ON THE DIVISIBILITY AMONG POWER LCM MATRICES ON GCD-CLOSED SETS

Abstract

Let $a,b$ and n be positive integers and let $S=\{x_1, \ldots , x_n\}$ be a set of n distinct positive integers. For ${x\in S}$ , define $G_{S}(x)=\{d\in S: d<x, \,d\mid x \ \mathrm {and} \ (d\mid y\mid x, y\in S)\Rightarrow y\in \{d,x\}\}$ . Denote by $[S^a]$ the $n\times n$ matrix having the ath power of the least common multiple of $x_i$ and $x_j$ as its $(i,j)$ -entry. We show that the bth power matrix $[S^b]$ is divisible by the ath power matrix $[S^a]$ if $a\mid b$ and S is gcd closed (that is, $\gcd (x_i, x_j)\in S$ for all integers i and j with $1\le i, j\le n$ ) and $\max _{x\in S} \{|G_S (x)|\}=1$ . This confirms a conjecture of Shaofang Hong [‘Divisibility properties of power GCD matrices and power LCM matrices’, Linear Algebra Appl. 428 (2008), 1001–1008].

1 Introduction

For arbitrary integers x and y, we denote by $(x,y)$ the greatest common divisor of x and y and by $[x,y]$ their least common multiple. Let $a, b$ and n be positive integers. Let $S = \{x_1, \ldots , x_n\}$ be a set of n distinct positive integers. Let $\xi _a$ be the arithmetic function defined by $\xi _a=x^a$ for any positive integer x. Let $(S^a)$ and $[S^a]$ stand for the $n\times n$ matrices whose $(i,j)$ -entry is $\xi _a((x_i, x_j))$ and $\xi _a([x_i, x_j])$ respectively. We call $(S^a)$ the $ath$ power GCD matrix and $[S^a]$ the $ath$ power LCM matrix. The set S is factor closed (FC) if $(x\in S, d\mid x)\Rightarrow d\in S$ and gcd closed if $(x_i, x_j)\in S$ for all integers i and j with $1\le i,j\le n$ . Obviously, an FC set must be gcd closed but the converse is not true. Nearly 150 years ago, Smith [15] proved that

(1.1) $$ \begin{align} \det([x_i,x_j])=\prod_{k=1}^n\varphi(x_k)\pi(x_k) \end{align} $$

if S is FC, where $\varphi $ is Euler’s totient function and $\pi $ is the multiplicative function defined for the prime power $p^r$ by $\pi (p^r)=-p$ . There are many generalisations of Smith’s determinant (1.1) and related results (see, for instance, [1–14, 16–21]). In particular, an elegant result was achieved by Hong et al. [8] stating that for any integer $n\ge 2$ ,

$$ \begin{align*}\det([i,j])_{2\le i,j\le n}=\bigg(\prod_{k=1}^n \varphi (k)\pi(k)\bigg)\sum_{\substack{t=1\\ t \text{ is square free}}}^n\frac{t\mu(t)}{\varphi (t)}, \end{align*} $$

where $\mu $ is the M $\ddot {\textrm {o}}$ bius function and an integer $x\ge 1$ is called square free if x is not divisible by the square of any prime number.

As usual, $\mathbb {Z}$ and $|S|$ denote the ring of integers and the cardinality of the set S. Hong [9] introduced the concept of greatest-type divisor when he solved the Bourque–Ligh conjecture. For any integer $x\in S$ , y is called a greatest-type divisor of x if

$$ \begin{align*}(y<x, \,y\mid z\mid x \ \mathrm{and} \ y,z\in S)\Rightarrow z\in\{y,x\}.\end{align*} $$

Let $ G_S(x):=\{y\in S: y \ \text {is a greatest-type divisor of} \ x \ \mathrm {in} \ S\} $ and let $M_n(\mathbb {Z})$ stand for the ring of $n\times n$ matrices over the integers. Bourque and Ligh [4] proved that $(S)$ divides $[S]$ in the ring $M_n(\mathbb {Z})$ (that is, $[S]=B(S)$ or $[S]=(S)B$ for some $B\in M_n(\mathbb {Z})$ ) if S is FC. Hong [10] showed that such a factorisation is not true when S is gcd closed and $\max _{x \in S}\{|{G_S(x)}|\}=2$ . The results of Bourque–Ligh and Hong were generalised by Korkee and Haukkanen [14] and by Chen et al. [6]. Feng et al. [7], Zhao [17], Altinisik et al. [1] and Zhao et al. [18] used the concept of greatest-type divisor to characterise the gcd-closed sets S with $\max _{x \in S}\{|{G_S(x)}|\}\le 3$ such that $(S^a)\mid [S^a]$ which partially solved an open problem of Hong [10].

