# If all Subgroups are Cyclic, is group Cylic? [duplicate]

I am having difficulty seeing if it is the case that if all subgroups of a group are cyclic, that the group itself is cyclic.

## marked as duplicate by Dietrich Burde, hardmath, user147263, Jonas Meyer, Jyrki LahtonenNov 1 '14 at 19:45

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• Do you mean to say "proper" subgroups? Because otherwise, since a group is a subgroup of itself, and every subgroup is cyclic, the group is cyclic. – Hayden Oct 29 '14 at 14:24
• There are many counterexamples, even with very small cardinalities (say... 4)... – Najib Idrissi Oct 29 '14 at 14:28
• Yes I meant proper subgroups. I had a feeling the reverse wasn't the case but I just started group theory so it was difficult to see a non cyclic group that has only cyclic proper subgroups. – Vivid Oct 29 '14 at 14:34

## 4 Answers

If you mean proper subgroups, then No! consider $G = \mathbf{Z}_2 \times \mathbf{Z}_2$.

• I guess any product of two finite cyclic groups does the trick ; the divisors of $pq$ are $1,p,q$ and $pq$ (even if $p=q$). By cardinality arguments and Cauchy's Theorem one concludes every proper subgroup is cyclic. – Patrick Da Silva Oct 29 '14 at 15:38

The group itself is a subgroup, so that it is cyclic as well, because all subgroups are cyclic. Did you maybe mean that all strict subgroups are cyclic ? In this case, it is wrong to conclude that the group itself is cyclic : for instance $$G = \left\{ \frac{x}{2^d}\;|\;x\in\mathbf{Z}, d\in\mathbf{n}\right\}$$ is not cyclic, and has only cyclic strict subgroups. (I am giving this example, as all given examples were finite groups.)

No this does not have to be the case - take $V_4$ or $S_3$, the Klein 4-group or the symmetric group on 3 symbols respectively. So the whole group does not have to be even abelian.

In addition to the finite groups mentioned above, there are also infinite examples. The construction is nontrivial, but a Tarski monster is an infinite group $G$ with every proper subgroup $H\subset G$ cyclic of order $p$ for some fixed prime $p$ (independent of $H$). They exist for sufficiently large $p$.