2. Preliminaries on the rescaled moment method

Given a partition $\lambda=(\lambda_1,\dots,\lambda_r)$, let

\begin{equation*}G_\lambda=\bigoplus_{i=1}^r \mathbb{Z}/p^{\lambda_i}\mathbb{Z}\end{equation*}

be the abelian $p$-group corresponding to $\lambda$, and let $\lambda'$ be the conjugate partition of $\lambda$. We also define $|\lambda|=\sum_{i=1}^r \lambda_i$.

We define $\operatorname{MC}(G_{\lambda'})$ as the number of chains of subgroups $\{0\}\subsetneq H_1\subsetneq \cdots\subsetneq H_{|\lambda|}= G_{\lambda'}$ of length $|\lambda|$, that is, of maximal length.

Given a positive integer $d$, let

\begin{align*}\operatorname{Sig}_d&=\{(\lambda_1,\lambda_2,\dots,\lambda_d)\in \mathbb{Z}^d\,:\, \lambda_1\ge\lambda_2\cdots\ge \lambda_d\}\,\text{and }\\ \operatorname{Sig}_d^{\ge 0}&=\{(\lambda_1,\lambda_2,\dots,\lambda_d)\in \mathbb{Z}^d\,:\, \lambda_1\ge\lambda_2\cdots\ge \lambda_d\ge 0\}. \end{align*}

Note that $\operatorname{Sig}_d^{\ge 0}$ is just the set of partitions with at most $d$ parts.

For $\chi \gt 0$, let $\mathcal{L}_{d,p^{-1},\chi}$ be a $\operatorname{Sig}_d\,$-valued random variable such that

\begin{equation*}{\mathbb{E} p^{\left\langle \mathcal{L}_{d,p^{-1},\chi}\,,\,\lambda \right\rangle}}=\frac{((p-1)\chi)^{|\lambda|}}{|\lambda|!}\operatorname{MC}(G_{\lambda'})\qquad \text{for all }\lambda\in\operatorname{Sig}_d^{\ge 0}.\end{equation*}

It was proved by Nguyen and Van Peski that such a random variable exists and its distribution is uniquely determined [Reference Nguyen and Van Peski25].

The convergence to $\mathcal{L}_{d,p^{-1},\chi}$ can be obtained using the so-called rescaled moment method given in the next theorem, which is a new variant of the widely used moment method of Wood [Reference Wood33, Reference Wood34].

Theorem 4. (Nguyen and Van Peski [Reference Nguyen and Van Peski25, Theorem 1.2 and Proposition 5.3.])

Fix a prime  $p$ and a positive integer $d$. Let $(\Gamma_j)_{j\ge 1}$ be a sequence of random finitely-generated abelian $p$-groups and let $(n_j)_{j\ge 1}$ be a sequence of positive real numbers such that the following holds:

1. (a) The fractional part $\{-\log_p(n_j)\}$ converges to $\zeta$.

2. (b) For all $\lambda\in \operatorname{Sig}_d^{\ge 0}$, we have

   \begin{equation*}\lim_{j\to\infty} \mathbb{E}\ \frac{|\operatorname{Hom}(\Gamma_j,G_{\lambda'})|}{n_j^{|\lambda|}}=\frac{\chi_0^{|\lambda|}\operatorname{MC}(G_{\lambda'})}{|\lambda|!}.\end{equation*}

Then

\begin{equation*}\left(\operatorname{rank}(p^{i-1}\Gamma_j)-\lfloor\log_p(n_{j})+\zeta\rceil\right)_{i=1}^d\end{equation*}

converges to $\mathcal{L}_{d,p^{-1},\chi}$ in distribution as $j\to\infty$, where $\chi=\frac{\chi_0 p^{-\zeta}}{p-1}$.

The quantities $\mathbb{E}|\operatorname{Hom}(\Gamma_j,G_{\lambda'})|$ are often referred to as the moments of $\Gamma_j$. If $\Gamma_n$ is defined as in the Introduction, then the following well-known lemma gives a useful expression for the moments of $\Gamma_n$, for the proof, see, for example, [Reference Mészáros20].

Lemma 5. For any deterministic finite abelian $p$-group $G$, we have

\begin{equation*}\mathbb{E}|\operatorname{Hom}(\Gamma_n,G)|=\mathbb{E}|\operatorname{Hom}(\operatorname{cok}(L_n),G)|=\mathbb{E}|\{v\in G^n\,:\,L_nv=0\}|.\end{equation*}

Combining Theorem 4 and Lemma 5, we see that Theorem 1 follows once we prove the following theorem.

Theorem 6 For any finite abelian $p$-group $G$ of size $p^\ell$, we have

\begin{equation*}\lim_{n\to\infty} n^{-\ell} \mathbb{E}|\{v\in G^n\,:\,L_nv=0\}|=\frac{\chi_0^\ell \operatorname{MC}(G)}{\ell!}.\end{equation*}

The proof of Theorem 6 relies on the following partition of $G^n$. Given a sequence of integers $0=n_0\le n_1 \lt n_2 \lt \cdots \lt n_k \lt n_{k+1}=n$, and a chain of subgroups $\{0\}=G_0\subsetneq G_1\subsetneq\cdots\subsetneq G_k$ of $G$, let

\begin{multline*} \bar{V}^{n_1,\dots,n_{k+1}}_{G_1,\dots,G_{k}}=\{v\in G^n\,:\,\text{for all }0\le i\le k\,\text{and }n_i \lt j\le n_{i+1},\\ \text{the subgroup generated by the first}\ j\ \text{components of}\ v\ \text{is}\ G_i\}. \end{multline*}

It is easy to see that the sets $\bar{V}^{n_1,\dots,n_{k+1}}_{G_1,\dots,G_{k}}$ form a partition of $G^n$. We prove that if $\{0\}=G_0\subsetneq G_1\subsetneq\cdots\subsetneq G_\ell$ is maximal chain of subgroups of $G$, moreover, $n_{i+1}-n_i \gt 3\log^2 n$ for all $i=1,\dots,\ell$, then

\begin{equation*}\mathbb{E}|\{v\in \bar{V}^{n_1,\dots,n_{\ell+1}}_{G_1,\dots,G_{\ell}}\,:\,L_nv=0\}|=(1+o(1))\chi_0^\ell.\end{equation*}