Hong [12] investigated divisibility among power GCD matrices and among power LCM matrices. It was proved in [12] that $(S^a)\mid (S^b), (S^a) \mid [S^b]$ and $[S^a] \mid [S^b]$ if $a \mid b$ and S is a divisor chain (that is, $x_{\sigma (1)}| \cdots |x_{\sigma (n)}$ for a permutation $\sigma $ of $\{1,\ldots , n\}$ ), and such factorisations are no longer true if $a\nmid b$ and $|S|\ge 2$ . Evidently, a divisor chain is gcd closed but not conversely. Recently, Zhu [19] confirmed two conjectures of Hong raised in [12] stating that if $a\mid b$ and S is a gcd-closed set with $\max _{x\in S}\{|G_S(x)|\}=1$ , then both the bth power GCD matrix $(S^b)$ and the bth power LCM matrix $[S^b]$ are divisible by the ath power GCD matrix $(S^a)$ . At the end of [12], Hong also conjectured that if $a \mid b$ and $S=\{x_1,\ldots , x_n \}$ is gcd closed and $\max _{x\in S}\{|G_S(x)|\}=1$ , then $[S^a] \mid [S^b]$ in the ring $M_n(\mathbb {Z})$ . Tan and Li [16] partially confirmed this conjecture by proving that $[S^a] \mid [S^b]$ in the ring $M_{|S|}(\mathbb {Z})$ if $a \mid b$ and S consists of finitely many coprime divisor chains with $1\in S$ and that such a divisibility relation is not true if $a\nmid b$ . However, the conjecture still remains open.

Our goal is to present a proof of Hong’s conjecture. The main result of the paper is the following theorem.

Theorem 1.1. If a and b are positive integers such that $a \mid b$ and S is a gcd-closed set such that $\max _{x\in S}\{|G_S(x)|\}=1$ , then the ath power LCM matrix $[S^a]$ divides the bth power LCM matrix $[S^b]$ in the ring $M_{|S|}(\mathbb {Z})$ .

The proof of Theorem 1.1 is similar to that of Feng et al. [7] in character, but it is more complicated. This paper is organised as follows. In Section 2, we supply several preliminary lemmas needed in the proof of Theorem 1.1. Section 3 is devoted to the proof of Theorem 1.1.

One can easily check that for any permutation $\sigma $ on the set $\{1, \ldots , n \}$ , $[S^a]\mid [S^b]\Leftrightarrow [S_{\sigma }^a]\mid [S_{\sigma } ^b]$ , where $S_\sigma :=\{x_{\sigma (1)},\ldots , x_{\sigma (n)}\}$ . Without loss of any generality, we can always assume that the set $S=\{x_1,\ldots , x_n \}$ satisfies $x_1<\cdots <x_n$ .

2 Auxiliary results

In this section, we provide several lemmas that will be needed in the proof of Theorem 1.1. We begin with a result due to Hong which gives the formula for the determinant of the power LCM matrix on a gcd-closed set.

Lemma 2.1 [11, Lemma 2.1].

If S is gcd closed, then

(2.1) $$ \begin{align} \det[S^a]=\prod\limits_{k=1}^n x_k^{2a}\alpha_{a,k}, \end{align} $$

where

(2.2) $$ \begin{align} \alpha_{a,k}:=\sum_{\substack{d|x_k\\ d\nmid x_t,\,x_t<x_k}} \bigg(\dfrac{1}{\xi_a}*\mu\bigg)(d) \end{align} $$

and ${1}/{\xi _a}$ is the arithmetic function defined for any positive integer x by ${({1}/{\xi _a})(x):= x^{-a}}$ .

Lemma 2.2 [5, Theorem 3].

