 Solution

We prove the unique-subgroup criterion for a finite abelian group to be cyclic using the invariant-factor classification from part (a). If $A$ is nontrivial, writewhere $C_m$ denotes a cyclic group of order $m$. If $s\ge 2$, choose a prime number $p$ dividing $d_1$; then $p$ also divides $d_2$. In the first cyclic factor, the element represented by $d_1/p$ has order of a group element $p$, and the corresponding element in the second factor has order $p$ as well.

The two elements supported in their separate coordinates generate distinct subgroups of order $p$: one has zero second coordinate, and the other has zero first coordinate. Both are subgroups of $A$, and $p\mid |A|$. This contradicts the assumed uniqueness for that divisor. Hence $s\le 1$, andIf $A$ is the trivial group, it is already cyclic, generated by its identity. Thus the argument covers every finite order.