intersect trivially pairwise. Thus the nonidentity elements in conjugates of and form disjoint subsets of . By this previous exercise, the number of nonidentity elements in conjugates of is and in conjugates of is , so that has at least nonidentity elements. Now note the following.
|(strictness since also contains 1)|
|=||(since is finite)|
|=||, so that|
|, and thus|
Then (without loss of generality) we have , since for natural numbers and with , .
Suppose . If is composite, then via Cauchy’s theorem we have a contradiction (as there exist proper nontrivial subgroups in ). Then is prime, but then is abelian, a contradiction. In particular, we may assume that .
Suppose . Then since is maximal, is in fact a Sylow 2-subgroup of . Hence where is odd, and by this previous exercise, is not simple, a contradiction.
Thus no such group exists.