Which of the following group has a proper subgroup that is not cyclic?

$1$. $\mathbb Z_{15} \times \mathbb Z_{17}$.

$2$. $S_3$

$3$. ($\mathbb Z$,+)

$4$. ($\mathbb Q$,+)

  • the proper subgroups of $\mathbb Z_{15} \times \mathbb Z_{17}$ have possible orders $3,5,15,17,51,85$ & all groups of orders $3,5,15,17,51,85$ are cyclic.So,all proper subgroups of $\mathbb Z_{15} \times \mathbb Z_{17}$ are cyclic.
  • Every proper subgroup of $S_3$ is cyclic.So,it is not the answer.
  • ($\mathbb Z$,+) is a cyclic group generated by $1$.And every proper subgroup of a cyclic group is cyclic.So,it is not the answer.
  • Any finitely generated subgroup of ($\mathbb Q$,+) is cyclic.So,it is not the answer.

The answer given in the answer key is $4$.

Please help me knowing which point i'm missing.

  • $\begingroup$ Why did you restrict yourself to finitely generated subgroups of $\mathbb{Q}$? $\endgroup$ – carmichael561 May 27 '17 at 4:17
  • $\begingroup$ @carmichael561:because i think any finitely generated subgroup will also be the subgroup.Is there difference between finitely generated subgroup and ordinary subgroups of a group? $\endgroup$ – P.Styles May 27 '17 at 4:20
  • $\begingroup$ Yes, because there could be subgroups which aren't finitely generated. $\endgroup$ – carmichael561 May 27 '17 at 4:21
  • $\begingroup$ @carmichael561:Will you please specify with the help of an example? $\endgroup$ – P.Styles May 27 '17 at 4:23
  • 1
    $\begingroup$ By the way, your argument for #1 isn't valid. The opposite of "has a proper subgroup that isn't cyclic" is "all proper subgroups are cyclic", You've only shown that it has at least one cyclic proper subgroup. $\endgroup$ – JonathanZ May 27 '17 at 4:27

Hint. As regards 4. what about $\left\{\frac{m}{2^n} : m, n \in \mathbb{Z}\right\}$.


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