1 Introduction
For arbitrary positive integers x and y, we denote by
$(x,y)$
the greatest common divisor of x and y and
$[x,y]$
their least common multiple. Throughout, let
$a,b$
and n be positive integers, and let f be an arithmetic function. Let
$S=\{x_1,\ldots ,x_n\}$
be a set of n distinct positive integers. Let
$(f(x_i, x_j))$
(abbreviated by
$(f(S))$
) and
$(f[x_i, x_j])$
(abbreviated by
$(f[S])$
) stand for the
$n\times n$
matrices whose
$(i,j)$
-entry is
$f((x_i, x_j))$
and
$f([x_i, x_j])$
, respectively. Evidently, the matrices
$(f(S))$
and
$(f[S])$
are symmetric. The set S is called factor closed (FC) if all positive divisors of any element of S are in S, and is called gcd closed if
$(x_i, x_j)\in S$
for all integers i and j with
$1\le i,j\le n$
. An FC set is clearly gcd closed but the converse is not true. For example, if
$a>1$
is an integer, then the set
$S=\{a, 2a, \ldots , na\}$
is gcd closed but not factor closed. In 1875, Smith [Reference Smith26] published his renowned result stating that if S is an FC set, then
$\det (f(S))=\prod _{k=1}^n(f*\mu )(x_k),$
where
$\mu $
is the Möbius function, and
$f*\mu $
is the Dirichlet convolution of f and
$\mu $
. There have been many generalisations of Smith’s determinant (see [Reference Apostol2–Reference Bourque and Ligh7, Reference Hong, Hu and Lin12, Reference Hong and Lin13, Reference Hong16, Reference Yamasaki28, Reference Zhu30]). The nonsingularity of Smith’s matrices was first considered by Bourque and Ligh in [Reference Bourque and Ligh4] and subsequently studied in [Reference Haukkanen, Mattila and Mantysalo10, Reference Hong14, Reference Hong15, Reference Hong19, Reference Hong20, Reference Korkee, Mattila and Haukkanen24]. The asymptotic behaviour of the eigenvalues of Smith’s matrices has also been investigated (see [Reference Altinisik1, Reference Hilberdink and Pushnitski11, Reference Hong and Enoch Lee22, Reference Hong and Loewy23]).
Let
${\mathbb Z}$
denote the ring of integers and let
$|T|$
stand for the cardinality of a finite set T. When
$f(x)=x$
, the matrices
$(S):=(f(S))$
and
$[S]:=(f[S])$
are called the GCD matrix and the LCM matrix, respectively. In 1992, Bourque and Ligh [Reference Bourque and Ligh4] showed that
$(S)\mid [S]$
in the ring
$M_n({\mathbb Z})$
of
$n\times n$
matrices over the integers if S is FC. That is, there is
$A \in M_n({\mathbb Z})$
such that
$[S]=(S)A$
or, equivalently,
$[S]=A^T(S)$
. Let
$x,y \in S$
with
$y<x$
. If
$y\mid x$
and the conditions
$y\mid z\mid x$
and
$z\in S$
imply that
$z\in \{y,x\}$
, then y is called a greatest-type divisor of x in S. We denote by
$G_S(x)$
the set of all the greatest-type divisors of x in S. Obviously,
$G_S(x)=G_{S(x)}(x)$
if
$S(x)$
is the set of all positive divisors of x in S. The concept of greatest-type divisor was introduced by Hong in [Reference Hong14] and played a key role in his solution [Reference Hong14, Reference Hong15] of the Bourque–Ligh conjecture [Reference Bourque and Ligh4]. In 2002, Hong [Reference Hong17] proved that there exist gcd-closed sets S with
$\max _{x\in S}\{|G_S(x)|\}=2$
such that
$(S)\nmid [S]$
in
$M_n({\mathbb Z})$
. Feng et al. [Reference Feng, Hong and Zhao9] used the concept of greatest-type divisor to study the divisibility of a power LCM matrix by a power GCD matrix defined on the gcd-closed set S in
$M_n({\mathbb Z})$
. Let
$C_S:=\{f: (f*\mu )(d)\in \mathbb {Z} \ \mathrm {whenever} \ d\mid \mathrm {lcm}(S)\}$
, where
$\mathrm {lcm}(S)$
means the least common multiple of all the elements of S. From [Reference Hong18, Lemma 4.2(i)], the conditions
$d\mid \mathrm {lcm}(S)$
and
$f\in C_S$
yield
$f(d)\in \mathbb {Z}$
. Li and Tan [Reference Li and Tan25] proved that
$(f(S))\mid (f[S])$
if S is gcd closed such that
$\max _{x\in S}\{|G_S(x)|\}=1$
and f is multiplicative with
$f\in D_S:=\{f\in C_S: f(u)\mid f(v) \ \text {whenever} \ u\mid v \ \text {and} \ u,v\in S\}$
and
$(f(S))$
is nonsingular.
