$\mathbb{Z}_{p}[t]/(P(t))$ is a DVR if $P(t)$ is irreducible in $\mathbb{F}_{p}[t]$

Let $p$ be a prime and $P(t) \in \mathbb{Z}_{p}[t]$ a monic polynomial whose image in $\mathbb{F}_{p}$ modulo $p$ (which we also denote by $P(t)$ if there is no ambiguity) is irreducible. In this posting, we give a proof of the following fact, which we learned from some lectures notes by Andrew Sutherland (Lemma 10.13 and Example 10.16).

Theorem. Keeping the notation as above, the quotient ring $R = \mathbb{Z}_{p}[t]/(P(t))$ is a DVR whose maximal ideal is generated by $p$.

Proof. We first show that $R$ is a local ring whose maximal ideal is $(p)$. First, note that $$R/(p) \simeq \mathbb{F}_{p}[t]/(P(t))$$ is a field because we assumed that $P(t)$ is irreducible in $\mathbb{F}_{p}[t].$ Hence $(p)$ is indeed a maximal ideal of $R.$ Let $\mathfrak{m}$ be any maximal ideal of $R.$ We want to show that $p \in \mathfrak{m}.$

For contradiction, suppose that $p \notin \mathfrak{m}.$ Then since $\mathfrak{m}$ is a maximal ideal, we must have $\mathfrak{m} + (p) = R.$ Since $R$ is Noetherian, we have $$\mathfrak{m} = (r_{1}, \dots, r_{l}) = Rr_{1} + \cdots + Rr_{l}$$ for some $r_{1}, \dots, r_{l} \in R.$ We have $$R = \mathbb{Z}_{p} \oplus \bar{t} \mathbb{Z}_{p} \oplus \cdots \oplus \bar{t}^{d-1} \mathbb{Z}_{p},$$ where $d = \deg(P),$ so $R$ is finite over $\mathbb{Z}_{p}.$ Since $\mathfrak{m}$ is finite over $R,$ we see that $\mathfrak{m}$ is finite over $\mathbb{Z}_{p}.$ That is, we have some $z_{1}, \dots, z_{n} \in \mathfrak{m}$ such that $$\mathfrak{m} = \mathbb{Z}_{p}z_{1} + \cdots + \mathbb{Z}_{p}z_{n}.$$ Since $\mathfrak{m} + (p) = R,$ it follows that the images $\bar{z}_{1}, \dots, \bar{z}_{n}$ of $z_{1}, \dots, z_{n} \in R$ generate $R/(p)$ as an $\mathbb{F}_{p}$-vector space. Since $\mathbb{Z}_{p}$ is a local ring with the maximal ideal $(p),$ by Nakayama's lemma, this implies that $R$ is generated by $z_{1}, \dots, z_{n}$ as a $\mathbb{Z}_{p}$-module, but then this implies that $$R = \mathbb{Z}_{p}z_{1} + \cdots + \mathbb{Z}_{p}z_{n} \subset Rz_{1} + \cdots + Rz_{n} = \mathfrak{m},$$ which is a contradiction. Thus, we conclude that $p \in \mathfrak{m},$ showing that $R$ is a local ring with the maximal ideal $(p).$ Since $R$ is a Noetherian local domain that is not a field, whose maximal ideal is principal, we conclude that $R$ is a DVR. $\Box$

Sandpile groups of random graphs: 2. Erdős–Rényi model

First, let's consider the following question:

Question. What is the distribution of the number $N_{\Gamma}$ of the spanning trees of a random graph $\Gamma$?

Of course, we need to make sense of what "random" means in the above question. One way to generate random graphs is to use the Erdős–Rényi model. That is, given $n \in \mathbb{Z}_{\geq 2}$ and a real number $0 < q < 1$, we first fix $n$ vertices and choose a graph with these vertices at random by requiring that for each pair of vertices, the probability that the two chosen vertices are connected by an edge is $q.$ We write $G(n, q)$ to mean the set of all graphs with $n$ vertices without self-loops and double edges with the probability measure given by the Erdős–Rényi model. We note that $G(n, q)$ is a finite set of graphs, each of which has the $n$ given vertices with

• no self-loops;
• no double edges;
• and $M$ edges ($0 \leq M \leq {n \choose 2}$),

and each graph in $G(n, q)$ with $M$ edges occurs with the probability $q^{M}(1 - q)^{{n \choose 2} - M}.$ Hence, the probability that the number of edges of an Erdős–Rényi random graph is equal to $M$ is ${{n \choose 2} \choose M}q^{M}(1 - q)^{{n \choose 2} - M}.$ That is, the distribution of the number of edges of a random $\Gamma \in G(n, q)$ follows the binomial distribution with parameters $q$ and ${n \choose 2}.$ In particular, the average number of the edges in a graph randomly chosen in $G(n, q)$ is ${n \choose 2}q.$

