The Sandpile Group of a Family of Nearly Complete Graphs

Abstract

Let $K_n-C_{k}$ denote the graph obtained from the complete graph $K_n$ by deleting k edges of a cycle $C_k$, that is, a nearly complete graph. In this paper, we completely determine the structure of the sandpile group of $K_{n}-C_{k}$ for $3\le k\le n-2$.

1 Introduction

The sandpile group is originated from the Abelian Sandpile Model in statistical physics [5]. In fact, the sandpile group pops up in many different fields under different names, such as the critical group in the chip-firing game [2,3,4], the Picard group or the Jacobian group in the divisor theory of graphs [1], the group of components on arithmetical graphs [8], etc.

In [7], Lorenzini studied the structure of the sandpile group of a complete graph with edges of two disjoint paths $P_a$ and $P_b$ deleted. He showed that if $a+b=n-1$ and $gcd(a,b)=1$, then $S(K_n-P_a-P_b)$ is cyclic, and at the same time, he suggested that the sandpile groups of $K_n-C_{n-1}$ and $K_n-C_{n}$ are never cyclic but with the problems open. Then, in 2011, Norine and Whalen [10] not only showed the other direction of the first result is also true, but settled the case $K_n-C_{n-1}$ by completely determining the structure of its sandpile group. Motivated by the above two papers, we shall consider the sandpile group of the graph $K_{n}-C_{k}$ for $3\le k\le n-2$ in this paper.

Let $G = (V, E)$ be a finite loopless graph with n vertices. Then, its Laplacian matrix $L(G) = D(G)-A(G)$, where $D(G) = \hbox {diag}(d_1, d_2, \ldots , d_n)$ and A(G) are the degree matrix and the adjacency matrix of G, respectively. Thinking of L(G) as a linear map ${\mathbb {Z}}^n\rightarrow {\mathbb {Z}}^n$, its cokernel has the form

$$\begin{aligned} coker L(G) = {\mathbb {Z}}^n/L(G){\mathbb {Z}}^n ={\mathbb {Z}}\oplus S(G), \end{aligned}$$

where S(G) is the sandpile group on G in the sense of isomorphism.

For $v\in V(G)$, we define the reduced Laplacian ${\tilde{L}}_v$ as the submatrix of L(G) obtained by omitting the row and column corresponding to v. Then, it is well known that for any $v\in V(G)$,

$$\begin{aligned} S(G)\cong {\mathbb {Z}}^{n-1}/{\tilde{L}}_v{\mathbb {Z}}^{n-1}=coker({\tilde{L}}_v). \end{aligned}$$

That is, $coker({\tilde{L}}_v)$ is independent of the choice of vertex v. Thus, we simply write ${\tilde{L}}_v$ as ${\tilde{L}}$.

Recall that two matrices $A,B\in {\mathbb {Z}}^{m\times n}$ are (unimodular) equivalent (written $A\sim B$) [9] if there exist $P\in GL(m,{\mathbb {Z}})$, $Q\in GL(n,{\mathbb {Z}})$, such that $B=PAQ$. Equivalently, B is obtained from A by a sequence of integer row and column operations that are invertible over the ring ${\mathbb {Z}}$ of integers. For any square integer matrix A, it is equivalent to a unique diagonal matrix $S(A) = \hbox {diag}(s_1, s_2,\ldots ,s_n)$ ( the Smith normal form) whose entries are nonnegative and $s_i$ divides $s_{i+1} (i=1,\ldots ,n-1)$. It can be seen easily that $A \sim B$ implies $coker A \cong coker B$. So if the Smith normal form of the reduced Laplacian ${\tilde{L}}$ is $\hbox {diag}(s_1, s_2,\ldots , s_{n-1})$, where $1=s_1=\cdots =s_{r-1}<s_{r}\le \cdots \le s_{n-1}$, then the sandpile group of G

$$\begin{aligned} S(G)={\mathbb {Z}}_{s_{r}}\oplus {\mathbb {Z}}_{s_{r+1}}\oplus \cdots \oplus {\mathbb {Z}}_{s_{n-1}}, \end{aligned}$$

where ${\mathbb {Z}}_{a}={\mathbb {Z}}/a{\mathbb {Z}}$. $s_r,\ldots ,s_n$ are called invariant factors of S(G), and $\mu (G)=n-r$ is the minimum number of generators of S(G).

The rest of the paper is arranged as follows: In Sect. 2, we give several lemmas which are needed for the main results. In Sect. 3, we determine the explicit structure of the sandpile group of $K_{n}-C_{k}$,$(3\le k\le n-2)$. In Sect. 4, as applications of the main results, we deduce more explicit results for the sandpile group of $K_{n}-C_{k}$ for $k=3,4,5,6,7$ and prime k.

2 Preliminaries

In this section, we shall first give several lemmas.

