---
abstract: |
    Let $h=\prod_{i=1}^{t}p_i^{s_i}$ be its decomposition into a product of powers of distinct primes, and $\mathbb{Z}_{h}$ be the residue class ring modulo $h$. Let $\mathbb{Z}_{h}^{n}$ be the $n$-dimensional row vector space over $\mathbb{Z}_{h}$. A generalized Grassmann graph for $\mathbb{Z}_{h}^n$, denoted by $G_r(m,n,\mathbb{Z}_{h})$ ($G_r$ for short), has all $m$-subspaces of $\mathbb{Z}_{h}^n$ as its vertices, and two distinct vertices are adjacent if their intersection is of dimension $>m-r$, where $2\leq r\leq m+1\leq n$. In this paper, we determine the clique number and geometric structures of maximum cliques of $G_r$. As a result, we obtain the Erdős-Ko-Rado theorem for $\mathbb{Z}_{h}^{n}$ and some bounds of the independence number of $G_r$.

    [*AMS classification*]{}: 05C50, 05D05

    [*Key words*]{}: Erdős-Ko-Rado theorem, Residue class ring, Grassmann graph, Clique number, Independence number
author:
- |
    Jun Guo[^1]\
    [Department of Mathematics, Langfang Normal University, Langfang 065000, China]{}
date:  
title: '**Erdős-Ko-Rado theorem and generalized Grassmann graphs for vector spaces over residue class rings**'
---

Introduction
============

Let $\mathbb{Z}$ denote the integer ring. For $a,b,h\in \mathbb{Z}$, integers $a$ and $b$ are said to be [*congruent modulo*]{} $h$ if $h$ divides $a-b$, and denoted by $a\equiv b \mod h$. Let $h=\prod_{i=1}^{t}p_i^{s_i}$ be its decomposition into a product of powers of distinct primes. Let $\mathbb{Z}_{h}$ denote the residue class ring modulo $h$ and $\mathbb{Z}_{h}^\ast$ denote its unit group. Then $\mathbb{Z}_{h}$ is a principal ideal ring and $|\mathbb{Z}_{h}^\ast|=h\prod_{i=1}^t(1-p_i^{-1}).$ By [@Ireland], $\mathbb{Z}_{h}\cong\mathbb{Z}_{p_1^{s_1}}\oplus\mathbb{Z}_{p_2^{s_2}}\oplus\cdots\oplus\mathbb{Z}_{p_t^{s_t}}$ and $\mathbb{Z}_{h}^\ast\cong\mathbb{Z}_{p_1^{s_1}}^\ast\times\mathbb{Z}_{p_2^{s_2}}^\ast\times\cdots\times\mathbb{Z}_{p_t^{s_t}}^\ast$. Note that $(p_i),i=1,2,\ldots,t$, are all the maximal ideals of $\mathbb{Z}_{h}$. Write $J_{(\alpha_1,\alpha_2,\ldots,\alpha_t)}=(\prod_{i=1}^{t}p_i^{\alpha_i})$, where $0\leq \alpha_i\leq s_i$ for $i=1,2,\ldots,t$. For brevity, we write $J_{(\alpha_1)}$ as $J_{\alpha_1}$ if $t=1$. For $a\in\mathbb{Z}$, we also denote by $a$ the congruence class of $a$ modulo $h$.

For a subset $S$ of $\mathbb{Z}_{h}$, let $S^{m\times n}$ be the set of all $m\times n$ matrices over $S$, and $S^{n}=S^{1\times n}$. A matrix in $S^n$ is also called an $n$-dimensional row vector over $S$. Let $I_r$ ($I$ for short) be the $r\times r$ identity matrix, and $0_{m,n}$ ($0$ for short) the $m\times n$ zero matrix. Let $\hbox{diag}(A_1,A_2,\ldots, A_k)$ denote the block diagonal matrix whose blocks along the main diagonal are matrices $A_1,A_2,\ldots, A_k$. The set of $n\times n$ invertible matrices forms a group under matrix multiplication, called the [*general linear group*]{} of degree $n$ over $\mathbb{Z}_{h}$ and denoted by $G\!L_n(\mathbb{Z}_{h})$. Let ${}^t\!A$ denote the transpose matrix of a matrix $A$ and $\det(X)$ the determinant of a square matrix $X$ over $\mathbb{Z}_{h}$. For $X\in \mathbb{Z}_{h}^{n\times n}$, by Corollary 2.21 in [@Brown], $X\in G\!L_n(\mathbb{Z}_{h})$ if and only if $\det(X)\in\mathbb{Z}_h^\ast$.

For $A\in\mathbb{Z}_{h}^{m\times n}$ and $B\in\mathbb{Z}_{h}^{n\times m}$, if $AB=I_m$, we say that $A$ has a [*right inverse*]{} and $B$ is a right inverse of $A$. Similarly, if $AB=I_m$, then $B$ has a [*left inverse*]{} and $A$ is a left inverse of $B$. $\mathbb{Z}_{h}^{n}$ is called the $n$-dimensional row vector space over $\mathbb{Z}_{h}$. Let $\alpha_i\in\mathbb{Z}_{h}^{n}$ for $i=1,2,\ldots,m$. The vector subset $\{\alpha_1,\alpha_2,\ldots,\alpha_m\}$ is called [*unimodular*]{} if the matrix ${}^t({}^t\alpha_1, {}^t\alpha_2, \ldots, {}^t\alpha_m)$ has a right inverse. By Lemma \[lem2.9\] below, a matrix $A\in\mathbb{Z}_{h}^{m\times n}$ has a right inverse if and only if all row vectors of $A$ are linearly independent in $\mathbb{Z}_{h}^{n}$.

Let $V\subseteq\mathbb{Z}_{h}^{n}$ be a [*linear subset*]{} (i.e., a $\mathbb{Z}_{h}$-module). A [*largest unimodular vector subset*]{} of $V$ is a unimodular vector subset of $V$ which has maximum number of vectors. The dimension of $V$, denoted by $\dim(V)$, is the number of vectors in a largest unimodular vector subset of $V$. Clearly, $\dim(V)=0$ if and only if $V$ does not contain a unimodular vector. If a linear subset $X$ of $\mathbb{Z}_{h}^{n}$ has a unimodular basis with $m$ vectors, then $X$ is called an $m$-[*dimensional vector subspace*]{} ($m$-[*subspace*]{} for short) of $\mathbb{Z}_{h}^{n}$. Every $m$-subspace of $\mathbb{Z}_{h}^{n}$ is isomorphic to $\mathbb{Z}_{h}^{m}$. Applying Lemma \[lem2.9\] below, it is easy to prove that every basis of a subspace of $\mathbb{Z}_{h}^{n}$ can be extended to a basis of $\mathbb{Z}_{h}^{n}$. We define the $0$-subspace to be $\{0\}$.

The Erdős-Ko-Rado theorem [@Erdos; @Wilson] is a classical result in extremal set theory which obtained an upper bound on the cardinality of a family of $m$-subsets of a set that every pairwise intersection has cardinality at least $r$ and describes exactly which families meet this bound. The results on Erdős-Ko-Rado theorem have inspired much research [@Frankl; @Godsi2; @Huang-T; @Tanaka; @Vanhove]. Let $0\leq r\leq m\leq n$ and ${\mathbb{Z}_{h}^{n}\brack m}$ be the set of all $m$-subspaces of $\mathbb{Z}_{h}^{n}$. A family ${\cal F}\subseteq{\mathbb{Z}_{h}^{n}\brack m}$ is called $r$-[*intersecting*]{} if $\dim(A\cap B)\geq r$ for all $A,B\in{\cal F}$. When $t=1$, Huang et al. [@Huang3] obtained an upper bound on the cardinality of an $r$-intersecting family in ${\mathbb{Z}_{h}^{n}\brack m}$ and described exactly which families meet this bound.

Let $0\leq2r\leq2m=n$ and $I\subseteq[t]:=\{1,2,\ldots,t\}$. Suppose that $\alpha_i=s_i,\pi_i(x)=1$ if $i\in I$, and $\alpha_i=0,\pi_i(x)=0$ if $i\in[t]\setminus I$. Define $$\label{equanew1}
{\cal F}_{(\alpha_1,\alpha_2,\ldots,\alpha_t)}^{(r,m,n,I)}
=\left\{\left(\begin{array}{cc}
X & 0\\
Y & xI_{r}
\end{array}\right)\in{\mathbb{Z}_h^n\brack m} :
X\in\mathbb{Z}_h^{(m-r)\times(n-r)},Y\in J_{(\alpha_1,\alpha_2,\ldots,\alpha_t)}^{r\times(n-r)} \right\}.$$ In this paper, we give the Erdőso-Rado theorem for $\mathbb{Z}_{h}^{n}$ as follows.

\[thm1.1\] Let $\lfloor n/2\rfloor\geq m\geq r\geq 0$ and ${\cal F}\subseteq{\mathbb{Z}_{h}^{n}\brack m}$ be an $r$-intersecting family. Then $$|{\cal F}|\leq \prod_{i=1}^tp_i^{(s_i-1)(n-m)(m-r)}{n-r\brack m-r}_{p_i},\;\hbox{where}\;
 {n-r\brack m-r}_{p_i}=\prod_{j=0}^{m-r-1}\frac{(p_i^{n-r}-p_i^j)}{(p_i^{m-r}-p_i^j)},$$ and equality holds if and only if either [(a)]{} ${\cal F}$ consists of all $m$-subspaces of $\mathbb{Z}_{h}^{n}$ which contain a fixed $r$-subspace of $\mathbb{Z}_{h}^{n}$, [(b)]{} $n=2m$ and ${\cal F}$ is the set of all $m$-subspaces of $\mathbb{Z}_{h}^{n}$ contained in a fixed $(n-r)$-subspace of $\mathbb{Z}_{h}^{n}$, or [(c)]{} $n=2m$ and there exists some $T\in G\!L_n(\mathbb{Z}_h)$ such that ${\cal F}={\cal F}_{(\alpha_1,\alpha_2,\ldots,\alpha_t)}^{(r,m,n,I)}T$ with $I\not=\emptyset,[t]$, where ${\cal F}_{(\alpha_1,\alpha_2,\ldots,\alpha_t)}^{(r,m,n,I)}$ is given by (\[equanew1\]).

To prove Theorem \[thm1.1\], we need to discuss generalized Grassmann graphs over $\mathbb{Z}_{h}$. The Grassmann graph over a finite field plays an important role in geometry and combinatorics, see [@Brouwer; @Pankov; @Wan2]. As a natural extension, the [*generalized Grassmann graph*]{} over $\mathbb{Z}_{h}$, denoted by $G_r(m,n,\mathbb{Z}_{h})$, has the vertex set ${\mathbb{Z}_{h}^n\brack m}$, and two distinct vertices are adjacent if their intersection is of dimension $>m-r$, where $2\leq r\leq m+1\leq n$. Note that $G_2(m,n,\mathbb{Z}_{h})$ is the Grassmann graph $G(m,n,\mathbb{Z}_{h})$.

Let $V(\Gamma)$ denote the vertex set of a graph $\Gamma$. For $A,B\in V(\Gamma)$, we write $A\sim B$ if vertices $A$ and $B$ are adjacent. A [*clique*]{} of a graph $\Gamma$ is a complete subgraph of $\Gamma$. A clique ${\cal C}$ is [*maximal*]{} if there is no clique of $\Gamma$ which properly contains ${\cal C}$ as a subset. A [*maximum clique*]{} of $\Gamma$ is a clique of $\Gamma$ which has maximum cardinality. The [*clique number*]{} $\omega(\Gamma)$ of $\Gamma$ is the number of vertices in a maximum clique. An [*independent set*]{} of a graph $\Gamma$ is a subset of vertices such that no two vertices are adjacent. A [*largest independent set*]{} of $\Gamma$ is an independent set of maximum cardinality. The [*independence number*]{} $\alpha(\Gamma)$ is the number of vertices in a largest independent set of $\Gamma$.

The paper is organized as follows. In Section 2, we introduce the basic properties of matrices over $\mathbb{Z}_{h}$ for later reference. In Section 3, we obtain the dimension formula and the Anzahl theorem for subspaces of $\mathbb{Z}_{h}^n$. In Section 4, we calculate the clique number and determine geometric structures of maximum cliques of $G_r(m,n,\mathbb{Z}_{h})$. As a result, Theorem \[thm1.1\] is proved. In Section 5, we obtain some bounds of the independence number of $G_r(m,n,\mathbb{Z}_{h})$ by using method in [@Huang3].