Cayley's formula says that the total number of spanning trees on given $n$ vertices is $n^{n-2},$ and the probability that each spanning tree occurs in a random graph in $G(n, q)$ is $q^{n-1}.$ Hence, given $0 \leq m \leq n^{n-2},$ we have $$\underset{\Gamma \in G(n, q)}{\mathrm{Prob}}(N_{\Gamma} = m) = {n^{n-2} \choose m} (q^{n-1})^{m}(1 - q^{n-1})^{n^{n-2} - m}.$$ That is, the number $N_{\Gamma}$ of a random graph $\Gamma \in G(n, q)$ follows the binomial distribution with parameters $q$ and $n^{n-2}.$ In particular, the average number of spanning trees of $\Gamma$ is $n^{n-2}q^{n-1}.$

Now, we know completely what the distribution of the number of spanning trees of $\Gamma$ randomly chosen from $G(n, q).$ In the last posting, we discussed how Kirchhoff's theorem tells us that the number $N_{\Gamma}$ of spanning trees of a graph $\Gamma$ is the product of the non-negative invariant factors of its sandpile group $S_{\Gamma}.$

Question. What is the distribution of the sandpile group $S_{\Gamma}$ of $\Gamma$ randomly chosen in $G(n, q)$?

It turns out from a paper of Wood (Corollary 9.3) that for any finite abelian group $A,$ we have $$\lim_{n \rightarrow \infty}\underset{\Gamma \in G(n, q)}{\mathrm{Prob}}(S_{\Gamma} \simeq A) = 0.$$ We already know that the expected number $$n^{n-2}q^{n-1} = \frac{(nq)^{n-1}}{n}$$ of $N_{\Gamma}$ is large (which is in particular nonzero) when $n$ is large, so we expect $S_{\Gamma}$ to be a finite group. More precisely, we have $$\underset{\Gamma \in G(n, q)}{\mathrm{Prob}}(S_{\Gamma} \text{ is infinite}) = \underset{\Gamma \in G(n, q)}{\mathrm{Prob}}(N_{\Gamma} = 0) = (1 - q^{n-1})^{n^{n-2}}.$$ If $0 < q < 1$ is fixed, then this implies that $$\lim_{n \rightarrow \infty}\underset{\Gamma \in G(n, q)}{\mathrm{Prob}}(S_{\Gamma} \text{ is infinite}) = 0.$$ Hence, the fact that the probability that $S_{\Gamma}$ is isomorphic to a specific finite abelian group is small for large $n$ has to do with that the events we are considering are too fine.

Remark. This is a strange phenomenon. Heuristically, one may say that there are too many candidates for $S_{\Gamma}$ when we take a random $\Gamma \in G(n, q),$ which makes it less likely if we try to specify the isomorphism class of $S_{\Gamma},$ but there are other similar instances, where we choose from many finitely abelian groups but the probability does not converge to $0.$ For example, see this paper of Nguyen and Wood (Theorem 1.1 with $u = 1$).

It also turns out that given a prime $p,$ the $p$-part $S_{\Gamma}[p^{\infty}]$ of the sanpile group of a random graph $\Gamma \in G(n, q)$ has a nontrivial distribution. The following theorem is from a paper of Wood (Theorem 1.1):

Theorem (Wood). Let $p$ be a prime and $G$ a finite abelian $p$-group so that $$G \simeq \mathbb{Z}/p^{\lambda_{1}}\mathbb{Z} \times \mathbb{Z}/p^{\lambda_{2}}\mathbb{Z} \times \cdots \mathbb{Z}/p^{\lambda_{l}}\mathbb{Z}$$ with $\lambda_{1} \geq \lambda_{2} \geq \cdots \geq \lambda_{l} \geq 0.$ Let $(\mu_{1}, \mu_{2}, \dots)$ be the transpose of the partition $(\lambda_{1}, \lambda_{2}, \dots, \lambda_{l}).$ Then $$\begin{align*}\lim_{n \rightarrow \infty} & \underset{\Gamma \in G(n, q)}{\mathrm{Prob}}(S_{\Gamma}[p^{\infty}] \simeq G) \\ &= \frac{p^{-\sum_{r \geq 1} \mu_{r}(\mu_{r}+1)/2}\prod_{i=1}^{\lambda_{1}}\prod_{j=1}^{\lfloor (\mu_{i} - \mu_{i+1})/2 \rfloor} (1 - p^{-2j})^{-1}}{|G||\mathrm{Aut}(G)|}\prod_{k \geq 0} (1 - p^{-2k-1}).\end{align*}$$

Sandpile groups of random graphs -- 1 Preliminaries

In this posting, we start discussing the topic of random sandpile groups.