Lemma 2.1

[9] Suppose that A is an $n\times n$ integer matrix, and its Smith normal form is $\hbox {diag}(s_1,s_2,\ldots ,s_n)$. Then,

$$\begin{aligned} s_i=\frac{\Delta _i}{\Delta _{i-1}},\quad \ \ i=1,2,\ldots ,n, \end{aligned}$$

where $\Delta _i$ (called i-th determinant divisor) is the greatest common divisor of the $i\times i$ minors of A. By convention $\Delta _0:=1$.

Now let $a_{k}(t), b_{k}(t), c_{k}(t)$ and $d_k(t)$ be four sequences satisfying the same linear recurrence relation: $y_{k}(t)= ty_{k-1}(t)-y_{k-2}(t)$ with initial values as follows, respectively,

$$\begin{aligned} \begin{aligned}&a_{0}(t)=1,\&a_{1}(t)=t;\\&b_{0}(t)=2,\&b_{1}(t)=t;\\&c_{0}(t)=1,\&c_{1}(t)=t+1;\\&d_{0}(t)=1,\&d_{1}(t)=t-1. \end{aligned} \end{aligned}$$

Solving the linear recurrence relations, it is easy to obtain that

$$\begin{aligned} \begin{aligned}&a_{k}(t)=\frac{1}{\sqrt{t^{2}-4}}(\lambda _1^{k+1}-\lambda _2^{k+1});\\&b_{k}(t)=\lambda _1^{k}+\lambda _2^{k};\\&c_{k}(t)=\frac{1}{t-2}(\lambda _1^{k}(\lambda _1-1)+\lambda _2^{k}(\lambda _2-1));\\&d_{k}(t)=\frac{1}{t+2}(\lambda _1^{k}(\lambda _1+1)+\lambda _2^{k}(\lambda _2+1)); \end{aligned} \end{aligned}$$

where

$$\begin{aligned} \lambda _1=\frac{t+\sqrt{t^{2}-4}}{2}, \ \ \lambda _2=\frac{t-\sqrt{t^{2}-4}}{2} \end{aligned}$$

are two roots of the characteristic equation $x^2-tx+1=0$.

Lemma 2.2

For $a_{k}(t), b_{k}(t), c_k(t)$ and $d_k(t)$ as defined above, we have

• (1) if $k=2m$, then

  $$\begin{aligned} \left\{ \begin{aligned}&a_{2m}(t)= c_{m}(t)d_m(t); \\&a_{2m}(t)+1= a_{m}(t)b_{m}(t). \end{aligned} \right. \end{aligned}$$

• (2) if $k=2m-1$, then

  $$\begin{aligned} \left\{ \begin{aligned}&a_{2m-1}(t)= a_{m-1}(t)b_{m}(t);\\&a_{2m-1}(t)-1= c_{m-1}(t)d_{m}(t). \end{aligned} \right. \end{aligned}$$
• (3)
  $$\begin{aligned} \begin{aligned}&tb_{m}(t)-2b_{m-1}(t)=(t^{2}-4)a_{m-1}(t);\\&tc_m(t)-2c_{m-1}(t)=(t+2)d_m(t). \end{aligned} \end{aligned}$$

Proof

Using the facts $\lambda _1\lambda _2=1, \lambda _1+\lambda _2=t, \lambda _1-\lambda _2=\sqrt{t^{2}-4}$. It is easy to check the identities. $\square$

Note that if t is an integer, then $a_{k}(t), b_{k}(t), c_k(t)$ and $d_k(t)$ are integer sequences. Let gcd(a, b) denote the greatest common divisor of a and b. Then, we have the following results.

Lemma 2.3

Let t be an integer, then for any $m>1$,

$$\begin{aligned} gcd(a_{m}(t),a_{m-1}(t))=gcd(c_{m}(t),c_{m-1}(t))=gcd(d_{m}(t),d_{m-1}(t))=1 \end{aligned}$$

and

$$\begin{aligned} gcd(b_{m}(t),b_{m-1}(t))=gcd(2,t). \end{aligned}$$

Proof

We only need to notice that for any sequence $x_n$ satisfying $x_n=tx_{n-1}-x_{n-2}$,

$$\begin{aligned} gcd(x_n, x_{n-1})=gcd(x_{n-1},x_{n-2})=\cdots =gcd(x_1,x_0). \end{aligned}$$

Thus, the results follow directly. $\square$

Using the fact

$$\begin{aligned} \begin{vmatrix} t&1&0&\cdots&0 \\ 1&t&1&\cdots&0 \\ 0&1&t&\cdots&0 \\ \vdots&\vdots&\vdots&\vdots \\ 0&0&0&\cdots&t \\ \end{vmatrix}_{k\times k}=a_k(t) \end{aligned}$$

for $k\ge 1$, we obtain the following result.

