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 hence | ||
, 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.
Comments
in the paragraph starting with “Let M in G be a maximal subgroup”, why do we have |G|=|MN| since MN may not be a subgroup(if it is, then we are done but what if not?)
Good point.
I think that the proof still works using the weaker (and certainly true) inequality .
We’re assuming M and N to be distinct maximal subgroups of G finite simple where every subgroup is abelian. So by the previous problem, and G = MN.
A (just) slightly different proof starting with G having at least nonidentity elements, so that G has at least elements. We want to show that this is > |G|. Rearranging, we want to show The first and last equalities follow because M and N are distinct maximal subgroups of G with trivial intersection. Then the inequality follows since for
This proof is wrong. The first and last inequalities don’t follow because G needs not be MN.
I can’t understand why has at least elements.
I can’t either… I came up with another argument which (I think) gets around that problem.