The indecomposable finite abelian groups are exactly the nontrivial cyclic groups of prime-power order

Given: The objects and hypotheses in the statement.

A nontrivial finite abelian group is indecomposable if it is not an internal direct product of two nontrivial subgroups in the sense of def-internal-direct-product-of-subgroups. It is decomposable if such a product exists. The trivial group is assigned neither label. (Indecomposable and decomposable nontrivial finite abelian groups).

Every finite abelian group is isomorphic to a finite direct product of cyclic groups of prime-power order. The multiset of their orders is uniquely determined by the group, up to permutation of the factors. (Fundamental theorem of finite abelian groups: elementary-divisor form).

Let $G$ be a finite abelian group and let $p$ be a prime dividing $\mathrm{\mid }G\mathrm{\mid }$. Then $G$ contains an element, and hence a subgroup, of order $p$. (Cauchy's theorem for finite abelian groups).

Every subgroup $H$ of a cyclic group $G=⟨g⟩$ is cyclic. If $H\mathrm{\ne }\{e\}$, then the least positive integer $d$ for which $g^d\in H$ satisfies $H=⟨g^d⟩$. (Every subgroup of a cyclic group is cyclic; the least positive exponent in a nontrivial subgroup supplies a generator).

Let $G$ be a finite group such that the positive integer $\mathrm{\mid }G\mathrm{\mid }$ is prime. Then every $g\mathrm{\ne }e$ has order $\mathrm{\mid }G\mathrm{\mid }$, satisfies $⟨g⟩=G$, and hence generates $G$. In particular, $G$ is cyclic. (A finite group of prime order is cyclic and every nonidentity element generates it).