If S is a gcd-closed set and $(f((x_i,x_j)))$ is invertible, then $ (f((x_i,x_j)))^{-1}=(a_{ij}), $ where

$$ \begin{align*}a_{ij}:=\sum_{\substack{x_i|x_k\\ x_j|x_k}} \dfrac{c_{ik}c_{jk}}{\delta_k}\end{align*} $$

with

(2.3) $$ \begin{align} \delta_k:= \sum_{\substack{d|x_k\\ d\nmid x_t,\, x_t<x_k}} (f*\mu)(d) \quad\mbox{and}\quad c_{ij}:=\sum _{\substack{dx_i|x_j\\ dx_i\nmid x_t,\, x_t<x_j}} \mu(d). \end{align} $$

Lemma 2.3 [11, Lemma 2.3].

Let m be a positive integer. Then

$$ \begin{align*}{\underset{d|m}{\sum}}\bigg(\dfrac{1}{\xi_a}*\mu \bigg)(d)=m^{-a}. \end{align*} $$

Lemma 2.4 [7, Lemma 2.2].

Let S be gcd closed and $\max _{x\in S}\{|G_S(x)|\}=1$ . Let $\alpha _{a,k}$ be defined as in (2.2). If $G_S(x_k)=\{x_{k_1}\}$ for $2\le k\le |S|$ , then $\alpha _{a,k}=x_k^{-a}-x_{k_1}^{-a}.$

Lemma 2.5. Let S be gcd closed and $\max _{x\in S}\{|G_S(x)|\}=1$ . Let $\alpha _{a,k}$ and $c_{ij}$ be defined as in (2.2) and (2.3), respectively. Then $[S^a]$ is nonsingular and $[S^a]^{-1}=(s_{ij})_{1\le i,j\le n}$ with

$$ \begin{align*} s_{ij}:=\dfrac{1}{x_i^a x_j^a} \sum_{\substack{x_i|x_k\\ x_j|x_k}} \dfrac{c_{ik}c_{jk}}{\alpha_{a,k}}. \end{align*} $$

Proof. Since ${[x_i,x_j]}^a={x_i^ax_j^a}/{{(x_i,x_j)}^a}$ ,

(2.4) $$ \begin{align} [S^a]=D\bigg(\dfrac{1}{\xi_a}(x_i,x_j)\bigg)D, \end{align} $$

where $D:=\mathrm {diag}(x_1^a,\ldots ,x_n^a)$ . By (2.1) and (2.4),

$$ \begin{align*} \det\bigg(\dfrac{1}{\xi_a}((x_i,x_j))\bigg) =\prod\limits_{k=1}^n\alpha_{a,k}. \end{align*} $$

By Lemma 2.3, $\alpha _{a,1}=x_1^{-a}$ . For $2\le k\le n$ , since $\max _{x\in S}\{|G_S(x)|\}=1$ , one may let $G_S(x_k)=\{x_{k_1}\}$ . By Lemma 2.4, $\alpha _{a,k}=x_k^{-a}-x_{k_1}^{-a}\neq 0$ . So the matrix $(({1/\xi _a})((x_i,x_j)))$ is nonsingular. Now applying Lemma 2.2 gives

(2.5) $$ \begin{align} \bigg(\dfrac{1}{\xi_a}((x_i,x_j))\bigg)^{-1}=(h_{ij}), \end{align} $$

where

$$ \begin{align*}h_{ij}:=\sum_{\substack{x_i|x_k\\ x_j|x_k}} \dfrac{c_{ik}c_{jk}}{\alpha_{a,k}}.\end{align*} $$

The desired result follows immediately from (2.4) and (2.5).

We next recall some basic results on gcd-closed sets.

Lemma 2.6 [7, Lemma 2.3].

Let S be a gcd-closed set with $|S|\ge 2$ . Let $c_{ij}$ be defined as in (2.3). Then

$$ \begin{align*}c_{w1}=\left\{\begin{aligned} {1}& \quad \mbox{if } w=1,\\ {0}&\quad \mbox{otherwise}. \end{aligned} \right. \end{align*} $$

Further, if $G_S(x_m)=\{x_{m_1}\}$ for $2\le m\le |S|$ , then

$$ \begin{align*}c_{wm}=\left\{ \begin{aligned} {-1}&\quad \mbox{if }w=m_1,\\ {1}&\quad \mbox{if } w=m,\\ {0}&\quad \mbox{otherwise}. \end{aligned} \right. \end{align*} $$

Lemma 2.7 [7, Lemma 3.1].