Matrices over $\mathbb{Z}_h$
============================

In this section, we introduce some basic properties of matrices over $\mathbb{Z}_h$.

\[lem2.1\][(See Lemma 2.3 in [@Guo3].)]{} Every non-zero element $x$ in $\mathbb{Z}_{h}$ can be written as $x=u\prod_{i=1}^tp_i^{\alpha_i}$, where $u$ is a unit, and $0\leq\alpha_i\leq s_i$ for $i=1,2,\ldots,t$. Moreover, the vector $(\alpha_1,\alpha_2,\ldots,\alpha_t)$ is unique and $u$ is unique modulo the ideal $J_{(s_1-\alpha_1,s_2-\alpha_2,\ldots,s_t-\alpha_t)}$.

\[lem2.2\][(See Theorem 2.10 in [@Guo3].)]{} Let $m\leq n$ and $A\in\mathbb{Z}_{h}^{m\times n}$. Then there are $S\in G\!L_m(\mathbb{Z}_h)$ and $T\in G\!L_n(\mathbb{Z}_h)$ such that $$\label{equa1}
A=S{\rm diag}\,\left(\prod_{i=1}^tp_i^{\alpha_{i1}},\prod_{i=1}^tp_i^{\alpha_{i2}},\ldots,\prod_{i=1}^tp_i^{\alpha_{im}}\right)T,$$ where $0\leq\alpha_{i1}\leq\alpha_{i2}\leq\cdots\leq\alpha_{im}\leq s_i$ for $i=1,2,\ldots,t$. Moreover, the array $$\Omega=((\alpha_{11},\alpha_{12},\ldots,\alpha_{1m}),(\alpha_{21},\alpha_{22},\ldots,\alpha_{2m}),\ldots,(\alpha_{t1},\alpha_{t2},\ldots,\alpha_{tm}))$$ is uniquely determined by $A$.

Let $A\in \mathbb{Z}_{h}^{m\times n}$ be a non-zero matrix. By Cohn’s definition [@Cohn2], the [*inner rank*]{} of $A$, denoted by $\rho(A)$, is the least integer $r$ such that $$\label{equa2}
A = BC\quad \hbox{where}\; B \in \mathbb{Z}_{h}^{m\times r} \;\hbox{and}\; C \in \mathbb{Z}_{h}^{r\times n}.$$ Let $\rho(0) =0$. Any factorization as (\[equa2\]) with $r=\rho(A)$ is called a [*minimal factorization*]{} of $A$. For $A\in \mathbb{Z}_{h}^{m\times n}$, it is obvious that $\rho(A)\leq\min\{m,n\}$ and $\rho(A)=0$ if and only if $A=0$.

For matrices over $\mathbb{Z}_{h}$, from [@Cohn; @Cohn2], we deduce that the following hold: $$\label{equa3}
 \rho(A)=\rho(SAT),\;\hbox{where}\;S\;\hbox {and}\;T\;\hbox{are invertible matrices}.$$ $$\label{equa4}
   \rho(AB) \leq\min\{\rho(A),\rho(B)\}.$$ $$\label{equa5}
   \rho\left(\begin{array}{cc}
   A_{11} & A_{12}\\
   A_{21}  & A_{22}
   \end{array}\right) \geq\max\{\rho(A_{ij}) :1\leq i,j\leq 2\}.$$

\[lem2.3\][(See Lemma 2.11 in [@Guo3].)]{} Let $A\in\mathbb{Z}_{h}^{m\times n}$ be as in (\[equa1\]). Then the inner rank of $A$ is $\max\{c: (\alpha_{1c},\alpha_{2c},\ldots,\alpha_{tc})\not=(s_1,s_2,\ldots,s_t)\}$.

For $i=1,2,\ldots,t$, let $\pi_i$ (resp. $\theta_i$) be the natural surjective homomorphism from $\mathbb{Z}_{h}$ to $\mathbb{Z}_{p_i^{s_i}}$ (resp. $\mathbb{Z}_{h/p_i^{s_i}}$), and $\pi_i(A)=(\pi_i(a_{uv}))$ (resp. $\theta_i(A)=(\theta_i(a_{uv}))$) for every $A=(a_{uv})\in\mathbb{Z}_{h}^{m\times n}$. For brevity, we write $\prod_{i=1}^{t}\mathbb{Z}_{p_i^{s_i}}^{m\times n}=\mathbb{Z}_{p_1^{s_1}}^{m\times n}\times\mathbb{Z}_{p_2^{s_2}}^{m\times n}\times\cdots\times\mathbb{Z}_{p_t^{s_t}}^{m\times n}$.

\[lem2.3.2\] For every $A\in\mathbb{Z}_{h}^{m\times n}$, let $\pi(A)=(\pi_1(A),\pi_2(A),\ldots,\pi_t(A))$ and $(\pi_i,\theta_i)(A)=(\pi_i(A),\theta_i(A))$ for $i=1,2,\ldots,t$. Then $\pi$ is a bijective map from $\mathbb{Z}_{h}^{m\times n}$ to $\prod_{i=1}^{t}\mathbb{Z}_{p_i^{s_i}}^{m\times n}$, and $(\pi_i,\theta_i)$ is a bijective map from $\mathbb{Z}_{h}^{m\times n}$ to $\mathbb{Z}_{p_i^{s_i}}^{m\times n}\times \mathbb{Z}_{h/p_i^{s_i}}^{m\times n}$ for $i=1,2,\ldots,t$.

[[*Proof.*]{}]{}For each $i$ with $1\leq i\leq t$ and $a,b\in\mathbb{Z}_h$, if $a=b$, then $(\pi_i,\theta_i)(a)=(\pi_i,\theta_i)(b)$. Suppose $(\pi_i,\theta_i)(a)=(\pi_i,\theta_i)(b)$. Then $\pi_i(a)=\pi_i(b)$ and $\theta_i(a)=\theta_i(b)$. Since $$a\equiv\pi_i(a)\mod p_i^{s_i}, b\equiv\pi_i(b)\mod p_i^{s_i},
a\equiv\theta_i(a) \mod h/p_i^{s_i},b\equiv\theta_i(b) \mod h/p_i^{s_i},$$ we obtain $p_i^{s_i}|(a-b)$ and $(h/p_i^{s_i})|(a-b)$, which imply that $a=b$. Therefore, $(\pi_i,\theta_i)$ is an injective map from $\mathbb{Z}_{h}$ to $\mathbb{Z}_{p_i^{s_i}}\times \mathbb{Z}_{h/p_i^{s_i}}$. By $|\mathbb{Z}_{h}|=|\mathbb{Z}_{p_i^{s_i}}\times \mathbb{Z}_{h/p_i^{s_i}}|$, $(\pi_i,\theta_i)$ is a bijective map from $\mathbb{Z}_{h}^{m\times n}$ to $\mathbb{Z}_{p_i^{s_i}}^{m\times n}\times \mathbb{Z}_{h/p_i^{s_i}}^{m\times n}$. Since $\mathbb{Z}_{h}\cong\mathbb{Z}_{p_1^{s_1}}\oplus\mathbb{Z}_{p_2^{s_2}}\oplus\cdots\oplus\mathbb{Z}_{p_t^{s_t}}$, $\pi$ is a bijective map from $\mathbb{Z}_{h}^{m\times n}$ to $\prod_{i=1}^{t}\mathbb{Z}_{p_i^{s_i}}^{m\times n}$. ${\hfill\Box\medskip}$

For matrices $A, B\in\mathbb{Z}_{h}^{m\times n}$ and $C\in\mathbb{Z}_{h}^{n\times k}$, it is easy to see that $$\label{equa6}
\pi_i(A+B)=\pi_i(A)+\pi_i(B)\quad\hbox{and}\quad\pi_i(AC)=\pi_i(A)\pi_i(C)\;\hbox{for}\;i=1,2,\ldots,t,$$ and $$\label{equa7}
\theta_i(A+B)=\theta_i(A)+\pi_i(B)\quad\hbox{and}\quad\theta_i(AC)=\theta_i(A)\pi_i(C)\;\hbox{for}\;i=1,2,\ldots,t.$$

\[lem2.4\][(See Lemma 2.12 in [@Guo3].)]{} Let $A\in\mathbb{Z}_{h}^{m\times n}$. Then $$\rho(A)=\max\{\rho(\pi_i(A)): i=1,2,\ldots,t\}=\max\{\rho(\theta_i(A)): i=1,2,\ldots,t\}.$$

Let $A\in\mathbb{Z}_{h}^{m\times n}$. Denote by $I_k(A)$ the ideal in $\mathbb{Z}_{h}$ generated by all $k\times k$ minors of $A$. Let $\hbox{Ann}_{\mathbb{Z}_{h}}(I_k(A)) =\{x\in \mathbb{Z}_{h}:xI_k(A)=0\}$ denote the annihilator of $I_k(A)$. The [*McCoy rank*]{} of $A$, denoted by $\hbox{rk}(A)$, is the following integer: $$\hbox{rk}(A) = \max\{k : \hbox{Ann}_{\mathbb{Z}_{h}}(I_k(A)) =(0)\} .$$ Note that $\hbox{rk}(A) =\hbox{rk}({}^t\!A)$; $\hbox{rk}(A) =\hbox{rk}(SAT)$ where $S$ and $T$ are invertible matrices; and $\hbox{rk}(A) =0$ if and only if $\hbox{Ann}_{\mathbb{Z}_{h}}(I_1(A)) \not=(0)$, see [@Brown].

\[lem2.5\] Let $A\in \mathbb{Z}_{h}^{m\times n}$ be as in (\[equa1\]). Then the McCoy rank of $A$ is $$\max\{c: (\alpha_{1c},\alpha_{2c},\ldots,\alpha_{tc})=0\}.$$

[[*Proof.*]{}]{}Let $\ell=\max\{c\in[m]:(\alpha_{1c},\alpha_{2c},\ldots,\alpha_{tc})=0\}$. Then $I_\ell(A)=\mathbb{Z}_{h}$ and $I_{\ell+1}(A)=(\prod_{i=1}^tp_i^{\alpha_{i,\ell+1}})$, which imply that $\hbox{Ann}_{\mathbb{Z}_{h}}(I_\ell(A))=(0)$ and $\hbox{Ann}_{\mathbb{Z}_{h}}(I_{\ell+1}(A))\not=(0)$. Similarly, $\hbox{Ann}_{\mathbb{Z}_{h}}(I_{k}(A))\not=(0)$ for $\ell+1 \leq k \leq \min\{m, n\}$. It follows that ${\rm rk}(A)=\ell$.${\hfill\Box\medskip}$

For matrices $A, B$ over $\mathbb{Z}_{h}$, by Lemmas \[lem2.2\], \[lem2.3\] and \[lem2.5\], if $\rho(A)={\rm rk}(A)$, then $$\label{equa8}
   \rho\left(\begin{array}{cc}
   A & 0\\ 0& B
   \end{array}\right)=\rho(A)+\rho(B)\quad\hbox{and}\quad
   \hbox{rk}\left(\begin{array}{cc}
   A &0 \\ 0 & B
   \end{array}\right)=\hbox{rk}(A)+\hbox{rk}(B).$$

\[lem2.6\] Let $A\in\mathbb{Z}_{h}^{m\times n}$. Then $${\rm rk}(A)=\min\{{\rm rk}(\pi_i(A)): i=1,2,\ldots,t\}=\min\{{\rm rk}(\theta_i(A)): i=1,2,\ldots,t\}.$$

[[*Proof.*]{}]{}By Lemma \[lem2.2\], we may assume that $A$ is as in (\[equa1\]). Let ${\rm rk}(A)=\ell$. For each $i$ with $1\leq i\leq t$, write $w_{ic}=\pi_i(\prod_{j=1}^tp_j^{\alpha_{jc}})$ for $c=\ell+1,\ldots,m$. By (\[equa6\]), $\pi_{i}(S)\in G\!L_m(\mathbb{Z}_{p_i^{s_i}}),\pi_{i}(T)\in G\!L_n(\mathbb{Z}_{p_i^{s_i}})$, and $$\pi_i(A)=\pi_{i}(S)\hbox{diag}(I_{\ell},w_{i,\ell+1},\dots,w_{im})\pi_{i}(T),$$ which imply that $${\rm rk}(\pi_i(A))={\rm rk}(\hbox{diag}(I_{\ell},w_{i,\ell+1},\dots,w_{im})\geq \ell={\rm rk}(A).$$ Therefore, ${\rm rk}(A)\leq\min\{{\rm rk}(\pi_i(A)): i=1,2,\ldots,t\}$. Since $(\alpha_{1,\ell+1},\alpha_{2,\ell+1},\ldots,\alpha_{t,\ell+1})\not=0$, there exists some $j$ such that $\alpha_{j,\ell+1}\not=0$, which implies that $w_{j,\ell+1}\not\in\mathbb{Z}_{p_j^{s_i}}^\ast$. It follows that ${\rm rk}(\pi_j(A))=\ell={\rm rk}(A)$.