1 Preliminaries

Let $n \in \mathbb{Z}_{\geq 1}$ and $\Gamma$ a graph (undirected) with verticies $1, 2, \dots, n.$ Given $1 \leq i \neq j \leq n,$ define $\deg(i, j)$ to be the number of edges between vertex $i$ and vertex $j.$ We define $\deg(i)$ to be the number of edges between vertex $i$ and any other vertices. Define the adjacency matrix $A(\Gamma)$ of $\Gamma$ to be the $n \times n$ matrix whose $(i, j)$ entry is $\deg(i, j).$ We define the Laplacian of $\Gamma$ as

$$L(\Gamma) := \mathrm{diag}(\deg(1), \deg(2), \dots, \deg(n)) - A(\Gamma).$$

Remark. Note that $L(\Gamma)$ cannot tell whether there are self-loops on any vertex of the graph $\Gamma.$

Because of the way we defined the degree of each vertex, it is always the case that

• each row of $L(\Gamma)$ sums to $0$ and
• each column of $L(\Gamma)$ sums to $0.$

Denote by $L_{ij}$ the $(i, j)$-entry of $L = L(\Gamma).$ Then

$$L\begin{bmatrix} k_{1} \\ \vdots \\ k_{n} \end{bmatrix} = \begin{bmatrix} L_{11}k_{1} + \cdots + L_{1n}k_{n} \\ \vdots \\ L_{n1}k_{1} + \cdots + L_{nn}k_{n} \end{bmatrix},$$

so if we add all the entries of the last $n \times 1$ matrix, we have

$$(L_{11} + \cdots + L_{n1})k_{1} + \cdots + (L_{1n} + \cdots + L_{nn})k_{n} = 0,$$

since each column of $L = L(\Gamma)$ sums to $0.$ That is, the image $\mathrm{im}(L)$ of $L = L(\Gamma)$ always sits inside

$$(\mathbb{Z}^{n})_{0} := \{(r_{1}, \dots, r_{n}) \in \mathbb{Z}^{n} : r_{1} + \cdots + r_{n} = 0\}.$$

The sandpile group of $\Gamma$ is defined as $S_{\Gamma} := (\mathbb{Z}^{n})_{0}/\mathrm{im}(L).$

Question. When we choose $\Gamma$ at random, how are the sandpile groups $S_{\Gamma}$ distributed?

Of course, this question needs to be polished a bit. 

Lemma. We have

$$\ker(L(\Gamma)) = \bigoplus\mathbb{Z}(e_{i_{1}} + \cdots + e_{i_{k}}),$$

where the direct sum is over the subsets $\{i_{1}, \dots, i_{k}\}$ of the vertex set $\{1, 2, \dots, n\}$ of $\Gamma$, each of which consists of vertices from a connected component of $\Gamma.$ In particular, we see that $\ker(L(\Gamma))$ is a free $\mathbb{Z}$-submodule of $\mathbb{Z}^{n}$ with rank equal to the number of connected components of $\Gamma.$

Remark. It follows from Lemma that $\mathrm{im}(L(\Gamma)) \simeq \mathbb{Z}^{n}/\ker(L(\Gamma))$ is a free $\mathbb{Z}$-module with rank $n - |\pi_{0}(\Gamma)|,$ where we denote by $\pi_{0}(\Gamma)$ the set of connected components of $\Gamma.$ Thus, we note that the sandpile group $S_{\Gamma} = (\mathbb{Z}^{n})_{0}/\mathrm{im}(L(\Gamma))$ is a free $\mathbb{Z}$-module with rank $|\pi_{0}(\Gamma)| - 1.$ In particular, we have

Corollary. The sandpile group $S_{\Gamma}$ of a graph $\Gamma$ is finite if and only if $\Gamma$ is connected.

Proof of Lemma. Note that the elements $e_{i_{1}} + \cdots + e_{i_{k}}$ we consider are $\mathbb{Z}$-linearly independent because the subsets $\{i_{1},  \dots, i_{k}\}$ of $\{1, 2, \dots, n\}$ we consider are disjoint.