Let S be gcd closed and $x,z\in S$ such that $x\nmid z$ . If ${G_S(x)=\{y\}}$ , then $(x, z)=(y, z)$ .

Lemma 2.8. Let S be gcd closed and $x, y\in S$ with $G_S(x)=\{y\}$ . If $a\mid b$ , then for any $z, r\in S$ with $r\mid x$ , $y^a[z,x]^b-x^a[z,y]^b$ is divisible by each of $x^a(y^a-x^a)$ and $r^a(y^a-x^a)$ .

Proof. We divide the proof into two cases.

Case 1: $x\nmid z$ . By Lemma 2.7, $(x, z)=(y, z)$ , which implies

(2.6) $$ \begin{align} y^a[z,x]^b-x^a[z,y]^b=y^a\dfrac{z^bx^b} {(z,x)^b}-x^a\dfrac{z^by^b}{(z,y)^b} =\dfrac{z^b}{(z,x)^b}x^ay^a(x^{b-a}-y^{b-a}). \end{align} $$

Since $a\mid b$ ,

$$ \begin{align*}x^{b-a}-y^{b-a}=(x^a-y^a)\sum _{i=0}^{{({b}/{a})}-2} (x^a)^{{({b}/{a})}-2-i}y^{ai} \quad\mbox{and}\quad \sum _{i=0}^{{({b}/{a})}-2} (x^a)^{{({b}/{a})}-2-i}y^{ai}\in \mathbb{Z}. \end{align*} $$

Hence, $(x^a-y^a)\mid (x^{b-a}-y^{b-a})$ . Then by (2.6), we deduce that $y^a[z,x]^b-x^a[z,y]^b$ is divisible by each of $x^a(y^a-x^a)$ and $r^a(y^a-x^a)$ .

Case 2: $x\mid z$ . Then $[x, z]=[y, z]=z$ . It follows that

$$ \begin{align*}y^a[z,x]^b-x^a[z,y]^b=y^az^b-x^az^b=z^b(y^a-x^a).\end{align*} $$

Since $a\mid b$ , the desired results follow immediately.

Lemma 2.9. Let S be gcd closed and $\max _{x\in S}\{|G_S(x)|\}=1$ . If $a\mid b$ , then all the elements of the nth column and the nth row of $[S^b][S^a]^{-1}$ are integers.

Proof. The proof of Lemma 2.9 is divided into two cases.

Case 1: $1\le i\le n$ and $j=n$ . By Lemmas 2.5 and 2.6,

$$ \begin{align*} {([S^b][S^a]^{-1})}_{in} =&\sum_{m=1}^n[x_i, x_m]^b\dfrac{1}{x_m^ax_n^a}\sum_{\substack{x_m|x_k\\ x_n|x_k}} \dfrac{c_{mk}c_{nk}}{\alpha _{a,k}}\\ =&\dfrac{1}{x_n^a}\sum_{m=1}^n\dfrac{[x_i,x_m]^bc_{mn}}{x_m^a\alpha_{a,n}} =\dfrac{1}{x_n^a\alpha_{a,n}}\sum_{m=1}^n\dfrac{[x_i,x_m]^bc_{mn}}{x_m^a}. \end{align*} $$

Since $\max _{x\in S}\{|G_S(x)|\}=1$ , we may let $G_S(x_n)=\{x_{n_1}\}$ . Then by Lemmas 2.4, 2.6 and 2.8,

$$ \begin{align*}{([S^b])[S^a]^{-1})}_{in} =\dfrac{x_{n_1}^a[x_i,x_n]^b-x_n^a[x_i,x_{n_1}]^b}{x_n^a(x_{n_1}^a-x_n^a)}\in \mathbb{Z} \end{align*} $$

as required.