When
$f=\xi _a$
with
$\xi _a(x):=x^a$
, the matrices
$(S^a):=(f(S))$
and
$[S^b]:=(f[S])$
are called the ath power GCD matrix and ath power LCM matrix, respectively. Hong [Reference Hong21] initiated the study of divisibility among power GCD matrices and among power LCM matrices and proved that
$(S^a)\mid (S^b)$
,
$(S^a)\mid [S^b]$
and
$[S^a]\mid [S^b]$
in the ring
$M_{n}({\mathbb Z})$
when
$a\mid b$
and S is a divisor chain (that is, there is a permutation
$\sigma $
of order n on
$\{1,2,\ldots,n\}$
such that
${x_{\sigma (1)}\mid \cdots \mid x_{\sigma (n)}}$
). Hong’s results were extended by Tan and Li [Reference Tan and Li27] to the case of finitely many coprime divisor chains. In 2022, Zhu [Reference Zhu29] and, in 2023, Zhu and Li [Reference Zhu and Li31] showed that if
$a\mid b$
and S is a gcd-closed set with
$\max _{x\in S}\{|G_S(x)|\}=1$
, then
$(S^a)\mid (S^b),\ (S^a)\mid [S^b]$
and
$(S^a)\mid [S^b]$
in the ring
$M_n({\mathbb Z})$
.
In this paper, we examine the divisibility among the power GCD matrices and the power LCM matrices defined on factor-closed sets. The main result of this paper can be stated as the following theorem.
Theorem 1.1. Let S be a factor-closed set and let a and b be positive integers with
$a\mid b$
. In the ring
$M_{|S|}({\mathbb Z})$
, the ath power GCD matrix
$(S^a)$
divides both the bth power GCD matrix
$(S^b)$
and the bth power LCM matrix
$[S^b]$
, and the ath power LCM matrix
$[S^a]$
divides the bth power LCM matrix
$[S^b]$
.
The proof of Theorem 1.1 is algebraic and p-adic. Setting
$a=b$
, Theorem 1.1 becomes the Bourque–Ligh theorem presented in [Reference Bourque and Ligh7].
Corollary 1.2 [Reference Bourque and Ligh7].
Let S be a factor-closed set and let a be a positive integer. Then, the ath power LCM matrix
$[S^a]$
is divisible by the ath power GCD matrix
$(S^a)$
in the ring
$M_{|S|}(\mathbb {Z})$
.
This paper is organised as follows. In Section 2, we establish the results on the divisibility among power GCD and among power LCM matrices on factor-closed sets. As a corollary, Theorem 1.1 follows immediately. Section 3 is devoted to some interesting remarks and conjectures.
2 Divisibility in the factor-closed case
In this section, we investigate the divisibility among power GCD and among power LCM matrices on factor-closed sets. As usual, for any given prime number p and any integer x, we denote by
$\nu _p(x)$
the p-adic valuation of x, that is, the largest nonnegative integer r such that
$p^r$
divides x. We write
$p^l\,||\,x$
if
$l=\nu _p(x)$
. We begin with a well-known result of Bourque and Ligh.
Lemma 2.1 [Reference Bourque and Ligh6].
Let
$S=\{x_1,\ldots ,x_n\}$
be an FC set. Then,
$(S^a)^{-1}=((x_i,x_j)^{a})^{-1}=(\alpha _{ij})$
, where
$$ \begin{align*}\alpha_{ij}=\sum_{\substack{x_i|x_k\\ x_j|x_k}}\frac{\mu({x_k}/{x_i}) \mu({x_k}/{x_j})}{\phi_{a}(x_k)}\end{align*} $$
and
$\phi _a$
is the Jordan totient function, that is,
$\phi _{a}(x_k)=(\xi _{a}\ast \mu )(x_k) =\sum _{d|x_k}d^{a}\mu ({x_k}/{d}).$
Lemma 2.2. Let a and b be positive integers with
$a\mid b$
. Then, for any positive integer m,
$$ \begin{align*} \frac{\phi_b(m)}{\phi_a(m)}=\prod_{p|m}p^{(b-a) (\nu_p(m)-1)}\cdot\frac{p^b-1}{p^a-1}\in{\mathbb Z}. \end{align*} $$
Proof. Let
$m\geq 1$
be an integer. If
$m=1$
, then
$\phi _{a}(m)=(\xi _{a}\ast \mu )(1)=1$
and
$\phi _{b}(m)=1$
. So Lemma 2.2 is true when
$m=1$
. In what follows, we let
$m\geq 2$
. Write
$m=p_{1}^{e_{1}}\cdots p_{r}^{e_{r}}$
, where
$r\geq 1$
,
$p_1,\ldots ,p_r$
are r distinct primes and
$e_1,\ldots ,e_r$
are positive integers. Since
$\phi _{a}=\xi _{a}\ast \mu $
and
$\phi _{b}=\xi _{b}\ast \mu $
, and
$\xi _{a}$
,
$\xi _{b}$
and
$\mu $
are multiplicative functions,
$\phi _{a}$
and
$\phi _{b}$
are multiplicative functions. It then follows that
$$ \begin{align*} \frac{\phi_{b}(m)}{\phi_{a}(m)} &= \prod_{i=1}^{r}\frac{\phi_{b}(p_{i}^{e_{i}})}{\phi_{a}(p_{i}^{e_{i}})} =\prod_{i=1}^{r}\frac{(\xi_{b}\ast\mu)(p_{i}^{e_{i}})} {(\xi_{a}\ast\mu)(p_{i}^{e_{i}})} = \prod_{i=1}^{r}\frac{\xi_{b}(p_{i}^{e_{i}})\mu(1)+\xi_{b}(p_{i}^{e_{i}-1}) \mu(p_{i})}{\xi_{a}(p_{i}^{e_{i}})\mu(1)+\xi_{a}(p_{i}^{e_{i}-1})\mu(p_{i})}\\ &=\prod_{i=1}^{r}\frac{(p_{i}^{e_{i}})^{b}-(p_{i}^{e_{i}-1})^{b}} {{(p_{i}^{e_{i}})}^{a}-{(p_{i}^{e_{i}-1})}^{a}} = \prod_{i=1}^{r}(p_{i}^{e_{i}-1})^{b-a}\cdot\frac{p_{i}^{b}-1}{p_{i}^{a}-1} =\prod_{p^{l}\|m}p^{(b-a)(l-1)}\cdot\frac{p^{b}-1}{p^{a}-1}. \end{align*} $$
Since
$a\mid b$
, we have
${(p^{b}-1)}/{(p^{a}-1)}\in \mathbb {Z}$
. Thus,
${\phi _{b}(m)}/{\phi _{a}(m)}\in \mathbb {Z}$
and Lemma 2.2 is proved.
