A distribution of finite abelian groups

Much of this posting is a brief regurgitation of (a corner of) a paper by Cohen and Lenstra. However, I will use some weird words that the authors do not use to feed my (and hopefully others') intuition.

Given a nonempty category $\mathcal{C}$ such that

• the automorphism group of each object is finite and
• there are finitely many isomorphism classes of $\mathcal{C},$

we define the number of points of $\mathcal{C}$ to be $$|\mathcal{C}| := \sum_{[G] \in \mathcal{C}/\simeq}\frac{1}{|\mathrm{Aut}(G)|},$$ where $\mathcal{C}/\simeq$ is the set of isomorphism classes of $\mathcal{C}.$ (This definition is not so "made-up"; see this Wikipedia page.) If $\mathscr{P}$ is a property defined on $\mathcal{C}/\simeq,$ we define $$\mathrm{Prob}_{G \in \mathcal{C}}(G \text{ satsifies } \mathscr{P}) := \frac{|\mathcal{C}_{\mathscr{P}}|}{|\mathcal{C}|},$$ where $\mathcal{C}_{\mathscr{P}}$ is the full subcategory of $\mathcal{C}$ given by the objects satisfying $\mathscr{P}.$ Note that we have $$|\mathcal{C}_{\mathscr{P}}| = \sum_{[G] \in \mathcal{C}/\simeq}\frac{\boldsymbol{1}_{\mathscr{P}}(G)}{|\mathrm{Aut}(G)|},$$ where $\boldsymbol{1}_{\mathscr{P}} : \mathcal{C}/\simeq \rightarrow \{0, 1\}$ is the indicator function for the property $\mathscr{P}.$ Replacing this indicator function to any other complex function $f : \mathcal{C}/\simeq \rightarrow \mathbb{C},$ we may define the average of $f$ in $\mathcal{C}$ as $$M_{\mathcal{C}}(f) := \frac{|\mathcal{C}|_{f}}{|\mathcal{C}|},$$ where $$|\mathcal{C}|_{f} = \sum_{[G] \in \mathcal{C}/\simeq}\frac{f(G)}{|\mathrm{Aut}(G)|}.$$ In particular, we have $$M_{\mathcal{C}}(\boldsymbol{1}_{\mathscr{P}}) = \mathrm{Prob}_{G \in \mathcal{C}}(G \text{ satisfies } \mathscr{P}).$$ What if the category $\mathcal{C}$ has infinitely many objects? An example we are going to focus on is the category $\textbf{Ab}^{<\infty}$ of finite abelian groups which was the focus of the paper of Cohen and Lenstra. Note that each object $\textbf{Ab}^{<\infty}$ has finite automorphism group, but the category itself has infinitely many isomorphism classes (e.g., take $\mathbb{Z}/(n)$ for infinitely many $n.$)

Limiting distributions. Now, let $\mathcal{C}$ be a category such that the automorphism group of each object is finite, but unlike before, suppose that the category has infinitely many isomorphism classes. How can we define the probability and the average? First off, we may hope that the number $$|\mathcal{C}| = \sum_{[G] \in \mathcal{C}/\simeq}\frac{1}{|\mathrm{Aut}(G)|}$$ of points on $\mathcal{C}$ is finite. In certain cases, this turns out to be true. For instance, consider the category $\textbf{Ab}_{p}^{<\infty}$ of finite abelian $p$-groups for a fixed prime $p.$ It turns out that $$|\textbf{Ab}_{p}^{<\infty}| = \sum_{[G] \in \textbf{Ab}_{p}^{<\infty}/\simeq}\frac{1}{|\mathrm{Aut}(G)|} = \prod_{i=1}^{\infty}(1 - p^{-i}),$$ and people seem to credit Hall for this computation. The first consequence of this computation is that we may give each $[G] \in \in \textbf{Ab}_{p}^{<\infty}/\simeq$ the mass $$\frac{1}{|\mathrm{Aut}(G)|}\prod_{i=1}^{\infty}(1 - p^{-i})$$ to make a discrete probability distribution on the set $\textbf{Ab}_{p}^{<\infty}/\simeq$ of isomorphism classes of fintie abelian $p$-groups. This distribution is now called Cohen-Lenstra distribution. On the other hand, note that $$|\textbf{Ab}^{<\infty}| \geq \sum_{p} |\textbf{Ab}_{p}^{<\infty}| = \infty,$$ where the sum is over all primes $p.$ Hence, we realize that even though we may easily build the notions of the probability and the average of a complex-valued function on the category $\textbf{Ab}_{p}^{<\infty},$ the same would not work for $\textbf{Ab}^{<\infty}.$

