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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.

Vivid
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    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
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    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 Answers4

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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.)

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

Human Learning
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    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
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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.

Nicky Hekster
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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 $p$.

anomaly
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