The first main result of this section is the following theorem.
Theorem 2.3. Let S be a factor-closed set, and let a and b be positive integers with
$a\mid b$
. Then,
$(S^b)=(S^a)A$
, where
$A=(a_{ij})$
is an integer matrix of order
$|S|$
with
$$ \begin{align*} a_{ij}:=\begin{cases} 0& {\mathit{if }} x_i\not\mid x_j, \\ \displaystyle{\sum_{x_i|x_k|x_j}\mu\bigg(\frac{x_k}{x_i}\bigg) \prod\limits_{p|x_k}p^{(b-a)(\nu_p(x_k)-1)} \sum_{t=0}^{{b}/{a}-1}p^{at}} & {\mathit{otherwise}.} \end{cases} \end{align*} $$
Proof. It is clear that
$\phi _{b}(p^l)=p^{bl}-p^{b(l-1)}=p^{b(l-1)}(p^{b}-1)$
for the prime power
$p^{l} \ (l\geq 1)$
. Set
$f_{x_{j},b}(x):=(x, x_{j})^{b} \mbox { for all } x\in \mathbb {Z}$
. Since S is factor closed, by Lemma 2.1,
$$ \begin{align} \!\!\! ((S^{a})^{-1} (S^{b}))_{ij} &= \sum_{t=1}^{n}(S^{a})^{-1}_{it}\cdot(S^{b})_{tj} =\sum_{t=1}^{n}(x_{t},x_{j})^{b}\alpha_{it} = \sum_{t=1}^{n}(x_{t},x_{j})^{b}\sum_{\substack{x_i|x_k\\ x_t|x_k}}\frac{\mu({x_k}/{x_i})\mu({x_k}/{x_t})} {\phi_{a}(x_k)}\notag\\[6pt] &= \sum_{x_i|x_k}\phi_{a}(x_k)^{-1} \mu\bigg(\frac{x_k}{x_i}\bigg)\sum_{\substack{t=1\\ x_t|x_k}}^{n} (x_{t},x_{j})^{b}\mu\bigg(\frac{x_k}{x_t}\bigg) = \sum_{x_i|x_k}\frac{\mu({x_k}/{x_i})}{\phi_{a}(x_k)} \sum_{d|x_k}(d,x_{j})^{b}\mu\bigg(\frac{x_k}{d}\bigg)\notag\\[6pt] &=\sum_{x_i|x_k}\frac{\mu({x_k}/{x_i})} {\phi_{a}(x_k)}\sum_{d|x_k}f_{x_{j},b}(d)\mu\bigg(\frac{x_k}{d}\bigg) =\sum_{x_i|x_k}\frac{\mu({x_k}/{x_i})}{\phi_{a}(x_k)} \cdot(f_{x_{j},b}\ast\mu)(x_k). \end{align} $$
Since
$\gcd (x,d_{1}d_{2})=\gcd (x,d_1)\cdot \gcd (x,d_2)$
whenever
$(d_1,d_2)=1$
, the function
$f_{x_{j},b}$
is a multiplication function. Computing
$(f_{x_{j},b}\ast \mu )(p^{l})$
for the prime power
$p^{l} \ (l\geq 1)$
,
$$ \begin{align*} (f_{x_{j},b}\ast\mu)(p^{l}) & = \sum_{d|p^{l}}(d,x_{j})^{b}\mu\bigg(\frac{p^l}{d}\bigg) =\sum_{e=0}^{l}(p^{e},x_j)^{b}\mu(p^{l-e})\\[6pt] & = (p^l,x_j)^{b}\mu(1)+(p^{l-1},x_j)^{b}\mu(p) = (p^l,x_j)^{b}-(p^{l-1},x_j)^{b}\\[6pt] & = \begin{cases} 0 & \mathrm{if}\ p^{l}\nmid x_j\Leftrightarrow \nu_p(x_j)\leq l-1, \\ \phi_{b}(p^l) & \mathrm{if}\ p^{l}|x_j\Leftrightarrow \nu_p(x_j)\geq l. \end{cases} \end{align*} $$
Thus,
$$ \begin{align} (f_{x_{j},b}\ast\mu)(x_k)= \begin{cases} \phi_{b}(x_k) & \mathrm{if}\ x_k\mid x_j, \\ 0 & \mathrm{if}\ x_k\nmid x_j. \end{cases} \end{align} $$
It then follows from (2.1) and (2.2) that
$$ \begin{align} ((S^{a})^{-1}(S^{b}))_{ij} & =\sum_{\substack{x_i|x_k\\ x_k|x_j}}\frac{\mu({x_k}/{x_i})} {\phi_{a}(x_k)}\cdot(f_{x_{j},b}\ast\mu)(x_k) +\sum_{\substack{x_i|x_k\\ x_k\nmid x_j}}\frac{\mu({x_k}/{x_i})} {\phi_{a}(x_k)}\cdot(f_{x_{j},b}\ast\mu)(x_k)\notag\\ & = \sum_{\substack{x_i|x_k\\ x_k|x_j}}\frac{\mu({x_k}/{x_i})} {\phi_{a}(x_k)}\phi_{b}(x_k) =\begin{cases} 0 & \mathrm{if}\ x_i\nmid x_j, \\ {\displaystyle \sum_{x_i|x_k|x_j}\frac{\phi_{b}(x_k)} {\phi_{a}(x_k)}\mu\bigg(\frac{x_k}{x_i}\bigg)} & \mathrm{if}\ x_i\mid x_j. \end{cases} \end{align} $$
$$ \begin{align*} ((S^{a})^{-1}(S^{b}))_{ij} & = \begin{cases} 0 & \mathrm{if}\ x_i\nmid x_j, \\ {\displaystyle \sum_{x_i|x_k|x_j} \mu\bigg(\frac{x_k}{x_i}\bigg)\prod_{p^{l}\|x_k} p^{(b-a)(l-1)}\sum_{t=0}^{b/a-1}(p^{a})^{t}} & \mathrm{if}\ x_i\mid x_j, \end{cases} \\ & = a_{ij}. \end{align*} $$
This concludes the proof of Theorem 2.3.