For each $i$ with $1\leq i\leq t$, write $x_{ic}=\theta_i(\prod_{j=1}^tp_j^{\alpha_{jc}})$ for $c=\ell+1,\ldots,m$. By (\[equa7\]), $\theta_{i}(S)\in G\!L_m(\mathbb{Z}_{h/p_i^{s_i}}),\theta_{i}(T)\in G\!L_n(\mathbb{Z}_{h/p_i^{s_i}})$, and $$\theta_i(A)=\theta_{i}(S)\hbox{diag}(I_{\ell},x_{i,\ell+1},\dots,x_{im})\theta_{i}(T),$$ which imply that $${\rm rk}(\theta_i(A))={\rm rk}(\hbox{diag}(I_{\ell},x_{i,\ell+1},\dots,x_{im}))\geq \ell={\rm rk}(A).$$ Therefore, ${\rm rk}(A)\leq\min\{{\rm rk}(\theta_i(A)): i=1,2,\ldots,t\}$. Since $(\alpha_{1,\ell+1},\alpha_{2,\ell+1},\ldots,\alpha_{t,\ell+1})\not=0$, there exists some $j$ such that $\alpha_{j,\ell+1}\not=0$, which implies that $x_{i,\ell+1}\not\in\mathbb{Z}_{h/p_j^{s_j}}^\ast$ for $i\not=j$. It follows that ${\rm rk}(\theta_i(A))=\ell={\rm rk}(A)$ for $i\not=j$. Hence, the desired result follows. ${\hfill\Box\medskip}$

\[lem2.7\][(See [@Rosen].)]{} Let $\pi_i$ (resp. $\pi$) also denote the restriction of $\pi_i$ (resp. $\pi$) on $G\!L_n(\mathbb{Z}_{h})$ for $i=1,2,\ldots,t$. Then the following hold:

-   $\pi_i$ is a natural surjective homomorphism from $G\!L_{n}(\mathbb{Z}_{h})$ to $G\!L_{n}(\mathbb{Z}_{p_i^{s_i}})$ for $i=1,2,\ldots,t$.

-   $\pi$ is an isomorphism from $G\!L_{n}(\mathbb{Z}_{h})$ to $\prod_{i=1}^tG\!L_{n}(\mathbb{Z}_{p_i^{s_i}})$.

\[lem2.8\] Let $X\in\mathbb{Z}_{h}^{m\times n}$. Then the following hold:

-   When $m\leq n$, $X$ has a right inverse if and only if ${\rm rk}(X)=m$.

-   When $n=m$, $X$ is invertible if and only if $\pi_i(X)$ is invertible for $i=1,2,\ldots,t$. Moreover, if $X$ is invertible, then $\pi_i(X^{-1})=(\pi_i(X))^{-1}$ for $i=1,2,\ldots,t$.

[[*Proof.*]{}]{}(i). By Lemma 2.4 in [@Huang3], $\pi_i(X)$ has a right inverse if and only if ${\rm rk}(\pi_i(X))=m$ for $i=1,2,\ldots,t$. If $X$ has a right inverse, then (\[equa6\]) implies that $\pi_i(X)$ has a right inverse for $i=1,2,\ldots,t$. It follows that ${\rm rk}(\pi_i(X))=m$ for $i=1,2,\ldots,t$. By Lemma \[lem2.6\], ${\rm rk}(X)=m$.

Conversely, suppose ${\rm rk}(X)=m$. By Lemma \[lem2.2\], there are $S\in G\!L_m(\mathbb{Z}_h)$ and $T\in G\!L_n(\mathbb{Z}_h)$ such that $X=S(I_m,0_{m,n-m})T$. Pick $Y=T^{-1}\left(\begin{array}{c}
I_{m}\\
0_{n-m,m}
\end{array}\right)S^{-1}$. Then $XY=I_m$, which implies that $X$ has a right inverse.

(ii). By Lemma \[lem2.7\], $X$ is invertible if and only if $\pi_i(X)$ is invertible for $i=1,2,\ldots,t$. Suppose that $X\in\mathbb{Z}_{h}^{n\times n}$ is invertible. Then $XX^{-1}=I_n$, it follows from (\[equa6\]) that $\pi_i(XX^{-1})=\pi_i(X)\pi_i(X^{-1})=I_n$ for $i=1,2,\ldots,t$. Therefore, we have $\pi_i(X^{-1})=(\pi_i(X))^{-1}$ for $i=1,2,\ldots,t$. ${\hfill\Box\medskip}$

\[lem2.9\] Let $m\leq n$, and $\alpha_1,\alpha_2,\ldots,\alpha_m\in\mathbb{Z}_{h}^{n}$. Then the following hold:

-   The matrix $A={}^t({}^t\alpha_1, {}^t\alpha_2, \ldots, {}^t\alpha_m)$ has a right inverse if and only if $\alpha_1,\alpha_2,\ldots,\alpha_m$ are linearly independent.

-   If $\alpha_1,\alpha_2,\ldots,\alpha_m$ are linearly independent, then $\alpha_1,\alpha_2,\ldots,\alpha_m$ can be extended to a basis of $\mathbb{Z}_{h}^{n}$.

[[*Proof.*]{}]{}(i). If $A$ has a right inverse, then there exists some $B\in\mathbb{Z}_h^{n\times m}$ such that $AB=I_m$. From $\sum_{i=1}^mx_i\alpha_i=0$, we deduce that $(x_1,x_2,\ldots,x_m)A=0$, which implies that $(x_1,x_2,\ldots,x_m)=(x_1,x_2,\ldots,x_m)AB=0$, and therefore $\alpha_1,\alpha_2,\ldots,\alpha_m$ are linearly independent.

Conversely, suppose that $\alpha_1,\alpha_2,\ldots,\alpha_m$ are linearly independent. Without loss of generality, by Lemma \[lem2.2\], we may assume that $A$ is as in (\[equa1\]). By Lemma \[lem2.5\], ${\rm rk}(A)=m$ if and only if $(\alpha_{1m},\alpha_{2m},\ldots,\alpha_{tm})=0$, which implies that $A$ has a right inverse if and only if $(\alpha_{1m},\alpha_{2m},\ldots,\alpha_{tm})=0$ by Lemma \[lem2.8\]. Assume that $(\alpha_{1m},\alpha_{2m},\ldots,\alpha_{tm})\not=0$, then $x_m=\prod_{i=1}^tp_i^{s_i-\alpha_{im}}\not=0$ such that $(0,\ldots,0,x_m)S^{-1}\not=0$ and $(0,\ldots,0,x_m)S^{-1}A=0$, which implies that $\alpha_1,\alpha_2,\ldots,\alpha_m$ are not linearly independent, a contradiction.

(ii). Since $\alpha_1,\alpha_2,\ldots,\alpha_m$ are linearly independent, $A$ has a right inverse by (i), which implies that ${\rm rk}(A)=m$ by Lemma \[lem2.8\]. By Lemma \[lem2.2\], there are $S\in G\!L_m(\mathbb{Z}_h)$ and $T\in G\!L_n(\mathbb{Z}_h)$ such that $A=S(I_m,0_{m,n-m})T$. Then $$\widetilde{A}=\left(\begin{array}{c}
A\\ (0_{n-m,m},I_{n-m})T
\end{array}\right)
=\left(\begin{array}{cc}
S & 0\\ 0 & I_{n-m}
\end{array}\right)T$$ is invertible, which implies that all row vectors of $\widetilde{A}$ are a basis of $\mathbb{Z}_{h}^{n}$, and therefore $\alpha_1,\alpha_2,\ldots,\alpha_m$ can be extended to a basis of $\mathbb{Z}_{h}^{n}$. ${\hfill\Box\medskip}$

Subspaces of $\mathbb{Z}_{h}^{n}$
=================================

In this section, we study subspaces of $\mathbb{Z}_{h}^{n}$ and obtain some useful results. Notation and terminology will be adopted from [@Huang3].

Suppose that $\alpha_1,\alpha_2,\ldots,\alpha_m \;(m\leq n)$ are $m$ linearly independent row vectors in $\mathbb{Z}_{h}^{n}$. Let $[\alpha_1,\alpha_2,\ldots,\alpha_m]$ be the $\mathbb{Z}_{h}$-module generated by $\alpha_1,\alpha_2,\ldots,\alpha_m$. Then $X=[\alpha_1,\alpha_2,\ldots,\alpha_m]$ is an $m$-subspace of $\mathbb{Z}_{h}^{n}$, and the matrix ${}^t({}^t\alpha_1, {}^t\alpha_2, \ldots, {}^t\alpha_m)$ is called a [*matrix representation*]{} of $X$. We use an $m\times n$ matrix $A$ with ${\rm rk}(A)=m$ to represent an $m$-subspace of $\mathbb{Z}_{h}^{n}$. For $A,B\in\mathbb{Z}_{h}^{m\times n}$ with ${\rm rk}(A)={\rm rk}(B)=m$, both $A$ and $B$ represent the same $m$-subspace if and only if there exists an $S\in G\!L_m(\mathbb{Z}_h)$ such that $B=SA$. For convenience, if $A$ is a matrix representation of an $m$-subspace $X$, then we write $X=A$.

For an $m$-subspace $X$ of $\mathbb{Z}_{h}^{n}\;(m\leq n)$, by Lemma \[lem2.2\], there exists a $T\in G\!L_n(\mathbb{Z}_h)$ such that $X$ has matrix representations $$X=(0,I_m)T=S(0,I_m)T\; \hbox{for all}\; S\in G\!L_m(\mathbb{Z}_h).$$ By Lemma \[lem2.1\], every $m$-subspace $X$ of $\mathbb{Z}_{h}^{n}\;(m\leq n)$ has the unique matrix representation $(A_1,I_m)T$, which is a [*row-reduced echelon form*]{}, where $T$ is a permutation matrix and $A_1\in\mathbb{Z}_{h}^{m\times(n-m)}$.

Let $X$ and $Y$ be two subspaces of $\mathbb{Z}_{h}^{n}$. A [*join*]{} of $X$ and $Y$ is a minimum dimensional subspace containing $X$ and $Y$ in $\mathbb{Z}_{h}^{n}$. In general, the join of two subspaces in $\mathbb{Z}_{h}^{n}$ is not unique, see [@Huang3] for example. Denoted by $X\vee Y$ the set of all joins of subspaces $X$ and $Y$ with the same minimum dimension $\dim(X\vee Y )$. Then $X\vee Y=Y\vee X$, and $X\vee Y=\{Y\}$ if $X\subseteq Y$. Clearly, $X\cap Y$ contains a subspace of dimension $\dim(X\cap Y)$. In general, $X\cap Y$ is a linear subset but it may not be a subspace, see [@Huang3] for example.

\[lem3.1\] [(See Lemma 3.2 in [@Huang3].)]{} Let $1\leq m\leq k<n$. Suppose that $A$ is a $k$-subspace of $\mathbb{Z}_{p^s}^{n}$ and $B$ is an $m$-subspace of $\mathbb{Z}_{p^s}^{n}$ with $B\not\subseteq A$. Then there is a $T\in G\!L_n(\mathbb{Z}_{p^s})$ such that $$A=(0,I_k)T\quad\hbox{and}\quad B=(D,I_m)T,$$ where $D={\rm diag}\left(p^{\alpha_{1}},p^{\alpha_{2}},\ldots,p^{\alpha_{r}},0_{m-r,n-m-r}\right),1\leq r\leq \min\{m,n-k\}$ and $0\leq\alpha_{1}\leq\alpha_{2}\leq\cdots\leq\alpha_{r}\leq s-1$.

\[lem3.2\] Let $1\leq m\leq k<n$. Suppose that $A$ is a $k$-subspace of $\mathbb{Z}_{h}^{n}$ and $B$ is an $m$-subspace of $\mathbb{Z}_{h}^{n}$ with $B\not\subseteq A$. Then there is a $T\in G\!L_n(\mathbb{Z}_h)$ such that $$\label{equa10}
A=(0,I_k)T\quad\hbox{and}\quad B=(D,I_m)T,$$ where $D={\rm diag}\left(\prod_{i=1}^tp_i^{\alpha_{i1}},\prod_{i=1}^tp_i^{\alpha_{i2}},\ldots,\prod_{i=1}^tp_i^{\alpha_{ir}},0_{m-r,n-m-r}\right),1\leq r\leq \min\{m,n-k\},\rho(D)=r$, and $0\leq\alpha_{i1}\leq\alpha_{i2}\leq\cdots\leq\alpha_{ir}\leq s_i$ for $i=1,2,\ldots,t$.