Lemma 2.4

For $k\ge 3$, let

$$\begin{aligned} A_k(t)=\begin{pmatrix} t &{} 1 &{} 0 &{} \cdots &{} 1 \\ 1 &{} t &{} 1 &{} \cdots &{} 0 \\ 0 &{} 1 &{} t &{} \cdots &{} 0 \\ \vdots &{} \vdots &{} \vdots &{} &{} \vdots \\ 1 &{} 0 &{} 0 &{} \cdots &{} t \\ \end{pmatrix}_{k\times k} \end{aligned}$$

then

$$\begin{aligned} \det A_k(t)=\left\{ \begin{array}{ll} (t^{2}-4)a_{m-1}^{2}(t),&{}\quad \ \ \text{ if }\ \ k=2m; \\ (t+2)d_{m}^{2}(t),&{}\quad \ \ \text{ if }\ \ k=2m+1. \end{array} \right. \end{aligned}$$

Proof

It is easy to see that

$$\begin{aligned} \det A_k(t)=ta_{k-1}(t)-2(a_{k-2}(t)+(-1)^k). \end{aligned}$$

(1) If $k=2m$, then by Lemma 2.2

$$\begin{aligned} \begin{aligned} \det A_k(t)&=ta_{2m-1}(t)-2(a_{2m-2}(t)+1)\\&=ta_{m-1}(t)b_{m}(t)-2a_{m-1}(t)b_{m-1}(t))\\&=a_{m-1}(t)(tb_{m}(t)-2b_{m-1}(t))\\&=(t^{2}-4)a_{m-1}^{2}(t). \end{aligned} \end{aligned}$$

(2) If $k=2m+1$, then by Lemma 2.2 again

$$\begin{aligned} \begin{aligned} \det A_k(t)&=ta_{2m}(t)-2(a_{2m-1}(t)-1)\\&=tc_{m}(t)d_{m}(t)-2c_{m-1}(t)d_{m}(t)\\&=d_{m}(t)(tc_m(t)-2c_{m-1}(t))\\&=(t+2)d_{m}^{2}(t). \end{aligned} \end{aligned}$$

Now the proof is completed. $\square$

3 The Sandpile Group of $K_{n}-C_{k}$

After the preparatory work we have done in the above section, in this section, we shall give the explicit structure of the sandpile group of the graph $G=K_{n}-C_{k}$ for $3\le k\le n-2$. Let $V(G)=\{1, \ldots , n\}$, by symmetry, we may assume that $d_1=d_2=\cdots =d_k=n-3$ and $d_{k+1}=\cdots =d_n=n-1$. Then, the Laplacian matrix of G has the form

$$\begin{aligned} L= \begin{pmatrix} n-3 &{} 0 &{} -1 &{} \cdots &{} 0 &{} -1 &{} \cdots &{} -1 \\ 0 &{} n-3 &{} 0 &{} \cdots &{} -1 &{} -1 &{} \cdots &{} -1 \\ -1 &{} 0 &{} n-3 &{} \cdots &{} -1 &{} -1 &{} \cdots &{} -1 \\ \vdots &{} \vdots &{} \vdots &{} &{} \vdots &{} \vdots &{} &{} \vdots \\ 0 &{} -1 &{} -1 &{} \cdots &{} n-3 &{} -1 &{} \cdots &{} -1 \\ -1 &{} -1 &{} -1 &{} \cdots &{} -1 &{} n-1 &{} \cdots &{} -1 \\ \vdots &{} \vdots &{} \vdots &{} &{} \vdots &{} \vdots &{} \ &{} \vdots \\ -1 &{} -1 &{} -1 &{} \cdots &{} -1 &{} -1 &{} \cdots &{} n-1 \\ \end{pmatrix}_{n\times n} \end{aligned}$$

Deleting the last row and column from L to obtain the reduced Laplacian matrix, ${\tilde{L}}$ of G. For ${\tilde{L}}$, first subtracting the last row of ${\tilde{L}}$ from all other rows, and then subtracting the last column from all other columns. We get

$$\begin{aligned} {\tilde{L}}\sim \begin{pmatrix} n-2 &{} 1 &{} 0 &{} \cdots &{} 1 &{} 0 &{} \cdots &{} 0 &{} 0\\ 1 &{} n-2 &{} 1 &{} \cdots &{} 0 &{} 0 &{} \cdots &{} 0 &{} 0\\ 0 &{} 1 &{} n-2 &{} \cdots &{} 0 &{} 0 &{} \cdots &{} 0 &{} 0\\ \vdots &{} \vdots &{} \vdots &{} &{} \vdots &{} \vdots &{} &{} \vdots &{} \vdots \\ 1 &{} 0 &{} 0 &{} \cdots &{} n-2 &{} 0 &{} \cdots &{} 0 &{} 0\\ 0 &{} 0 &{} 0 &{} \cdots &{} 0 &{} n &{} \cdots &{} 0 &{} 0\\ \vdots &{} \vdots &{} \vdots &{} &{} \vdots &{} \vdots &{} &{} \vdots &{} \vdots \\ 0 &{} 0 &{} 0 &{} \cdots &{} 0 &{} 0 &{} \cdots &{} n &{} 0\\ 0 &{} 0 &{} 0 &{} \cdots &{} 0 &{} 0 &{} \cdots &{} 0 &{} 1\\ \end{pmatrix}_{(n-1)\times (n-1)} = \begin{pmatrix} A_k(n-2) &{} 0\\ 0 &{} B \end{pmatrix}, \end{aligned}$$

