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If every proper subgroup of a finite abelian group $G$ is cyclic, then $G$ is cyclic.

I think the statement is true for some cases.

My attempt :

Case 1 : $o(G) = p_1^{n_1}p_2^{n_2} \ldots p_k^{n_k}$ , where $ p _1 , p_ 2, \ldots, p_k$ are prime numbers. Now I will take two elements $s , v$ (say) whose orders are prime to each other. So $o(sv) = o(G)$. So $G$ is cyclic.

Case 2 : $o(G) = p^{n}$, where $p$ is a prime. I think the statement would not be true for this case. A counterexample is the non-cyclic abelian group of order $4$.

Can anyone please tell me If I have gone wrong anywhere?

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    $\begingroup$ It isn't. Take $G = C_2 \times C_2$. $\endgroup$ – the_fox Nov 10 '18 at 6:29
  • $\begingroup$ Can you please take a look the question once again.. I have made some edits..@the_fox $\endgroup$ – cmi Nov 10 '18 at 6:36
  • $\begingroup$ The statement remains the same though, as does your claim. Perhaps you mean to say that if $G$ is abelian and of composite order and all of its proper subgroups are cyclic then $G$ itself is cyclic. That's certainly true, though your attempt at a proof does not make much sense to me (why is $o(sv) = o(G)$ and what does that tell you?) $\endgroup$ – the_fox Nov 10 '18 at 6:43
  • $\begingroup$ If $G$ has prime-power order then it is still true, except in some cases I have hinted at. Have a look at the fundamental theorem for the structure of finite abelian groups. $\endgroup$ – the_fox Nov 10 '18 at 6:45
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    $\begingroup$ Case 1 is OK. But you could also say that Case 2 is only wrong when $n=2$. $\endgroup$ – Derek Holt Nov 10 '18 at 16:01
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First you need to know that $C_m \times C_n \cong C_{mn}$ when $\gcd(m,n)=1$ and also know the fundamental theorem for abelian groups.

Now, suppose that $|G|$ has at least two distinct prime divisors, say $|G| = p_1^{a_1} \cdots p_k^{a_k}$ , $k \geq 2$. Let $p_i$ be one of the primes and let $P_i$ be the subgroup of order $p_i^{a_i}$ of $G$. Since $|P_i| < |G|$, $P_i$ is a proper subgroup of $G$ so it is cyclic. The same holds for all $i$ though, and since $G \cong P_1 \times \cdots \times P_k$, it follows that $G$ is the direct product of cyclic groups of coprime orders, so it is cyclic itself by the first observation.

Suppose now that $|G|=p^m$ for some positive integer $m$. If $m>2$, I argue that $G$ must be cyclic. By the fundamental theorem, there are positive integers $a_1, a_2, \ldots, a_n$ such that $G = C_{p^{a_1}} \times \ldots \times C_{p^{a_n}}$, where $a_1 + \ldots a_n = m$. Suppose for a contradiction that $G$ is not cyclic, but every proper subgroup of $G$ is. Then certainly $n>1$. Let $H$ be the subgroup of order $p$ of $C_{p^{a_1}}$, $K$ the subgroup of order $p$ of $C_{p^{a_2}}$. Consider the subgroup $H \times K$ of $G$. Since $|H \times K| = p^2 <|G|$, it follows that $H \times K = C_p \times C_p$ is a proper subgroup of $G$ and thus must be cyclic by assumption. But this is a contradiction, so $G$ is cyclic, just as we wanted to show.

Now, if $m=1$ there is nothing to do ($C_p$ is cyclic and its only proper subgroup is the trivial group). If $m=2$, the structure theorem tells you that either $G \cong C_{p^2}$ or $G \cong C_p \times C_p$. Note that, in this case, the claim does not hold, since $C_p \times C_p$ is not cyclic but every proper subgroup of $C_p \times C_p$ is.

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    $\begingroup$ not comprehensive..Please go through the question once again.. $\endgroup$ – INDIAN Nov 10 '18 at 7:13
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    $\begingroup$ Sorry, I don't see what I'm missing. Is there something wrong with the proof? $\endgroup$ – the_fox Nov 10 '18 at 7:14
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    $\begingroup$ @cmi do you at least understand? $\endgroup$ – the_fox Nov 10 '18 at 7:28
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    $\begingroup$ Your argument is basically correct so I do not understand the negative votes. Perhaps you should have stated the conclusion more clearly. $\endgroup$ – Derek Holt Nov 10 '18 at 8:24
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    $\begingroup$ Who knows? They probably don't understand some part. $\endgroup$ – the_fox Nov 10 '18 at 19:33

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