We can now give the second main result of this section.
Theorem 2.4. Let S be a factor-closed set, and let a and b be positive integers with
$a\mid b$
. Then,
$[S^b]=(S^a) B,$
where
$B=(b_{ij})$
is an integer matrix of order
$|S|$
with
$$ \begin{align*} b_{ij}:=\sum_{x_i|x_k}\bigg(\frac{x_j}{(x_k,x_j)}\bigg)^{b} \mu\bigg(\frac{x_k}{x_i}\bigg)\frac{\phi_{b}(x_k)} {\phi_{a}(x_k)}\delta(x_k,x_j) \end{align*} $$
and
$$ \begin{align*} \delta(x_k,x_j):=\begin{cases} 0 & \mathit{if}\ p^{l}\mid x_j\,\ \mathit{for\ some}\ p^{l}\,\|\,x_k, \\ 1 & \mathit{otherwise}. \end{cases} \end{align*} $$
Proof. Set
$g_{x_{j},b}(x):={x^b}/{(x,x_{j})^{b}}$
for all
$x\in \mathbb {Z}$
. Since S is factor closed, Lemma 2.1 yields
$$ \begin{align} ((S^{a})^{-1}[S^{b}])_{ij} &= \sum_{t=1}^{n}((S^{a})^{-1})_{it}[S^{b}]_{tj} =\sum_{t=1}^{n}\alpha_{it}[x_{t},x_{j}]^{b} =\sum_{t=1}^{n}\frac{x_{t}^{b} x_{j}^{b}} {(x_{t},x_{j})^{b}}\cdot\alpha_{it}\notag\\ &= x_{j}^{b}\sum_{t=1}^{n}\frac{x_{t}^{b}}{(x_{t},x_{j})^{b}} \sum_{\substack{x_i|x_k\\ x_t|x_k}}\frac{1}{\phi_{a}(x_k)} \mu\bigg(\frac{x_k}{x_i}\bigg)\mu\bigg(\frac{x_k}{x_t}\bigg)\notag\\ &= x_{j}^{b}\sum_{x_i|x_k}\frac{\mu({x_k}/{x_i})} {\phi_{a}(x_k)}\sum_{t=1\atop x_t|x_k}^{n}\frac{x_{t}^{b}} {(x_{t},x_{j})^{b}}\mu\bigg(\frac{x_k}{x_t}\bigg)\notag\\ &= x_{j}^{b}\sum_{x_i|x_k}\frac{\mu({x_k}/{x_i})} {\phi_{a}(x_k)}\sum_{d|x_k}\frac{d^b} {(d,x_{j})^{b}}\mu\bigg(\frac{x_k}{d}\bigg)\notag\\ &=x_{j}^{b}\sum_{x_i|x_k}\frac{\mu({x_k}/{x_i})} {\phi_{a}(x_k)}\sum_{d|x_k}g_{x_{j},b}(d) \mu\bigg(\frac{x_k}{d}\bigg) \notag\\ &=x_{j}^{b}\sum_{x_i|x_k}\frac{\mu({x_k}/{x_i})} {\phi_{a}(x_k)}(g_{x_{j},b}\ast\mu)(x_k). \end{align} $$
For any coprime positive integers
$d_1$
and
$d_2$
,
$$ \begin{align*}g_{x_{j},b}(d_1d_2) =\frac{d_{1}^{b}d_{2}^{b}}{(d_{1}d_{2},x_j)^{b}} =\frac{d_{1}^{b}}{(d_{1},x_j)^{b}}\cdot\frac{d_{2}^{b}} {(d_{2},x_j)^{b}}=g_{x_{j},b}(d_1)g_{x_{j},b}(d_2),\end{align*} $$
so
$g_{x_{j},b}$
is a multiplicative function. Now let
$l=\nu _p(x_k)$
. Calculating
$(g_{x_{j},b}\ast \mu )(p^{l})$
,
$$ \begin{align*} (g_{x_{j},b}\ast\mu)(p^{l}) =& \sum_{e=0}^{l}g_{x_{j},b}(p^{e})\mu(p^{l-e}) = g_{x_{j},b}(p^{l})-g_{x_{j},b}(p^{l-1}) =\frac{p^{lb}}{ (p^l,x_j)^{b}}-\frac{p^{(l-1)b}}{(p^{l-1},x_j)^{b}}. \end{align*} $$
If
$p^{l}\mid x_j$
, then
$\nu _p(x_j)\geq l$
and
$$ \begin{align*}\frac{p^{lb}}{ (p^l,x_j)^{b}}-\frac{p^{(l-1)b}}{ (p^{l-1},x_j)^{b}} =\frac{p^{lb}}{p^{lb}}-\frac{p^{(l-1)b}}{p^{(l-1)b}}=1-1=0.\end{align*} $$
If
$p^{l}\nmid x_j$
, then
$\nu _p(x_j)\leq l-1$
and so