The number of ways we can choose a chain $\{0\}=G_0\subsetneq G_1\subsetneq\cdots\subsetneq G_\ell$ and $0=n_0\le n_1 \lt n_2 \lt \cdots \lt n_\ell \lt n_{\ell+1}=n$ satisfying the conditions above is $(1+o(1))\frac{n^\ell \operatorname{MC}(G)}{\ell!}$. Therefore, the contribution of the vectors contained in the union of the corresponding sets $\bar{V}^{n_1,\dots,n_{\ell+1}}_{G_1,\dots,G_{\ell}}$ to the expectation $\mathbb{E}|\{v\in G^n\,:\,L_nv=0\}|$ is $(1+o(1))\frac{n^\ell \chi_0^\ell \operatorname{MC}(G)}{\ell!}$. Theorem 6 will follow once we prove that the contribution of all the other vectors is negligible.

3. Calculating the moments

This section is devoted to the proof of Theorem 6.

Let $H$ be a finite abelian $p$-group. Throughout this section, constants are allowed to depend on $H$ and $\xi$, but not on anything else.

For a $v\in H^m$ and a character $\varrho\in \widehat{H}=\operatorname{Hom}(H,\mathbb{C}^*)$, we define

\begin{equation*}\|v\|_\varrho=|\{1\le i\le m\,:\,v_i\notin \ker \varrho\}|.\end{equation*}

We also define

\begin{equation*}\tau(v)=\mathbb{P}\left(\sum_{i=1}^m \xi_i v_i=0\right),\end{equation*}

where $\xi_1,\xi_2,\dots,\xi_m$ are i.i.d. copies of $\xi$.

We use $1$ to denote the trivial character in $\widehat{H}$.

Lemma 7. There is a constant $c \gt 0$, such that for any $m$ and any $v\in H^m$, if $g\in H$ is a random variable independent from $\xi_1,\dots,\xi_m$, then

\begin{equation*}\left||H|\mathbb{P}\left(\sum_{i=1}^m \xi_i v_i=g\right)-1\right|\le \sum_{1\neq \varrho \in \widehat{H}} \exp\left(-c\|v\|_\varrho\right).\end{equation*}

In particular,

\begin{equation*}\Big||H|\tau(v)-1\Big|\le \sum_{1\neq \varrho \in \widehat{H}} \exp\left(-c\|v\|_\varrho\right).\end{equation*}

Proof. Combining the inversion formula for the discrete Fourier transform and the fact that $g,\xi_1,\xi_2,\dots,\xi_m$ are independent, we obtain that

(2)\begin{equation} |H|\mathbb{P}\left(\sum_{i=1}^m \xi_i v_i=g\right)=\sum_{\varrho\in \widehat{H}}\mathbb{E}\varrho\left(\sum_{i=1}^m \xi_i v_i-g\right)=1+\sum_{1\neq\varrho\in \widehat{H}}\left(\mathbb{E}\varrho(-g)\right)\prod_{i=1}^m\mathbb{E}\varrho\left( \xi_i v_i\right). \end{equation}

By [Reference Wood32, Lemma 2.2.] and (1), we have a $c \gt 0$ such that

\begin{equation*}\left|\mathbb{E}\varrho\left( \xi_i v_i\right)\right|\le \exp(-c)\end{equation*}

whenever $v_i\notin\ker \varrho$, and in all other cases $\mathbb{E}\varrho\left( \xi_i v_i\right)=1$. Combining this with (2), the statement follows.

Let $w\in H^h$. We assume that the components of $w$ generate $H$.

Let $X=(X_1,X_2,\dots,)\in H^{\mathbb{N}}$ be a random vector such that $X_i=w_i$ for all $1\le i\le h$ and $X_{h+1},X_{h+2},X_{h+3},\dots$ are independent uniform random elements of $H$.

For a vector $v$ in $H^m$ or $H^{\mathbb{N}}$, let $v_{\le i}$ be the prefix of $v$ of length $i$.

We define

\begin{equation*}Z_i=\prod_{j=h+1}^{h+i} |H| \tau(X_{\le j})\quad\text{and }\quad Z_\infty=\prod_{j=h+1}^{\infty} |H| \tau(X_{\le j}).\end{equation*}

Proof. Let us choose $\varepsilon \gt 0$ small enough. Let $1\neq \varrho\in \widehat{H}$. We define

\begin{equation*}T_\varrho=\sup\left(\left\{i\ge 1\,:\,\left\|X_{\le h+i}\right\|_\varrho\le \varepsilon i\right\}\cup \{0\}\right).\end{equation*}

Since $\mathbb{P}(X_{h+j}\notin \ker \varrho)\ge \frac{p-1}p$, it follows from the law of large numbers that $T_\varrho$ is almost surely finite provided that $\varepsilon$ is small enough.

We define

\begin{equation*}H_i=\bigcap_{\substack{1\neq \varrho\in \widehat{H}\\T_\varrho\ge i}} \ker\varrho,\end{equation*}

where the intersection over the empty set is defined to be $H$.

By Lemma 7, we have

(3)\begin{align} |H|&\tau(X_{\le h+i})\cr &\le 1+\sum_{1\neq \varrho \in \widehat{H}} \exp\left(-c\|X_{\le h+i}\|_\varrho\right)\cr \nonumber\\ &\le \left(1+\sum_{\substack{1\neq \varrho \in \widehat{H}\ H_i\subseteq\ker \varrho}} \exp\left(-c\|X_{\le h+i}\|_\varrho\right)\right)\nonumber\\ & \quad \times \left(1+\sum_{\substack{1\neq \varrho \in \widehat{H}\ H_i\not\subseteq\ker \varrho}} \exp\left(-c\|X_{\le h+i}\|_\varrho\right)\right). \end{align}

Note that if $H_i\not\subseteq\ker \varrho$, then $T_\varrho \lt i$, so $\|X_{\le h+i}\|_\varrho \gt \varepsilon i$. Therefore,

