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I've had some trouble with this problem since I can't seem to know how to continue from certain point of my proof.

First of all, using Cauchy's theorem, we can assure that for every $p\mid |H|$ that is prime there exists an element of order $p$ in $H$, let that be $x$. Therefore, the subgroup $\langle x\rangle$ has order prime $p$, and so, every element different from $1$ of that group has to have order $p$ (since they all generate it). These are $p-1$ elements, and since our group has exactly that number of elements with order $p$, then these are all. Equivalently, $\forall p$ that is prime and divides $|H|, \exists!L\leq H$ of order $p$, and since every group of order prime is cyclic, it's the one we just described, and it is cyclic.

From this point, I've tried to see if for every divisor of the order of the group there exists only one subgroup of that order that is cyclic (and so, that the group is cyclic), but I can't seem to know how to prove it. I've also tried to see if this condition forces $H$ to have an element of order $H$, and so $H$ is cyclic, but I don't know how to prove that either. If anyone has any ideas on how to tackle this problem I would be really grateful, thanks.

Arturo Magidin
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Pabloo
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    What theorems are you allowed to use? Sylow’s theorems? The problem becomes trivial when you’re allowed to use the classification of finitely generated abelian groups, of course. – Jan Oct 29 '24 at 19:02
  • Yes, Sylow's theorems can be used. Also, I don't see why the classification of finitely generated abelian groups makes the problem trivial, could you dumb it down for me a little? – Pabloo Oct 29 '24 at 19:19
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    Decompose $G$ into a direct sum of cyclic groups of prime power order. If there are two factors with the same prime, say order $p^r$ and $p^s$, then you get at least $p^2-1$ elements of order $p$ between them. So there is at most one factor for each prime, which means $G$ is a direct sum of cyclic groups of relatively prime orders, which means it is cyclic. – Arturo Magidin Oct 29 '24 at 19:26

1 Answers1

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A finite abelian group $G$ is not cyclic if and only if there exists a prime $p$ such that $G$ contains a subgroup isomorphic to $C_p\times C_p$; see, e.g., here for a proof that does not use the classification of finite abelian groups, here for one that uses the Chinese Remainder Theorem, and here for links to other duplicates.

If $G$ has a subgroup isomorphic to $C_p\times C_p$, then $G$ has at least $p^2-1$ elements of order $p$. Thus, if $G$ has exactly $p-1$ elements of order $p$ for every prime $p$, then $G$ does not contain a subgroup isomorphic to $C_p\times C_p$ for any prime $p$, hence $G$ is cyclic.

Arturo Magidin
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  • Possibly dumb question but why does this imply $G$ has at least $p^2-1$ elements of order $p$? – Ishigami Oct 30 '24 at 09:41
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    Because every nontrivial element of $C_p$ has oder $p$ and hence, denoted $C_p\times C_p=\langle x\rangle \times \langle y\rangle$, the element $(x^k,y^l)$ has order $p$ for every $k,l\in{1,\dots,p-1}$, as well as every element $(x^k,1)$ and $(1,y^l)$. Therefore, overall $(p-1)^2+(p-1)+(p-1)=$ $p^2-2p+1+2p-2=$ $p^2-1$ elements of order $p$, at least, @Ishigami. – Kan't Oct 30 '24 at 11:37
  • @Kan't simpler: $C_p\times C_p$ has $p^2$ elements, all of order dividing $p$, and only one, $(1,1)$ of order $1$. So $p^2-1$. – Arturo Magidin Oct 30 '24 at 12:33
  • @Ishigami Every element of $C_p\times C_p$ has order dividing $p$, hence order 1 (which only the identity has) or $p$. So all its elements except one. – Arturo Magidin Oct 30 '24 at 12:35
  • For some reason, I assumed that the questioner hadn't Lagrange at hands, @ArturoMagidin. – Kan't Oct 30 '24 at 12:46