Infinite non cyclic group with proper subgroups being cyclic

Question

I was thinking about the question regarding infinite groups with all of its proper subgroups being cyclic but the group itself is not cyclic. The finite case is a well known fact $(K_4)$. I have found this beautiful counterexample for infinite groups here about $$\left\langle \frac{a}{2^k} \mid a\in \mathbb{Z}, k \in \mathbb{N} \right\rangle.$$ Any prime $p$ would have done the job besides $2$. Now my question is the following-

Are there any other groups satisfying this property?

See here Is there some sort of classification of all minimal non-cyclic groups? and the answer of the same author. — kabenyuk, Nov 01 '23 at 14:35
Some authors have mentioned "Tarski monsters". But there are also torsion-free 2-generated groups with the same property (due to Olshanskii around 1980), and the construction of such groups is a much easier application of Gromov-hyperbolicity that the finite exponent version. — YCor, Dec 13 '23 at 11:18
By the way the group you mention, usually known as $\mathbf{Z}[1/2]$, does not satisfy the claimed property (unlike the claim in the linked post). Indeed $3\mathbf{Z}[1/2]$ is a proper, non-cyclic subgroup (of index 3). — YCor, Jan 10 '24 at 00:04

score 4 · Answer 1 · answered Nov 01 '23 at 13:38

4

Yes. Look up Tarski monster group. Given a prime $p$, this is an infinite simple group all of whose proper subgroups are finite cyclic of order $p$. They exist for all large enough $p$.

answered Nov 01 '23 at 13:38

Dave Benson

536

score 2 · Answer 2 · edited Nov 01 '23 at 14:48

2

For instance the Tarski monster group has the property that it is finitely generated, infinite and every proper subgroup is a cyclic group of a fixed prime order $p$.

edited Nov 01 '23 at 14:48

Shaun

47,747

answered Nov 01 '23 at 13:38

Moishe Kohan

111,854

I don't think there is a particular group that is called "the" Tarski monster. There is a construction, making use of plenty of choices, providing many groups. Choices can be made so as to produce a recursive presentation, but this is not canonical at all. – YCor Dec 13 '23 at 11:21
@YCor: True, is is a class of groups, I should have called it a "group of the Tarski monster class." But OP does not seem to care one way or another... – Moishe Kohan Dec 13 '23 at 19:46
No, just "a Tarski monster group" is fine, rather than "the Tarski monster group". – YCor Dec 13 '23 at 23:10

Infinite non cyclic group with proper subgroups being cyclic

2 Answers2