(4)\begin{equation} 1+\sum_{\substack{1\le \varrho \in \widehat{H}\\ H_i\not\subseteq\ker \varrho}} \exp\left(-c\|X_{\le h+i}\|_\varrho\right)\le 1+|H|\exp(-c\varepsilon i)\le \exp\left(|H|\exp(-c\varepsilon i)\right). \end{equation}

On the event $H_i=H$, we trivially have

(5)\begin{equation}1+\sum_{\substack{1\neq \varrho \in \widehat{H}\\ H_i\subseteq\ker \varrho}} \exp\left(-c\|X_{\le h+i}\|_\varrho\right)=1.\end{equation}

Now assume that $H_i\neq H$. In this case, there is a one-to-one correspondence between the sets $\{\varrho \in \widehat{H}\,:\,H_i\subseteq\ker \varrho\}$ and $\widehat{H/H_i}$. Thus,

(6)\begin{equation}|\{\varrho \in \widehat{H}\,:\,H_i\subseteq\ker \varrho\}|=|H|/|H_i|.\end{equation}

Since the components of $w$ generate $H$, we have a $j\in\{1,2,\dots,h\}$ such that $w_j\notin H_i$. Then, we can find a $\varrho\in \widehat{H}$ such that $H_i\subseteq \ker \varrho$ and $w_j\notin \ker \varrho$. For such a $\varrho$, we have $\|X_{\le h+i}\|_\varrho\ge 1$. Combining this with (6), we obtain that

(7)\begin{equation}1+\sum_{\substack{1\neq \varrho \in \widehat{H}\\ H_i\subseteq\ker \varrho}} \exp\left(-c\|X_{\le h+i}\|_\varrho\right)\le \frac{|H|}{|H_i|}-1+\exp(-c). \end{equation}

Combining (3), (4), (5) and (7), we obtain that

(8)\begin{equation}|H|\tau(X_{\le h+i})\le \left(\frac{|H|}{|H_i|}-\mathbb{1}(H_i\neq H)(1-\exp(-c))\right) \exp\left(|H|\exp(-c\varepsilon i)\right).\end{equation}

Introducing the notation $T_{\max}=\max_{1\neq \varrho\in \widehat{H}} T_\varrho$, it follows from (8) that

\begin{equation*}\sup_{i\ge 1}Z_i\le K_0 \prod_{i=1}^{T_{\max}} \left(\frac{|H|}{|H_i|}-1+\exp(-c)\right),\end{equation*}

where $K_0$ is a constant defined as

\begin{equation*}K_0=\prod_{i=1}^\infty \exp\left(|H|\exp(-c\varepsilon i)\right)=\exp\left(\frac{|H|\exp(-c\varepsilon )}{1-\exp(-c\varepsilon)}\right).\end{equation*}

If we choose $\kappa_0$ small enough then there is a constant $0 \lt \gamma \lt 1$ such that for all $1\le \kappa\le \kappa_0$ and for all subgroup $H'\neq H$ of $H$, we have

\begin{equation*}\left(\frac{|H|}{|H'|}-1+\exp(-c)\right)^\kappa\le \gamma^2 \frac{|H|}{|H'|}. \end{equation*}

Thus, for all $1\le \kappa\le \kappa_0$, we have

(9)\begin{equation}\left(\sup_{i\ge 1}Z_i\right)^\kappa\le K_0^{\kappa_0} \gamma^{2T_{\max}} \prod_{i=1}^{T_{\max}} \frac{|H|}{|H_i|}. \end{equation}

Note that the choice of $\gamma$ only depends on $H$ and $\xi$, but not on $\varepsilon$.

Let $t$ be a positive integer and let $C_1\subseteq C_2\subseteq\cdots\subseteq C_t\neq H$ be a chain of subgroups of $H$.

Lemma 9. Let $\mathcal{A}$ be the event that $T_{\max}=t$ and $H_i=C_i$ for all $1\le i\le t$. Provided that $\varepsilon$ is small enough, we have

\begin{equation*}\mathbb{P}(\mathcal{A})\le \gamma^{-t}\prod_{i=1}^t \frac{|C_i|}{|H|}.\end{equation*}

Moreover, for all $1\le \kappa\le \kappa_0$, we have

\begin{equation*}\mathbb{E}\mathbb{1}(\mathcal{A}) \left(\sup_{i\ge 1} Z_i\right)^\kappa\le \gamma^{t}K_0^{\kappa_0}.\end{equation*}

Proof. For $1\neq \varrho\in \widehat{H}$, let

\begin{equation*}B_{\varrho}=\{1\le j\le T_\varrho\,:\,X_{h+j}\notin \ker \varrho \}.\end{equation*}

On the event $\mathcal{A}$, if $X_{h+j}\notin C_j$ for some $1\le j\le t$, then there is a $1\neq \varrho\in \widehat{H}$ such that $T_\varrho\ge j$ and $X_{h+j}\notin \ker \varrho$. In particular, $j\in B_\varrho$.

Therefore, we just proved that on the event $\mathcal{A}$, we have

\begin{equation*}X_{h+j}\in C_j\text{for all }j\in \{1,2,\dots,t\}\setminus \bigcup_{1\neq\varrho \in \widehat{H}} B_\varrho.\end{equation*}

Observe that on the event $\mathcal{A}$, we have

\begin{equation*}\left|\bigcup_{1\neq\varrho \in \widehat{H}} B_\varrho\right|\le \sum_{1\neq\varrho \in \widehat{H}} \varepsilon T_\varrho\le \varepsilon|H|t. \end{equation*}

Note that assuming that $\varepsilon$ small enough the right hand side is less than $t$.

Below, $\sum_B$ will stand for a summation over all $\lfloor \varepsilon|H|t \rfloor$ element subsets $B$ of $\{1,2,\dots,t\}$.