We write $L = L(\Gamma)$ for convenience. Let $i_{1} < \cdots < i_{k}$ be the vertices of a connected component of $\Gamma.$ We have 

$$L(e_{i_{1}} + \cdots + e_{i_{k}}) = L(e_{i_{1}}) + \cdots + L(e_{i_{k}}) = \begin{bmatrix} L_{1,i_{1}} + \cdots + L_{1,i_{k}} \\ L_{2,i_{1}} + \cdots + L_{2,i_{k}} \\ \vdots \\ L_{n,i_{1}} + \cdots + L_{n,i_{k}}\end{bmatrix}.$$

Now, for each $1 \leq i \leq n,$ we know that if $i$ does not belong to the connected component, then $L_{i,i_{1}} = \cdots = L_{i,i_{k}} = 0.$ Otherwise (if $i$ belongs to the connected component), we have $i = i_{j}$ for some $1 \leq j \leq k$ and $$L_{i_{j},i_{1}} + \cdots + L_{i_{j},i_{k}} = L_{i_{j},1} + \cdots + L_{i_{j},n} = 0.$$ Hence, we must have that $L(e_{i_{1}} + \cdots + e_{i_{k}}) = 0$ in $\mathbb{Z}^{n},$ so $$e_{i_{1}} + \cdots + e_{i_{k}} \in \ker(L).$$ This shows that

$$\ker(L) \supset \bigoplus\mathbb{Z}(e_{i_{1}} + \cdots + e_{i_{k}}),$$

so it remains to show the opposite inclusion.

Let $x = \begin{bmatrix}x_{1} \\ \vdots \\ x_{n}\end{bmatrix} \in \ker(L) \subset \mathbb{Z}^{n}.$ We have $$0 = x^{T}Lx = \sum_{1 \leq i, j \leq n}L_{ij}x_{i}x_{j} = \sum (x_{i} - x_{j})^{2},$$ where the last sum is over $1 \leq i, j \leq n$ such that $v_{i}$ and $v_{j}$ are connected by an edge. (This is by definition of $L = L(\Gamma).$) This implies that for any connected component $\{i_{1}, \dots, i_{k}\},$ we have $x_{i_{1}} = \cdots = x_{i_{k}}$, and these are only relations among $x_{1}, \dots, x_{n}.$ Hence, we see that $x$ is a $\mathbb{Z}$-linear combination of elements of the form $e_{i_{1}} + \cdots + e_{i_{k}}$ in $\mathbb{Z}^{n},$ where $\{i_{1}, \dots, i_{k}\}$ vary in connected components of $\Gamma.$ This finishes the proof. $\Box$

Lemma. Let $n \in \mathbb{Z}_{\geq 2}.$ Given any graph $\Gamma$ with $n$ vertices, its sandpile group $S_{\Gamma}$ is isomorphic to the cokernel of any $(n - 1) \times (n - 1)$ submatrix of $L(\Gamma).$ In particular, the cokernels of $(n - 1) \times (n - 1)$ submatrice of $L(\Gamma)$ are isomorphic to each other.

Proof. We write $L = L(\Gamma).$ Since all the entries of any row or column sum to $0,$ given any $1 \leq i, j \leq n,$ we may add all the other rows to the $i$-th row and then all the other columns to the $j$-th column and replace all the entries of the $i$-th row and the $j$-th column into $0$'s. Since these operations are elementary row operations over $\mathbb{Z},$ this does not change the cokernel of $L$ up to isomorphisms. Denote by $L^{(i,j)},$ the new $n \times n$ matrix. Then we have $$\mathrm{cok}(L) \simeq \mathbb{Z}^{n}/\mathrm{im}(L^{(i,j)}).$$ Note that $$\mathrm{im}(L^{(i,j)}) \subset \mathbb{Z}e_{1} \oplus \cdots \oplus \mathbb{Z}e_{j-1} \oplus \mathbb{Z}e_{j+1} \oplus \cdots \oplus \mathbb{Z}e_{n}.$$ This implies that $$S_{\Gamma} = (\mathbb{Z}^{n})_{0}/\mathrm{im}(L) \simeq (\mathbb{Z}e_{1} \oplus \cdots \oplus \mathbb{Z}e_{j-1} \oplus \mathbb{Z}e_{j+1} \oplus \cdots \oplus \mathbb{Z}e_{n})/\mathrm{im}(L^{(i,j)}).$$ Let $L_{(i,j)}$ be the $(n-1) \times (n-1)$ matrix given by erasing the $i$-th row and the $j$-th column of $L.$ Then we have $$\begin{align*}(\mathbb{Z}^{n})_{0}/\mathrm{im}(L) &\simeq (\mathbb{Z}e_{1} \oplus \cdots \oplus \mathbb{Z}e_{j-1} \oplus \mathbb{Z}e_{j+1} \oplus \cdots \oplus \mathbb{Z}e_{n})/\mathrm{im}(L^{(i,j)}) \\ &\simeq \mathbb{Z}^{n-1}/\mathrm{im}(L_{(i,j)}) \\ &= \mathrm{cok}(L_{(i,j)}),\end{align*}$$ so $S_{\Gamma} \simeq \mathrm{cok}(L_{(i,j)}),$ as desired. $\Box$

Theorem (Kirchhoff). Let $n \in \mathbb{Z}_{\geq 2}.$ The number of spanning trees of a graph $\Gamma$ with $n$ vertices is equal to the determinant of any $(n-1) \times (n-1)$ submatrix of $L(\Gamma).$ In particular, if $\Gamma$ is connected, then this number is equal to the size of the sandpile group $S_{\Gamma}.$