where

$$\begin{aligned} B= \begin{pmatrix} n &{} \cdots &{} 0 &{} 0 \\ \vdots &{} &{} \vdots &{} \vdots \\ 0 &{} \cdots &{} n &{} 0 \\ 0 &{} \cdots &{} 0 &{} 1 \\ \end{pmatrix}_{(n-1-k)\times (n-1-k)}. \end{aligned}$$

That is,

$$\begin{aligned} S(K_n-C_k)\cong ({\mathbb {Z}}_{n})^{n-k-2}\oplus coker(A_k(n-2)). \end{aligned}$$

(1)

Remark

Note that $K_n-C_k$ can be viewed as the join of the graph $K_k-C_k$ with the complete graph $K_{n-k}$. The structure of the sandpile group of this family of graphs had been addressed in [6, 11]. (1) is just the same result of Theorem 1 (1) in [6]. In the following, we first determine the explicit structure of $coker(A_k(n-2))$. Then, we determine all invariant factors of $S(K_n-C_k)$.

Theorem 3.1

For the Smith normal form of $A_k(n-2)$, we have

$$\begin{aligned} A_k(n-2)= \begin{pmatrix} n-2 &{} 1 &{} 0 &{} \cdots &{} 1 \\ 1 &{} n-2 &{} 1 &{} \cdots &{} 0 \\ 0 &{} 1 &{} n-2 &{} \cdots &{} 0 \\ \vdots &{} \vdots &{} \vdots &{} &{} \vdots \\ 1 &{} 0 &{} 0 &{} \cdots &{} n-2 \\ \end{pmatrix}_{k\times k} \sim \begin{pmatrix} 1 &{} 0 &{} \cdots &{} 0 &{} 0 \\ 0 &{} 1 &{} \cdots &{} 0 &{} 0 \\ \vdots &{} \vdots &{} &{} \vdots &{} \vdots \\ 0 &{} 0 &{} \cdots &{} s_{1} &{} 0 \\ 0 &{} 0 &{} \cdots &{} 0 &{} s_{2} \\ \end{pmatrix}_{k\times k} \end{aligned}$$

where

$$\begin{aligned} s_{1}= & {} \left\{ \begin{array}{ll} a_{m-1}(n-2)gcd (2, n),&{}\quad \ \ \text{ if }\ \ k=2m; \\ d_{m}(n-2),&{}\quad \ \ \text{ if }\ \ k=2m+1. \end{array} \right. \\ s_{2}= & {} \left\{ \begin{array}{ll} \frac{n(n-4)}{gcd^{2}(2,n)}s_{1},&{}\quad \ \ \text{ if }\ \ k=2m; \\ ns_{1},&{}\quad \ \ \text{ if }\ \ k=2m+1. \end{array} \right. \end{aligned}$$

Proof

Note that there exists a $(k-2)\times (k-2)$ minor in $A_k(n-2)$ with value 1 (deleting the first and the $k-$th columns, the $k-1$-th and k-th rows, respectively). So $\Delta _1=\cdots =\Delta _{k-2}=1$. Also $\Delta _k=\det A_k(n-2)$ is given in Lemma 2.4. The only remain is to compute $\Delta _{k-1}$. Since $A_k(n-2)$ is a circulant matrix, so does its adjoint matrix $adj(A_k(n-2))$. Hence,

$$\begin{aligned} \Delta _{k-1}=gcd(A_{ij};i,j=1,\ldots ,k)=gcd(A_{1j}; j=1,\ldots ,k), \end{aligned}$$

where $A_{ij}(1 \le j \le k)$ is the algebraic cofactor of $ij-$element of $A_k(n-2)$. For simplicity, let $(x_{1},x_{2},\ldots ,x_{k})=(A_{11},A_{12},\ldots ,A_{1k})$. Note that $x_{i}=x_{k+2-i}(i=2,3,\ldots ,k$) since $A_k(n-2)$ is a symmetric circulant matrix. From

$$\begin{aligned} \begin{pmatrix} n-2 &{} 1 &{} 0 &{} \cdots &{} 1 \\ 1 &{} n-2 &{} 1 &{} \cdots &{} 0 \\ 0 &{} 1 &{} n-2 &{} \cdots &{} 0 \\ \vdots &{} \vdots &{} \vdots &{} &{} \vdots \\ 1 &{} 0 &{} 0 &{} \cdots &{} n-2 \\ \end{pmatrix}_{k\times k} \begin{pmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \vdots \\ x_{k} \end{pmatrix} =\begin{pmatrix} \det A_k(n-2) \\ 0 \\ 0 \\ \vdots \\ 0 \end{pmatrix}, \end{aligned}$$