Case 2: $i=n,\ 1\le j\le n-1$ . Then

$$ \begin{align*} {( [S^b][S^a]^{-1})}_{nj} =\sum_{m=1}^n[x_n, x_m]^b\dfrac{1}{x_m^ax_j^a}\sum_{\substack{x_m|x_k\\ x_j|x_k}} \frac {c_{mk}c_{jk}}{\alpha _{a,k}} =\sum_{x_j|x_k}\frac{c_{jk}}{x_j^a\alpha _{a,k}}\sum_{x_m|x_k}\dfrac{1}{x_m^a}c_{mk}[x_m,x_n]^b. \end{align*} $$

We claim that

$$ \begin{align*}\gamma_k:={\frac{1}{x_j^a\alpha _{a,k}}\sum_{x_m|x_k}\dfrac{1}{x_m^a}c_{mk}[x_m,x_n]^b}\in \mathbb{Z}\end{align*} $$

for any positive integer k with $x_j\mid x_k$ .

If $k=1$ , then $m=j=1$ . In this case,

$$ \begin{align*}\gamma_1=\dfrac{1}{\alpha _{a,1}}\cdot\dfrac{1} {x_1^{2a}}\cdot c_{11}\cdot[x_1,x_n]^b =\dfrac{[x_1,x_n]^b}{x_1^a} =\dfrac{x_1^{b-a}x_n^b}{{(x_1,x_n)}^b}\in \mathbb{Z}.\end{align*} $$

Now let $k>1$ . We can set $G_S(x_k)=\{x_{k_1}\}$ since $|G_S(x_k)|=1$ . By Lemmas 2.4, 2.6 and 2.8,

$$ \begin{align*}\gamma_k={\frac{1}{x_j^a\alpha _{a,k}}\sum_{x_m|x_k} \dfrac{1}{x_m^a}c_{mk}[x_m,x_n]^b} ={\frac{x^a_{k_1}[x_k,x_n]^b-x_k^a[x_{k_1},x_n]^b} {x_j^a(x_{k_1}^a-x_k^a)}}\in \mathbb{Z} \end{align*} $$

as desired. This concludes the proof of the claim and of Lemma 2.9.

Finally, we can use Lemma 2.9 to establish the main result of this section.

Lemma 2.10. Let S be gcd closed and $\max _{x\in S}\{|G_S(x)|\}=1$ . Let $S_1:=S\setminus \{x_n\}=\{x_1, \ldots , x_{n-1}\}$ . If $a\mid b$ , then $[S^b][S^a]^{-1}\in M_n(\mathbb {Z})$ if and only if $[S_1^b][S_1^a]^{-1}\in M_{n-1}(\mathbb {Z}).$

Proof. First, it follows from the hypothesis and Lemma 2.9 that all the elements of the nth column and the nth row of $[S^b][S^a]^{-1}$ are integers. So it suffices to show that

(2.7) $$ \begin{align} \mathcal{A}_{ij}:={([S^b][S^a]^{-1})}_{ij} -{([S_1^b][S_1^a]^{-1})}_{ij}\in \mathbb{Z} \end{align} $$

for all integers i and j with $1\le i,j\le n-1$ .

To see this, define

$$ \begin{align*} e_{uv}:=\left\{ \begin{aligned} 1&\quad \mbox{if} \ x_v\mid x_u,\\ 0&\quad \mbox{if} \ x_v\nmid x_u, \end{aligned} \right. \end{align*} $$

for all integers u and v between 1 and n. Then $e_{nj}=1$ if $x_j\mid x_n$ and $e_{nj}=0$ otherwise. Furthermore, for any integer m with $1\le m\le n-1$ , one has $e_{nm}=1$ if $x_m\mid x_n$ and $e_{nm}=0$ otherwise. We then deduce that

(2.8) $$ \begin{align} \mathcal{A}_{ij} & =\sum_{m=1}^n[x_i,x_m]^b\sum_{\substack{x_m|x_k\\ x_j|x_k}} \frac{c_{mk}c_{jk}}{x_m^ax_j^a\alpha _{a,k}} -\sum_{m=1}^{n-1}[x_i,x_m]^b\sum _{\substack{x_m|x_k\\ x_j|x_k,\, x_k\neq x_n}} \frac{c_{mk}c_{jk}}{x_m^ax_j^a\alpha _{a,k}}\notag\\ & =\frac{c_{nn}c_{jn}}{x_n^ax_j^a\alpha _{a,n}}[x_i,x_n]^b e_{nj}+ \sum_{m=1}^{n-1}\frac{c_{mn}c_{jn}}{x_m^ax_j^a\alpha _{a,n}} [x_i,x_m]^b e_{nj}e_{nm} \notag\\ & =e_{nj}\frac{c_{jn}}{x_j^a\alpha _{a,n}} \bigg(\dfrac{[x_i,x_n]^b}{x_n^a}+\sum_{m=1}^{n-1} \dfrac{[x_i,x_m]^bc_{mn}e_{nm}}{x_m^a} \bigg) :=e_{nj} A_{ij}. \end{align} $$