Remark. Let $L$ be the Laplacian of a graph $\Gamma$ with $n$ vertices with $n \geq 2,$ and $L'$ the $(n-1) \times (n-1)$ submatrix of $L$ given by erasing the last row and the last columns. Then there are always $P, Q \in \mathrm{GL}_{n-1}(\mathbb{Z})$ such that $$PL'Q = \mathrm{diag}(\lambda_{1}, \dots, \lambda_{n-1})$$ with $\lambda_{1} \geq \cdots \geq \lambda_{n-1} \geq 0$ in $\mathbb{Z}.$ Hence, we have $$S_{\Gamma} \simeq \mathrm{cok}(L') \simeq \mathbb{Z}/\lambda_{1}\mathbb{Z} \times \cdots \times \mathbb{Z}/\lambda_{n-1}\mathbb{Z},$$ and Kirchohoff's theorem tells us that $$\det(L') = \lambda_{1} \cdots \lambda_{n-1}$$ is the number of spanning trees of $\Gamma.$ That is, Kirchohoff's theorem says that the number $N_{\Gamma}$ of spanning trees of $\Gamma$ is equal to the product of (non-negative) invariant factors of its sandpile group $S_{\Gamma}.$

A vague question. If we choose $\Gamma$ at random, what does the distribution of $N_{\Gamma}$ look like? More generally, what does the distribution of $S_{\Gamma}$ look like?

Surjection moment method

In a paper by M.M.Wood (Theorem 3.1), we find the following theorem:

Theorem (Wood). Let 

• $(X_{n})_{n \geq 1}$ and $(Y_{n})_{n \geq 1}$ be sequences of random finitely generated abelian groups and
• $\mathcal{A}_{m}$ be the set of isomorphism classes of finite abelian groups with exponent dividing $m \in \mathbb{Z}_{\geq 1}$ (i.e., $mG = 0$ for all $G \in \mathcal{A}_{m}$).

Suppose that for every $G \in \mathcal{A}_{m},$ the limit  $$\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, G)).$$ exists and $$\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, G)) \leq |\wedge^{2}G|.$$

Then for every $H \in \mathcal{A}_{m},$ the limit $$\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H)$$ exists, and $$\sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Sur}(H, G) = \lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, G)).$$

If we also have $$\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, G)) = \lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(Y_{n}, G))$$ for all $G \in \mathcal{A}_{m},$ then $$\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) = \lim_{n \rightarrow \infty}\mathrm{Prob}(Y_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H).$$

Remarks. 

1. Note that it is implicit in the statement of the theorem that if we denote by $\mathcal{S}$ the probability space of finitely generated abelian groups for $X_{n},$ then for any $H \in \mathcal{A}_{m},$ the subset $$\{X \in \mathcal{S} : X \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H\}$$ is measurable, and similarly for $Y_{n}.$
2. It is also implicit in the statement that the expected value $\mathbb{E}(\#\mathrm{Sur}(X_{n}, G))$ exists for every $n \geq 1,$ and similarly for $Y_{n}.$
3. In the references we follow, it is not mentioned that the abelian groups appearing as representatives of classes in $\mathcal{A}_{m}$ are of finite size. It seems that the proof of Theorem requires it, so we assume this. (This does not affect the main results in the references.)

Rest of this posting. We cover the proof of the above theorem, which we learn from another paper by Wood (Theorem 8.3).

Step 1. We first assume that for every $H \in \mathcal{A}_{m}$ the limit $$\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H)$$ exists and deduce $$\sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z})\#\mathrm{Sur}(H, G) = \lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, G))$$ from this assumption.

Claim. For every $G \in \mathcal{A}_{m},$ there exists $G' \in \mathcal{A}_{m}$ such that the infinite sum $$\sum_{H \in \mathcal{A}_{m}}\frac{\#\mathrm{Hom}(H, G)}{\#\mathrm{Hom}(H, G')}$$ converges.