we obtain that

$$\begin{aligned} \begin{aligned}&\det A_k(n-2)=(n-2)x_{1}+2x_{2}\\&x_{1}+(n-2)x_{2}+x_{3}=0 \Rightarrow x_{3}=-x_{1}-(n-2)x_{2}\\&x_{2}+(n-2)x_{3}+x_{4}=0 \Rightarrow x_{4}=-x_{2}-(n-2)x_{3}\\&\vdots \\&x_{k-2}+(n-2)x_{k-1}+x_{k}=0 \Rightarrow x_{k}=-x_{k-2}-(n-2)x_{k-1}. \end{aligned} \end{aligned}$$

That is, $x_{3},\ldots ,x_{k}$ can be represented linearly by $x_{1},x_{2}$.

So

$$\begin{aligned} \Delta _{k-1}= & {} gcd(x_{1},x_{2},\ldots ,x_{k})= gcd(x_{1},x_{2}).\\ x_{1}= & {} A_{11}= \begin{vmatrix} n-2&1&\cdots&0\\ 1&n-2&\cdots&0\\ \vdots&\vdots&\vdots \\ 0&0&\cdots&n-2\\ \end{vmatrix}_{(k-1)\times (k-1)} =a_{k-1}(n-2)\\ x_{2}= & {} A_{12}=A_{1k}=(-1)^{k+1}\cdot \begin{vmatrix} 1&n-2&1&\cdots&0\\ 0&1&n-2&\cdots&0\\ \vdots&\vdots&\vdots&\vdots \\ 1&0&0&\cdots&1\\ \end{vmatrix}_{(k-1)\times ((k-1)}\\= & {} (-1)^{k+1}\cdot (1+(-1)^{k}a_{k-2}(n-2))\\= & {} (-1)^{k+1}-a_{k-2}(n-2), \end{aligned}$$

So by Lemma 2.1,

$$\begin{aligned} \left\{ \begin{aligned}&s_{1}=\Delta _{k-1}=gcd(a_{k-1}(n-2),a_{k-2}(n-2)+(-1)^{k});\\&s_{2}=\frac{\det A_{k}(n-2)}{s_{1}}. \end{aligned} \right. \end{aligned}$$

(1) If $k=2m$, then by Lemmas 2.2, 2.3 and 2.4

$$\begin{aligned} \begin{aligned} s_{1}&=gcd(a_{2m-1}(n-2),a_{2m-2}(n-2)+1)\\&=gcd(a_{m-1}(n-2)b_{m}(n-2),a_{m-1}(n-2)b_{m-1}(n-2))\\&=a_{m-1}(n-2)gcd(b_{m}(n-2),b_{m-1}(n-2))\\&=a_{m-1}(n-2)gcd(2,n-2)\\&=a_{m-1}(n-2)gcd(2,n),\\ s_{2}&=\frac{(n^{2}-4n)a_{m-1}^{2}(n-2)}{a_{m-1}(n-2)gcd(2,n)}\\&=\frac{n(n-4)}{gcd^{2}(2,n)}s_{1}. \end{aligned} \end{aligned}$$

(2) If $k=2m+1$, then by Lemmas 2.2, 2.3 and 2.4

$$\begin{aligned} \begin{aligned} s_{1}&=gcd(a_{2m}(n-2),a_{2m-1}(n-2)-1)\\&=gcd(c_{m}(n-2)d_{m}(n-2),c_{m-1}(n-2)d_{m}(n-2))\\&=d_{m}(n-2)gcd(c_{m}(n-2),c_{m-1}(n-2))\\&=d_{m}(n-2),\\ s_{2}&=\frac{nd_{m}^{2}(n-2)}{d_{m}(n-2)}\\&=ns_{1}. \end{aligned} \end{aligned}$$

$\square$

In order to give all invariant factors of $S(K_n-C_k)$, we need the following result about $a_{k}(n-2)$.

Lemma 3.2

$$\begin{aligned} a_{k}(n-2)=nf_{k}(n)+(-1)^{k}(k+1), \end{aligned}$$

where $f_{k}(n)$’s are all integers.

Proof

We prove the result by induction on k.

First $a_{0}(n-2)=1, a_{1}(n-2)=n-2$, so $f_0(n)=0, f_1(n)=1$ are integers.

