Let $G$ be a group such that if $H$ is a subset of $G\setminus Z(G)$ and any two element of $H$ commute, then $H$ is finite. Is it true that the set of all the sizes of such $H$ has a maximum element? Thanks in advance.
Let me try.
First, the center should be finite, otherwise the coset of any non-central element $xZ(G)$ is an infinite abelian subset in $G-Z(G)$.
As any subset with pairwise commuting elements generates an abelian subgroup, and conversely the elements of any abelian subgroup are pairwise commuting, we are actually asking about a group in which every abelian subgroup is finite (this is caused by the finiteness of the center), with no bound on the order of these subgroups.
Such a question is already asked in MathOverflow ( see https://mathoverflow.net/questions/80998/groups-with-no-bounds-on-the-size-of-abelian-subgroups-without-infinite-ones) , and such a group is already constructed in Olshanskii's book "The Geometry of defining relations in groups".
I reproduce the answer here : There exists a countable $2$-generated simple group $G$, that contains a copy of any cyclic group of odd order, moreover every proper subgroup of $G$ is either cyclic (of order dividing some integer $n$) or a conjugate to one of our embedded copies of the cyclic groups.