Proof of Claim. If $m = 1,$ then the only isomorphism class in $\mathcal{A}_{m}$ is that of the trivial group, so the claim immediately holds. Hence, assume that $m > 1.$ Consider the prime factorization $m = p_{1}^{e_{1}} \cdots p_{k}^{e_{k}},$ and write $$H = H_{1} \oplus \cdots \oplus H_{k}$$ with $p_{i}^{e_{i}}H_{i} = 0$ for $H \in \mathcal{A}_{m},$ which implies that $$\sum_{H \in \mathcal{A}_{m}}\frac{\#\mathrm{Hom}(H, G)}{\#\mathrm{Hom}(H, G')} = \sum_{H_{1} \in \mathcal{A}_{p_{1}^{e_{1}}}}\frac{\#\mathrm{Hom}(H_{1}, G_{1})}{\#\mathrm{Hom}(H_{1}, G'_{1})} \cdots \sum_{H_{k} \in \mathcal{A}_{p_{k}^{e_{k}}}}\frac{\#\mathrm{Hom}(H_{k}, G_{k})}{\#\mathrm{Hom}(H_{k}, G'_{k})}$$ for any $G' \in \mathcal{A}_{m}.$ This reduces the problem to the case where $m = p^{e}$ where $p$ is a prime and $e \in \mathbb{Z}_{\geq 1}.$ Without loss of generality, assume that $G$ is nontrivial. In this case, we have $$G \simeq \mathbb{Z}/p^{\lambda_{1}}\mathbb{Z} \times \mathbb{Z}/p^{\lambda_{2}}\mathbb{Z} \cdots \times \mathbb{Z}/p^{\lambda_{l}}\mathbb{Z}$$ with $e \geq \lambda_{1} \geq \cdots \geq \lambda_{l} \geq 1,$ so if we let $\lambda = (\lambda_{1}, \dots, \lambda_{l})$ be the corresponding partition, then we choose $$G' = \mathbb{Z}/p^{\mu_{1}}\mathbb{Z} \times \mathbb{Z}/p^{\mu_{2}}\mathbb{Z} \cdots \times \mathbb{Z}/p^{\mu_{l'}}\mathbb{Z},$$ where $\mu_{i} = 2\lambda'_{i} + 1$ for $1 \leq i \leq e.$ (Here, we have $\lambda'_{i} = 0$ for $\lambda_{1} < i \leq e.$) By using a standard formula that counts the number of group maps from a finite abelian $p$-groups to another (Lemma 7.1 of Wood), we can compute that $$\sum_{H \in \mathcal{A}_{p^{e}}}\frac{\#\mathrm{Hom}(H, G)}{\#\mathrm{Hom}(H, G')} = \sum_{c_{1} \geq \cdots \geq c_{e} \geq 0}p^{-c_{1}(\lambda_{1} + 1) \cdots - c_{e}(\lambda_{e} + 1)},$$ which is convergent as it is bounded by a convergent geometric series. This proves the claim. $\Box$

Back to Step 1, we have $$\begin{align*}\sum_{B \in \mathcal{A}_{m}}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq B) \#\mathrm{Hom}(B, G') &= \mathbb{E}( \#\mathrm{Hom}(X_{n}, G')) \\ &= \sum_{H \subset G'}\mathbb{E}(\#\mathrm{Sur}(X_{n}, H)),\end{align*}$$ where $H$ ranges over subgroups of (a representative of) fixed $G' \in \mathcal{A}_{m}.$ The second identity is evident, but the first identity needs an explanation. By definition, we have $$\begin{align*}\mathbb{E}(\#\mathrm{Hom}(X_{n}, G')) &= \int_{X \in \mathcal{S}} \#\mathrm{Hom}(X, G') \mu_{n}(X), \end{align*}$$ where $\mu_{n}$ means the given probability measure on the probability space $\mathcal{S}$ of finitely generated abelian groups, where $X_{n}$ takes place. Since $mG' = 0,$ we have $$\#\mathrm{Hom}(X, G') = \#\mathrm{Hom}(X \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z}, G')$$ for any $X \in \mathcal{S}$ so that $$\begin{align*}\mathbb{E}(\#\mathrm{Hom}(X_{n}, G')) &= \int_{X \in \mathcal{S}} \#\mathrm{Hom}(X \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z}, G') \mu_{n}(X) \\ &= \sum_{B \in \mathcal{A}_{m}}\int_{\substack{X \in \mathcal{S}: \\ (X \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z}) \simeq B}}\#\mathrm{Hom}(B, G')\mu_{n}(X) \\ &= \sum_{B \in \mathcal{A}_{m}}\int_{\substack{X \in \mathcal{S}: \\ (X \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z}) \simeq B}}\mu_{n}(X) \#\mathrm{Hom}(B, G') \\ &= \sum_{B \in \mathcal{A}_{m}}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq B) \#\mathrm{Hom}(B, G'),\end{align*}$$ as desired.