One situation that we may deal with is when we have a sequence of full subcategories $$\mathcal{C}_{1} \hookrightarrow \mathcal{C}_{2} \hookrightarrow \mathcal{C}_{3} \hookrightarrow \cdots$$ of $\mathcal{C}$ such that

• each $\mathcal{C}_{n}$ has finitely many isomorphism classes and
• $\bigcup_{n=1}^{\infty}\mathcal{C}_{n} = \mathcal{C}.$

If $|\mathcal{C}|$ is finite, then writing $\mathcal{C}_{0} := \emptyset,$ we have $$|\mathcal{C}| = \sum_{j=1}^{\infty}|\mathcal{C}_{j} \smallsetminus \mathcal{C}_{j-1}| = \lim_{n \rightarrow \infty}|\mathcal{C}_{n}|.$$ Thus, for any function $f : \mathcal{C}/\simeq \rightarrow \mathbb{C},$ we must have $$M_{\mathcal{C}}(f) = \lim_{n \rightarrow \infty}M_{\mathcal{C}_{n}}(f),$$ if $|\mathcal{C}|_{f}$ (defined above) is finite, which will necessarily yield $$|\mathcal{C}|_{f} = \lim_{n \rightarrow \infty}|\mathcal{C}_{n}|_{f}.$$ This justifies the following definition.

Definition. Let $\mathcal{C}$ be a category each of whose object has finitely many automorphisms. Suppose that there is a sequence of full subcategories $$\mathcal{C}_{1} \hookrightarrow \mathcal{C}_{2} \hookrightarrow \mathcal{C}_{3} \hookrightarrow \cdots$$ of $\mathcal{C}$ such that 

• each $\mathcal{C}_{n}$ has finitely many isomorphism classes and
• $\bigcup_{n=1}^{\infty}\mathcal{C}_{n} = \mathcal{C}.$

Given any complex-valued function on the set $\mathcal{C}/\simeq$ of isomorphism classes of $\mathcal{C},$ we define the average of $f$ with respect to the sequence to be $$M_{\mathcal{C}}(f) := \lim_{n \rightarrow \infty}M_{\mathcal{C}_{n}}(f),$$ if the limit on the right-hand side exists. Given any property $\mathscr{P}$ on $\mathcal{C}/\simeq,$ we define $$\mathrm{Prob}_{G \in \mathcal{C}}(G \text{ satisfies } \mathscr{P}) := M_{\mathcal{C}}(\boldsymbol{1}_{\mathscr{P}}),$$ if the limit in the definition of the right-hand side exists.

Remark/Conjecture. The above definition a priori depends on the given sequence of full subcategories with finitely many isomorphism classes. My guess is that any such two sequences should give the same limit (if exists, under perhaps some mild conditions), but I cannot immediately come up with a proof nor an enlightening counterexample.

Finite abelian groups. Now, we focus on the category $\mathcal{C} = \textbf{Ab}^{<\infty}$ of finite abelian groups. Our goal is to compute $$M_{\textbf{Ab}^{<\infty}}(f) = \lim_{n \rightarrow \infty}\frac{\sum_{[G] \in \textbf{Ab}^{\leq n}/\simeq}\frac{f(G)}{|\mathrm{Aut}(G)|}}{\sum_{[G] \in \textbf{Ab}^{\leq n}/\simeq}\frac{1}{|\mathrm{Aut}(G)|}}$$ for some "reasonable" complex-valued function $f$ defined on the set $\textbf{Ab}^{< \infty}/\simeq$ of isomorphism classes of finite abelian groups with respect to the sequence $$\textbf{Ab}^{\leq 1} \hookrightarrow  \textbf{Ab}^{\leq 2} \hookrightarrow \textbf{Ab}^{\leq 3} \hookrightarrow \cdots,$$ where $\textbf{Ab}^{\leq n}$ denotes the full subcategory of $\textbf{Ab}^{< \infty},$ consisting of finite abelian groups of size $\leq n.$

