Let $G$ be a finite, commutative group. Is
$$\text{$G$ is cyclic.}$$ equivalent to
$$\text{For each $n$, there are at most $n$ pairwise different elements of $G$,}$$ $$\text{such that the order of such an element divides $n$.}$$ ?
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2"$n$ pairwise different elements" is a long-winded way of saying "$n$ elements"... Anyway, yes. This should be proved in most intro texts that cover abelian groups, no? – anon Jan 11 '18 at 22:23
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2I think both directions have already been answered here: https://math.stackexchange.com/questions/346936/finite-group-for-which-xxm-e-leq-m-for-all-m-is-cyclic and https://math.stackexchange.com/questions/517113/can-we-conclude-that-this-group-is-cyclic?noredirect=1&lq=1 – user408858 Jan 11 '18 at 22:31
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1I guess that's just the backward direction. The forward direction is not hard. – Caleb Stanford Jan 12 '18 at 01:29
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Ok. This is simply again using $\sum_{d|n}\phi(d)=n$ and $\phi(d)$ is the amount of elements with order $d$. (Since $G$ is cyclic there is only one subgroup of order $d$) – user408858 Jan 13 '18 at 18:35