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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.

Styles
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  • Why did you restrict yourself to finitely generated subgroups of $\mathbb{Q}$? – carmichael561 May 27 '17 at 04:17
  • @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? – Styles May 27 '17 at 04:20
  • Yes, because there could be subgroups which aren't finitely generated. – carmichael561 May 27 '17 at 04:21
  • @carmichael561:Will you please specify with the help of an example? – Styles May 27 '17 at 04:23
  • Robert Z's answer is a nice example. – carmichael561 May 27 '17 at 04:24
  • @carmichael561:Yeah i got it.thank you – Styles May 27 '17 at 04:25
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    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. – JonathanZ May 27 '17 at 04:27
  • @user275313:My bad,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. – Styles May 27 '17 at 06:43
  • @user275313:thanks a lot for pointing this. – Styles May 27 '17 at 06:46
  • No problem. (You've actually left me wondering what the precise condition is for a finite Abelian group to have all of its subgroups be cyclic, but you don't need that for your particular problem.) – JonathanZ May 28 '17 at 02:03
  • @user275313:good,i did'nt get that thought.So,have you reached to some conclusion? – Styles May 28 '17 at 06:30
  • I came up with some ideas, but then I noticed this question in the "Related" column on the right which seems to contain the core of my solution – JonathanZ May 28 '17 at 22:24

1 Answers1

4

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

Robert Z
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