In Isaac's Finite Group Theory Page 28, it states:

There exist infinite groups in which the abelian subgroups have bounded order.

I fail to construct such group. In fact, I'm only able to deduce that the order of every element is bounded by an constant, which makes me feel hard to construct an example.

Hope for an answer!

  • $\begingroup$ Does the group with countable many generators each of which has square the identity work? I think all the abelian subgroups have two elements (or one). (Comment not an answer since I haven't thought it through.) $\endgroup$ – Ethan Bolker Mar 22 at 17:17
  • $\begingroup$ @EthanBolker I think you mean, $$G = \langle x_i \mid x_i^2 = 1, i \in \mathbb N \rangle.$$ I was typing this comment to argue that it works, but I think it doesn't. The cyclic subgroup generated by $x_1 x_2$ is infinite. $\endgroup$ – M. Vinay Mar 22 at 17:23
  • $\begingroup$ @EthanBolker I wonder how you define the product of each two generators. $\endgroup$ – Wembley Inter Mar 22 at 17:24
  • $\begingroup$ @M.Vinay You're right. What if you declare every square to be the identity (so infinitely many relations)? $\endgroup$ – Ethan Bolker Mar 22 at 17:38
  • $\begingroup$ @EthanBolker Such a group is Abelian [$(gh)^2 = 1 \implies gh = (gh)^{-1} = h^{-1}g^{-1} = hg$]. $\endgroup$ – M. Vinay Mar 22 at 17:39

To complement Arturo's answer:

A famous theorem of Ph. Hall is that every infinite locally finite group has an infinite abelian subgroup.

As a consequence, an infinite group whose abelian subgroups are finite of bounded order, has an infinite finitely generated subgroup, clearly sharing the same property. It's, in particular, of finite exponent. So, there's no easy approach to your question.

Adian in 1979 proved that for odd $n\ge 665$ and all $m\ge 2$, the Burnside group $B(m,n)$ (free group of exponent $n$ on $m$ generators) is infinite and all its abelian subgroups are cyclic (hence of order $\le n$).

Note also that finitely generated groups of finite exponent can have infinite abelian subgroups: for instance, if $G$ is infinite, finitely generated of exponent $n$ then $(\mathbf{Z}/p\mathbf{Z})\wr G$ is finitely generated of exponent $pn$ and has an infinite abelian subgroup of exponent $p$.


Tarski monsters provide examples. These are (infinite) groups $G$ in which every proper subgroup $H$ is either trivial, or cyclic of order a fixed prime $p$. In particular, the only abelian subgroups are of order $1$ or $p$.

Such groups exist for every sufficiently large prime, as shown by Olshanski'i. They are 2-generated, nonabelian, simple, and a rich source of all sorts of counterexamples.

  • $\begingroup$ Thanks! May I ask whether exists an example easier to construct, as Olshanski'i's proof of existence might be a little bit too sophisticated? $\endgroup$ – Wembley Inter Mar 23 at 2:59
  • 1
    $\begingroup$ @WembleyInter: I doubt there is anything elementary. After all, that is why Isaacs just mentions them and does not even hint at what they look like. $\endgroup$ – Arturo Magidin Mar 23 at 3:17

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