$$ \begin{align*} \frac{p^{lb}}{(p^l,x_j)^{b}}-\frac{p^{(l-1)b}}{ (p^{l-1},x_j)^{b}} =\frac{p^{lb}}{p^{b\cdot \nu_p(x_j)}}-\frac{p^{(l-1)b}} {p^{b\cdot \nu_p(x_j)}} =\frac{p^{lb}}{p^{b\cdot \nu_p(x_j)}}\Big(1-\frac{1}{p^b}\Big) =\frac{\phi_{b}(p^l)}{(p^l,x_j)^{b}}. \end{align*} $$
Hence,
$$ \begin{align*} (g_{x_{j},b}\ast\mu)(p^{l})=\begin{cases} 0 & \mathrm{if}\ p^{l}\mid x_j\Leftrightarrow \nu_p(x_j)\geq l=\nu_p(x_k), \\ {\displaystyle \frac{\phi_{b}(p^l)}{(p^l,x_j)^{b}}} & \mathrm{if}\ p^{l}\nmid x_j\Leftrightarrow \nu_p(x_j)\leq l-1=\nu_p(x_k)-1. \end{cases} \end{align*} $$
It then follows from
$$ \begin{align*}(g_{x_{j},b}\ast\mu)(x_k)=\prod\limits_{p^{l}\|x_k} (g_{x_{j},b}\ast\mu)(p^{l})\end{align*} $$
that
$(g_{x_{j},b}\ast \mu )(x_k)=0$
if there exists a prime power
$p^l$
with
$p^{l}\,\|\,x_k$
such that
$p^l\mid x_j$
. Furthermore, if
$p^{l}\nmid x_j$
for any prime power
$p^l$
with
$p^{l}\,\|\,x_k$
, then
$$ \begin{align*} (g_{x_{j},b}\ast\mu)(x_k) =\prod_{\substack{p^{l}\|x_k \\ p^{l}\nmid x_j}}\frac{p^{lb}} {p^{b \nu_p(x_j)}}\bigg(1-\frac{1}{p^b}\bigg) =\frac{1}{\prod\limits_{\substack{p^l\|x_k\\ p^l\nmid x_j}}p^{b\nu_p(x_j)}} \prod\limits_{\substack{p^ l\| x_k \\ p^l \not\mid x_j}}p^{lb}\bigg(1-\frac{1}{p^b}\bigg) = \frac{\phi_{b}(x_k)}{\gcd(x_k,x_j)^{b}}. \end{align*} $$
Therefore,
$$ \begin{align} (g_{x_{j},b}\ast\mu)(x_k)=\begin{cases} 0 & \mathrm{if}\ p^{l}|x_j \ \mathrm{for \ some \ prime \ power}\ p^{l}\|x_k, \\ {\displaystyle \frac{\phi_{b}(x_k)}{\gcd(x_k, x_j)^{b}}} & \mathrm{otherwise}. \end{cases} \end{align} $$
Substituting (2.5) into (2.4) yields
$$ \begin{align*}((S^{a})^{-1}[S^{b}])_{ij}=\sum_{x_i|x_k} \bigg(\frac{x_j}{\gcd(x_k,x_j)}\bigg)^{b} \mu\bigg(\frac{x_k}{x_i}\bigg) \frac{\phi_{b}(x_k)}{\phi_{a}(x_k)} \delta(x_k,x_j)=b_{ij},\end{align*} $$
where
$$ \begin{align*} \delta(x_k,x_j):=\begin{cases} 0 & \mathrm{if}\ p^{l}\mid x_j \ \mathrm{for \ some \ prime \ power}\ p^{l}\,\|\,x_k, \\ 1 & \mathrm{otherwise}. \end{cases} \end{align*} $$
This finishes the proof of Theorem 2.4.
The following result is also due to Bourque and Ligh.
Lemma 2.5 [Reference Bourque and Ligh7].
Let S be an FC set. Then,
$[S^a]^{-1}=(\beta _{ij})$
, where
$$ \begin{align*}\beta_{ij}=\frac{1}{(x_{i}x_{j})^{a}}\sum_{\substack{x_i|x_k\\ x_j|x_k}} \frac{x_{k}^{2a}v(x_k)}{r(x_k)^{a}\phi_{a}(x_k)} \mu\bigg(\frac{x_k}{x_i}\bigg)\mu\bigg(\frac{x_k}{x_j}\bigg), \end{align*} $$
with v and r being multiplicative functions defined for a prime power
$p^e$
by
$v(p^e)=-1$
and
$r(p^e)=p$
.
The third main result of this section is the following theorem.