It follows from the discussion above that

\begin{align*} \mathbb{P}(T_{\max}&=t\,\,\,\text{and }H_i=C_i\,\,\,\text{for all }1\le i\le t)\\ &\le \sum_B \mathbb{P}(X_{h+j}\in C_j\,\,\,\text{for all }j\in \{1,\dots,t\}\setminus B)\\ & = \sum_B \prod_{j\in \{1,\dots,t\}\setminus B} \frac{|C_j|}{|H|}\\ & \le {{t}\choose{\lfloor\varepsilon|H|t\rfloor}} |H|^ {\varepsilon|H|t} \prod_{j=1}^t \frac{|C_j|}{|H|}. \end{align*}

By choosing $\varepsilon$ small enough we can achieve that ${{t}\choose{\lfloor\varepsilon|H|t\rfloor}} |H|^ {\varepsilon|H|t}\le \gamma^{-t}$. Thus, the first statement follows. The second statement follows by combining the first one with (9).

It is easy to see that there is a constant $c_2$ such that the number of chains of subgroups $C_1\subseteq C_2\subseteq\cdots\subseteq C_t$ is at most $c_2(t+1)^{\log_p|H|}$. Combining this estimate with Lemma 9, we get that

\begin{equation*}\mathbb{E}\mathbb{1}(T_{\max}=t)\left(\sup_{i\ge 1} Z_i\right)^\kappa\le c_2 (t+1)^{\log_p|H|}\gamma^{t}K_0^{\kappa_0}.\end{equation*}

Note that this estimate also true for $t=0$. Thus,

\begin{equation*}\mathbb{E}\left(\sup_{i\ge 1} Z_i\right)^\kappa=\sum_{t=0}^\infty\mathbb{E}\mathbb{1}(T_{\max}=t)\left(\sup_{i\ge 1} Z_i\right)^\kappa\le\sum_{t=0}^\infty c_2 (t+1)^{\log_p|H|}\gamma^{t}K_0^{\kappa_0} \lt \infty.\end{equation*}

Therefore, part (a) of the lemma follows.

Let $i_0$ be the smallest $i$ such that $1-|H|\exp(-c\varepsilon i) \gt 0$.

Assuming that $\ell \gt i \gt \max(T_{\max},i_0)$, by Lemma 7, we have

\begin{equation*}\prod_{j=i+1}^\ell(1-|H|\exp(-c\varepsilon j))\le \frac{Z_\ell}{Z_i}\le \prod_{j=i+1}^\ell (1+|H|\exp(-c\varepsilon j)).\end{equation*}

Thus, it follows easily that with probability $1$, $Z_\infty$ is defined and finite. Moreover, $Z_i$ converges to $Z_\infty$ almost surely. Thus, part (b) follows.

Finally, by parts (a) and (b), we can apply the dominated convergence theorem to obtain that

\begin{equation*}\lim_{i\to \infty}\mathbb{E}Z_i=\mathbb{E}Z_\infty.\end{equation*}

Let $G$ be a finite abelian $p$-group. Let $0\le n_1 \lt n_2 \lt \cdots \lt n_{k+1}$ be a sequence of integers, and let $g_1,g_2,\dots,g_{k}\in G$. We define $G_0$ as $\{0\}$ and $G_i$ as the subgroup of $G$ generated by $g_1,g_2,\dots,g_i$. Let $V^{n_1,\dots,n_{k+1}}_{g_1,\dots,g_{k}}$ be the set of vectors $v\in G^{n_{k+1}}$ satisfying

• • $v_{n_i+1}=g_i$ for all $i=1,2,\dots,k$;

• • $v_j\in G_i$ for all $i=0,1,\dots,k$ and $n_{i}+1 \lt j\le n_{i+1}$.

Here we defined $n_0$ to be $-1$.

Note that

\begin{equation*}\left|V^{n_1,\dots,n_{k+1}}_{g_1,\dots,g_{k}}\right|=\prod_{i=0}^{k}|G_i|^{n_{i+1}-n_i-1}.\end{equation*}

Let $X=(X_1,\dots,X_{n_{k+1}})$ be a uniform random element of $V^{n_1,\dots,n_{k+1}}_{g_1,\dots,g_{k}}$. Let

\begin{equation*}R=\left(\prod_{i=0}^{k}|G_i|^{n_{i+1}-n_i-1}\right)\prod_{i=1}^{n_{k+1}}\tau(X_{\le i}).\end{equation*}

Note that for any $v\in G^n$, we have

(10)\begin{equation}\mathbb{P}(L_nv=0)=\prod_{i=1}^n \tau(v_{\le i}).\end{equation}

Therefore,

(11)\begin{equation}\mathbb{E}R=\mathbb{E}\left|\{v\in V^{n_1,\dots,n_{k+1}}_{g_1,\dots,g_{k}}\,:\,L_{n_{k+1}}v=0\}\right|.\end{equation}

Let

\begin{equation*}M=\max_{H\le G} K_{H,\xi}\quad\text{and }\quad\kappa_1= \min_{H\le G} \kappa_{0,H,\xi},\end{equation*}

where the constants $K_{H,\xi}$ and $\kappa_{0,H,\xi}$ are provided by Lemma 8.

Lemma 10. For any $1\le\kappa\le \kappa_1$, we have

\begin{equation*}\mathbb{E}R^{\kappa}\le M^{k}.\end{equation*}

Proof. We prove by induction on $k$. For $k=0$, the statement is trivial. Assume that $k \gt 0$. Let

\begin{equation*}R_0=\prod_{j=n_{k}+2}^{n_{k+1}} |G_{k}|\tau(X_{\le j}).\end{equation*}

Let $v\in V^{n_1,\dots,n_{k}}_{g_1,\dots,g_{k-1}}$. We set $h=n_{k}+1$, $H=G_{k}$. Let $w\in H^h$ be defined by appending $v$ with $g_{k}$. Let $Z_n$ be defined as in Lemma 8.

Note that $R_0$ conditioned on the event that $X_{\le n_{k}}=v$ has the same distribution as $Z_{n_{k+1}-n_{k}-1}$.