Let us now show that $A_{ij}\in \mathbb {Z}$ . Since $\max _{x\in S}\{|G_S(x)|\}=1$ , one may let $G_S(x_n)=\{x_{n_1}\}$ . From Lemma 2.4, $\alpha _{a,n}=x_n^{-a}-x_{n_1}^{-a}.$ However, by Lemma 2.6, for any integer m with $1\le m\le n-1$ , $c_{mn}=-1$ if $m=n_1$ and $c_{mn}=0$ otherwise. It follows from (2.8) and Lemma 2.8 that

(2.9) $$ \begin{align} {A_{ij}=\frac{x_{n_1}^a[x_i,x_n]^b-x_n^a[x_i,x_{n_1}]^b} {x_j^a(x_{n_1}^a-x_n^a)}\cdot c_{jn}}\in \mathbb{Z}. \end{align} $$

Since $e_{nj}\in \{0,1\}$ , (2.8) and (2.9) yield (2.7).

The proof of Lemma 2.10 is complete.

3 Proof of Theorem 1.1

We prove Theorem 1.1 by using induction on $n=|S|$ .

For $n=1$ , the statement is clearly true.

Let $n=2$ . Since $S=\{x_1, x_2\}$ is gcd closed, $(x_1, x_2)=x_1$ and $x_1\mid x_2$ . It follows that

$$ \begin{align*}\begin{aligned} {[S^b]}[S^a]^{-1}&= \begin{pmatrix} x_1^b&x_2^b\\[3pt] x_2^b&x_2^b \end{pmatrix} \cdot\frac {1}{x_2^a(x_1^a-x_2^a)} \begin{pmatrix} x_2^a&-x_2^a\\[3pt] -x_2^a&x_1^a \end{pmatrix} =\begin{pmatrix} \mathcal{B}&-x_1^a\mathcal{C}\\[3pt] 0&x_2^{b-a} \end{pmatrix}, \end{aligned} \end{align*} $$

where

$$ \begin{align*}\mathcal{B}:=\dfrac {x_2^b-x_1^b}{x_2^a-x_1^a} \quad\mathrm{and}\quad \mathcal{C}:=\dfrac {x_2^{b-a}-x_1^{b-a}}{x_2^a-x_1^a}.\end{align*} $$

Since $a\mid b$ , implying that $a\mid (b-a)$ , it follows that $\mathcal {B}\in \mathbb {Z}$ and $\mathcal {C}\in \mathbb {Z}$ , that is, ${[S^b]}[S^a]^{-1}\in M_2(\mathbb {Z})$ . The statement is true for this case.

Let $n=3$ . Since $S=\{x_1,x_2,x_3\}$ is gcd closed, we have $x_1\mid x_i\ (i=2, 3)$ and $(x_2,x_3)=x_1$ or $x_2$ . Consider the following two cases.

Case 1: $(x_2,x_3)=x_1$ . Then one computes

$$ \begin{align*} \begin{split} {[S^b]}[S^a]^{-1} &=\begin{pmatrix} x_1^b&x_2^b&x_3^b\\[4pt] x_2^b&x_2^b&\dfrac{x_2^bx_3^b}{x_1^b}\\[4pt] x_3^b&\dfrac{x_2^bx_3^b}{x_1^b}&x_3^b \end{pmatrix} \cdot \frac {x_1^a}{x_2^ax_3^a(x_2^a-x_1^a)(x_3^a-x_1^a)}\\[4pt] &\quad \times \begin{pmatrix} \dfrac{x_1^{2a}x_2^ax_3^a-x_2^{2a}x_3^{2a}}{x_1^{2a}} &\dfrac{x_2^ax_3^{2a}-x_1^ax_2^ax_3^a}{x_1^a} &\dfrac{x_2^{2a}x_3^a-x_1^ax_2^ax_3^a}{x_1^a}\\[4pt] \dfrac{x_2^ax_3^{2a}-x_1^ax_2^ax_3^a}{x_1^a}&x_1^ax_3^a-x_3^{2a}&0\\[4pt] \dfrac{x_2^{2a}x_3^a-x_1^ax_2^ax_3^a}{x_1^a}&0&x_1^ax_2^a-x_2^{2a} \end{pmatrix}\\ &=\begin{pmatrix} \mathcal{B}+x_3^a\mathcal{F}&-x_1^a\mathcal{C}&-x_1^a\mathcal{F}\\[4pt] x_3^a\mathcal{D}\mathcal{F}&x_2^{b-a}&-x_1^a\mathcal{D}\mathcal{F}\\[4pt] x_2^a\mathcal{E}\mathcal{C}&-x_1^a\mathcal{E}\mathcal{C}&x_3^{b-a} \end{pmatrix}, \end{split} \end{align*} $$

