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.

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    $\begingroup$ The quaternion group is not a semi-direct product, but all its proper subgroups are abelian. $\endgroup$ – t.b. Jun 28 '11 at 12:03
  • $\begingroup$ @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 ;-) $\endgroup$ – Thomas Connor Jun 28 '11 at 12:07
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    $\begingroup$ The existence of (continuum-many) Tarski monsters suggests to me this even wider class will be pretty hard to deal with. $\endgroup$ – Chris Eagle Jun 28 '11 at 12:20
  • $\begingroup$ 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 $2$. $\endgroup$ – Amitesh Datta Jun 28 '11 at 12:26
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    $\begingroup$ @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). $\endgroup$ – Jack Schmidt Jun 29 '11 at 5:36

The finite case was settled by Miller–Moreno (1903) and described again in Redei (1950). The infinite case is substantially different, due to the existence of Tarski monsters. However, the case of non-perfect groups is reasonably similar to finite groups and is handled in Beljaev–Sesekin (1975), along with some more general conditions. Generally speaking, infinite groups are not very similar to finite groups in these sorts of questions, but if you restrict to (nearly) solvable groups, then things are a bit better. Nearly solvable in this case means "not perfect" and for quotient properties, often means "non-trvial Fitting/Hirsch-Plotkin radical". In other words, for subgroup properties you want an abelian quotient, and for quotient properties you want a (locally nilpotent or) abelian normal subgroup. More current research along the lines I enjoy generalize "abelian" rather than "finite"; a reasonable framework for this is given in Beidleman–Heineken (2009).

  • Miller, G. A.; Moreno, H. C. Non-abelian groups in which every subgroup is abelian. Trans. Amer. Math. Soc. 4 (1903), no. 4, 398–404. MR1500650 JFM34.0173.01 DOI:10.2307/1986409

  • Rédei, L. Die Anwendung des schiefen Produktes in der Gruppentheorie. J. Reine Angew. Math. 188, (1950). 201–227. MR48432 DOI:10.1515/crll.1950.188.201

  • Beljaev, V. V.; Sesekin, N. F. Infinite groups of Miller-Moreno type. Acta Math. Acad. Sci. Hungar. 26 (1975), no. 3-4, 369–376. MR404457 DOI:10.1007/BF01902346

  • Beidleman, J. C.; Heineken, H. Minimal non-F-groups. Ric. Mat. 58 (2009), no. 1, 33–41. MR2507791 DOI:10.1007/s11587-009-0044-2

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    $\begingroup$ Thank you for this nice answer! $\endgroup$ – Thomas Connor Jun 28 '11 at 16:49
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    $\begingroup$ +1 An excellent answer (as usual). It really is great to have knowledgeable people like you on math.stackexchange.com. $\endgroup$ – Amitesh Datta Jun 29 '11 at 6:19
  1. If $G$ is a group with all its proper subgroups abelian, then $G$ itself may not be abelian. A perfect counter example is group $D_6$, i.e. $S_3$.

  2. If $G$ is a group with all its subgroups abelian, for sure including itself and trivial one, then yes, $G$ itself is also abelian. You can easily check commutative of arbitrary 2 elements by generating them into a subgroup of $G$.

Hope that helps.


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