[[*Proof.*]{}]{}By Lemmas \[lem2.6\] and \[lem2.8\], $\pi_i(A)$ is a $k$-subspace of $\mathbb{Z}_{p_i^{s_i}}^{n}$ and $\pi_i(B)$ is an $m$-subspace of $\mathbb{Z}_{p_i^{s_i}}^{n}$ for $i=1,2,\ldots,t$. For each $i$ with $1\leq i\leq t$, by Lemma \[lem3.1\], there is a $T_i\in G\!L_n(\mathbb{Z}_{p_i^{s_i}})$ such that $$\pi_i(A)=(0,I_k)T_i\quad\hbox{and}\quad \pi_i(B)=(D_i,I_m)T_i,$$ where $D_i={\rm diag}\,\left(p_i^{\alpha_{i1}},p_i^{\alpha_{i2}},\ldots,p_i^{\alpha_{ir_i}},0_{m-r_i,n-m-r_i}\right),1\leq r_i\leq \min\{m,n-k\}$ and $0\leq\alpha_{i1}\leq\alpha_{i2}\leq\cdots\leq\alpha_{ir_i}\leq s_i-1$.

Let $r=\max\{r_i:i=1,2,\ldots,t\}$ and $\alpha_{ij}=s_i$ for $i=1,2,\ldots,t$ and $j=r_i+1,\ldots,r$. Then $1\leq r\leq \min\{m,n-k\}$ and $0\leq\alpha_{i1}\leq\alpha_{i2}\leq\cdots\leq\alpha_{ir}\leq s_i$ for $i=1,2,\ldots,t$. Since $\mathbb{Z}_{h}\cong\mathbb{Z}_{p_1^{s_1}}\oplus\mathbb{Z}_{p_2^{s_2}}\oplus\cdots\oplus\mathbb{Z}_{p_t^{s_t}}$, there exist $d_c\in\mathbb{Z}_{h}$ such that $\pi_i(d_c)=p_i^{\alpha_{ic}}$ for $i=1,2,\ldots,t$ and $c=1,2,\ldots,r$. Let $\widetilde{D}={\rm diag}(d_1,d_2,\ldots,d_r,0_{m-r,n-m-r})$. By Lemma \[lem2.4\], $\rho(\widetilde{D})=r$ and $\pi_i(\widetilde{D})=D_i$ for $i=1,2,\ldots,t$.

Let $\pi_i(\prod_{j=1}^tp_j^{\alpha_{jc}})=d_{ic}p_i^{\alpha_{ic}}$, where $d_{ic}=\pi_i(\prod_{j\not=i}p_j^{\alpha_{jc}})\in\mathbb{Z}_{p_i^{s_i}}^\ast$, for $i=1,2,\ldots,t$ and $c=1,2,\ldots,r$. Since $\mathbb{Z}_{h}^\ast\cong\mathbb{Z}_{p_1^{s_1}}^\ast\times\mathbb{Z}_{p_2^{s_2}}^\ast\times\cdots\times\mathbb{Z}_{p_t^{s_t}}^\ast$, there exist $u_{c}\in\mathbb{Z}_h^\ast$ such that $\pi_i(u_c)=d_{ic}$ for $i=1,2,\ldots,t$ and $c=1,2,\ldots,r$. Since $\pi_i(u_cd_c)=\pi_i(\prod_{j=1}^tp_j^{\alpha_{jc}})$ for $i=1,2,\ldots,t$ and $c=1,2,\ldots,r$, we have $u_cd_c=\prod_{j=1}^tp_j^{\alpha_{jc}}$ for $c=1,2,\ldots,r$.

By Lemma \[lem2.7\], there exists the unique $\widetilde{T}\in G\!L_n(\mathbb{Z}_h)$ such that $\pi_i(\widetilde{T})=T_i$ for $i=1,2,\ldots,t$. By (\[equa6\]), $$\pi_i(A)=\pi_i((0,I_k)\widetilde{T})\quad\hbox{and}\quad \pi_i(B)=\pi_i((\widetilde{D},I_m)\widetilde{T})\;\hbox{for}\;i=1,2,\ldots,t,$$ which imply that $\pi(A)=\pi((0,I_k)\widetilde{T})$ and $\pi(B)=\pi((\widetilde{D},I_m)\widetilde{T}).$ Since $\pi$ is a bijective map, $A=(0,I_k)\widetilde{T}$ and $B=(\widetilde{D},I_m)\widetilde{T}.$ Let $D={\rm diag}(u_1d_1,u_2d_2,\ldots,u_rd_r,0_{m-r,n-m-r})$ and $T={\rm diag}\,(u_1^{-1},u_2^{-1},\ldots,u_r^{-1},I_{n-r})\widetilde{T}$. Since $r\leq n-k$, we have $$A=(0,I_k)T\quad\hbox{and}\quad B=(D,I_m)T.$$ Therefore, we complete the proof of this lemma. ${\hfill\Box\medskip}$

\[lem3.3\][(Dimensional formula.)]{} Let $A$ and $B$ be two subspaces of $\mathbb{Z}_{h}^n$. Then $$\label{equa11}
\dim(A\vee B)=\dim(A) + \dim(B) - \dim(A\cap B) = \rho\left(\begin{array}{c}A\\ B \end{array}\right).$$ Moreover, $\dim(A\cap B)=\min\{\dim(\pi_i(A)\cap\pi_i(B)):i=1,2,\ldots,t\}=\min\{\dim(\theta_i(A)\cap\theta_i(B)):i=1,2,\ldots,t\}.$

[[*Proof.*]{}]{}Let $k=\dim(A)$ and $m=\dim(B)$. Without loss of generality, we may assume that $1\leq m\leq k<n$ and $B\not\subseteq  A$. By Lemma \[lem3.2\], there is a $T\in G\!L_n(\mathbb{Z}_h)$ such that $A$ and $B$ are as in (\[equa10\]). Let $D_1={\rm diag}\left(\prod_{i=1}^tp_i^{\alpha_{i1}},\prod_{i=1}^tp_i^{\alpha_{i2}},\ldots,\prod_{i=1}^tp_i^{\alpha_{ir}}\right)$. Without loss of generality, we may assume that $T=I_n$ and $m\geq2$. Hence we can assume further that $A$ and $B$ have matrix representations $$\label{equa12}
A=(0,I_k)\quad\hbox{and}\quad
B=\left(\begin{array}{cccc}
D_1 & 0_{r,n-m-r} & I_r & 0\\
0 & 0 & 0 & I_{m-r}
\end{array}\right),$$ respectively. Clearly, the $(m-r)$-subspace $(0_{m-r,n-m+r},I_{m-r})$ is contained in $A\cap B$, and the $(k+r)$-subspace $\left(\begin{array}{ccc}
I_r & 0_{r,n-k-r}  & 0\\
0 &  0 & I_{k}
\end{array}\right)$ contains $A$ and $B$. It follows that $\dim(A\cap B)\geq m-r$ and $\dim(A\vee B)\leq k+r$.

Let $\alpha\in A\cap B$ be an $n$-dimensional row vector. Then there are matrices $C_1\in\mathbb{Z}_{h}^k$ and $C_2=(C_{21},C_{22})\in\mathbb{Z}_{h}^m$, where $C_{21}\in\mathbb{Z}_{h}^r$ and $C_{22}\in\mathbb{Z}_{h}^{m-r}$, such that $\alpha=C_1A=(0_{1,n-k},C_1)$ and $\alpha=C_2B=(C_{21}D_1,0,C_{21},C_{22})$. By $r\leq n-k$, we have that $C_{21}D_1=0$ and therefore $\alpha=(0_{1,n-m},C_{21},C_{22})$. From $C_{21}D_1=0$ and $\rho(D_1)=r$, we deduce that $C_{21}\in\bigcup_{i=1}^t(p_i)^{r}$, where $(p_i),i=1,2,\ldots,t,$ are the maximal ideals of $\mathbb{Z}_h$, which implies that $\dim(A\cap B)\leq m-r$. Hence, we have $\dim(A\cap B)=m-r$.

Let $d=\dim(A\vee B)$. Then there exists a $d$-subspace $W\in A\vee B$. Since $B\not\subseteq A$, one obtains $d>k$. By Lemma \[lem2.9\], $W$ has a matrix representation $W={\rm diag}(W_{11},I_k)$, where $W_{11}\in\mathbb{Z}_{h}^{(d-k)\times(n-k)}$ has a right inverse. By $B\subseteq W$, the $r$-subspace $(D_1,0_{r,n-m-r},I_r,0)\subseteq W$. Therefore, there is a matrix $C=(C_1,C_2)\in\mathbb{Z}_{h}^{r\times d}$, where $C_1\in\mathbb{Z}_{h}^{r\times(d-k)}$ and $C_2\in\mathbb{Z}_{h}^{r\times k}$, such that $(D_1,0_{r,n-m-r},I_r,0)=CW=(C_1W_{11},C_2)$. Hence $(D_1,0_{r,n-k-r})=C_1W_{11}$. It follows that $r=\rho(D_1)=\rho(C_1W_{11})\leq\rho(W_{11})\leq d-k$, which implies that $\dim(A\vee B)=k+r$.

From (\[equa8\]) and (\[equa12\]), we deduce that $$\rho\left(\begin{array}{c}A\\ B \end{array}\right)
=\rho\left(\begin{array}{ccc}
I_k&0&0\\
0&D_1 &  0\\
0 & 0 & 0
\end{array}\right)=k+r.$$ Therefore, we complete the proof of (\[equa11\]).

Let us continue to assume that $A$ and $B$ have matrix representations as in (\[equa12\]). From $r=\rho(D_1)$, we deduce that $\dim(A\cap B)=m-\rho(D_1)$. By Lemmas \[lem2.6\] and \[lem2.8\], $\pi_i(A)$ and $\pi_i(B)$ are two subspaces of $\mathbb{Z}_{p_i^{s_i}}^n$ for $i=1,2,\ldots,t$. Similarly, we can prove that $\dim(\pi_i(A)\cap \pi_i(B))=m-\rho(\pi_i(D_1))$ for $i=1,2,\ldots,t$. By Lemma \[lem2.4\], we have $$\begin{aligned}
\dim(A\cap B)&=&m-\rho(D_1)=m-\max\{\rho(\pi_i(D_1)): i=1,2,\ldots,t\}\\
&=&\min\{\dim(\pi_i(A)\cap\pi_i(B)):i=1,2,\ldots,t\}.\end{aligned}$$ By Lemmas \[lem2.4\], \[lem2.6\] and \[lem2.8\], we have similarly that $\dim(A\cap B)=\min\{\dim(\theta_i(A)\cap\theta_i(B)):i=1,2,\ldots,t\}$. ${\hfill\Box\medskip}$

\[lem3.4\][(See Theorem 3.5 in [@Huang3].)]{} Let $1\leq k<n$. Then the number of $k$-subspaces of $\mathbb{Z}_{p^s}^n$ is $p^{(s-1)k(n-k)}{n\brack k}_p$.

\[lem3.5\] Let $1\leq m\leq k<n$. Then the following hold:

-   The number of $k$-subspaces of $\mathbb{Z}_{h}^n$ is $\prod_{i=1}^tp_i^{(s_i-1)k(n-k)}{n\brack k}_{p_i}.$

-   In $\mathbb{Z}_{h}^n$, the number of $m$-subspaces in a given $k$-subspace is $\prod_{i=1}^tp_i^{(s_i-1)m(k-m)}{k\brack m}_{p_i}.$

-   In $\mathbb{Z}_{h}^n$, the number of $k$-subspaces containing a fixed $m$-subspace is\
    $\prod_{i=1}^tp_i^{(s_i-1)(k-m)(n-k)}{n-m\brack k-m}_{p_i}.$

[[*Proof.*]{}]{}(i). Let $\pi$ be the bijective map from $\mathbb{Z}_{h}^{k\times n}$ to $\prod_{i=1}^t\mathbb{Z}_{p_i^{s_i}}^{k\times n}$. By Lemma \[lem2.6\] and matrix representations of subspaces, $\pi$ induces a bijective map from ${\mathbb{Z}_{h}^{n}\brack k}$ to $\prod_{i=1}^t{\mathbb{Z}_{p_i^{s_i}}^{n}\brack k}$ such that $\pi(X)=(\pi_1(X),\pi_2(X),\ldots,\pi_t(X))$ for a matrix representation $X$ of every $k$-subspace of $\mathbb{Z}_h^n$. By Lemma \[lem3.4\], the number of $k$-subspaces of $\mathbb{Z}_{p_i^{s_i}}^n$ is $p_i^{(s_i-1)k(n-k)}{n\brack k}_{p_i}$ for $i=1,2,\ldots,t$, which imply that the number of $k$-subspaces of $\mathbb{Z}_{h}^n$ is $\prod_{i=1}^tp_i^{(s_i-1)k(n-k)}{n\brack k}_{p_i}.$

(ii). Since every $k$-subspace of $\mathbb{Z}_{h}^n$ is isomorphic to $\mathbb{Z}_{h}^k$, the desired result follows by (i).

