# What can we say of a group all of whose proper subgroups are abelian?

Let $G$ be a group (not necessarily finite). Can we say something about its structure if we suppose that all of its proper subgroups are abelian? Is there a difference between the finite case and the infinite case?

To put it in another way, is the class of such groups wild or do we control it? Naturally, abelian groups are part of it, but I am interested in the nonabelian case.

This question may sound quite open but I think it should be interesting to investigate it.

• The quaternion group is not a semi-direct product, but all its proper subgroups are abelian. – t.b. Jun 28 '11 at 12:03
• @Theo: You're right, I edited it. I wrote it to give examples of the kind of answers I am expecting, but I should have thought more about it before writing it ;-) – Thomas Connor Jun 28 '11 at 12:07
• The existence of (continuum-many) Tarski monsters suggests to me this even wider class will be pretty hard to deal with. – Chris Eagle Jun 28 '11 at 12:20
• A finite group with this property is necessarily solvable. In fact, if $G$ is a finite group with an abelian maximal subgroup, then $G$ is solvable. I think this is fairly easy to prove using character theory. Hence $G$ must have derived length $2$. – Amitesh Datta Jun 28 '11 at 12:26
• @Amitesh Datta, Steve: I think you guys misunderstood each other. Steve means a group can have an abelian maximal subgroup and still have derived length 3 (like SL(2,3) does). You just meant a minimal-non-abelian group has an abelian maximal subgroup so must be solvable, and hence its commutator subgroup is proper, so abelian, so the original is derived length 2. A similar idea happens in the infinite case, as long as the group is not perfect (which is no longer impossible). – Jack Schmidt Jun 29 '11 at 5:36