where $\mathcal {B}$ and $\mathcal {C}$ are as given earlier in this section, $\mathcal {D}:={x_2^b}/{x_1^b}$ , $\mathcal {E}:={x_3^b}/{x_1^b}$ and $\mathcal {F}:=({x_3^{b-a}-x_1^{b-a})}/({x_3^a-x_1^a}).$ Since $x_1\mid x_2$ , $x_1\mid x_3$ and $a\mid (b-a)$ , all of $\mathcal {B}, \mathcal {C}, \mathcal {D}, \mathcal {E}$ and $\mathcal {F}$ are integers. Hence, ${[S^b]}[S^a]^{-1}\in M_3(\mathbb {Z})$ . The statement holds in this case.

Case 2: $(x_2,x_3)=x_2$ . Then $x_2\mid x_3$ . We compute

$$ \begin{align*}\begin{aligned} {[S^b]}[S^a]^{-1}&= \begin{pmatrix} x_1^b&x_2^b&x_3^b\\[4pt] x_2^b&x_2^b&x_3^b\\[4pt] x_3^b&x_3^b&x_3^b \end{pmatrix} \cdot \frac {1}{x_3^a(x_2^a-x_1^a)(x_3^a-x_2^a)}\\ &\quad \times \begin{pmatrix} x_3^a(x_2^a-x_3^a)&x_3^a(x_3^a-x_2^a)&0\\[4pt] x_3^a(x_3^a-x_2^a)&x_3^a(x_1^a-x_3^a)&x_3^a(x_2^a-x_1^a)\\[4pt] 0&x_3^a(x_2^a-x_1^a)&x_2^a(x_1^a-x_2^a) \end{pmatrix}\\ &=\begin{pmatrix} \mathcal{B}&-\mathcal{B}+\mathcal{G}&-x_2^a\mathcal{H}\\[4pt] 0&\mathcal{G}&-x_2^a\mathcal{H}\\[4pt] 0&0&x_3^{b-a} \end{pmatrix}, \end{aligned} \end{align*} $$

where $\mathcal {B}$ is as before, $\mathcal {G}:={({x_3^b-x_2^b)}/{(x_3^a-x_2^a})}$ and $\mathcal {H}:={(x_3^{b-a}-x_2^{b-a})}/{(x_3^a-x_2^a)}$ . Since $a\mid b$ and $a\mid (b-a)$ imply that $\mathcal {G}\in \mathbb {Z}$ and $\mathcal {H}\in \mathbb {Z}$ , it follows immediately that ${[S^b]}[S^a]^{-1}\in M_3(\mathbb {Z})$ . The statement is true for this case.

Now let $n\ge 4$ . Assume that the statement is true for the $n-1$ case. In what follows, we show that the statement is true for the n case. Since S is gcd closed and $\max _{x\in S}\{|G_S(x)|\}=1$ , it follows that $S_1:=\{x_1,\ldots , x_{n-1}\}$ is also gcd closed and $\max _{x\in S_1}\{|G_{S_1}(x)|\}=1$ . Hence by the inductive hypothesis, $[S_1^b][S_1^a]^{-1}\in M_{n-1}( \mathbb {Z})$ . Finally, from Lemma 2.10, $[S^b][S^a]^{-1}\in M_n(\mathbb {Z})$ as desired.

This finishes the proof of Theorem 1.1. $\Box $