(iii). Let $${\cal S}=\left\{(A,B):A\in{\mathbb{Z}_{h}^{n}\brack m},B\in{\mathbb{Z}_{h}^{n}\brack k},A\subseteq B\right\}.$$ We compute the cardinality of ${\cal S}$ in two ways. By (i) and (ii), the number of $k$-subspaces containing a fixed $m$-subspace is $$\frac{|{\cal S}|}{\left|{\mathbb{Z}_{h}^{n}\brack m}\right|}=\frac{\left|{\mathbb{Z}_{h}^{n}\brack k}\right|\cdot\left|{\mathbb{Z}_{h}^{k}\brack m}\right|}{\left|{\mathbb{Z}_{h}^{n}\brack m}\right|}
=\prod_{i=1}^tp_i^{(s_i-1)(k-m)(n-k)}{n-m\brack k-m}_{p_i},$$ as desired. ${\hfill\Box\medskip}$

Let $A$ be an $m$-subspace of $\mathbb{Z}_{h}^n$. Let $$A^\perp=\{y\in\mathbb{Z}_{h}^n: y{}^tx=0\;\hbox{for all}\;x\in A\}.$$ Then $A^\perp$ is a linear subset of $\mathbb{Z}_{h}^n$. By Lemma \[lem2.2\], $A$ has a matrix representation $A=(0,I_m)T$, where $T\in G\!L_n(\mathbb{Z}_h)$. It is easy to prove that $A^\perp$ has a matrix representation $$\label{equa13}
A^\perp=(I_{n-m},0){}^tT^{-1}\; \hbox{if}\; A=(0,I_m)T.$$ Hence $A^\perp$ is an $(n-m)$-subspace of $\mathbb{Z}_{h}^n$. The subspace $A^\perp$ is called the [*dual subspace*]{} of $A$. Note that $\dim(A)+\dim(A^\perp)=n$ and $(A^{\perp})^\perp=A.$ If $A_1$ and $A_2$ are two subspaces of $\mathbb{Z}_{h}^n$, then $A_1\subseteq A_2$ if and only if $A_2^\perp\subseteq A_1^\perp.$

\[lem3.6\] Let $m<n$ and let $A$ and $B$ be two $m$-subspaces of $\mathbb{Z}_{h}^n$. Then $$m-\dim(A\cap B)=n-m-\dim(A^\perp\cap B^\perp).$$

[[*Proof.*]{}]{}Without loss of generality, by Lemma \[lem3.2\], we may assume that $A=(0,I_m)$ and $B=(D,I_m)$, where $D={\rm diag}\left(\prod_{i=1}^tp_i^{\alpha_{i1}},\prod_{i=1}^tp_i^{\alpha_{i2}},\ldots,\prod_{i=1}^tp_i^{\alpha_{ir}},0_{m-r,n-m-r}\right),1\leq r\leq \min\{m,n-m\},\rho(D)=r$, and $0\leq\alpha_{i1}\leq\alpha_{i2}\leq\cdots\leq\alpha_{ir}\leq s_i$ for $i=1,2,\ldots,t$. Then $\dim(A\cap B)=m-r$.

Note that $A^\perp$ and $B^\perp$ are $(n-m)$-subspaces of $\mathbb{Z}_{h}^n$. By (\[equa13\]), $A^\perp$ has a matrix representation $A^\perp=(I_{n-m},0)$. Let $B^\perp=(B_1,B_2)$ be a matrix representation of $B^\perp$, where $B_1\in\mathbb{Z}_{h}^{(n-m)\times(n-m)}$ and $B_2\in\mathbb{Z}_{h}^{(n-m)\times m}$. Then $B_1{}^tD+B_2=0$, which implies that $\rho(B_2)=\rho(-B_1{}^tD)\leq r$. By Theorem \[lem3.3\], we have $$\begin{aligned}
n-m-\dim(A^\perp\cap B^\perp)&=&\rho\left(\begin{array}{c}
A^\perp\\ B^\perp
\end{array}\right)-(n-m)
=\rho\left(\begin{array}{cc}
I_{n-m} & 0\\ B_1 & B_2
\end{array}\right)-(n-m)\\
&=&\rho(B_2)\leq r=m-\dim(A\cap B).\end{aligned}$$ On the other hand, we have similarly $$m-\dim(A\cap B)=m-\dim((A^\perp)^\perp\cap(B^\perp)^\perp)
\leq n-m-\dim(A^\perp\cap B^\perp).$$ Therefore, we obtain $m-\dim(A\cap B)=n-m-\dim(A^\perp\cap B^\perp).$ ${\hfill\Box\medskip}$

Proof of Theorem \[thm1.1\]
===========================

In this section, we discuss the generalized Grassmann graph $G_r(m,n,\mathbb{Z}_{h})$ over $\mathbb{Z}_{h}$ and prove Theorem \[thm1.1\]. We always assume that $2\leq r\leq m+1\leq n$.

For convenience, if $X$ is a vertex of $G_r(m,n,\mathbb{Z}_{h})$, then the matrix representation of $X$ also is denoted by $X$. By Lemma \[lem2.3.2\], $\pi$ induces a bijective map from $V(G_r(m,n,\mathbb{Z}_{h}))$ to $\prod_{i=1}^{t}V(G_r(m,n,\mathbb{Z}_{p_i^{s_i}}))$, and $(\pi_i,\theta_i)$ induces a bijective map from $V(G_r(m,n,\mathbb{Z}_{h}))$ to $V(G_r(m,n,\mathbb{Z}_{p_i^{s_i}}))\times V(G_r(m,n,\mathbb{Z}_{h/p_i^{s_i}}))$ for $i=1,2,\ldots,t$.

\[lem4.1\] Let $n\leq 2m$ and $2\leq r\leq \max\{m+1,n-m+1\}$. Then $G_r(m,n,\mathbb{Z}_{h})$ is a connected vertex transitive graph, and $G_r(m,n,\mathbb{Z}_{h})\cong G_r(n-m,n,\mathbb{Z}_{h})$.

[[*Proof.*]{}]{}If $X$ is a vertex of $G_r(m,n,\mathbb{Z}_{h})$, then there is a $T_X\in G\!L_n(\mathbb{Z}_h)$ such that $X=(0,I_m)T_X$ by Lemma \[lem2.2\]. Let $\varphi(X)=XT_X^{-1}$. Then $\varphi$ is an automorphism of $G_r(m,n,\mathbb{Z}_{h})$ and $\varphi(X)=(0,I_m)$. Hence $G_r(m,n,\mathbb{Z}_{h})$ is vertex transitive. By Lemma \[lem3.2\] and Theorem \[lem3.3\], it is easy to prove that $G_r(m,n,\mathbb{Z}_{h})$ is a connected graph.

By Theorem \[lem3.5\], we have $|V(G_r(m,n,\mathbb{Z}_{h}))|=|V(G_r(n-m,n,\mathbb{Z}_{h}))|.$ By Lemma \[lem3.6\], the map $X\mapsto X^\perp$ is an isomorphism from $G_r(m,n,\mathbb{Z}_{h})$ to $G_r(n-m,n,\mathbb{Z}_{h})$. Therefore, we obtain $G_r(m,n,\mathbb{Z}_{h})\cong G_r(n-m,n,\mathbb{Z}_{h})$. ${\hfill\Box\medskip}$

By Theorem \[lem4.1\], we may assume that $n\geq2m$ in our discussion on $G_r(m,n,\mathbb{Z}_{h})$.

\[lem4.2\][(See Lemma 4.7 and Theorem 4.9 in [@Huang3].)]{} Let $n\geq 2m$. Then the clique number of $G_r(m,n,\mathbb{Z}_{p^s})$ is $p^{(s-1)(n-m)(r-1)}{n-m+r-1\brack r-1}_{p}.$ Moreover, ${\cal F}$ is a maximum clique of $G_r(m,n,\mathbb{Z}_{p^s})$ if and only if either [(a)]{} ${\cal F}$ consists of all $m$-subspaces of $\mathbb{Z}_{p^s}^n$ which contain a fixed $(m-r+1)$-subspace of $\mathbb{Z}_{p^s}^n$, or [(b)]{} $n=2m$ and ${\cal F}$ is the set of all $m$-subspaces of $\mathbb{Z}_{p^s}^n$ contained in a fixed $(m+r-1)$-subspace of $\mathbb{Z}_{p^s}^n$.

\[lem4.3\] Let $n\geq 2m$, and $\pi_i$ be the surjective homomorphism from $V(G_r(m,n,\mathbb{Z}_{h}))$ to $V(G_r(m,n,\mathbb{Z}_{p_i^{s_i}}))$ for $i=1,2\ldots,t$. Suppose that ${\cal F}$ is a maximum clique of $G_r(m,n,\mathbb{Z}_{h})$. Then the following hold:

-   The clique number of $G_r(m,n,\mathbb{Z}_{h})$ is $\prod_{i=1}^tp_i^{(s_i-1)(n-m)(r-1)}{n-m+r-1\brack r-1}_{p_i}.$

-   $\pi_i({\cal F})$ is a maximum clique of $G_r(m,n,\mathbb{Z}_{p_i^{s_i}})$ for $i=1,2,\ldots,t$.

[[*Proof.*]{}]{}Let ${\cal F}$ be a maximum clique of $G_r(m,n,\mathbb{Z}_{h})$. For each $i$ with $1\leq i\leq t$, let $\{\pi_i(A_{i1}),\pi_i(A_{i2}),\ldots,\pi_i(A_{ik_i})\}$ be the set of all different elements in $\pi_i({\cal F})$. Then $\pi_i({\cal F})=\{\pi_i(A_{i1}),\pi_i(A_{i2}),\ldots,\pi_i(A_{ik_i})\}$. By Lemma \[lem2.6\] and Theorem \[lem3.3\], $\pi_i({\cal F})$ is a clique of $G_r(m,n,\mathbb{Z}_{p_i^{s_i}})$. By Lemma \[lem4.2\], we have $$\label{equa14}
k_i\leq\omega(G_r(m,n,\mathbb{Z}_{p_i^{s_i}}))=p_i^{(s_i-1)(n-m)(r-1)}{n-m+r-1\brack r-1}_{p_i}.$$ Moreover, ${\cal F}$ has a partition into $k_i$ cliques: ${\cal F}=\bigcup_{j=1}^{k_i}{\cal C}_j$, where ${\cal C}_j$ is a clique with $\pi_i({\cal C}_j)=\{\pi_i(A_{ij})\}$ for $j=1,2,\ldots,k_i$, and ${\cal C}_u\cap{\cal C}_v=\emptyset$ for all $u\not=v$. It follows that $\omega(G_r(m,n,\mathbb{Z}_{h}))=\sum_{j=1}^{k_i}|{\cal C}_j|$.

(i). By Lemma \[lem4.2\], the result is trivial for $t=1$. Suppose that $t\geq2$ and the result is true for $t-1$. Let $n_j=|{\cal C}_j|$ for $j=1,2,\ldots,k_i$. Then there exists $\{B_{1j},B_{2j},\ldots,B_{n_jj}\}\subseteq (p_i^{s_i})^{m\times n}$ such that ${\cal C}_j=\{A_{ij}+B_{1j},A_{ij}+B_{2j},\ldots,A_{ij}+B_{n_jj}\}$. By Lemmas \[lem2.3.2\], \[lem2.6\] and Theorem \[lem3.3\], $|(\pi_i,\theta_i)({\cal C}_j)|=|\theta_i({\cal C}_j)|=|{\cal C}_j|$ and $\theta_i({\cal C}_j)$ is a clique of $G_r(m,n,\mathbb{Z}_{h/p_i^{s_i}})$. By induction, we have $$n_j\leq\omega(G_r(m,n,\mathbb{Z}_{h/p_i^{s_i}}))=\prod_{j\not=i}p_j^{(s_j-1)(n-m)(r-1)}{n-m+r-1\brack r-1}_{p_j}\;\hbox{for}\;j=1,2,\ldots,k_i.$$ Therefore, by (\[equa14\]), we have $$\label{equa15}
\omega(G_r(m,n,\mathbb{Z}_{h}))=\sum_{j=1}^{k_i}n_j\leq  \prod_{i=1}^tp_i^{(s_i-1)(n-m)(r-1)}{n-m+r-1\brack r-1}_{p_i}.$$

On the other hand, $${\cal C}=\left\{\left(\begin{array}{cc}
X & 0\\
0 & I_{m-r+1}
\end{array}\right): X\in V(G_r(r-1,n-m+r-1,\mathbb{Z}_{h}))\right\}$$ is a clique of $G_r(m,n,\mathbb{Z}_{h})$ and $|{\cal C}|=\prod_{i=1}^tp_i^{(s_i-1)(n-m)(r-1)}{n-m+r-1\brack r-1}_{p_i}$. It follows that $\omega(G_r(m,n,\mathbb{Z}_{h}))\geq\prod_{i=1}^tp_i^{(s_i-1)(n-m)(r-1)}{n-m+r-1\brack r-1}_{p_i}$. Therefore, the clique number of $G_r(m,n,\mathbb{Z}_{h})$ is $\prod_{i=1}^tp_i^{(s_i-1)(n-m)(r-1)}{n-m+r-1\brack r-1}_{p_i}$.