Then

\begin{align*}\mathbb{E}[R^\kappa\,|\,X_{\le n_{k}}=v]&=\left(\left(\prod_{i=0}^{k-1}|G_i|^{n_{i+1}-n_i-1}\right)\prod_{i=1}^{n_{k}+1}\tau(v_{\le i})\right)^{\kappa} \mathbb{E}[R_0^{\kappa}\,|\,X_{\le n_{k}}=v]\\ &\le \left(\left(\prod_{i=0}^{k-1}|G_i|^{n_{i+1}-n_i-1}\right)\prod_{i=1}^{n_{k}}\tau(v_{\le i})\right)^{\kappa} \mathbb{E}Z_{n_{k+1}-n_{k}-1}^\kappa\\ & \le M\left(\left(\prod_{i=0}^{k-1}|G_i|^{n_{i+1}-n_i-1}\right)\prod_{i=1}^{n_{k}}\tau(v_{\le i})\right)^{\kappa},\end{align*}

where the last step we used Lemma 8.

Combining this with the induction hypothesis, we obtain that

\begin{equation*}\mathbb{E}R^{\kappa}\le M \mathbb{E}\left(\left(\prod_{i=0}^{k-1}|G_i|^{n_{i+1}-n_i-1}\right)\prod_{i=1}^{n_{k}}\tau(X_{\le i})\right)^{\kappa}\le M^k.\end{equation*}

Let

\begin{equation*}A_n=\left\{v\in \mathbb{F}_p^n\,:\,v_1\neq 0, |\{1\le i \le n\,: v_i\neq 0\}|\ge \frac{n}3\right\}.\end{equation*}

The following lemma is clearly stronger than Lemma 2.

Proof. Part (a): Let $0\neq g\in \mathbb{F}_p$. Combining equation (10) and Lemma 8 with the choice $H=\mathbb{F}_p$, $h=1$, $w_1=g$, we obtain that,

\begin{equation*}\lim_{n\to\infty} \mathbb{E}|\{v\in\mathbb{F}_p^n\,:\,v_1= g, L_nv=0\}|=\lim_{n\to\infty} \tau(w)\mathbb{E}Z_{n-1}\end{equation*}

exists and finite. Summing this over all the possible choices of $0\neq g\in \mathbb{F}_p$, we get the statement.

Part (b): We already seen that $\chi_0 \lt \infty$. To show that $\chi_0 \gt 0$, let $0\neq g\in \mathbb{F}_p$. We will rely on the proof of Lemma 8 with the choice $H=\mathbb{F}_p$, $h=1$, $w_1=g$. Let $X$ be defined as in that lemma. Let $q$ be the probability that $\xi$ is divisible by $p$. By (1), we have $q \gt 0$. Note that

\begin{equation*}\tau(X_{\le i})\ge q^i.\end{equation*}

Choose a positive integer $t$ such that $1-p\exp(-c\varepsilon t) \gt 0$ and $\mathbb{P}(T_{\max}\le t) \gt 0$. Then following the argument of the proof of Lemma 8, we obtain that

\begin{equation*}\mathbb{E} \mathbb{1}(T_{\max}\le t) Z_\infty\quad{\ge}\quad\mathbb{P}(T_{\max}\le t) \left(\prod_{i=1}^{t}pq^{i+1}\right) \prod_{i=t+1}^\infty \left(1-p\exp(-c\varepsilon i)\right) \gt 0.\end{equation*}

Since by Lemma 8 and (10), we have

\begin{equation*}\lim_{n\to\infty}\mathbb{E}|\{v\in \mathbb{F}_p^n\,:\, v_1=g,L_nv=0\}|=q\mathbb{E}Z_\infty,\end{equation*}

part (b) follows.

Part (c): let $0\neq g\in \mathbb{F}_p$. We apply Lemma 10 with the choice $G=\mathbb{F}_p$, $k=1$, $g_1=g$, $n_1=0$ and $n_2=n$. Let $X$ be defined as in that lemma. Then combining Lemma 10 with (10), we have

\begin{equation*}\mathbb{E} \left(p^{n-1} \mathbb{P}(L_nX=0)\right)^{\kappa_1}\le M.\end{equation*}

It follows from the law of large numbers that

\begin{equation*}\lim_{n\to \infty}\mathbb{P}(X\notin A_n)=0.\end{equation*}

Choose $\kappa_2$ such that $\frac{1}{\kappa_1}+\frac{1}{\kappa_2}=1$. By Hölder’s inequality,

\begin{align*}\lim_{n\to\infty}&\mathbb{E} |\{v\in\mathbb{F}_p^n\setminus A_n, v_1=g,L_nv=0\}|\\ &=\lim_{n\to\infty}\mathbb{E} \mathbb{1}(X\notin A_n)p^{n-1} \mathbb{P}(L_nX=0)&\\ &\le \lim_{n\to\infty}\mathbb{P}(X\notin A_n)^{1/\kappa_2}\left(\mathbb{E} \left(p^{n-1} \mathbb{P}(L_nX=0)\right)^{\kappa_1}\right)^{1/\kappa_1}\\ &\le \lim_{n\to\infty}\mathbb{P}(X\notin A_n)^{1/\kappa_2} M^{1/\kappa_1}\\ &=0.\end{align*}

Summing these over $0\neq g\in \mathbb{F}_p$ and using part (a), the statement follows.

Let us consider $n_1,n_2,\dots,n_{k+1},g_1,g_2,\dots, g_{k}$ and $G_0,G_1,\dots,G_{k}$ as before. We say that $(n_1,n_2,\dots,n_{k+1})$ and $(g_1,g_2,\dots, g_{k})$ form an $(n,k)$-skeleton, if $n_{k+1}=n$ and $G_0, G_1,\dots, G_{k}$ are pairwise distinct. Moreover, by an $n$-skeleton we mean an $(n,k)$-skeleton for some $k$. The proof of the following lemma is straightforward.

We call the $n$-skeleton provided by the previous lemma the skeleton of $v$, and we use $k(v), n_1(v),n _2(v),\dots$ to denote the $k,n_1,n_2,\dots$ above, respectively.

Let $\ell=\log_p|G|$. Note that $k(v)\le \ell$.