Now suppose $a_{k-1}(n-2)=nf_{k-1}(n)+(-1)^{k-1}k$, $a_{k-2}(n-2)=nf_{k-2}(n)+(-1)^{k-2}(k-1)$, where $f_{k-1}(n)$ and $f_{k-2}(n)$ are integers. Then, by $a_{k}(n-2)= (n-2)a_{k-1}(n-2)-a_{k-2}(n-2)$,

$$\begin{aligned} \begin{aligned} a_{k}(n-2)&=(n-2)(nf_{k-1}(n)+(-1)^{k-1}k)-(nf_{k-2}(n)+(-1)^{k-2}(k-1))\\&=n((n-2)f_{k-1}(n)-f_{k-2}(n)+(-1)^{k-1}k)+(-1)^{k}2k-(-1)^{k}(k-1)\\&=n((n-2)f_{k-1}(n)-f_{k-2}(n)+(-1)^{k-1}k)+(-1)^{k}(k+1)\\&=nf_{k}(n)+(-1)^{k}(k+1), \end{aligned} \end{aligned}$$

where $f_{k}(n)=(n-2)f_{k-1}(n)-f_{k-2}(n)+(-1)^{k-1}k$ is an integer since $f_{k-1}(n)$ and $f_{k-2}(n)$ are integers. This completes the proof. $\square$

Now we are ready to give the main result.

Theorem 3.3

For the nearly complete graph $G=K_{n}-C_{k}$ $(3\le k\le n-2)$, we have

• (1) if $k=n-2$, then

  $$\begin{aligned} S(G)\cong {\mathbb {Z}}_{s_{1}}\oplus {\mathbb {Z}}_{s_{2}}; \end{aligned}$$

• (2) if $3\le k\le n-3$, then

  $$\begin{aligned} S(G)\cong \left\{ \begin{array}{ll} \underbrace{{\mathbb {Z}}_{n}\oplus \cdots \oplus {\mathbb {Z}}_{n}}_{n-3-k}\oplus {\mathbb {Z}}_{ns_{1}}\oplus {\mathbb {Z}}_{s_{2}},&{}\quad \ \ \mathrm{if}\ \gcd (n,k)=1;\\ {\mathbb {Z}}_{i}\oplus \underbrace{{\mathbb {Z}}_{n}\oplus \cdots \oplus {\mathbb {Z}}_{n}}_{n-3-k}\oplus {\mathbb {Z}}_{\frac{n}{i}\cdot s_{1}}\oplus {\mathbb {Z}}_{s_{2}},&{}\ \ \mathrm{if}\ \gcd (n,k)=i,(1<i\le k); \end{array} \right. \end{aligned}$$

where

$$\begin{aligned} s_{1}=\left\{ \begin{array}{ll} a_{m-1}(n-2)gcd (2, n),&{}\quad \ \ \text{ if }\ \ k=2m; \\ d_{m}(n-2),&{}\quad \ \ \text{ if }\ \ k=2m+1; \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} s_{2}=\left\{ \begin{array}{ll} \frac{n(n-4)}{gcd^{2}(2,n)}s_{1},&{}\quad \ \ \text{ if }\ \ k=2m; \\ ns_{1},&{}\quad \ \ \text{ if }\ \ k=2m+1. \end{array} \right. \end{aligned}$$

Proof

By Theorem 3.1, we have

$$\begin{aligned} {\tilde{L}}\sim \begin{pmatrix} 1 &{} \\ &{} \ddots &{} \\ &{} &{} 1 &{} \\ &{} &{} &{} s_{1} &{} \\ &{} &{} &{} &{} s_{2} &{}\\ &{} &{} &{} &{} &{} n &{} \\ &{} &{} &{} &{} &{} &{} \ddots &{}\\ &{} &{} &{} &{} &{} &{} &{} n \\ \end{pmatrix}_{(n-1)\times (n-1)}. \end{aligned}$$

Obviously, $S(G)\cong {\mathbb {Z}}_{s_{1}}\oplus {\mathbb {Z}}_{s_{2}}$ for $k=n-2$.

For $3\le k\le n-3$, let

$$\begin{aligned} A= \begin{pmatrix} s_{1} &{} \\ &{} s_{2} &{}\\ &{} &{} n &{} \\ &{} &{} &{} \ddots &{}\\ &{} &{} &{} &{} n \\ \end{pmatrix}_{(n-k)\times (n-k)}. \end{aligned}$$

Note that both $s_1$ and n divide $s_2$, so it is not difficult to get that

$$\begin{aligned}&\Delta _{i}(A)=n^{i-1}gcd(s_{1},n),\quad \ \ i=1,\ldots ,n-k-2;\\&\Delta _{n-k-1}(A)=n^{n-k-2}s_1,\quad \ \ \Delta _{n-k}(A)=n^{n-k-2}s_1s_2. \end{aligned}$$

Now we shall show that $gcd(s_{1},n)=gcd(n,k)$. First, for $k=2m$, by Lemma 3.2

$$\begin{aligned} \begin{aligned} s_{1}&=a_{m-1}(n-2)gcd(2,n)\\&=(nf_{m-1}(n)+(-1)^{m-1}m)gcd(2,n)\\&=nf_{m-1}(n)gcd(2,n)+(-1)^{m-1}gcd(2,n)m, \end{aligned} \end{aligned}$$

so $gcd(s_{1},n)=gcd(gcd(2,n)m,n)=gcd(2m,n)=gcd(n,k)$. Then, for $k=2m+1$, by Lemma 3.2 again

$$\begin{aligned} \begin{aligned} s_{1}&=d_{m}(n-2)\\&=a_{m}(n-2)-a_{m-1}(n-2)\\&=nf_{m}(n)+(-1)^{m}(m+1)-(nf_{m-1}(n)+(-1)^{m-1}m)\\&=n(f_{m}(n)-f_{m-1}(n))+(-1)^{m}(2m+1), \end{aligned} \end{aligned}$$

so $gcd(s_{1},n)=gcd(2m+1,n)=gcd(n,k)$. That is, for any case, $gcd(s_{1},n)=gcd(n,k)$.