(ii). By (\[equa14\]) and (\[equa15\]), $k_i=p_i^{(s_i-1)(n-m)(r-1)}{n-m+r-1\brack r-1}_{p_i}$ for $i=1,2,\ldots,t$. By Lemma \[lem4.2\], $\pi_i(\cal F)$ is a maximum clique of $G_r(m,n,\mathbb{Z}_{p_i^{s_i}})$ for $i=1,2,\ldots,t$. ${\hfill\Box\medskip}$

\[lem4.4\] Let $4\leq2r\leq2m=n$ and $I\subseteq[t]$. Then ${\cal F}_{(\alpha_1,\alpha_2,\ldots,\alpha_t)}^{(m-r+1,m,n,I)}$ is a maximum clique of $G_r(m,n,\mathbb{Z}_{h})$, where ${\cal F}_{(\alpha_1,\alpha_2,\ldots,\alpha_t)}^{(m-r+1,m,n,I)}$ is given by (\[equanew1\]).

[[*Proof.*]{}]{}Since $\pi$ is a bijective map from $V(G_r(m,n,\mathbb{Z}_{h}))$ to $\prod_{i=1}^{t}V(G_r(m,n,\mathbb{Z}_{p_i^{s_i}}))$, $\pi$ induces a bijective map from ${\cal F}_{(\alpha_1,\alpha_2,\ldots,\alpha_t)}^{(m-r+1,m,n,I)}$ to $\pi\left({\cal F}_{(\alpha_1,\alpha_2,\ldots,\alpha_t)}^{(m-r+1,m,n,I)}\right)$. Therefore, we have $$\left|{\cal F}_{(\alpha_1,\alpha_2,\ldots,\alpha_t)}^{(m-r+1,m,n,I)}\right|=\left|\pi\left({\cal F}_{(\alpha_1,\alpha_2,\ldots,\alpha_t)}^{(m-r+1,m,n,I)}\right)\right|
=\prod_{i=1}^{t}\left|\pi_i\left({\cal F}_{(\alpha_1,\alpha_2,\ldots,\alpha_t)}^{(m-r+1,m,n,I)}\right)\right|.$$

For any $$U=\left(\begin{array}{cc}
X & 0\\
Y & xI_{m-r+1}
\end{array}\right)\in{\cal F}_{(\alpha_1,\alpha_2,\ldots,\alpha_t)}^{(m-r+1,m,n,I)},$$ we have that $X$ is an $(r-1)$-subspace of $\mathbb{Z}_h^{m+r-1}$ since $U\in{\mathbb{Z}_h^n\brack m}$. By Theorem \[lem3.5\], one obtains $$\left|{\cal F}_{(\alpha_1,\alpha_2,\ldots,\alpha_t)}^{(m-r+1,m,n,I)}\right|
\geq\left|{\mathbb{Z}_h^{m+r-1}\brack r-1}\right|=\prod_{i=1}^tp_i^{(s_i-1)m(r-1)}{m+r-1\brack r-1}_{p_i}.$$

If $i\in I$, from Lemmas \[lem2.6\] and \[lem2.8\], we deduce that $\pi_i\left({\cal F}_{(\alpha_1,\alpha_2,\ldots,\alpha_t)}^{(m-r+1,m,n,I)}\right)\subseteq{\cal A}_i$, where ${\cal A}_i$ is the set of all $m$-subspaces of $\mathbb{Z}_{p_i^{s_i}}^n$ containing the fixed $(m-r+1)$-subspace $(0,I_{m-r+1})$. If $i\in[t]\setminus I$, from Lemmas \[lem2.6\] and \[lem2.8\], we deduce that $\pi_i\left({\cal F}_{(\alpha_1,\alpha_2,\ldots,\alpha_t)}^{(m-r+1,m,n,I)}\right)\subseteq{\cal A}_i$, where ${\cal A}_i$ is the set of all $m$-subspaces of $\mathbb{Z}_{p_i^{s_i}}^n$ contained in the fixed $(m+r-1)$-subspace $(I_{m+r-1},0)$. Therefore, by Theorem \[lem3.5\], we have $$\prod_{i=1}^{t}\left|\pi_i\left({\cal F}_{(\alpha_1,\alpha_2,\ldots,\alpha_t)}^{(m-r+1,m,n,I)}\right)\right|\leq\prod_{i=1}^t|{\cal A}_i|
=\prod_{i=1}^tp_i^{(s_i-1)m(r-1)}{m+r-1\brack r-1}_{p_i}.$$ Hence, we obtain $\left|{\cal F}_{(\alpha_1,\alpha_2,\ldots,\alpha_t)}^{(m-r+1,m,n,I)}\right|=\prod_{i=1}^tp_i^{(s_i-1)m(r-1)}{m+r-1\brack r-1}_{p_i}$.

Let $U$ and $W$ be two distinct subspaces in ${\cal F}_{(\alpha_1,\alpha_2,\ldots,\alpha_t)}^{(m-r+1,m,n,I)}$. If $i\in I$, then $\pi_i(U)\cap \pi_i(W)\supseteq(0,I_{m-r+1})$, which implies that $\dim(\pi_i(U)\cap \pi_i(W))\geq m-r+1$. If $i\in[t]\setminus I$, then $\pi_i(U),\pi_i(W)\subseteq(I_{m+r-1},0)$, which imply that $\dim(\pi_i(U)\cap \pi_i(W))\geq m-r+1$ by Theorem \[lem3.3\] and $\dim(\pi_i(U)\vee\pi_i(W))\leq m+r-1$. Therefore, we have $\dim(\pi_i(U)\cap \pi_i(W))\geq m-r+1$ for $i=1,2,\ldots,t$. By Theorem \[lem3.3\] again, we have $\dim(U\cap W)\geq m-r+1$, and therefore ${\cal F}_{(\alpha_1,\alpha_2,\ldots,\alpha_t)}^{(m-r+1,m,n,I)}$ is a clique of $G_r(m,n,\mathbb{Z}_{h})$. By Lemma \[lem4.3\] and $\left|{\cal F}_{(\alpha_1,\alpha_2,\ldots,\alpha_t)}^{(m-r+1,m,n,I)}\right|=\prod_{i=1}^tp_i^{(s_i-1)m(r-1)}{m+r-1\brack r-1}_{p_i}$, ${\cal F}_{(\alpha_1,\alpha_2,\ldots,\alpha_t)}^{(m-r+1,m,n,I)}$ is a maximum clique of $G_r(m,n,\mathbb{Z}_{h})$. ${\hfill\Box\medskip}$

\[lem4.5\] Let $n\geq 2m$. Then ${\cal F}$ is a maximum clique of $G_r(m,n,\mathbb{Z}_{h})$ if and only if either [(a)]{} ${\cal F}$ consists of all $m$-subspaces of $\mathbb{Z}_{h}^n$ which contain a fixed $(m-r+1)$-subspace of $\mathbb{Z}_{h}^n$, [(b)]{} $n=2m$ and ${\cal F}$ is the set of all $m$-subspaces of $\mathbb{Z}_{h}^n$ contained in a fixed $(m+r-1)$-subspace of $\mathbb{Z}_{h}^n$, or [(c)]{} $n=2m$ and there exists some $T\in G\!L_n(\mathbb{Z}_h)$ such that ${\cal F}={\cal F}_{(\alpha_1,\alpha_2,\ldots,\alpha_t)}^{(m-r+1,m,n,I)}T$ with $I\not=\emptyset,[t]$, where ${\cal F}_{(\alpha_1,\alpha_2,\ldots,\alpha_t)}^{(m-r+1,m,n,I)}$ is given by (\[equanew1\]).

[[*Proof.*]{}]{}By the definition of $G_r(m,n,\mathbb{Z}_{h})$, we have $2\leq r\leq m+1$. When $r=m+1$ or $m=1$, $G_r(m,n,\mathbb{Z}_{h})$ is a clique, which implies that this theorem holds by Theorem \[lem3.5\]. From now on, we assume that $4\leq2r\leq2m\leq n$.

Suppose that either [(a)]{} ${\cal F}$ consists of all $m$-subspaces of $\mathbb{Z}_{h}^n$ which contain a fixed $(m-r+1)$-subspace of $\mathbb{Z}_{h}^n$, [(b)]{} $n=2m$ and ${\cal F}$ is the set of all $m$-subspaces of $\mathbb{Z}_{h}^n$ contained in a fixed $(m+r-1)$-subspace of $\mathbb{Z}_{h}^n$, or [(c)]{} $n=2m$ and there exists some $T\in G\!L_n(\mathbb{Z}_h)$ such that ${\cal F}={\cal F}_{(\alpha_1,\alpha_2,\ldots,\alpha_t)}^{(m-r+1,m,n,I)}T$ with $I\not=\emptyset,[t]$. By the definition of $G_r(m,n,\mathbb{Z}_{h})$, Theorem \[lem3.5\] and Lemma \[lem4.4\], ${\cal F}$ is a maximum clique of $G_r(m,n,\mathbb{Z}_{h})$.

Let ${\cal F}$ be a maximum clique of $G_r(m,n,\mathbb{Z}_{h})$. We will prove that either [(a)]{} ${\cal F}$ consists of all $m$-subspaces of $\mathbb{Z}_{h}^n$ which contain a fixed $(m-r+1)$-subspace of $\mathbb{Z}_{h}^n$, [(b)]{} $n=2m$ and ${\cal F}$ is the set of all $m$-subspaces of $\mathbb{Z}_{h}^n$ contained in a fixed $(m+r-1)$-subspace of $\mathbb{Z}_{h}^n$, or [(c)]{} $n=2m$ and there exists some $T\in G\!L_n(\mathbb{Z}_h)$ such that ${\cal F}={\cal F}_{(\alpha_1,\alpha_2,\ldots,\alpha_t)}^{(m-r+1,m,n,I)}T$ with $I\not=\emptyset,[t]$. There are the following two cases to be considered.

