Finite subgroups of the multiplicative group of a skew field It is well known that any finite subgroup of the multiplicative group of a field is cyclic, but this certainly fails for skew fields (look in the quaternions for an easy example). Yet there are constraints on what finite subgroups of multiplicative groups of a skew fields can be: for example, every all their abelian subgroups must be cyclic (this follows readily from the above result).
Is there a nice characterization of such subgroups, or at least some interesting constraints on them? I'd be surprised if any finite group with all its abelian subgroups cyclic occurred as such a group, though I don't have a counterexample. I'm also curious about the simplest example of such a group which doesn't come from the quaternions, and results that extend for infinite subgroups.
 A: Let $G$ be a group such that all abelian subgroups are cyclic. Then in particular, $G$ has no subgroup of the form $C_p\times C_p$ for any prime $p$. This means that the Sylow $p$-subgroups of $G$ are cyclic for $p$ odd, and cyclic or generalized quaternion for $p=2$. That's all you can say about the structure, since if $H$ is an abelian subgroup it factorizes as the direct product of its Sylow $p$-subgroups, and is cyclic if and only if all Sylow $p$-subgroups are cyclic.
So what is the structure of such groups? If the Sylow $2$-subgroup is also cyclic then $G$ is soluble, and in fact has a normal subgroup of odd order and index a power of $2$. Even more, it is actually supersoluble, which you can see once you know it's soluble. If $G$ has generalized quaternion Sylow $2$-subgroups then more examples can occur, such as $\mathrm{SL}_2(p)$.
A paper by Suzuki, On finite groups with cyclic Sylow subgroups for all odd primes, Amer. J. Math. 77 (1955) 657–691, examines this case. He finds that if $G$ is not soluble then it possesses a subgroup $G_1$ of index at most $2$, and $G_1$ is a direct product of a group of odd order and a subgroup $\mathrm{SL}_2(p)$.