Theorem 2.6. Let S be a factor-closed set and let a and b be positive integers with
$a\mid b$
. Then,
$[S^b]=[S^a]C,$
where
$C=(c_{ij})$
is an integer matrix of order
$|S|$
with
$$ \begin{align*} c_{ij}\ &{:=}\ \sum_{x_i|x_k}(-1)^\alpha\bigg(\frac{x_k}{x_i}\bigg)^{a} \mu\bigg(\frac{x_k}{x_i}\bigg) \prod_{\substack{p|x_k\\ \nu_p(x_j) \geq \nu_p(x_k)}}p^{\nu_p(x_j) b-\nu_p(x_k) a}\\& \quad \times \prod_{\substack{p|x_k\\ \nu_p(x_j)< \nu_p(x_k)}} \frac{p^{(b-a)(\nu_p(x_k)-1)}(p^{b-a}-1)}{p^{a}-1} \prod_{\substack{p|x_j\\ p\not\mid x_k}} p^{\nu_p(x_j)b}\end{align*} $$
and
$\alpha ={\sum _{p|x_k,\, \nu _p(x_j)<\nu _p(x_k)}1}$
.
Proof. By Lemma 2.5,
$$ \begin{align*} c_{ij} &= ([S^{a}]^{-1}[S^{b}])_{ij}=\sum_{t=1}^{n} ([S^{a}]^{-1})_{it}\cdot[S^{b}]_{tj}\\ & = \sum_{t=1}^{n}\beta_{it}[x_t,x_j]^{b}=\sum_{t=1}^{n} \frac{x_{t}^{b}x_{j}^{b}}{(x_t,x_j)^{b}}\beta_{it}\\ & = x_{j}^{b}\sum_{t=1}^{n}\frac{x_{t}^{b}}{(x_t,x_j)^{b}} \frac{1}{x_{i}^{a}x_{t}^{a}}\sum_{\substack{x_i|x_k\\ x_t|x_k}} \frac{x_{k}^{2a} v(x_k)}{r(x_k)^{a}\phi_{a}(x_k)} \mu\bigg(\frac{x_k}{x_i}\bigg)\mu\bigg(\frac{x_k}{x_t}\bigg)\\ & = \frac{x_{j}^{b}}{x_{i}^{a}}\sum_{x_i|x_k}\frac{x_{k}^{2a} v(x_k) \mu({x_k}/{x_i})}{r(x_k)^{a}\phi_{a}(x_k)}\sum_{\substack{t=1\\ x_t|x_k}}^{n}\frac{x_t^{b-a}}{(x_t,x_j)^{b}}\mu\bigg(\frac{x_k}{x_t}\bigg). \end{align*} $$
Define the arithmetic function
$h_{x_{j},a,b}$
by
$h_{x_{j},a,b}(x):=x^{b-a}/(x,x_{j})^{b}$
for all positive integers x. Since S is factor closed,
$$ \begin{align*} c_{ij} & = \frac{x_{j}^{b}}{x_{i}^{a}}\sum_{x_i|x_k}\frac{x_{k}^{2a} v(x_k)\mu({x_k}/{x_i})}{r(x_k)^{a}\phi_{a}(x_k)} \sum_{d|x_k}\frac{d^{b-a}}{(d,x_{j})^{b}}\mu\bigg(\frac{x_k}{d}\bigg)\\ & = \frac{x_{j}^{b}}{x_{i}^{a}}\sum_{x_i|x_k}\frac{x_{k}^{2a} v(x_k) \mu({x_k}/{x_i})}{r(x_k)^{a}\phi_{a}(x_k)} \sum_{d|x_k}h_{x_{j},a,b}(d)\mu\bigg(\frac{x_k}{d}\bigg)\\ & = \frac{x_{j}^{b}}{x_{i}^{a}}\sum_{x_i|x_k}\frac{x_{k}^{2a} v(x_k) \mu({x_k}/{x_i})}{r(x_k)^{a}\phi_{a}(x_k)} (h_{x_{j},a,b}\ast\mu)(x_k). \end{align*} $$
For all positive integers
$d_1$
and
$d_2$
with
$(d_1,d_2)=1$
,
$$ \begin{align*}h_{x_{j},a,b}(d_1d_2) =\frac{(d_{1}d_{2})^{b-a}}{(d_{1}d_{2},x_j)^{b}} =\frac{d_{1}^{b-a}}{(d_{1},x_j)^{b}} \cdot\frac{d_{2}^{b-a}}{(d_{2},x_j)^{b}} =h_{x_{j},a,b}(d_1) h_{x_{j},a,b}(d_2),\end{align*} $$
so
$h_{x_{j},a,b}$
is a multiplicative function. Computing
$(h_{x_{j},a,b}\ast \mu )(p^{l})$
for
$l=\nu _p(x_k)$
,
$$ \begin{align*} (h_{x_{j},a,b}\ast\mu)(p^{l})=&h_{x_{j},a,b}(p^l)-h_{x_{j},a,b}(p^{l-1}) =\frac{p^{l(b-a)}}{ (p^l,x_j)^{b}}-\frac{p^{(l-1)(b-a)}}{(p^{l-1},x_j)^{b}}. \end{align*} $$
If
$p^{l}\mid x_j$
, then
$\nu _{p}(x_j)\geq l=\nu _p(x_k)$
, and so
$$ \begin{align*} \frac{p^{l(b-a)}}{(p^l,x_j)^{b}}-\frac{p^{(l-1)(b-a)}}{(p^{l-1},x_j)^{b}} =\frac{p^{l(b-a)}}{p^{lb}}-\frac{p^{(l-1)(b-a)}}{p^{(l-1)b}} =\frac{1}{p^{la}}-\frac{1}{p^{(l-1)a}}=\frac{1-p^a}{p^{la}}. \end{align*} $$
If
$p^{l}\nmid x_j$
, then