So by Lemma 2.1, we have

$$\begin{aligned} A\sim \hbox {diag}\left( gcd(n,k), n, \ldots , n, \frac{n}{gcd(n,k)}\cdot s_{1},s_2\right) . \end{aligned}$$

Now the results about S(G) follow directly. $\square$

Recall that the minimum number of generators of the sandpile group S(G) is denoted by $\mu (G)$. From the above theorem, we immediately derive the following corollary.

Corollary 3.4

Let $G=K_{n}-C_{k}$ $(3\le k\le n-2)$, we have

• (1) if $k=n-2$, then $\mu (G)=2$;

• (2) if $3\le k\le n-3$, then

  $$\begin{aligned} \mu (G)=\left\{ \begin{array}{ll} n-k-1,&{}\quad \ \ \mathrm{if}\ \gcd (n,k)=1; \\ n-k,&{}\quad \ \ \mathrm{if}\ \gcd (n,k)=i,(1<i\le k). \end{array} \right. \end{aligned}$$

Remark

It is worthy of pointing out that our method can be used to determine the sandpile group of any graph as $K_n-\cup _{i=1}^tC_{k_i}$, where $C_{k_{i}}$ are mutual disjoint cycles and satisfy $3\le k_1+k_2+\cdots +k_t\le n-2$. It is also not difficult to deduce that the sandpile groups of these graphs are never cyclic.

4 More Explicit Results for Some Special k

In this section, as applications of Theorem 3.3, we give more explicit results for the sandpile group of $K_{n}-C_{k}$ for $k=3,4,5,6,7$ and prime k.

Theorem 4.1

For the nearly complete graph $G=K_{n}-C_{3}$, we have

• (1) if $G=K_{5}-C_{3}$, then

  $$\begin{aligned} S(G)\cong {\mathbb {Z}}_{2}\oplus {\mathbb {Z}}_{10}; \end{aligned}$$

• (2) if $n\ge 6$, then

  $$\begin{aligned} S(G)\cong \left\{ \begin{array}{ll} {\mathbb {Z}}_{3}\oplus ({\mathbb {Z}}_{n})^{n-6}\oplus {\mathbb {Z}}_{p(n-3)} \oplus {\mathbb {Z}}_{n(n-3)},&{}\quad \ \ \ \mathrm{if}\ \ n=3p;\\ ({\mathbb {Z}}_{n})^{n-6}\oplus ({\mathbb {Z}}_{n(n-3)})^{2},&{}\quad \ \ \ \mathrm{if}\ \ n=3p+1\ or\ 3p+2. \end{array} \right. \end{aligned}$$

Theorem 4.2

For the nearly complete graph $G=K_{n}-C_{4}$, we have

• (1) if $G=K_{6}-C_{4}$, then

  $$\begin{aligned} S(G)\cong {\mathbb {Z}}_{8}\oplus {\mathbb {Z}}_{24}; \end{aligned}$$

• (2) if $n\ge 7$, then

  $$\begin{aligned} S(G)\cong \left\{ \begin{array}{llll} {\mathbb {Z}}_{4}\oplus ({\mathbb {Z}}_{n})^{n-7}\oplus {\mathbb {Z}}_{2p(n-2)} \oplus {\mathbb {Z}}_{2p(n-2)(n-4)},&{}\quad \ \ \ \mathrm{if}\ \ n=4p;\\ {\mathbb {Z}}_{2}\oplus ({\mathbb {Z}}_{n})^{n-7}\oplus {\mathbb {Z}}_{n(n-2)} \oplus {\mathbb {Z}}_{(2p+1)(n-2)(n-4)},&{}\quad \ \ \ \mathrm{if}\ \ n=4p+2;\\ ({\mathbb {Z}}_{n})^{n-7}\oplus {\mathbb {Z}}_{n(n-2)}\oplus {\mathbb {Z}}_{n(n-2)(n-4)},&{}\quad \ \ \ \mathrm{if}\ \ n=4p+1\ or\ 4p+3. \end{array} \right. \end{aligned}$$

Theorem 4.3

For the nearly complete graph $G=K_{n}-C_{5}$, we have

• (1) if $G=K_{7}-C_{5}$, then

  $$\begin{aligned} S(G)\cong {\mathbb {Z}}_{19}\oplus {\mathbb {Z}}_{133}; \end{aligned}$$