Lemma 13. We have

\begin{equation*}\lim_{n\to\infty}n^{-\ell}\mathbb{E}|\{v\in G^n\,:\,k(v) \lt \ell,\, L_nv=0\}|=0.\end{equation*}

Proof. The number of $(n,k)$-skeletons is at most $(n|G|)^k$. Combining this with Lemma 10 and (11), we see that for any $k$, we have

\begin{equation*}\mathbb{E}|\{v\in G^n\,:\,k(v)=k,\, L_nv=0\}|\le (n|G|M)^k.\end{equation*}

Thus, the statement follows easily.

From now on we mostly restrict our attention to $(n,\ell)$-skeletons. Note that for an $(n,\ell)$-skeleton $(n_1,\dots,n_{\ell+1}),(g_1,\dots, g_{\ell})$, the chain $G_0\subsetneq G_1\subsetneq\cdots\subsetneq G_\ell$ is a maximal chain of subgroups in $G$. Thus, $G_\ell=G$ and $G_{i}/G_{i-1}\cong \mathbb{F}_p$ for all $i=1,2,\dots,\ell$.

Let $(n_1,\dots,n_{\ell+1}),(g_1,\dots, g_{\ell})$ be an $(n,\ell)$-skeleton. We say that this skeleton is well separated if

(12)\begin{equation}n_{i+1}-n_{i} \gt 3\log^2 n\quad\text{for all }i=1,\dots,\ell.\end{equation}

Lemma 14. We have

\begin{multline*} \lim_{n\to\infty}n^{-\ell}\mathbb{E}|\{v\in G^n\,:\,k(v)=\ell,\, L_nv=0,\text{the skeleton of }v\text{is not well separated}\}|\\=0. \end{multline*}

Let $(n_1,\dots,n_{\ell+1}),(g_1,\dots, g_{\ell})$ be a well separated skeleton. We partition $V^{n_1,\dots,n_{\ell+1}}_{g_1,\dots, g_{\ell}}$ into two subsets $U^{n_1,\dots,n_{\ell+1}}_{g_1,\dots, g_{\ell}}$ and $W^{n_1,\dots,n_{\ell+1}}_{g_1,\dots, g_{\ell}}$ such that $U^{n_1,\dots,n_{\ell+1}}_{g_1,\dots, g_{\ell}}$ consists of those elements $v$ of $V^{n_1,\dots,n_{\ell+1}}_{g_1,\dots, g_{\ell}}$ which satisfy that

\begin{equation*}|\{n_{i}+1\le j\le n_{i+1}\,:\,v_j\notin G_{i-1}\}|\ge\frac{n_{i+1}-n_i}3\quad \text{for all }i=1,2,\dots,\ell.\end{equation*}

In the lemma below $\sum_{ws(n)}$ denotes a sum over all well separated $(n,\ell)$-skeletons.

Lemma 15. We have

\begin{equation*}\lim_{n\to\infty}n^{-\ell}\sum_{ws(n)}\mathbb{E}\left|\left\{v\in W^{n_1,\dots,n_{\ell+1}}_{g_1,\dots, g_{\ell}}\,:\, L_nv=0\right\}\right|=0.\end{equation*}

Proof. Let $(n_1,\dots,n_{\ell+1}),(g_1,\dots, g_{\ell})$ be a well separated $(n,\ell)$-skeleton, and let $X$ be a uniform random element of $V^{n_1,\dots,n_{\ell+1}}_{g_1,\dots, g_{\ell}}$, and define $R$ as in Lemma 10. Let $\varepsilon \gt 0$. It follows from the law of large numbers that provided that $n$ is large enough, we have

\begin{equation*}\mathbb{P}\left(X\in W^{n_1,\dots,n_{\ell+1}}_{g_1,\dots, g_{\ell}}\right) \lt \varepsilon\end{equation*}

for all choices of well separated $(n,\ell)$-skeletons.

Choose $\kappa_2$ such that $\frac{1}{\kappa_1}+\frac{1}{\kappa_2}=1$. Combining (10) with Hölder’s inequality and Lemma 10, we obtain that

\begin{align*} \mathbb{E}\left|\left\{v\in W^{n_1,\dots,n_{\ell+1}}_{g_1,\dots, g_{\ell}}\,:\, L_nv=0\right\}\right|&=\mathbb{E} \mathbb{1}\left(X\in W^{n_1,\dots,n_{\ell+1}}_{g_1,\dots, g_{\ell}}\right) R\\ & \le \left(\mathbb{E} R^{\kappa_1}\right)^{1/\kappa_1} \left(\mathbb{P}\left(X\in W^{n_1,\dots,n_{\ell+1}}_{g_1,\dots, g_{\ell}}\right)\right)^{1/\kappa_2}\\ &\le M^{\ell/\kappa_1} \varepsilon^{1/\kappa_2}. \end{align*}

Since the number of $(n,\ell)$-skeletons is at most $(|G|n)^\ell$, and $\varepsilon$ can be arbitrary the statement follows.

Let $(n_1,\dots,n_{\ell+1}),(g_1,\dots, g_{\ell})$ be a well separated $(n,\ell)$-skeleton, and let $v\in V^{n_1,\dots,n_{\ell+1}}_{g_1,\dots, g_{\ell}}$. For $i=1,\dots,\ell$, we define

\begin{equation*}v^{(i)}=(v_{n_i+1}+G_{i-1},\,v_{n_i+2}+G_{i-1},\dots,\,v_{n_{i+1}}+G_{i-1})\in \left(G_i/G_{i-1}\right)^{n_{i+1}-n_i}.\end{equation*}

Lemma 16. Assume that $n$ is large enough. Let $(n_1,\dots,n_{\ell+1}),(g_1,\dots, g_{\ell})$ be a well separated $(n,\ell)$-skeleton and let $v\in U^{n_1,\dots,n_{\ell+1}}_{g_1,\dots, g_{\ell}}$.

Then for $i=2,3,\dots, \ell$ and $1\le j\le n_{i+1}-n_{i}$, we have

\begin{align*} &(1-(p-1)\exp(-c\log^2 n))^{i-1}\frac{\tau(v^{(i)}_{\le j})}{p^{i-1}}\le\tau(v_{\le n_{i}+j})\\ &\le (1+(p-1)\exp(-c\log^2 n))^{i-1} \frac{\tau(v^{(i)}_{\le j})}{p^{i-1}} .\end{align*}

Proof. Obviously $\sum_{k=1}^{n_i+j} \xi_k v_k=\sum_{k=n_1+1}^{n_i+j} \xi_k v_k$.