$\nu _{p}(x_j)\leq l-1=\nu _p(x_k)-1$
and so
$$ \begin{align*} \frac{p^{l(b-a)}}{(p^l,x_j)^{b}}-\frac{p^{(l-1)(b-a)}}{(p^{l-1},x_j)^{b}} =\frac{p^{l(b-a)}-p^{(l-1)(b-a)}}{p^{b \nu_p(x_j)}} = \frac{p^{l(b-a)}(1-{1}/{p^{b-a}})}{p^{b \nu_p(x_j)}}. \end{align*} $$
So,
$$ \begin{align*} (h_{x_{j},a,b}\ast\mu)(p^{l})=\begin{cases} \displaystyle \frac{1-p^a}{p^{la}} & \mathrm{if}\ p^{l}\mid x_j \Leftrightarrow \nu_p(x_j)\geq l=\nu_p(x_k),\\[8pt] \displaystyle \frac{p^{l(b-a)}(1-{1}/{p^{b-a}})}{p^{b\cdot \nu_p(x_j)}} & \mathrm{if}\ p^{l}\nmid x_j\Leftrightarrow \nu_p(x_j)\leq l-1=\nu_p(x_k)-1. \end{cases} \end{align*} $$
Thus,
$$ \begin{align*} (h_{x_{j},a,b}\ast\mu)(x_k) & = \prod_{p^{l}\|x_k}(h_{x_{j},a,b}\ast\mu)(p^{l}) = \prod_{\substack{p^{l}\|x_k\\ \nu_p(x_j)\geq l}}(h_{x_{j},a,b}\ast\mu)(p^{l})\cdot \prod_{\substack{p^{l}\|x_k\\ \nu_p(x_j)<l}}(h_{x_{j},a,b}\ast\mu)(p^{l})\\ & = \prod_{\substack{p^{l}\|x_k\\ \nu_p(x_j)\geq l}}\frac{1-p^a}{(p^l)^{a}}\cdot \prod_{\substack{p^{l}\|x_k\\ \nu_p(x_j)< l}}\frac{p^{(l-1)(b-a)} (p^{b-a}-1)}{(p^{\nu_p(x_j)})^{b}}. \end{align*} $$
Moreover,
$$ \begin{align*} \phi_{a}(x_k)= x_{k}^{a}\prod_{p|x_k}\bigg(1-\frac{1}{p^a}\bigg) =x_{k}^{a}\prod_{p^{l}\|x_k}\bigg(1-\frac{1}{p^a}\bigg). \end{align*} $$
Hence,
$$ \begin{align*} c_{ij} & =\frac{x_{j}^{b}}{x_{i}^{a}}\sum_{x_i|x_k}\frac{x_{k}^{2a} v(x_k)\mu({x_k}/{x_i})}{r(x_k)^{a} x_{k}^{a} \cdot\prod_{p^{l}\|x_k}(1-{1}/{p^a})} \prod_{\substack{p^{l}\|x_k\\ \nu_p(x_j)\geq l}} \frac{1-p^a}{(p^l)^{a}} \prod_{\substack{p^{l}\|x_k\\ \nu_p(x_j)<l}}\frac{p^{(l-1)(b-a)}(p^{b-a}-1)}{(p^{\nu_p(x_j)})^{b}}\\ & = x_{j}^{b}\sum_{x_i|x_k}\bigg(\frac{x_k}{x_i}\bigg)^{a} \mu\bigg(\frac{x_k}{x_i}\bigg) \frac{v(x_k)} {\prod_{p|x_k}(p^a-1)} \prod_{\substack{p^{l}\|x_k\\ \nu_p(x_j)\geq l}} \frac{1-p^a}{(p^l)^{a}}\prod_{\substack{p^{l}\|x_k\\ \nu_p(x_j) <l}}\frac{p^{(l-1)(b-a)}(p^{b-a}-1)}{(p^{\nu_p(x_j)})^{b}}. \end{align*} $$
It then follows that
$$ \begin{align*} c_{ij} & = \sum_{x_i|x_k} \Big(\frac{x_k}{x_i}\Big)^{a} \mu\Big(\frac{x_k}{x_i}\Big)\, v(x_k) \prod_{\substack{p|x_k\\ \nu_p(x_j) \geq \nu_p(x_k)}}\frac{p^{\nu_p(x_j) b}}{p^{\nu_p(x_k) a}} \prod_{\substack{p|x_k\\ \nu_p(x_j)< \nu_p(x_k)}} \frac{p^{\nu_p(x_j) b}}{p^{\nu_p(x_j) b}} \\ &\quad \times \prod_{\substack{p|x_k\\ \nu_p(x_j)\geq \nu_p(x_k)}} \frac{1-p^a}{p^{a}-1}\cdot\prod_{\substack{p|x_k\\ \nu_p(x_j)<\nu_p(x_k)}} \frac{p^{(b-a)(\nu_p(x_k)-1)}(p^{b-a}-1)}{p^{a}-1}\prod_{\substack{p|x_j\\ p\not\mid x_k}} p^{\nu_p(x_j)b}\\ & = \sum_{x_i|x_k} \bigg(\frac{x_k}{x_i}\bigg)^{a}\mu\bigg(\frac{x_k}{x_i}\bigg)\,(-1)^{\sum_{p|x_k, v_p(x_j)<v_p(x_k)1}} \prod_{\substack{p|x_k\\ \nu_p(x_j) \geq \nu_p(x_k)}}p^{\nu_p(x_j) b-\nu_p(x_k) a} \\ &\quad \times \prod_{\substack{p|x_k\\ \nu_p(x_j)< \nu_p(x_k)}} \frac{p^{(b-a)(\nu_p(x_k)-1)}(p^{b-a}-1)}{p^{a}-1}\prod_{\substack{p|x_j\\ p\not\mid x_k}}p^{\nu_p(x_j)b} \end{align*} $$
as required. This completes the proof of Theorem 2.6.