[**Case**]{} 1: $n>2m$. For each $i$ with $1\leq i\leq t$, by Lemmas \[lem4.2\] and \[lem4.3\], $\pi_i({\cal F})$ consists of all $m$-subspaces of $\mathbb{Z}_{p_i^{s_i}}^n$ which contain a fixed $(m-r+1)$-subspace $P_i$ of $\mathbb{Z}_{p_i^{s_i}}^n$. By Lemma \[lem2.2\], there is a $T_i\in G\!L_n(\mathbb{Z}_{p_i^{s_i}})$ such that $P_i=(0,I_{m-r+1})T_i$ for $i=1,2,\ldots,t$. So, we have $$\pi_i({\cal F})=\left\{\left(\begin{array}{cc}
X_i & 0\\
0 & I_{m-r+1}
\end{array}\right)T_i\in{\mathbb{Z}_{p_i^{s_i}}^{n}\brack m} : X_i\in{\mathbb{Z}_{p_i^{s_i}}^{n-m+r-1}\brack r-1}\right\}
\;\hbox{for}\;i=1,2,\ldots,t.$$ By Lemma \[lem2.7\], there is the unique $T\in G\!L_n(\mathbb{Z}_{h})$ such that $\pi_i(T)=T_i$ for $i=1,2,\ldots,t$. Since $\pi$ is a bijective map from $V(G_r(r-1,n-m+r-1,\mathbb{Z}_{h}))$ to $\prod_{i=1}^{t}V(G_r(r-1,n-m+r-1,\mathbb{Z}_{p_i^{s_i}}))$, we have $$\pi_i({\cal F})=\left\{\pi_i\left(\left(\begin{array}{cc}
X & 0\\
0 & I_{m-r+1}
\end{array}\right)T\right)\in{\mathbb{Z}_{p_i^{s_i}}^{n}\brack m} : X\in{\mathbb{Z}_{h}^{n-m+r-1}\brack r-1}\right\}
\;\hbox{for}\;i=1,2,\ldots,t,$$ which imply that $$\pi({\cal F})=\left\{\pi\left(\left(\begin{array}{cc}
X & 0\\
0 & I_{m-r+1}
\end{array}\right)T\right)\in\prod_{i=1}^t{\mathbb{Z}_{p_i^{s_i}}^{n}\brack m} : X\in{\mathbb{Z}_{h}^{n-m+r-1}\brack r-1}\right\}.$$ Since $\pi$ is a bijective map from $V(G_r(m,n,\mathbb{Z}_{h}))$ to $\prod_{i=1}^{t}V(G_r(m,n,\mathbb{Z}_{p_i^{s_i}}))$, we have $${\cal F}=\left\{\left(\begin{array}{cc}
X & 0\\
0 & I_{m-r+1}
\end{array}\right)T\in{\mathbb{Z}_{h}^{n}\brack m} : X\in{\mathbb{Z}_{h}^{n-m+r-1}\brack r-1}\right\},$$ which implies that ${\cal F}$ consists of all $m$-subspaces of $\mathbb{Z}_{h}^n$ which contain a fixed $(m-r+1)$-subspace of $\mathbb{Z}_{h}^n$.

[**Case**]{} 2: $n=2m$. By Lemmas \[lem4.2\] and \[lem4.3\], there exists some subset $I$ of $[t]$ such that, $\pi_i({\cal F})$ consists of all $m$-subspaces of $\mathbb{Z}_{p_i^{s_i}}^n$ which contain a fixed $(m-r+1)$-subspace of $\mathbb{Z}_{p_i^{s_i}}^n$ if $i\in I$, and $\pi_i({\cal F})$ is the set of all $m$-subspaces of $\mathbb{Z}_{p_i^{s_i}}^n$ contained in a fixed $(m+r-1)$-subspace of $\mathbb{Z}_{p_i^{s_i}}^n$ if $i\in[t]\setminus I$. Then there are the following three cases to be considered.

[*Case*]{} 2.1: $I=[t]$. Similar to the proof of Case 1, we may obtain that ${\cal F}$ consists of all $m$-subspaces of $\mathbb{Z}_{h}^n$ which contain a fixed $(m-r+1)$-subspace of $\mathbb{Z}_{h}^n$.

[*Case*]{} 2.2: $I=\emptyset$. For each $i$ with $1\leq i\leq t$, $\pi_i({\cal F})$ is the set of all $m$-subspaces of $\mathbb{Z}_{p_i^{s_i}}^n$ contained in a fixed $(m+r-1)$-subspace $Q_i$ of $\mathbb{Z}_{p_i^{s_i}}^n$. By Lemma \[lem2.2\], there is a $T_i\in G\!L_n(\mathbb{Z}_{p_i^{s_i}})$ such that $Q_i=(I_{m+r-1},0)T_i$ for $i=1,2,\ldots,t$. So, we have $$\pi_i({\cal F})=\left\{(X_i, 0)T_i\in{\mathbb{Z}_{p_i^{s_i}}^{n}\brack m} : X_i\in{\mathbb{Z}_{p_i^{s_i}}^{m+r-1}\brack m}\right\}\;\hbox{for}\;i=1,2,\ldots,t.$$ By Lemma \[lem2.7\], there is the unique $T\in G\!L_n(\mathbb{Z}_{h})$ such that $\pi_i(T)=T_i$ for $i=1,2,\ldots,t$. Since $\pi$ is a bijective map from $V(G_r(m,m+r-1,\mathbb{Z}_{h}))$ to $\prod_{i=1}^{t}V(G_r(m,m+r-1,\mathbb{Z}_{p_i^{s_i}}))$, we have $$\pi_i({\cal F})=\left\{\pi_i((X,0)T)\in{\mathbb{Z}_{p_i^{s_i}}^{n}\brack m} : X\in{\mathbb{Z}_{h}^{m+r-1}\brack m}\right\}\;\hbox{for}\;i=1,2,\ldots,t,$$ which imply that $$\pi({\cal F})=\left\{\pi((X,0)T)\in\prod_{i=1}^t{\mathbb{Z}_{p_i^{s_i}}^{n}\brack m} : X\in{\mathbb{Z}_{h}^{m+r-1}\brack m}\right\}.$$ Since $\pi$ is a bijective map from $V(G_r(m,n,\mathbb{Z}_{h}))$ to $\prod_{i=1}^{t}V(G_r(m,n,\mathbb{Z}_{p_i^{s_i}}))$, we have $${\cal F}=\left\{(X,0)T\in{\mathbb{Z}_{h}^{n}\brack m} : X\in{\mathbb{Z}_{h}^{m+r-1}\brack m}\right\},$$ which implies that ${\cal F}$ is the set of all $m$-subspaces of $\mathbb{Z}_{h}^n$ contained in a fixed $(m+r-1)$-subspace of $\mathbb{Z}_{h}^n$.

[*Case*]{} 2.3: $I\not=\emptyset,[t]$. If $i\in I$, $\pi_i({\cal F})$ consists of all $m$-subspaces of $\mathbb{Z}_{p_i^{s_i}}^n$ which contain a fixed $(m-r+1)$-subspace $P_i$ of $\mathbb{Z}_{p_i^{s_i}}^n$. By Lemma \[lem2.2\], there is a $T_i\in G\!L_n(\mathbb{Z}_{p_i^{s_i}})$ such that $P_i=(0,I_{m-r+1})T_i$ for each $i\in I$. So, we have $$\pi_i({\cal F})=\left\{\left(\begin{array}{cc}
X_i & 0\\
0 & I_{m-r+1}
\end{array}\right)T_i\in{\mathbb{Z}_{p_i^{s_i}}^{n}\brack m} : X_i\in{\mathbb{Z}_{p_i^{s_i}}^{m+r-1}\brack r-1}\right\}\;\hbox{for each}\;i\in I.$$

If $i\in[t]\setminus I$, $\pi_i({\cal F})$ is the set of all $m$-subspaces of $\mathbb{Z}_{p_i^{s_i}}^n$ contained in a fixed $(m+r-1)$-subspace $Q_i$ of $\mathbb{Z}_{p_i^{s_i}}^n$. By Lemma \[lem2.2\], there is a $T_i\in G\!L_n(\mathbb{Z}_{p_i^{s_i}})$ such that $Q_i=(I_{m+r-1},0)T_i$ for each $i\in [t]\setminus I$. So, we have $$\pi_i({\cal F})=\left\{\left(\begin{array}{cc}
X_i & 0\\
Y_i & 0
\end{array}\right)T_i\in{\mathbb{Z}_{p_i^{s_i}}^{n}\brack m} :
\left(\begin{array}{c}
X_i\\ Y_i
\end{array}\right)\in{\mathbb{Z}_{p_i^{s_i}}^{m+r-1}\brack m}\right\}\;\hbox{for each}\;i\in[t]\setminus I.$$

Let $T\in G\!L_n(\mathbb{Z}_{h})$ satisfy $\pi_i(T)=T_i$ for $i=1,2,\ldots,t$. Since $\pi$ is a bijective map from $V(G_r(m,n,\mathbb{Z}_{h}))$ to $\prod_{i=1}^{t}V(G_r(m,n,\mathbb{Z}_{p_i^{s_i}}))$ and $\pi_i(J_{(\alpha_1,\alpha_2,\ldots,\alpha_t)})=J_{\alpha_i}\subseteq\mathbb{Z}_{p_i^{s_i}}$ for $i=1,2,\ldots,t$, we have $$\begin{aligned}
\pi_i({\cal F})&\subseteq&\left\{\pi_i\left(\left(\begin{array}{cc}
X & 0\\
Y & xI_{m-r+1}
\end{array}\right)T\right)\in{\mathbb{Z}_{p_i^{s_i}}^n \brack m} :\right.\\
&&\left.X\in\mathbb{Z}_h^{(r-1)\times(m+r-1)},Y\in J_{(\alpha_1,\alpha_2,\ldots,\alpha_t)}^{(m-r+1)\times(m+r-1)}\right\}\;\hbox{for}\;i=1,2,\ldots,t,\end{aligned}$$ where $\alpha_i=s_i,\pi_i(x)=1$ if $i\in I$, and $\alpha_i=0,\pi_i(x)=0$ if $i\in[t]\setminus I$, which imply that $\pi({\cal F})\subseteq\pi\left({\cal F}_{(\alpha_1,\alpha_2,\ldots,\alpha_t)}^{(m-r+1,m,n,I)}T\right)$. Since $\pi$ is a bijective map, one obtains ${\cal F}\subseteq{\cal F}_{(\alpha_1,\alpha_2,\ldots,\alpha_t)}^{(m-r+1,m,n,I)}T$. Since ${\cal F}$ is a maximum clique, ${\cal F}={\cal F}_{(\alpha_1,\alpha_2,\ldots,\alpha_t)}^{(m-r+1,m,n,I)}T$. ${\hfill\Box\medskip}$

[**Proof of Theorem \[thm1.1\].**]{} Let $\lfloor n/2\rfloor\geq m\geq r\geq 0$ and ${\cal F}\subseteq{\mathbb{Z}_{h}^{n}\brack m}$ be an $r$-intersecting family. Without loss of generality, we assume that $1\leq r\leq m-1$. By Theorem \[lem3.3\], ${\cal F}$ is a clique of the generalized Grassmann graph $G_{m-r+1}(m,n,\mathbb{Z}_{h})$. By Lemma \[lem4.3\], $|{\cal F}|\leq\prod_{i=1}^tp_i^{(s_i-1)(n-m)(m-r)}{n-r\brack m-r}_{p_i}$ and equality holds if and only if ${\cal F}$ is a maximum clique of $G_{m-r+1}(m,n,\mathbb{Z}_{h})$. By Theorem \[lem4.5\], this theorem holds. ${\hfill\Box\medskip}$

Independence numbers of $G_{r}(m,n,\mathbb{Z}_{h})$
===================================================

In this section, we discuss the independence number of $G_{r}(m,n,\mathbb{Z}_{h})$ and give some estimations of the independence number $\alpha(G_{r}(m,n,\mathbb{Z}_{h}))$.

If $\Gamma$ is a vertex transitive graph, by Lemma 2.7.2 in [@Roberson], we obtain the following inequality $$\label{equa17}
|V(\Gamma)|\geq\alpha(\Gamma)\omega(\Gamma).$$

\[lem5.1\][(See Theorem 5.1 in [@Huang3].)]{} If $n\geq 2m$, then $$p^{(s-1)(n-m)(m-r+1)}\alpha(G_{r}(m,n,\mathbb{Z}_{p}))\leq\alpha(G_{r}(m,n,\mathbb{Z}_{p^s}))\leq p^{(s-1)(n-m)(m-r+1)}\frac{{n\brack m}_p}{{n-m+r-1\brack r-1}_p}.$$

\[lem5.2\] If $n\geq 2m$, then $$\prod_{i=1}^tp_i^{\omega_i}\alpha(G_{r}(m,n,\mathbb{Z}_{p_i}))\leq\alpha(G_{r}(m,n,\mathbb{Z}_{h}))\leq
\prod_{i=1}^t\frac{p_i^{\omega_i}{n\brack m}_{p_i}}{{n-m+r-1\brack r-1}_{p_i}},$$ where $\omega_i=(s_i-1)(n-m)(m-r+1)$ for $i=1,2,\ldots,t$.

