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.

  • $\begingroup$ A little observation: any element $x \in G \setminus Z(G)$ must have finite order. Suppose not. Note $H = \langle x \rangle \setminus Z(G)$ has pairwise commuting elements and is therefore finite. In particular, we have $x^n , x^{n+1} \in Z(G)$ for large enough $n$. But then, $x \in Z(G)$, contrary to assumption. $\endgroup$ – Mike F Sep 24 '13 at 7:46
  • $\begingroup$ Maximal elements exist among $H$ by Zorn's lemma, but of course this is not enough to conclude that the set of sizes of such $H$ contains a maximum element. Also, any maximal element is $A - Z(G)$ where $A$ is an abelian subgroup of $G$ and $Z(G) \leq A$. So it suffices to consider abelian subgroups of $G$ that contain $Z(G)$. In a counterexample, $A - Z(G)$ would be finite for any abelian subgroup $A$ containing $Z(G)$ and $A - Z(G)$ could be arbitrarily large. $\endgroup$ – spin Sep 24 '13 at 8:06
  • $\begingroup$ Your question is a dual of a famous question of Erdos : If $G$ is a group such that any subset $H$ consisting of mutually non commuting elements is finite, Then are they boundedly finite?, the answer was given in the afirmative by B. H. Neumann (1976) 'A problem of Paul Erdos on groups' J .Austral. Math. Soc. $\endgroup$ – Yassine Guerboussa Sep 24 '13 at 9:35
  • $\begingroup$ Note also (in the sprit of Neumann), that you can reforlmulate your question as Graph theoretic one. Consider the elements of $G\Z(G)$ as vertices, and connect to vertices $x$ and $y$ by an edge if they commute; your condition is equivalent to that every complete subgraph is finite. I will not be surprised if an answer follows from a pure graph theoretic argument. $\endgroup$ – Yassine Guerboussa Sep 24 '13 at 9:55
  • $\begingroup$ I am a little confused by this question - do you mean "H is a finite set of pairwise commuting, non-central elements". Can you not just take $C_2\ast C_3\ast C_4\ast\ldots$? What am I missing? $\endgroup$ – user1729 Sep 24 '13 at 10:05

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.


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