Note that $\sum_{k=n_1+1}^{n_i} \xi_k v_k\in G_{i-1}$. Thus, on the event that $\sum_{k=n_1+1}^{n_i+j} \xi_k v_k=0$, we must have $\sum_{k=n_i+1}^{n_i+j} \xi_k v_k\in G_{i-1}$. Therefore,

\begin{multline*}\mathbb{P}\left(\sum_{k=n_1+1}^{n_i+j} \xi_k v_k=0\right)\\=\mathbb{P}\left(\sum_{k=n_i+1}^{n_i+j} \xi_k v_k\in G_{i-1}\right)\mathbb{P}\left(\sum_{k=n_1+1}^{n_i+j} \xi_k v_k=0\,\Big|\,\sum_{k=n_i+1}^{n_i+j} \xi_k v_k\in G_{i-1}\right).\end{multline*}

Continuing in a similar manner

\begin{align*} \mathbb{P}&\left(\sum_{k=n_1+1}^{n_i+j} \xi_k v_k=0\right)\\ &=\mathbb{P}\left(\sum_{k=n_i+1}^{n_i+j} \xi_k v_k\in G_{i-1}\right) \prod_{h=1}^{i-1} \mathbb{P}\left(\sum_{k=n_h+1}^{n_i+j} \xi_k v_k\in G_{h-1}\,\Big|\,\sum_{k=n_{h+1}+1}^{n_{i}+j} \xi_k v_k\in G_{h}\right) \\ &=\mathbb{P}\left(\sum_{k=n_i+1}^{n_i+j} \xi_k (v_k+G_{i-1})=0+G_{i-1}\right) \\ &\cdot\prod_{h=1}^{i-1} \mathbb{P}\left(\sum_{k=n_h+1}^{n_{h+1}} \xi_k (v_k+G_{h-1})=-\sum_{k=n_{h+1}+1}^{n_{i}+j} \xi_k (v_k+G_{h-1})\,\Big|\,\sum_{k=n_{h+1}+1}^{n_{i}+j} \xi_k v_k\in G_{h}\right). \end{align*}

Here

\begin{equation*}\mathbb{P}\left(\sum_{k=n_i+1}^{n_i+j} \xi_k (v_k+G_{i-1})=0+G_{i-1}\right)=\tau(v^{(i)}_{\le j}).\end{equation*}

Let $h\in\{1,2,\dots,i-1\}$. Since $(n_1,\dots,n_{\ell+1}),(g_1,\dots, g_{\ell})$ is a well separated $(n,\ell)$-skeleton and $v\in U^{n_1,\dots,n_{\ell+1}}_{g_1,\dots, g_{\ell}}$, we see that

\begin{equation*}\|v^{(h)}\|_\varrho \gt \log^2 n\quad\text{for all }1\neq \varrho\in \widehat{G_h/G_{h-1}}.\end{equation*}

Thus, by Lemma 7, we obtain that

\begin{align*} & \frac{1}p\left(1-(p-1)\exp(-c\log^2 n)\right)\\ &\le \mathbb{P}\left(\sum_{k=n_h+1}^{n_{h+1}} \xi_k (v_k+G_{h-1})=-\sum_{k=n_{h+1}+1}^{n_{i}+j} \xi_k (v_k+G_{h-1})\,\Big|\,\sum_{k=n_{h+1}+1}^{n_{i}+j} \xi_k v_k\in G_{h}\right)\\ &\le \frac{1}p\left(1+(p-1)\exp(-c\log^2 n)\right). \end{align*}

Therefore, the statement follows.

Lemma 17. We have

\begin{equation*}\lim_{n\to\infty} \max \left|\frac{\prod_{i=1}^\ell p^{-(i-1)(n_{i+1}(v)-n_i(v))}\mathbb{P}(L_{n_{i+1}(v)-n_i(v)} v^{(i)}=0)}{\mathbb{P}(L_nv=0)}-1\right|=0,\end{equation*}

where the max is over all $v$ such that $v\in U^{n_1,\dots,n_{\ell+1}}_{g_1,\dots, g_{\ell}}$ for some well separated $(n,\ell)$-skeleton $(n_1,\dots,n_{\ell+1}),(g_1,\dots, g_{\ell})$.

Proof. Let $(n_1,\dots,n_{\ell+1}),(g_1,\dots, g_{\ell})$ be a well separated $(n,\ell)$-skeleton, and let $v\in U^{n_1,\dots,n_{\ell+1}}_{g_1,\dots, g_{\ell}}$. By Lemma 16, we have

\begin{align*}&\mathbb{P}(L_n v=0)\\ &=\prod_{i=1}^\ell\prod_{j=1}^{n_{i+1}-n_i}\tau(v_{\le n_i+j})\\ &\le \prod_{i=1}^\ell p^{-(i-1)(n_{i+1}-n_i)}\mathbb{P}(L_{n_{i+1}-n_i}v^{(i)}=0)(1+(p-1)\exp(-c\log^2 n))^{(i-1)(n_{i+1}-n_i)}\\ &\le (1+(p-1)\exp(-c\log^2 n))^{\ell n}\prod_{i=1}^\ell p^{-(i-1)(n_{i+1}-n_i)}\mathbb{P}(L_{n_{i+1}-n_i}v^{(i)}=0). \end{align*}

Note that here $\lim_{n\to\infty} (1+(p-1)\exp(-c\log^2 n))^{\ell n}=1.$ We can obtain a matching lower bound in a similar manner.

Consider a sequence $0\le n_1 \lt \dots \lt n_{\ell+1}$ satisfying (12). For a maximal chain of subgroups $\{0\}=G_0\subsetneq G_1\subsetneq \cdots\subsetneq G_\ell=G$, let $\bar{U}^{n_1,\dots,n_{\ell+1}}_{G_1,\dots,G_\ell}=\bigcup U^{n_1,\dots,n_{\ell+1}}_{g_1,\dots,g_\ell}$ where the union is over all the choices of $g_i=G_i\setminus G_{i-1}$.