[[*Proof.*]{}]{}Let ${\cal S}_i=\{S_{i1},S_{i2},\ldots,S_{i\xi_i}\}$ be a largest independent set of $G_{r}(m,n,\mathbb{Z}_{p_i^{s_i}})$, where $S_{i\eta_i}\in {\mathbb{Z}_{p_i^{s_i}}^{n}\brack m}$ for $1\leq i\leq t$ and $1\leq \eta_i\leq \xi_i$. By the definition of $G_{r}(m,n,\mathbb{Z}_{p_i^{s_i}})$, we obtain $\dim(S_{ij}\cap S_{i\ell})\leq m-r$ for $1\leq i\leq t$ and $1\leq j,\ell\leq \xi_i$ with $j\not=\ell$. By Lemma \[lem5.1\], we have $$\xi_i=|{\cal S}_i|\geq p_i^{(s_i-1)(n-m)(m-r+1)}\alpha(G_{r}(m,n,\mathbb{Z}_{p_i}))\;\hbox{for}\;i=1,2,\ldots,t.$$ Let $${\cal S}=\{\pi^{-1}(S_{1\eta_1},S_{2\eta_2},\ldots,S_{t\eta_t}):1\leq \eta_i\leq \xi_i\;\hbox{for}\;i=1,2,\ldots,t\}.$$ Since $\pi$ is a bijective map from $V(G_r(m,n,\mathbb{Z}_{h}))$ to $\prod_{i=1}^{t}V(G_r(m,n,\mathbb{Z}_{p_i^{s_i}}))$, we have $$|{\cal S}|=\xi_1\xi_2\cdots\xi_t\geq \prod_{i=1}^tp_i^{(s_i-1)(n-m)(m-r+1)}\alpha(G_{r}(m,n,\mathbb{Z}_{p_i})).$$

For any two distinct vertices $A,B\in{\cal S}$, we have $\pi_i(A),\pi_i(B)\in{\cal S}_i$ for $i=1,2,\ldots,t$, which imply that $\dim(A\cap B)=\min\{\dim(\pi_i(A)\cap\pi_i(B)): i=1,2,\ldots,t\}\leq m-r$ by Theorem \[lem3.3\]. Therefore, ${\cal S}$ is a independent set of $G_{r}(m,n,\mathbb{Z}_{h})$, which implies that $\alpha(G_{r}(m,n,\mathbb{Z}_{h}))\geq |{\cal S}|\geq \prod_{i=1}^tp_i^{(s_i-1)(n-m)(m-r+1)}\alpha(G_{r}(m,n,\mathbb{Z}_{p_i}))$.

Since $G_{r}(m,n,\mathbb{Z}_{h})$ is vertex transitive, by (\[equa17\]) and Theorem \[lem3.5\], we have $$\alpha(G_{r}(m,n,\mathbb{Z}_{h}))\leq \prod_{i=1}^tp_i^{(s_i-1)(n-m)(m-r+1)}\frac{{n\brack m}_{p_i}}{{n-m+r-1\brack r-1}_{p_i}}.$$ Therefore, the desired result follows. ${\hfill\Box\medskip}$

For a vertex transitive graph $\Gamma$ and $\Delta\subseteq V(\Gamma)$, let $\alpha(\Delta)$ denote the independence number of the subgraph of $\Gamma$ induced by $\Delta$.

\[lem5.3\][(See Lemma 2.1 in [@Zhang].)]{} Let $\Gamma$ be a vertex transitive graph. Then $\frac{\alpha(\Gamma)}{|V(\Gamma)|}\leq\frac{\alpha(\Delta)}{|\Delta|}$ for any $\Delta\subseteq V(\Gamma)$, and equality implies that $|S\cap\Delta|=\alpha(\Delta)$ for every largest independent set $S$ of $\Gamma$.

\[lem5.4\] Let $n>m\geq2$. Then the following inequalities hold: $$\label{equa18}
  \alpha(G_{r}(m,n,\mathbb{Z}_{h}))\leq\left\lfloor\beta_{0}\alpha(G_{r}(m-1,n-1,\mathbb{Z}_{h}))\right\rfloor,$$ $$\label{equa19}
\alpha(G_{r}(m,n,\mathbb{Z}_{h}))\leq\left\lfloor\beta_{0}
\left\lfloor\beta_{1}\cdots
\left\lfloor\beta_{m-r}\right\rfloor\cdots\right\rfloor\right\rfloor,$$ where $$\beta_{j}=\prod_{i=1}^tp_i^{(s_i-1)(n-m)}\frac{p_i^{n-j}-1}{p_i^{m-j}-1}\quad\hbox{for}\quad j=0,1,\ldots,m-r.$$

[[*Proof.*]{}]{}Let $$\Delta=\left(\begin{array}{cc}
1 & 0\\
0 & V(G_{r}(m-1,n-1,\mathbb{Z}_{h}))
\end{array}\right).$$ Then $\alpha(\Delta)=\alpha(G_{r}(m-1,n-1,\mathbb{Z}_{h}))$. By Lemma \[lem5.3\] and Theorem \[lem3.5\], we have $$\alpha(G_{r}(m,n,\mathbb{Z}_{h}))\leq\frac{|V(G_{r}(m,n,\mathbb{Z}_{h}))|}{|\Delta|}\alpha(\Delta)=\beta_{0}\alpha(G_{r}(m-1,n-1,\mathbb{Z}_{h})).$$ Since $\alpha(G_{r}(m,n,\mathbb{Z}_{h}))$ is a positive integer, we obtain (\[equa18\]). Since $G_{r}(r-1,n-m+r-1,\mathbb{Z}_{h})$ is a complete graph, $\alpha(G_{r}(r-1,n-m+r-1,\mathbb{Z}_{h}))=1$. By (\[equa18\]), we obtain (\[equa19\]). ${\hfill\Box\medskip}$

\[lem5.5\][(See Theorem 5.4 in [@Huang3].)]{} If $n>m\geq2$ and $m$ divides $n$, then $$\alpha(G_{m}(m,n,\mathbb{Z}_{p^s}))=p^{(s-1)(n-m)}\frac{p^{n}-1}{p^{m}-1}.$$

\[lem5.6\] If $n>m\geq2$ and $m$ divides $n$, then $$\alpha(G_{m}(m,n,\mathbb{Z}_{h}))=\prod_{i=1}^tp_i^{(s_i-1)(n-m)}\frac{p_i^{n}-1}{p_i^{m}-1}.$$

[[*Proof.*]{}]{}Since $G_{m}(m-1,n-1,\mathbb{Z}_{h})$ is a complete graph, $\alpha(G_{m}(m-1,n-1,\mathbb{Z}_{h}))=1$. By (\[equa18\]), when $n>m\geq 2$, we have $$\alpha(G_{m}(m,n,\mathbb{Z}_{h}))\leq\left\lfloor\prod_{i=1}^tp_i^{(s_i-1)(n-m)}\frac{p_i^{n}-1}{p_i^{m}-1}\right\rfloor.$$ If $n>m\geq2$ and $m$ divides $n$, by Lemma \[lem5.5\], $\alpha(G_{m}(m,n,\mathbb{Z}_{p_i}))=(p_i^n-1)/(p_i^m-1)$ for $i=1,2,\ldots,t$. By Theorem \[lem5.2\], we have $$\prod_{i=1}^tp_i^{(s_i-1)(n-m)}\frac{p_i^{n}-1}{p_i^{m}-1}\leq\alpha(G_{m}(m,n,\mathbb{Z}_{h})).$$ Therefore, the desired result follows. ${\hfill\Box\medskip}$

\[lem5.7\][(See Theorem 5.5 in [@Huang3].)]{} If $n>m\geq2$ and $n\equiv\; \ell\mod\; m$, then $$\alpha(G_{m}(m,n,\mathbb{Z}_{p^s}))\geq p^{(s-1)(n-m)}\frac{p^{n}-p^m(p^\ell-1)-1}{p^{m}-1}.$$

\[lem5.8\] If $n>m\geq2$ and $n\equiv\; \ell \mod\; m$, then $$\label{equa20}
\alpha(G_{m}(m,n,\mathbb{Z}_{h}))\geq\prod_{i=1}^tp_i^{(s_i-1)(n-m)}\frac{p_i^{n}-p_i^m(p_i^\ell-1)-1}{p_i^{m}-1}.$$

[[*Proof.*]{}]{}By Lemma \[lem5.7\], we have $$\alpha(G_{m}(m,n,\mathbb{Z}_{p_i}))\geq \frac{p_i^{n}-p_i^m(p_i^\ell-1)-1}{p_i^{m}-1}\;\hbox{for}\;i=1,2,\ldots,t.$$ By Theorem \[lem5.2\], we obtain (\[equa20\]). ${\hfill\Box\medskip}$

For two subspaces $A$ and $B$ of $\mathbb{Z}_h^n$, the [*subspace distance*]{} between $A$ and $B$ is $d(A,B):=\dim(A)+\dim(B)-2\dim(A\cap B)$. It is easy to prove that $d(A,B)\geq0,d(A,B)=0\Leftrightarrow A=B$, $d(A,B)=d(B,A)$ and $d(A,C)\leq d(A,B)+d(B,C)$ for subspaces $A,B$ and $C$ of $\mathbb{Z}_h^n$.

Huang et al. [@Huang3] discussed the Grassmannian code for $\mathbb{Z}_{p^s}^n$. As a natural extension, a [*Grassmannian code*]{} ${\cal C}$ for $\mathbb{Z}_h^n$ is a subset of ${\mathbb{Z}_h^n\brack m}$. An $(n,2r,m)_{\mathbb{Z}_h}$ [*code*]{} for $\mathbb{Z}_h^n$ is a Grassmannian code for $\mathbb{Z}_h^n$ with minimum subspace distance $2r$. An $(n,2r,m)_{\mathbb{Z}_h}$ code with the maximum cardinality is called an [*optimal*]{} $(n,2r,m)_{\mathbb{Z}_h}$ code for $\mathbb{Z}_h^n$.

Note that $d(A,B)\geq 2r$ if and only if $\dim(A\cap B)\leq m-r$ for subspaces $A,B$ in ${\mathbb{Z}_h^n\brack m}$. It follows that an $(n,2r,m)_{\mathbb{Z}_h}$ code for $\mathbb{Z}_h^n$ is an independent set of $G_r(m,n,\mathbb{Z}_h)$ and vice versa. Therefore, an optimal $(n,2r,m)_{\mathbb{Z}_h}$ code for $\mathbb{Z}_h^n$ is a largest independent set of $G_r(m,n,\mathbb{Z}_h)$ and vice versa.

Acknowledgment {#acknowledgment .unnumbered}
==============

This research is supported by National Natural Science Foundation of China (11971146).

[00]{}

A.E. Brouwer, A.M. Cohen, A. Neumaier, Distance-Regular Graphs, Springer-Verlag, Berlin, Heidelberg, New York, 1989.

W.C. Brown, Matrices over Commutative rings, Marcel Dekker, New York/Basel/Hong Kong, 1993.

P.M. Cohn, Free Ideal Rings and Localization in General Rings, Cambridge University Press, Cambridge, 2006.

P.M. Cohn, Free Ring and Their Relations, second edition, Academic Press, London, 1985.

P. Erdős, C. Ko, R. Rado, Intersection theorems for systems of finite sets, Quart. J. Math. Oxford Ser. 12 (1961) 313–320.

P. Frankl, R.M. Wilson, The Erdős-Ko-Rado theorem for vector spaces, J. Combin. Theory Ser. A 43 (1986) 228–236.

C. Godsil, K. Meagher, Erdős-Ko-Rado Theorem: Algebraic Approaches, Cambridge University Press, Cambridge, 2016.

J. Guo, Erdős-Ko-Rado theorem and bilinear forms graphs for matrices over residue class rings, arXiv:2002.03560.

L. Huang, B. Lv, K. Wang, Erdős-Ko-Rado theorem, Grassmann graphs and $p^s$-Kneser graphs for vector spaces over a residue class ring, J. Comb. Theory Ser. A 164 (2019) 125–158.

T. Huang, An analogue of the Erdős-Ko-Rado theorem for the distance-regular graphs of bilinear forms, Discrete Math. 64 (1987) 191–198.

K. Ireland, M. Rosen, A Classical Introduction to Modern Number Theory, Springer-Verlag, 1982.

M. Pankov, Geometry of Semilinear Embeddongs, World Scientific, New Jersey/London/Singapore, 2015.

D.E. Roberson, Variations on a Theme: Graph Homomorphisms, Ph.D. dissertation, University of Waterloo, 2013.

M. Rosen, Direct products of cyclic groups, Amer. Math. Monthly 73 (1966) 55–57.

H. Tanaka, Classification of subsets with minimal width and dual width in Grassmann, bilinear forms and dual polar graphs, J. Comb. Theory Ser. A 113 (2006) 903–910.

F. Vanhove, V. Pepe, L. Storme, Theorems of Erdős-Ko-Rado-type in polar spaces, J. Combin. Theory Ser. A 118 (2011) 1291–1312.

Z. Wan, Geometry of Classical Groups over Finite Fields, second edition, Science Press, Beijing/New York, 2002.

R.M. Wilson, The exact bound in the Erdős-Ko-Rado theorem, Combinatorica 4 (1984) 247–257.

H. Zhang, Primitivity and independent sets in direct products of vertex-transitive graphs, J. Graph Theory 67 (2011) 218–225.

[^1]: Corresponding author. guojun$_-$lf@163.com
