The number of finite abelian groups of order n is the product of the partition numbers of the prime exponents of n

Given: The objects and hypotheses in the statement.

If $G$ is finite abelian and $\mathrm{\mid }G\mathrm{\mid }={\prod }_{i<r}{p}_i^{a_i}$ is its prime factorisation, then the subgroups $G(p_i)$ form an internal direct product of $G$. Thus $G\cong {\prod }_{i<r}G(p_i)\mathrm{.}$ For the trivial group, this is the empty product. (A finite abelian group is the internal direct product of its primary components).

For a prime $p$ and $n>0$, isomorphism classes of abelian groups of order $p^n$ are in bijection with partitions of $n$. For $n=0$, the unique group is the trivial group and corresponds separately to the empty partition. (Isomorphism classes of abelian groups of order p^n are counted by partitions of n).

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).

Let $(M,\cdot ,e)$ be a monoid (def-semigroup-and-monoid) and let $g:\mathbb{N}\to M$ be a family of elements of $M$, written $g_i:=g(i)$. There is exactly one function $P_g:\mathbb{N}\to M$ satisfying $P_g(0)=e,P_g(\sigma (n))=P_g(n)\cdot g_n(n\in \mathbb{N}),$ and we write ${\prod }_{i<n}{g}_i:=P_g(n),\text{also written }{g}_0{g}_1\cdots g_{n-1}\mathrm{.}$ In particular the empty product is ${\prod }_{i<0}{g}_i=e$, and ${\prod }_{i<1}{g}_i=e\cdot g_0=g_0$. Why the recursion is legitimate. The clause $P_g(\sigma (n))=P_g(n)\cdot g_n$ consults $n$ as well as $P_g(n)$, so thm-recursion does not apply to it directly. Apply that theorem instead with the set $A=\mathbb{N}\times M$, the element $a=(0,e)$, and the function $F:A\to A$ given by $F(n,x)=(\sigma (n),x\cdot g_n)$: it yields a unique $H:\mathbb{N}\to \mathbb{N}\times M$ with $H(0)=(0,e)$ and $H(\sigma (n))=F(H(n))$. Writing $H(n)=(H_1(n),H_2(n))$, induction (thm-induction-principle) gives $H_1(n)=n$ for every $n$, since $H_1(0)=0$ and $H_1(\sigma (n))=\sigma (H_1(n))$. Hence $H(\sigma (n))=(\sigma (n),H_2(n)\cdot g_n)$, so $P_g:=H_2$ satisfies the two displayed equations. It is the only such function: if $Q$ satisfies them too, then $\{n:P_g(n)=Q(n)\}$ contains $0$ and is closed under $\sigma$, hence is all of $\mathbb{N}$ by induction. The value depends only on $g_0,\dots ,g_{n-1}$. If $g,g^{\mathrm{\prime }}:\mathbb{N}\to M$ satisfy $g_i=g_i^{\mathrm{\prime }}$ for every $i<n$, then $P_g(n)=P_{g^{\mathrm{\prime }}}(n)$. Indeed the set of $n$ for which this implication holds contains $0$, both products then being $e$; and if it holds at $n$, and $g,g^{\mathrm{\prime }}$ agree at every $i<\sigma (n)$, then they agree at every $i<n$ and also at $n$ itself, because $i<\sigma (n)$ is equivalent to $i\le n$ (lem-nat-order-is-membership), so $P_g(\sigma (n))=P_g(n)\cdot g_n=P_{g^{\mathrm{\prime }}}(n)\cdot g_n^{\mathrm{\prime }}=P_{g^{\mathrm{\prime }}}(\sigma (n))$. Induction finishes it. This is what makes the notation $g_0{g}_1\cdots g_{n-1}$ unambiguous: it names a value determined by the first $n$ terms alone, and a finite list $u$ of length $n$, that is a function $u:n\to M$ on the von Neumann natural $n=\{0,\dots ,n-1\}$ (def-natural-numbers), determines the product ${\prod }_{i<n}{u}_i:=P_{\tilde{u}}(n)$ computed from any extension $\tilde{u}:\mathbb{N}\to M$ of $u$. (The product $g_0{g}_1\cdots g_{n-1}$ of a finite list in a monoid, by recursion, with the empty product ($n=0$) equal to the identity).