10
$\begingroup$

Is there a group $G$ in which every abelian subgroup is finite and there is no upper bound on the sizes of its abelian subgroups?

Let me say that a counterpart of the question above has posed by Paul Erd\"os in 1975 as follows: Is there a group $G$ in which every subset consisting of pairwise non-commuting elements is finite and there is no upper bound on the sizes of such subsets?

The asnwer was given by B. H. Neumann in 1976 in negative. It was proved that groups $G$ satisfying the first assumption of the latter question must be center-by-finite and so the index of the center is a upper bound for sizes of subsets consisting of pairwise non-commuting elements.

Improve this question
$\endgroup$
1
  • 3
    $\begingroup$ This contains lots of $\mathbb{Z}$'s. $\endgroup$
    – seldom seen
    Commented Nov 15, 2011 at 19:15

2 Answers 2

Reset to default
10
$\begingroup$

Yes, such groups exist.

Consider the disjoint union of cyclic groups of odd order $\mathcal{C}:=\{ \mathbb{Z}/(2n+1)\mathbb{Z} \mid n \in \mathbb{N} \}$. By a theorem of A. Ol'shanskii (see Thm 35.1 from his book "Geometry of defining relations in groups"), there exists an odd integer $n_0$ for which this collection of groups can be embedded into a countable $2$-generated simple group $G$ such that every proper subgroup of $G$ either is cyclic of order dividing $n_0$ or is conjugate to a subgroup of a group from $\mathcal C$.

Clearly $G$ is non-abelian, but every proper subgroup of $G$ is cyclic of finite odd order. Since each group from $\mathcal C$ is a subgroup of $G$, one sees that there is no bound on the orders of the cyclic subgroups.

Improve this answer
$\endgroup$
7
$\begingroup$

One can repeat the construction of the free Burnside group from Olshanskii "Geometry of defining relations in groups" imposing relations $w^{n_w}=1$ where $n_w\gg 1$ odd, depending on $w$ instead of $w^n=1$. All the auxiliary properties of the free Burnside groups except finite exponent will hold. In particular all Abelian subgroups of it will be cyclic, but the orders of them will be unbounded. The proof carries verbatim. I am pretty sure somebody (a student of Olshanskii) has done it explicitly already: one can construct torsion groups satisfying many "strange" properties. A more general way is to use lacunary hyperbolic groups (see our paper with Olshanskii and Osin in arXiv). There are many torsion lacunary hyperbolic groups, all Abelian subgroups there are cyclic but their orders unbounded.

Improve this answer
$\endgroup$
8
  • 3
    $\begingroup$ Tarski Monsters originally constructed by Olshanskii satisfy the requirements. All abelian subgroups there are cyclic, but the exponent is unbounded. The construction can be found in the book, of course. $\endgroup$
    – Denis Osin
    Commented Nov 15, 2011 at 20:26
  • $\begingroup$ @Denis: Yes, that is the oldest example. Also one does not need to refer to a proof, only to the result. $\endgroup$
    – user6976
    Commented Nov 15, 2011 at 21:51
  • $\begingroup$ I think that there might also be infinitely generated examples, which should be easier to construct. Something like an iterated semidirect product $((A1\rtimes A2)\rtimes A3)\dots$ of finite groups $A_i$, such that $A_n$ acts on the semidirect product $B_{n−1}$, before it, so that the induced action on the conjugacy classes of $B_{n−1}$ is free. However, I do not know how to construct such a sequence... $\endgroup$
    – Ashot Minasyan
    Commented Nov 16, 2011 at 11:07
  • $\begingroup$ @Ashot: Locally finite groups tend to have infinite Abelian subgroups. All answers so far are lacunary hyperbolic groups. This is almost certainly not the only way to construct an example. One can probably take the "lacunary large" construction of Olshanskii and Osin (impose relations $w^{n_w}$ keeping the group large), Golod-Shafarevich groups may not contain infinite Abelian subgroups. Also I do not know if Grigorchuk group contains an infinite Abelian subgroup (that must be known because the centralizers of elements in Grigorchuk groups are known). $\endgroup$
    – user6976
    Commented Nov 16, 2011 at 15:08
  • $\begingroup$ See: jlms.oxfordjournals.org/content/s1-39/1/235.extract $\endgroup$
    – user6976
    Commented Nov 16, 2011 at 15:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .