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If an Abelian group $G$ has order $n$ and at most one subgroup of order $d$ for all $d$ dividing $n$ then $G$ is cyclic.

I am trying to use the structure theorem for finitely generated abelian groups.

So I write $n=p_1^{\alpha_1}\ldots p_n^{\alpha_n}$.

I am hoping to show each of the alpha's must =1 then I will have that $G$ is isomorphic to $\prod_i^n \mathbb{Z}/p_i \mathbb{Z}$, which is cyclic.

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    Structure theorem is a 'big' theorem and its use as far as possible must be avoided. So go through the answers below. And, more on this line, look at problem U212 in http://awesomemath.org/wp-content/uploads/reflections/2011_6/MR6.pdf2011-12-22

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Just the most basic group theory and a simple counting argument will suffice. By Lagrange's theorem $G$ only has elements of orders $d$ dividing $n$. If such an element exists, then the cyclic group it generates has a number $\phi(d)$ of elements of order $d$ that only depends on $d$ (this is Euler's totient function but we need to know nothing about it). By the hypothesis $G$ cannot contain other elements of order $d$, so it has either $\phi(d)$ such elements or none; set $\chi_G(d)=1$ if it does have such elements and $\chi_G(d)=0$ if it doesn't. Then counting element of $G$ by their order one has $ n=\#G=\sum_{d\mid n}\chi_G(d)\phi(d). $

But a cyclic group of order $n$ has exactly one cyclic subgroup of order $d$ for every $d$ that divides $n$. This means that $ \sum_{d\mid n}\phi(d)=n. $ Thus the formula for $G$ can only be satisfied if all values $\chi_G(d)$ are equal to $1$. In particular $\chi_G(n)=1$: the group $G$ contains elements of its own order $n$ and is therefore cyclic.

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    @MarcvanLeeuwenThanks,especially to respond so close to midnight in France.2012-01-07
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Let $\phi$ be Euler's totient function.

The assumption implies that $G$ has at most $\phi(d)$ elements of order $d$ whenever $d$ divides $n$.

But then it has at least $\phi(d)$ elements of order $d$ whenever $d$ divides $n$.

So it has elements of order $n$.

EDIT. As Thomas suggests, I should add that the following fact has been implicitly used: We have $ \sum_{d|n}\ \phi(d) = n $ by considering the case when $G$ is cyclic.

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    Dear @ThomasAndrews: Thank you very much!2011-12-22
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I would prove this along the lines of J Rotman - An introduction to the theory of Groups, Graduate Texts in Mathematics, Springer-Verlag, 4th Edition. Pg:28

Observe that the converse of what you'd like to prove is quite a trivial issue.

For what you'd like to prove, J R has the following Recipe:

Lemma: Let $\phi$ be the Euler's totient function that counts the number of numbers less than $n$ and coprime to it, $ \sum_{d|n} \phi (d) = n$

Proof:

The proof of this fact follows from the following: Let G be a cyclic group of order $n$. I'll assume you can prove that there are $\phi(n)$ generators for this $G$ and in general $\phi(k)$ generators for a cyclic group of order $k$.

$G=\bigcup gen(C)$ where $gen(C)$ stands for generators of cyclic subgroups $C$ of G.(So $C$ ranges over all cyclic subgroups of $G$) And this union is a disjoint union, from the definition of a generator.

Note that the disjoint union holds for all $G$ and there is nothing sacrosanct about $G$ being cyclic, except from here: I claim now that, as the subgroups are unique for each divisor, the disjoint union translates into a sum that is exactly the lemma we seek to prove.

Your Question:

As pointed out before, I'll make use of the disjoint union.

$ n=|G| = \sum |gen(C)| \leq \sum_{d|n}{\phi(d)}$

So, now I use my lemma to conclude that, I have cyclic subgroups for each divisor of $n$ and in particular for $n$. So, that must be $G$. So, G must be cyclic.

I would like to emphasise, that this classifies all finite cyclic groups.

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    Will the downvoter care to explain why he downvoted this answer?2012-01-23
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It may be easier if you use the form of the structure theorem that says that $G$ is a product of cyclic groups where the order of each divides the order of the next.

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    @Zev Sorry about the language, not intentional at all. All I wanted to convey was in much lighter sense the fact that, this post doesn't answer the question2011-12-22
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By Cayley Theorem $G$ has an element $x$ of order $p_i$. It follows from your requirement that the subgroup $\langle x\rangle$ generated by $\langle x\rangle$ contains all the elements of order $p_i$.

Exercise for you: Prove that $p_i$ cannot divide the order of the factor group $G/\langle x\rangle$ (Why?)

and you are done...

P.S. If you covered this topic already, looking to the $p$-Sylow subgroups also helps, but I think this is more complicated. :)

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    Wouldn't that imply that $n$ is squarefree, although any cyclic group satisfies the requirement?2011-12-22