The additive rationals do not decompose as a product of finite cyclic prime-power groups

Given: The objects and hypotheses in the statement refuted.

On the set of pairs $(a,b)$ with $a, b \in \mathbb{Z}$ and $b \ne 0$, define $(a,b) \sim (c,d) \iff ad = cb \quad \text{in } \mathbb{Z}.$ This is an equivalence relation (lem-rat-equivalence). The rationals are the quotient $\mathbb{Q}$, and $[(a,b)]$ is written $a/b$. (The rationals as equivalence classes of pairs of integers).

Natural exponents, in a monoid. Let $(M,\cdot,e)$ be a monoid (def-semigroup-and-monoid) and $g \in M$. By the recursion theorem (thm-recursion), applied with the set $M$, the element $e$ and the function $x \mapsto x \cdot g$ from $M$ to $M$, there is exactly one function $\mathbb{N} \to M$, written $n \mapsto g^{n}$, with $g^{0} = e, \qquad g^{\sigma(n)} = g^{n} \cdot g \quad (n \in \mathbb{N}).$ In particular $g^{0} = e$ for every $g$, including $g = e$, and $g^{1} = g^{\sigma(0)} = e \cdot g = g$. Since $\mathbb{N}$ contains $0$ (def-natural-numbers), the exponent $0$ is a genuine value of the definition and not a separate convention. Integer exponents, in a group. Let $G$ be a group (def-group) and $g \in G$. Write $\iota : \mathbb{N} \to \mathbb{Z}$ for the embedding $k = [(k,0)]$ of lem-nat-embeds-int, which is injective, preserves addition, multiplication and order, and has as image exactly the nonnegative integers. For $x \in \mathbb{Z}$ define - $g^{x} := g^{k}$, the natural power, when $0 \le x$ and $x = k$; - $g^{x} := (g^{k})^{-1}$ when $x < 0$ and $-x = k$. Why this is well defined. The order on $\mathbb{Z}$ is total and antisymmetric (thm-int-ordered-ring, def-int-order), so exactly one of $0 \le x$ and $x < 0$ holds and the two clauses never both apply. In the first clause $x$ is nonnegative, so $x = k$ for some $k \in \mathbb{N}$, and $k$ is unique because $\iota$ is injective. In the second clause $x < 0$ gives $0 = x + (-x) < 0 + (-x) = -x$ by compatibility of the order with addition (thm-int-ordered-ring, def-int-operations), so $-x$ is a positive integer and again $-x = k$ for a unique $k$. The inverse $(g^{k})^{-1}$ is a single determined element by lem-inverse-unique and def-invertible-element. Finally the two readings of $g^{k}$, as a natural power and as an integer power, agree by construction, so no ambiguity is introduced. Abbreviation. In an exponent we write $k$ for the integer $k$ when a natural number $k$ is used where an integer is expected; this is unambiguous because $\iota$ is injective and preserves the arithmetic and the order, and because the two readings of $g^{k}$ agree as just noted. Additive notation. When the group is written additively the same object is written $n g$ or $n \cdot g$ rather than $g^{n}$, with $0 g = 0$ and $\sigma(n) g = n g + g$; the definitions are identical, only the symbols differ. (Powers $g^{n}$: natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$).

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).