Ideas of Cohen-Lenstra. Cohen and Lenstra approached the computation of $M_{\textbf{Ab}^{<\infty}}(f)$ as follows: with some "reasonable" conditions on $f$, they showed that $$M_{\textbf{Ab}^{<\infty}}(f) = \lim_{s \rightarrow 0}\frac{\sum_{[G] \in \textbf{Ab}^{<\infty}/\simeq}\frac{f(G)|G|^{-s}}{|\mathrm{Aut}(G)|}}{\sum_{[G] \in \textbf{Ab}^{<\infty}/\simeq}\frac{|G|^{-s}}{|\mathrm{Aut}(G)|}}.$$ It turns out that manipulating the sum $$\sum_{[G] \in \textbf{Ab}^{<\infty}/\simeq}\frac{f(G)|G|^{-s}}{|\mathrm{Aut}(G)|}$$ with complex variable $s$ is quite doable for certain $f$, so computing the above limit on the right-hand side became doable. It turns out that $$\zeta_{\textbf{Ab}^{<\infty}}(s) := \sum_{[G] \in \textbf{Ab}^{<\infty}/\simeq}\frac{|G|^{-s}}{|\mathrm{Aut}(G)|} = \prod_{j=1}^{\infty}\zeta(s + j),$$ where $s \mapsto \zeta(s)$ denotes the Riemann zeta function. We know that $$\zeta(s) = 1^{-s} + 2^{-s} + 3^{-s} + \cdots$$ is well-defined for $\mathrm{Re}(s) > 1,$ and this ensures that $\zeta_{\textbf{Ab}^{<\infty}}(s)$ is well-defined for $\mathrm{Re}(s) > 0.$ Using the fact that $\zeta(s)$ has analytic continuation to $\mathbb{C}$ except at the unique pole at $s = 1$ of order $1,$ we see that $\zeta_{\textbf{Ab}^{<\infty}}(s)$ has analytic continuation to $\mathbb{C}$ except the poles at $s = 0, 1, 2, \dots,$ each of whose order is $1.$

Writing $$\zeta_{\textbf{Ab}^{<\infty}, f}(s) := \sum_{[G] \in \textbf{Ab}^{<\infty}/\simeq}\frac{f(G)|G|^{-s}}{|\mathrm{Aut}(G)|},$$ Cohen and Lenstra showed (in Corollary 5.5 of their paper) that if $\zeta_{\textbf{Ab}^{<\infty}, f}(s)$ has a unique pole at $s = 0$ (for $\mathrm{Re}(s) \geq 0$) then we may compute $$M_{\textbf{Ab}^{<\infty}}(f) = \lim_{s \rightarrow 0}\frac{\zeta_{\textbf{Ab}^{<\infty}, f}(s)}{\zeta_{\textbf{Ab}^{<\infty}}(s)},$$ just as we desired above.

More flexible definition. For any full subcategory $\mathcal{C}$ of $\textbf{Ab}^{<\infty},$ we define $$\zeta_{\mathcal{C}}(s) := \sum_{[G] \in \mathcal{C}/\simeq}\frac{|G|^{-s}}{|\mathrm{Aut}(G)|},$$ just a notation we will use later for a computation.

Independence among primes. Let $T$ be any nonempty set of primes. Suppose that

• $f$ is multiplicative among primes in $T$ (i.e., $f(G) = \prod_{p \in T}f(G_{(p)})$).

This incorporates that $f(\text{trivial group}) = 1.$ Write $\textbf{Ab}_{T}^{<\infty}$ to be the category of finite abelian groups that are supported by the primes in $T,$ and $f|_{\textbf{Ab}_{T}^{<\infty}}$ the function given by $$f|_{\textbf{Ab}_{T}^{<\infty}}(G) := f(G_{T}),$$ where $$G_{T} := \bigoplus_{p \in T}G_{(p)}.$$

We have $$\zeta_{\textbf{Ab}^{<\infty}, f|_{\textbf{Ab}_{T}^{<\infty}/\simeq}}(s) = \left( \prod_{p \notin T}\sum_{[G] \in \textbf{Ab}_{p}^{<\infty}/\simeq}\frac{|G|^{-s}}{|\mathrm{Aut}(G)|} \right) \left( \prod_{p \in T}\sum_{[G] \in \textbf{Ab}_{p}^{<\infty}/\simeq}\frac{f(G)|G|^{-s}}{|\mathrm{Aut}(G)|} \right).$$ Thus, we have $$\frac{\zeta_{\textbf{Ab}^{<\infty}, f|_{\textbf{Ab}_{T}^{<\infty}/\simeq}}(s)}{\zeta_{\textbf{Ab}^{<\infty}}(s)} = \prod_{p \in T} \left( \frac{\sum_{[G] \in \textbf{Ab}_{p}^{<\infty}/\simeq}\frac{f(G)|G|^{-s}}{|\mathrm{Aut}(G)|}}{\sum_{[G] \in \textbf{Ab}_{p}^{<\infty}/\simeq}\frac{|G|^{-s}}{|\mathrm{Aut}(G)|}} \right).$$ When we are in a situation where we can exchange the limit of taking $s \rightarrow 0$ with the product (e.g., when $T$ is a finite set), then this implies that 

