Elementary divisors regroup uniquely into invariant factors

Given: The objects and hypotheses in the statement.

An elementary-divisor decomposition of a finite abelian group $G$ is an isomorphism $G\cong C_{q_0}\times\cdots\times C_{q_{r-1}},$ where every $q_i>1$ is a prime power. The unordered multiset of the $q_i$, counted with multiplicity, is the elementary-divisor data. The cyclic factors and product use thm-classification-of-cyclic-groups and def-external-direct-product-of-groups. The data records factor isomorphism types, not distinguished internal subgroups; the trivial group has empty data. (Elementary-divisor data for a finite abelian group).

An invariant-factor list for a finite abelian group $G$ is a finite list of integers $1<n_1\mid n_2\mid\cdots\mid n_r$ together with an isomorphism $G\cong C_{n_1}\times\cdots\times C_{n_r}$. The cyclic factors and product use thm-classification-of-cyclic-groups and def-external-direct-product-of-groups. Unit factors are omitted. The trivial group has the empty list. (Invariant-factor data for a finite abelian group).

Let $n_0,\ldots,n_{r-1}$ be a finite pairwise-coprime list of positive integers and let $N:=\prod_{i<r}n_i$. The map $\Phi:\mathbb Z/N\longrightarrow\prod_{i<r}\mathbb Z/n_i,\qquad[x]_N\longmapsto([x]_{n_i})_{i<r},$ is a bijection. It preserves addition, multiplication, $[0]$, and $[1]$ componentwise. For the empty list, $N=1$ and both sides have one element. (Chinese remainder theorem for a finite pairwise-coprime list: simultaneous residues determine one class modulo the product, and the resulting bijection preserves addition and multiplication).

Let $\iota:\mathbb N\to\mathbb Z$ be the canonical embedding. If $g\in G$ and $h\in H$ have finite orders $m,n\ge1$, then in the external direct product $\operatorname{ord}(g,h)=\operatorname{lcm}(m,n).$ (If $g$ and $h$ have finite orders $m$ and $n$, then $\iota(\operatorname{ord}(g,h))=\operatorname{lcm}(\iota(m),\iota(n))$ in $G\times H$).

If $G=\langle g\rangle$ 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 $n\ge1$, then $G\cong(\mathbb Z/n,+)$. (Every cyclic group is isomorphic to $(\mathbb Z,+)$ or to $(\mathbb Z/n,+)$ for its finite order $n\ge1$).

Powers are the natural powers of def-group-power and finite products those of def-monoid-finite-product, both taken in the commutative monoid $(\mathbb{Z},\cdot,1)$ of lem-units-of-z. Call $p : r \to \mathbb{Z}$ an injective list of primes when every $p_i$ is prime (def-prime) and $p_i = p_j$ forces $i = j$ (def-injection-surjection-bijection). Let $n \in \mathbb{Z}$ with $n \ge 1$ and let $p : r \to \mathbb{Z}$ be an injective list of primes such that every prime divisor of $n$ equals $p_i$ for some $i < r$. Then, with $v_q$ as in def-p-adic-valuation: 1. $\displaystyle n \;=\; \prod_{i<r} p_i^{\,v_{p_i}(n)}$; 2. $v_q(n) = 0$ for every prime $q$ that is not among $p_0,\dots,p_{r-1}$; 3. the exponents are determined by $n$: if $e : r \to \mathbb{N}$ and $n = \prod_{i<r} p_i^{\,e_i}$, then $e_j = v_{p_j}(n)$ for every $j < r$. Clause 3 needs only injectivity of the list, not the covering hypothesis. (For $n \ge 1$ and any injective list $p : r \to \mathbb{Z}$ of primes containing every prime divisor of $n$, one has $n = \prod_{i<r} p_i^{\,v_{p_i}(n)}$; the exponents are determined by $n$, and $v_q(n) = 0$ for every prime $q$ outside the list).