By hypothesis, the limit $$\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, H))$$ exists for each subgroup $H$ of $G'$, so $$\lim_{n \rightarrow \infty}\sum_{B \in \mathcal{A}_{m}}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq B) \#\mathrm{Hom}(B, G') = \sum_{H \subset G'}\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, H))$$ exists as well. In particular, there exists a bound $D_{G'} > 0$ independent of $n$ such that $$\sum_{B \in \mathcal{A}_{m}}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq B) \#\mathrm{Hom}(B, G') \leq D_{G'}.$$ Thus, for any fixed $H \in \mathcal{A}_{m},$ we have $$\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Hom}(H, G') \leq D_{G'}.$$ This implies that $$\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Hom}(H, G) \leq D_{G'}\frac{\#\mathrm{Hom}(H, G)}{\#\mathrm{Hom}(H, G')},$$ so $$\sum_{H \in \mathcal{A}_{m}}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Hom}(H, G) \leq D_{G'}\sum_{H \in \mathcal{A}_{m}}\frac{\#\mathrm{Hom}(H, G)}{\#\mathrm{Hom}(H, G')},$$ and by Claim the right-hand side converges as $n \rightarrow \infty.$ Thus, by the Lebesgue dominated convergence theorem, the left-hand side also converges and $$\begin{align*}&\lim_{n \rightarrow \infty}\sum_{H \in \mathcal{A}_{m}}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Hom}(H, G) \\ &= \sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Hom}(H, G),\end{align*}$$ where we used our assumption that the limit $$\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H)$$ exists (which will be proven below). Since $$\#\mathrm{Hom}(H, G) = \sum_{B \subset G}\#\mathrm{Sur}(H, B)$$ where the sum is over subgroups $B$ of $G,$ subtracting the corresponding quantities for all subgroups of $G,$ we obtain $$\begin{align*}&\lim_{n \rightarrow \infty}\sum_{H \in \mathcal{A}_{m}}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Sur}(H, G) \\ &= \sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Sur}(H, G).\end{align*}$$ A similar computation as above gives $$\mathbb{E}(\#\mathrm{Sur}(X_{n}, G)) = \sum_{H \in \mathcal{A}_{m}}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Sur}(H, G),$$ so $$\sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Sur}(H, G) = \lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, G))$$ by the hypothesis. This finishes Step 1.

Step 2. We assume that $\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H)$ and $\mathrm{Prob}(Y_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H)$ both converge as $n \rightarrow \infty$ for every $H \in \mathcal{A}_{m}.$  and $$\begin{align*}&\sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Sur}(H, G) \\ &= \sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(Y_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Sur}(H, G)\end{align*}$$ for every $G \in \mathcal{A}_{m}.$ By Step 1, both sides are equal to $$\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, G)) = \lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(Y_{n}, G)).$$

Then we show that $$\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) = \lim_{n \rightarrow \infty}\mathrm{Prob}(Y_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H)$$

Using a similar argument as in Step 1, we have $$\begin{align*}&\sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Hom}(H, G) \\ &= \sum_{B \subset G}\sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Sur}(H, G) \\ &= \sum_{B \subset G} \lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, G)),\end{align*}$$ where $B$ ranges over subgroups of $G.$  We can also deduce a similar identity for $(Y_{n})_{n \geq 1}.$

We need two lemmas to finish Step 2, and this is where we need $$\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, G)) \leq |\wedge^{2}G|.$$

Lemma 1. Let $\lambda = (\lambda_{1} \geq \cdots \geq \lambda_{l})$ be a partition and $$G = \mathbb{Z}/p^{\lambda_{1}}\mathbb{Z} \times \cdots \times \mathbb{Z}/p^{\lambda_{l}}\mathbb{Z}.$$ Then $$\sum_{B \subset G}|\textstyle\bigwedge^{2}G| \leq F^{\lambda_{1}}p^{\sum_{i=1}^{\infty}\lambda'_{i}(\lambda'_{i}-1)/2},$$ where $B$ ranges over subgroups of $G$ and $$F = \frac{2}{1 - 2^{-1/8}} \prod_{i=1}^{\infty}\frac{1}{1 - 2^{-i}}.$$

Remark. For a later use, we note that the number of parts of $\lambda$ in Lemma 1 is at most $\lambda_{1}.$

For the next lemma, we need more notation. Fix any distinct primes $p_{1}, \dots, p_{s}$ and integers $e_{1}, \dots, e_{k} \geq 1.$ Let $M_{j}$ be the set of partitions with at most $e_{j}$ parts, and set $M_{(e_{1}, \dots, e_{k})} = M := M_{1} \times \cdots \times M_{k}.$ Given any $\mu \in M,$ we may write $$\mu = (\mu^{(1)}, \dots, \mu^{(k)}).$$ Since $\mu^{(j)} \in M_{j}$, we write $\mu^{(j)} = (\mu^{(j)}_{1} \geq \cdots \geq \mu^{(j)}_{e_{j}}),$ where $\mu^{(j)}_{i} \geq 0.$