$$\begin{align*}M_{\textbf{Ab}^{<\infty}}(f|_{\textbf{Ab}_{T}^{<\infty}/\simeq}) &= \lim_{s \rightarrow 0}\frac{\zeta_{\textbf{Ab}^{<\infty}, f|_{\textbf{Ab}_{T}^{<\infty}/\simeq}}(s)}{\zeta_{\textbf{Ab}^{<\infty}}(s)} \\ &= \prod_{p \in T} \left( \frac{\sum_{[G] \in \textbf{Ab}_{p}^{<\infty}/\simeq}\frac{f(G)}{|\mathrm{Aut}(G)|}}{\sum_{[G] \in \textbf{Ab}_{p}^{<\infty}/\simeq}\frac{1}{|\mathrm{Aut}(G)|}} \right) \\ &= \prod_{p \in T}M_{\textbf{Ab}_{p}^{<\infty}}(f) \\ &= M_{\textbf{Ab}_{T}^{<\infty}}(f).\end{align*}$$ However, note that (assuming  that $f(\text{trivial group}) = 1$) we get the identity $$\zeta_{\textbf{Ab}^{<\infty}, f|_{\textbf{Ab}_{T}^{<\infty}/\simeq}}(s) = \zeta_{\textbf{Ab}_{T}^{<\infty}, f}(s)\zeta_{\textbf{Ab}_{|\mathrm{Spec}(\mathbb{Z})| \smallsetminus T}^{<\infty}}(s),$$ where $|\mathrm{Spec}(\mathbb{Z})|$ denotes the set of all primes, even without assuming that $f$ is multiplicative. That is, we are only using that $f|_{\textbf{Ab}_{T}^{<\infty}/\simeq}$ is multiplicative between the complements $T$ and $|\mathrm{Spec}(\mathbb{Z})| \smallsetminus T$. Therefore, we must have $$M_{\textbf{Ab}^{<\infty}}(f|_{\textbf{Ab}_{T}^{<\infty}/\simeq}) = M_{\textbf{Ab}_{T}^{<\infty}}(f)$$ for more general family of $f.$ It is known (due to Hall) that $$\sum_{[G] \in \textbf{Ab}_{p}^{<\infty}/\simeq}\frac{1}{|\mathrm{Aut}(G)|} = \prod_{i=1}^{\infty}(1 - p^{-i}),$$ so $$M_{\textbf{Ab}^{<\infty}}(f|_{\textbf{Ab}_{T}^{<\infty}/\simeq}) = \frac{\sum_{[G] \in \textbf{Ab}_{T}^{<\infty}/\simeq}\frac{f(G)}{|\mathrm{Aut}(G)|}}{\prod_{p \in T}\prod_{i=1}^{\infty}(1 - p^{-i})}.$$ If $f$ is multiplicative over each element of $T,$ then we also have $$\sum_{[G] \in \textbf{Ab}_{p}^{<\infty}/\simeq}\frac{1}{|\mathrm{Aut}(G)|} = \prod_{i=1}^{\infty}(1 - p^{-i})^{-1},$$ so $$M_{\textbf{Ab}^{<\infty}}(f|_{\textbf{Ab}_{T}^{<\infty}/\simeq}) = \prod_{p \in T} \left( \sum_{[G] \in \textbf{Ab}_{p}^{<\infty}/\simeq}\frac{f(G)}{|\mathrm{Aut}(G)|}  \prod_{i=1}^{\infty}(1 - p^{-i})\right).$$

Computing some distributions. By the last paragraph, for any fixed prime $p,$ we may compute $$\mathrm{Prob}_{G \in \textbf{Ab}^{<\infty}}(G_{(p)} = 0) = \prod_{i=1}^{\infty}(1 - p^{-i}),$$ and for any nonzero finite abelian $p$-group $H,$ we have $$\mathrm{Prob}_{G \in \textbf{Ab}^{<\infty}}(G_{(p)} \simeq H \text{ or } 0) = \left(1 + \frac{1}{|\mathrm{Aut}(G)|}\right) \prod_{i=1}^{\infty}(1 - p^{-i}).$$ This implies that $$\mathrm{Prob}_{G \in \textbf{Ab}^{<\infty}}(G_{(p)} \simeq H) = \frac{1}{|\mathrm{Aut}(G)|} \prod_{i=1}^{\infty}(1 - p^{-i}).$$ That is, Cohen and Lenstra showed that the $p$-part of a random finite abelian group follows the Cohen-Lenstra distribution.