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Let $A$ be a finite abelian group. Prove that $A$ is cyclic iff for each $n \in \mathbb{N}$ $$\#\{a \in A : na = 0\}\le n.$$

Any help or hint will be appreciated.

Noa Even
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user101010
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  • Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – A.P. Dec 29 '15 at 21:55
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    I don't think that this is true for finitely generated abelian groups. Take $\Bbb Z\times \Bbb Z$, for instance. Then the given set has cardinality $1$ for any $n$, but the group is not cyclic. – Arthur Dec 29 '15 at 22:03
  • thank you. It is Finite abelian group , not finitely generated. I'm editing the question – user101010 Dec 29 '15 at 22:21

2 Answers2

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As pointed out in the comment section by Arthur, your statement is partially false. Let us show your statement for finite abelian groups.

Let $G$ be a finite abelian group, then there exists $n\in\mathbb{N}$ and $(d_i)_{i\in\{1,\ldots,n\}}\in{\mathbb{N}_{\geqslant 2}}^n$ such that: $$\forall i\in\{2,\ldots,n\},d_{i-1}\vert d_i\textrm{ and }G\cong\bigoplus_{i=1}^n\mathbb{Z}/(d_i).$$

  • Assume that for all $m\in\mathbb{N}$, $|\{g\in G\textrm{ s.t. }mg=0\}|\leqslant m$. One has to show that $n\in\{0,1\}$. Assume by contradiction that $n\geqslant 2$, let us take $m:=d_2$ and notice that for all $(x,y)\in\mathbb{Z}/(d_1)\oplus\mathbb{Z}/(d_2)$, one has: $$m(x,y,0,\ldots,0)=0.$$ Which is a contradiction, since we have found $d_1d_2>m$ elements in $\{g\in G\textrm{ s.t. }mg=0\}$.

  • Assume that $G$ is cyclic, since $G$ is finite, $G\cong\mathbb{Z}/(|G|)$ and I let you show that the given condition holds.

Community
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C. Falcon
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Here is a proof that appears in Weil's gem Number theory for beginners and in several other books.

Let $m$ be the order of $A$ and let $\psi(n)$ be the number of elements of order $n$ in $A$.

Then, by Lagrange's theorem, $m=\sum_{d\mid m} \psi(d)$ because every element has an order that is a divisor of $m$.

If $\psi(d)>0$, then there is an element of order $d$ and the subgroup generated by it has order $d$ and so the hypothesis on $A$ implies that $\psi(d) = \phi(d)$.

Therefore, for all $d \mid m$, we have $\phi(d) \ge \psi(d)$.

Now $m = \sum_{d\mid m} \phi(d) \ge \sum_{d\mid m} \psi(d) = m$. (*)

This means that $\psi(d)=\phi(d)$ for all $d \mid m$.

In particular, $\psi(m)>0$, which means that $A$ has an element of order $m$ and so is cyclic.

(*) The first equality comes from counting the fractions $1/m,\dots,m/m$ when they are reduced.

lhf
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    This proof does not assume that $A$ is abelian. – lhf Feb 23 '18 at 14:21
  • In the fourth paragraph, why say the hypothesis implies that $\psi(d)=\phi(d)$ please? –  Feb 25 '18 at 14:03
  • @Benny, if $g$ has order $d$ then the hypothesis on $A$ imply that ${a \in A : a^d = 1} = \langle g \rangle$, which has $\phi(d)$ elements of order $d$. – lhf Feb 25 '18 at 15:28
  • Could you elaborate more on the (*)? I neither understood how you got that from counting the fractions nor why you counted the fractions. – LoneStar Nov 19 '19 at 05:13
  • @StarPlatinumZAWARUDO, see https://math.stackexchange.com/a/194713/589 – lhf Nov 19 '19 at 10:01