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.
 A: 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
A: *

*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. $S_3$.

*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.
