TITLE: Multiplicative groups of skew fields
QUESTION [5 upvotes]: Is every group isomorphic to a subgroup of the multiplicative group of some skew field?

REPLY [15 votes]: No. Indeed, every finite abelian subgroup is cyclic. Indeed, passing to the subfield it generates, we boil down to the standard easy fact that in a field every finite abelian subgroup of the multiplicative group is cyclic (a non-abelian finite group contains $C_p^2$ for some $p$, where $C_p$ means a cyclic group of order $p$, and we would get $p^2$ $p$-roots of $1$ in a field).