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

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?

• It isn't. Take $G = C_2 \times C_2$. – the_fox Nov 10 '18 at 6:29
• Can you please take a look the question once again.. I have made some edits..@the_fox – cmi Nov 10 '18 at 6:36
• 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?) – the_fox Nov 10 '18 at 6:43
• 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. – the_fox Nov 10 '18 at 6:45
• Case 1 is OK. But you could also say that Case 2 is only wrong when $n=2$. – Derek Holt Nov 10 '18 at 16:01

## 1 Answer

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.

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