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We have simple examples of finite groups (that may be abelian or not) that are not cyclic but all their proper subgroups are cyclic (e.g. Klein's $4$-group and $S_3$ respectively for abelian and non-abelian). In recent times, I have been able to produce a few examples of infinite abelian groups that are not cyclic but all their proper subgroups are cyclic.

But currently, I am pondering whether the same can also be said for some infinite non-abelian group or not, precisely, does there exist an infinite non-abelian group such that all of its proper subgroups become cyclic?

And if there do exist such groups, what can be an example?

And if possible, it will be very much helpful if someone can give a general algorithm for constructing such a group.

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  • $\begingroup$ If $a,b$ are non-commuting members of such a group, the subgroup $<a,b>$ generated by the set $\{a,b\}$ is non-Abelian, hence non-cyclic, hence must be the whole group. $\endgroup$ – DanielWainfleet Jun 17 '17 at 14:31
  • $\begingroup$ (continued from my prior comment) : This also applies to (e.g) $<a,ab>$ and to $<a,ba>$ and to $<a^2,b>$.... I dk the answer to whether such groups exist. $\endgroup$ – DanielWainfleet Jun 17 '17 at 14:39
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    $\begingroup$ @DanielWainfleet please give a precise and rigorous argument about what you mean to say. $\endgroup$ – reflexive Jun 21 '17 at 4:03
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    $\begingroup$ Tarski monsters: en.wikipedia.org/wiki/Tarski_monster_group $\endgroup$ – Matthew Towers Jul 30 '17 at 23:59
  • $\begingroup$ Can you give an example of an infinite abelian group where all proper subgroups are cyclic? $\endgroup$ – Myridium Jan 5 '18 at 5:05
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Response to the original question:

Such non-abelian groups, that all their proper subgroups are cyclic (and moreover, such that all their proper subgroups are isomorphic to $C_p$ for a fixed prime $p$) actually do exist. Such groups are called Tarski Monster Groups and there are continuum many of them for each prime $p > 10^{75}$ (this fact was proved by Alexander Ol'shansky in 1979).

Response to the comment by @Myridium:

The infinite abelian groups, that all their proper subgroups are cyclic can be described in the following way:

If all proper subgroups of an infinite abelian group $G$ are cyclic, then $G$ is isomorphic either to $C_\infty$ (infinite cyclic), or to $C_{p^\infty}$ (quasicyclic for some prime $p$), or to $\mathbb{Q}_p$ (the subgroup of $\mathbb{Q}_+$, which consists of fractions with powers of a prime $p$ in denominators).

To prove this statement we will need two lemmas:

Lemma 1:

If all proper subgroups of an infinite abelian group $G$ are cyclic, then $G$ is locally cyclic

If $G$ is finitely genersted, then by classification of finitely generated abelian groups it is isomorphic to $C_\infty$.

If $G$ is infinitely generated then all its finitely generated subgroups are proper, and thus cyclic.

Lemma 2:

A group is locally cyclic iff it is a subquotient of $\mathbb{Q}_+$

The complete proof of this fact can be found there

Proof of the main statement:

If $G$ is finitely generated, then it is $C_\infty$

If $G$ is infinitely generated periodic, then it has a quasicyclic subgroup (by classification of locally cyclic groups). And as quasicyclic groups are not cyclic, this subgroup is the whole group.

If $G$ is infinitely generated aperiodic, then its quotient by an infinite cyclic subgroup is infinitely generated periodic and also satisfies the required property. So $G$ is a locally cyclic extension of $C_\infty$ by $C_{p^\infty}$ for some prime $p$, which can be only $\mathbb{Q}_p$.

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