Lemma 18. We have

\begin{equation*}\lim_{n\to\infty}\max \left|\mathbb{E}\left|\left\{v\in \bar{U}^{n_1,\dots,n_{\ell+1}}_{G_1,\dots,G_\ell}\,:\,L_nv=0\right\}\right|-\chi_0^\ell\right|=0,\end{equation*}

where the maximum is over all the choices $0\le n_1 \lt \dots \lt n_{\ell+1}=n$ satisfying (12) and maximal chains of subgroups $\{0\}=G_0\subsetneq G_1\subsetneq \cdots\subsetneq G_\ell=G$.

Proof. Consider a sequence $0\le n_1 \lt \dots \lt n_{\ell+1}=n$ satisfying (12) and a maximal chain of subgroups $\{0\}=G_0\subsetneq G_1\subsetneq \cdots\subsetneq G_\ell=G$. Let us consider the map

\begin{equation*}v\mapsto (v^{(1)},\dots,v^{(\ell)})\end{equation*}

restricted to $\bar{U}^{n_1,\dots,n_{\ell+1}}_{G_1,\dots,G_\ell}$. After fixing an isomorphism between $\mathbb{F}_p$ and $G_i/G_{i-1}$, the image of this map is $A_{n_2-n_1}\times A_{n_3-n_2}\times \cdots\times A_{n_{\ell+1}-n_\ell}$. Moreover, the preimage of each element of this set has size $\prod_{i=1}^\ell p^{(i-1)(n_{i+1}-n_i)}$. Combining this observation with Lemma 17, we have

\begin{equation*} \lim_{n\to\infty}\max \left|\frac{\mathbb{E}|\{v\in \bar{U}^{n_1,\dots,n_{\ell+1}}_{G_1,\dots,G_\ell}\,:\,L_nv=0\}|}{\prod_{i=1}^{\ell} \mathbb{E}|\{v\in A_{n_{i+1}-n_i}\,:\,L_{n_{i+1}-n_i}v=0\}|}-1\right|=0,\end{equation*}

where the maximum is over all the choices $0\le n_1 \lt \dots \lt n_{\ell+1}=n$ satisfying (12) and a maximal chain of subgroups $\{0\}=G_0\subsetneq G_1\subsetneq \cdots\subsetneq G_\ell=G$. Combining this with part (c) of Lemma 11, the statement follows.

Clearly, the number of tuples $0\le n_1 \lt n_2 \lt \cdots \lt n_{\ell+1}=n$ satisfying (12) is $(1+o(1))\frac{n^\ell}{\ell!}$. Thus, Theorem 6 follows by combining Lemma 13, Lemma 14, Lemma 15 and Lemma 18.

4. Values of $\chi_0$ in special cases

In this section, we prove Lemma 3.

Let $1\neq\varrho\in \widehat{\mathbb{F}_p}$. If $0\neq g\in \mathbb{F}_p$, then

\begin{equation*}\mathbb{E}\varrho(\xi g)=\frac{p\alpha-1}{p-1}\varrho(0)+\sum_{h\in \mathbb{F}_p}\frac{1-\alpha}{p-1}\varrho(h)=\frac{p\alpha-1}{p-1}.\end{equation*}

Given a vector $v\in \mathbb{F}_p^m$, let

\begin{equation*}\|v\|=|\{1\le i\le m\,:\,v_i\neq 0\}|.\end{equation*}

Then it follows from (2) that,

(13)\begin{equation}p\tau(v)=1+(p-1)\left(\frac{p\alpha-1}{p-1}\right)^{\|v\|}=\beta_{\|v\|}.\end{equation}

Let us define $X=(X_1,X_2,\dots)$, where $X_1,X_2,\dots$ are independent, $X_1$ is a uniform random element of $\mathbb{F}_p\setminus \{0\}$, and $X_2,X_3,\dots$ are uniform random elements of $\mathbb{F}_p$.

Let

\begin{equation*}Y_n=\prod_{i=1}^n p\tau(X_{\le i})\quad\text{and }\quad Y_\infty=\prod_{i=1}^n p\tau(X_{\le i}).\end{equation*}

Then

\begin{equation*}\mathbb{E}|\{v\in \mathbb{F}_p^n\,:\,v_1\neq 0, L_nv=0\}|=\frac{p-1}p\mathbb{E} Y_n.\end{equation*}

By conditioning on $X_1$ and applying Lemma 8, we obtain

(14)\begin{equation}\lim_{n\to\infty}\mathbb{E}|\{v\in \mathbb{F}_p^n\,:\,v_1\neq 0, L_nv=0\}|=\lim_{n\to\infty}\frac{p-1}p\mathbb{E} Y_n=\frac{p-1}p\mathbb{E} Y_\infty.\end{equation}

For $i\ge 1$, let us define $T_i=\min\{j\,:\,\|X_{\le j}\|=i\}$. By (13), we see that

\begin{equation*}p \tau(X_{\le j})=\beta_i \quad \text{for all }T_i\le j \lt T_{i+1}.\end{equation*}

Thus,

\begin{equation*}Y_\infty=\prod_{i=1}^\infty \beta_i^{T_{i+1}-T_i}.\end{equation*}

Note that $T_2-T_1,T_3-T_2,T_4-T_3,\dots$ are i.i.d. random variables such that for all $k\ge 1$, we have

\begin{equation*}\mathbb{P}(T_{i+1}-T_i=k)=\frac{p-1}{p^k}.\end{equation*}

Thus,

\begin{equation*}\mathbb{E}\beta_i^{T_{i+1}-T_i}=\sum_{k=1}^\infty (p-1)\left(\frac{\beta_i}p\right)^k=\frac{(p-1)\beta_i}{p-\beta_i}.\end{equation*}

By the independence of $T_2-T_1,T_3-T_2,T_4-T_3,\dots$, we obtain

\begin{equation*}\mathbb{E} Y_\infty=\prod_{i=1}^\infty \frac{(p-1)\beta_i}{p-\beta_i}.\end{equation*}

Combining this with (14), Lemma 3 follows.