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?

  • 7
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

  • 1
    $\begingroup$ not comprehensive..Please go through the question once again.. $\endgroup$ – INDIAN Nov 10 '18 at 7:13
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @cmi do you at least understand? $\endgroup$ – the_fox Nov 10 '18 at 7:28
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ Who knows? They probably don't understand some part. $\endgroup$ – the_fox Nov 10 '18 at 19:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.