Every finite abelian p-group is a direct product of cyclic p-groups

Given: The objects and hypotheses in the statement.

Let $G$ be a finite abelian $p$-group and let $a\in G$ have maximal element order. Then there is a subgroup $H\le G$ such that $G=⟨a⟩\oplus H\mathrm{.}$ (A maximal-order cyclic subgroup splits off a finite abelian p-group).

Let $P$ be a property of naturals such that for every $n\in \mathbb{N}$, if $P(m)$ holds for all $m<n$ then $P(n)$. Then $P(n)$ holds for all $n\in \mathbb{N}$. (At $n=0$ the hypothesis is vacuous, so $P(0)$ is forced.) (Strong (complete) induction).

Let $N_0,\dots ,N_{r-1}⊴G$. The following are equivalent: the $N_i$ form an internal direct product of $G$; every $g\in G$ has a unique expression $g=n_0\cdots n_{r-1}$ with $n_i\in N_i$; and the multiplication map $\mu :{\prod }_{i<r}{N}_i\to G$ is an isomorphism. These statements include the empty family and the one-factor case. (Internal direct products are external direct products, equivalently every element has a unique factorisation).

If $G=⟨g⟩$ is cyclic, then exactly one of the following applies: - if $g$ has infinite order, $G\cong (\mathbb{Z},+)$; - if $g$ has finite order $n$, necessarily $n\ge 1$, then $G\cong (\mathbb{Z}\mathrm{/}n,+)$. (Every cyclic group is isomorphic to $(\mathbb{Z},+)$ or to $(\mathbb{Z}\mathrm{/}n,+)$ for its finite order $n\ge 1$).