• (2) if $n\ge 8$, then

  $$\begin{aligned} S(G)\cong \left\{ \begin{array}{ll} {\mathbb {Z}}_{5}\oplus ({\mathbb {Z}}_{n})^{n-8}\oplus {\mathbb {Z}}_{p(n-2)(n-3)-p} \oplus {\mathbb {Z}}_{n(n-2)(n-3)-n},&{}\quad \ \ \ \mathrm{if}\ \ n=5p;\\ ({\mathbb {Z}}_{n})^{n-8}\oplus ({\mathbb {Z}}_{n(n-2)(n-3)-n})^{2},&{}\quad \ \ \ \mathrm{otherwise.} \end{array} \right. \end{aligned}$$

Theorem 4.4

For the nearly complete graph $G=K_{n}-C_{6}$, we have

• (1) if $G=K_{8}-C_{6}$, then

  $$\begin{aligned} S(G)\cong {\mathbb {Z}}_{70}\oplus {\mathbb {Z}}_{560}; \end{aligned}$$

• (2) if $n\ge 9$, then

  $$\begin{aligned} S(G)\cong \left\{ \begin{array}{ll} {\mathbb {Z}}_{6}\oplus ({\mathbb {Z}}_{n})^{n-9}\oplus {\mathbb {Z}}_{2p(n-1)(n-3)}\oplus {\mathbb {Z}}_{3p(n-1)(n-3)(n-4)},&{}\quad \ \ \ \mathrm{if}\ \ n=6p;\\ {\mathbb {Z}}_{2}\oplus ({\mathbb {Z}}_{n})^{n-9}\oplus {\mathbb {Z}}_{n(n-1)(n-3)}\oplus {\mathbb {Z}}_{(3p+1)(n-1)(n-3)(n-4)},&{}\quad \ \ \ \mathrm{if}\ \ n=6p+2;\\ {\mathbb {Z}}_{3}\oplus ({\mathbb {Z}}_{n})^{n-9}\oplus {\mathbb {Z}}_{(2p+1)(n-1)(n-3)}\oplus {\mathbb {Z}}_{n(n-1)(n-3)(n-4)},&{}\quad \ \ \ \mathrm{if}\ \ n=6p+3;\\ ({\mathbb {Z}}_{n})^{n-9}\oplus {\mathbb {Z}}_{n(n-1)(n-3)}\oplus {\mathbb {Z}}_{n(n-1)(n-3)(n-4)},&{}\quad \ \ \ \mathrm{if}\ \ n=6p+1\ or\ 6p+5. \end{array} \right. \end{aligned}$$

Theorem 4.5

For the nearly complete graph $G=K_{n}-C_{7}$, we have

• (1) if $G=K_{9}-C_{7}$, then

  $$\begin{aligned} S(G)\cong {\mathbb {Z}}_{281}\oplus {\mathbb {Z}}_{2529}; \end{aligned}$$

• (2) if $n\ge 10$, then

  $$\begin{aligned} S(G)\cong \left\{ \begin{array}{lll} {\mathbb {Z}}_{7}\oplus ({\mathbb {Z}}_{n})^{n-10}\oplus {\mathbb {Z}}_{p(n-1)(n-2)(n-4)+p} \oplus {\mathbb {Z}}_{n(n-1)(n-2)(n-4)+n},&{}\quad \ \ \ \mathrm{if}\ \ n=7p;\\ ({\mathbb {Z}}_{n})^{n-10}\oplus ({\mathbb {Z}}_{n(n-1)(n-2)(n-4)+n})^{2},&{}\quad \ \ \ \mathrm{otherwise}. \end{array} \right. \end{aligned}$$

In fact, for any odd prime $k=2m+1$, we have the following result.

Theorem 4.6

For the nearly complete graph $G=K_{n}-C_{k}$. If $k=2m+1$ is a prime, then

• (1) if $G=K_{k+2}-C_{k}$, then

  $$\begin{aligned} S(G)\cong {\mathbb {Z}}_{d_m(k)}\oplus {\mathbb {Z}}_{(k+2)d_m(k)}; \end{aligned}$$

• (2) if $n\ge k+3$, then

  $$\begin{aligned} S(G)\cong \left\{ \begin{array}{lllll} {\mathbb {Z}}_{k}\oplus ({\mathbb {Z}}_{n})^{n-k-3}\oplus {\mathbb {Z}}_{pd_m(n-2)} \oplus {\mathbb {Z}}_{nd_m(n-2)},&{}\quad \ \ \ \mathrm{if}\ \ n=kp;\\ ({\mathbb {Z}}_{n})^{n-k-3}\oplus ({\mathbb {Z}}_{nd_m(n-2)})^{2},&{}\quad \ \ \ \mathrm{otherwise.} \end{array} \right. \end{aligned}$$