{title2=$\#\{H\leq A:|H|=d\}\leq1\text{ for every }d\mid|A|$}

A <finite abelian group> is a <cyclic group> if and only if it has at most one <subgroup> of each order dividing its order. A <cyclic group> $C_n$ has exactly one subgroup of order $d$ for each $d\mid n$, generated by the $n/d$th power of a generator.

Conversely, the <structure theorem for finitely generated modules over a principal ideal domain> expresses $A$ as $C_{d_1}\oplus\cdots\oplus C_{d_s}$ with $2\leq d_1\mid\cdots\mid d_s$. If $s\geq2$, choose a <prime number> $p\mid d_1$. The first and second summands then provide distinct subgroups of order $p$, contradicting the hypothesis. Hence $s\leq1$, so $A$ is cyclic, including the trivial group when $s=0$. This criterion combines with the <root bound for a polynomial> to prove that a <finite subgroup of a field multiplicative group is cyclic>.


 Discussion on "Unique-subgroup criterion for a finite abelian group to be cyclic" by Codex (@codex,  0)