Lemma 2. Assume the above notation. For each $\lambda \in M,$ suppose that we have $x_{\lambda}, y_{\lambda} \in \mathbb{R}_{\geq 0}$ such that $$\sum_{\mu \in M}x_{\mu}\prod_{j=1}^{k}p_{j}^{\sum_{i=1}^{\infty}\lambda_{i}^{(j)}\mu_{i}^{(j)}} = \sum_{\mu \in M}y_{\mu}\prod_{j=1}^{k}p_{j}^{\sum_{i=1}^{\infty}\lambda_{i}^{(j)}\mu_{i}^{(j)}} \leq \prod_{j=1}^{s}F^{e_{j}}p_{j}^{\sum_{i=1}^{\infty}\lambda_{i}^{(j)}(\lambda_{i}^{(j)}-1)/2},$$ for some constant $F > 0$ that does not depend on $\lambda.$ Then we have $x_{\lambda} = y_{\lambda}$ for all $\lambda \in M.$

Going back to Step 2, prime factorize $m = p_{1}^{e_{1}} \cdots p_{k}^{e_{k}}$ with $e_{j} \geq 1.$ Then there is a bijection between $M_{(e_{1}, \dots, e_{k})} = M$ and $\mathcal{A}_{m}$ by mapping each $$\lambda = (\lambda^{(1)}, \dots, \lambda^{(k)}) \in M$$ to a finite abelian group, whose $p_{j}$-part corresponds to the conjugate of the partition $\lambda^{(j)}$ for $1 \leq j \leq k.$ (This is where we need abelian groups in $\mathcal{A}_{m}$ to be finite.) What's important here is that if $G, H \in \mathcal{A}_{m}$ correspond to $\lambda, \mu \in M$ respectively under this bijection, then $$\prod_{j=1}^{k}p_{j}^{\sum_{i=1}^{\infty}\lambda_{i}^{(j)}\mu_{i}^{(j)}} = \#\mathrm{Hom}(H, G).$$ Now, if we let $$x_{\lambda} := \lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H)$$ and $$y_{\lambda} := \lim_{n \rightarrow \infty}\mathrm{Prob}(Y_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H),$$ then to finish Step 2 with the help of Lemma 2, we only need to show that $$\sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Hom}(H, G) \leq \prod_{j=1}^{s}F^{e_{j}}p_{j}^{\sum_{i=1}^{\infty}\lambda_{i}^{(j)}(\lambda_{i}^{(j)}-1)/2},$$ for some constant $F > 0$ that does not depend on $G \in \mathcal{A}_{m}.$ The left-hand side is $$\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Hom}(X_{n}, G)) = \sum_{B \subset G}\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, B)),$$ where $B$ varies over subgroups of $G,$ and we now use the assumption that $$\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, B)) \leq |\wedge^{2} B|,$$ which gives us $$\begin{align*}\sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Hom}(H, G) &= \sum_{B \subset G}\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, B)) \\ &\leq \sum_{B \subset G} |\wedge^{2}B| \\ &\leq \prod_{j=1}^{k} F^{e_{j}}p^{\sum_{i=1}^{\infty}\lambda'_{i}(\lambda'_{i}-1)/2},\end{align*}$$ where $$F = \frac{2}{1 - 2^{-1/8}} \prod_{i=1}^{\infty}\frac{1}{1 - 2^{-i}}.$$ We used Lemma 1 along with how direct sums work with wedge products. This finishes Step 2.

Remark. Step 2 is the most nontrivial step, and Lemma 1 and Lemma 2 are where interesting bounds are analyzed. Lemma 2 seems to be where the nontriviality lies. We will cover these two lemmas in a different post, and think about whether the bound given here can be improved.

 Step 3. We show that $\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H)$ converges as $n \rightarrow \infty.$

Since the sequence in question is bounded in $[0, 1],$ by the Bolzano--Weierstrass theorem, there must be a subsequence $X_{n_{1}}, X_{n_{2}}, X_{n_{3}}, \dots$ such that the limit $$L = \lim_{i \rightarrow \infty}\mathrm{Prob}(X_{n_{i}} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H)$$ exists for any $H \in \mathcal{A}_{m}.$

For contradiction, suppose that the sequence in question does not converge for some $H \in \mathcal{A}_{m}.$ This means that there exists $\epsilon > 0$ and a subsequence $X_{m_{1}}, X_{m_{2}}, X_{m_{3}}, \dots$ such that $$|\mathrm{Prob}(X_{m_{i}} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) - L| \geq \epsilon.$$ We may refine this subsequence by the Bolzano--Weirstrass theorem to assume that the limit $$\lim_{i \rightarrow \infty}\mathrm{Prob}(X_{m_{i}} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H)$$ exists. The contradiction comes from the fact that $$\lim_{i \rightarrow \infty}\mathrm{Prob}(X_{m_{i}} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \neq L,$$ because if we go through Step 1 and Step 2 with these two subsequences, Step 2 tells us that the two limits must be equal. This finishes the proof of Step 3.