Finally, it is easy to see that the conclusion of Theorem 1.1 follows immediately from Theorems 2.3, 2.4 and 2.6.
3 Remarks and conjectures in the gcd-closed case
For any set R of positive integers and for any
$x\in R$
, we define
$g_R(x):=|G_{R}(x)|$
to be the number of greatest-type divisors of x in R. For brevity, we write
$g(x)$
for
$g_R(x)$
.
Definition 3.1. Let S be a set of positive integers. Let
$x\in S$
with
$g(x)\ge 2$
.
-
(i) We say that the two distinct greatest-type divisors
$y_1$
and
$y_2$
of x in S satisfy the condition
$\mathcal {G}$
if
$[y_1,y_2]=x$
and
$(y_1,y_2)\in G_S(y_1)\cap G_S(y_2)$
. -
(ii) We say that x satisfies the condition
$\mathcal {G}$
if any two distinct greatest-type divisors of x in S satisfy the condition
$\mathcal {G}$
.
It should be pointed out that if S is gcd closed, then the definition of condition
$\mathcal {G}$
given above coincides with that of condition
$\mathcal {C}$
given in [Reference Feng, Hong and Zhao9].
Definition 3.2. Let S be a set of positive integers. We say that the set S satisfies the condition
$\mathcal {G}$
if, for any element x in S, either
$g(x)\le 1$
, or
$g(x)\ge 2$
and x satisfies the condition
$\mathcal {G}$
.
For instance, let
$S=\{2,3,4,5,9,11,12,15,33,36,55,165\}$
. Then, S is not gcd closed and S does not satisfy the condition
$\mathcal {G}$
. In fact,
${g(2)=g(3)=g(5)=g(11)=0}$
,
$g(4)=g(9)=1$
,
$g(12)=g(15)=g(33)=g(36)=g(55)=2$
and
$g(165)=3$
. Since
$G_S(12)=\{3,4\}$
and
$(3,4)=1\notin G_S(3)\cap G_S(4)$
, it follows that 12 does not satisfy the condition
$\mathcal {G}$
. Likewise, each of 15, 33 and 55 does not satisfy the condition
$\mathcal {G}$
. Since
$G_S(36)=\{9,12\}$
and
$[9,12]=36$
as well as
$(9,12)=3\in G_{S}(9)\cap G_{S}(12)$
,
$9$
and
$12$
satisfy the condition
$\mathcal {G}$
and so
$36$
satisfies the condition
$\mathcal {G}$
. Since
$G_{S}(165)=\{15,33,55\}$
,
$[15,33]=165$
and
$(15,33)=3\in G_S(15)\cap G_S(33)$
, 15 and 33 satisfy the condition
$\mathcal {G}$
. Similarly, one can check that 15 and 55 satisfy the condition
$\mathcal {G}$
, and 33 and 55 satisfy the condition
$\mathcal {G}$
. Therefore, 165 satisfies the condition
$\mathcal {G}$
. In a similar way, one can easily check that the set
$T_1=\{1,3,4,5,9,11,12,15,33,36,55,165\}$
is a gcd-closed set satisfying the condition
$\mathcal {G}$
, the set
$T_2=\{1,3,5,6,10,11,15,33,55,165\}$
is not gcd closed but satisfies the condition
$\mathcal {G}$
since
$\gcd (6,10)=2\not \in S$
, and the set
$T_3=\{1,2,3,5,11,12,15,33,55,165\}$
is gcd closed but does not satisfy the condition
$\mathcal {G}$
.
Theorem 3.3. Every factor-closed set is a gcd-closed set satisfying the condition
$\mathcal {G}$
.
Proof. Let S be a factor-closed set. First of all, we show that S is gcd closed. For any
$x,y\in S$
, since
$(x,y)\mid x$
and
$x\in S$
and S is factor closed, we must have
$(x,y)\in S$
. Hence, S is gcd closed as desired.
Next, we show that S satisfies the condition
$\mathcal {G}$
. Clearly,
$1\in S$
and
$g(1)=0$
. In the following, we let
$x\in S$
and
$x>1$
. If x is a prime power, say,
$p^i$
with
$i\ge 1$
, then
$\{p,p^2,\ldots ,p^{i-1}\}\subseteq S$
and
$G_S(x)=\{p^{i-1}\}$
, which implies that
$g(x)=1$
. In what follows, we assume that x contains at least two prime divisors, say,
$x=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$
, where the
$e_i$
are positive integers and
$p_1,\ldots , p_k$
are
$k\ge 2$
distinct primes. Since S is factor closed, it follows that for any k-tuple
$(r_1,r_2,\ldots , r_k)\in \mathbb {Z}^k$
with
$0\le r_i\le e_i$
for
$1\le i\le k$
, one has
$p_1^{r_1}p_2^{r_2}\cdots p_k^{r_k}\in S$
. It then follows that
$$ \begin{align*}G_S(x)=\{p_1^{e_1-1}p_2^{e_2}\cdots p_k^{e_k}, p_1^{e_1}p_2^{e_2-1}\cdots p_k^{e_k},\ldots,p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k-1}\} =\bigg\{\frac{x}{p_1},\frac{x}{p_2},\ldots,\frac{x}{p_k}\bigg\}.\end{align*} $$
Then,
$g(x)=k\ge 2$
, and for any two distinct integers i and j with
$1\le i< j\le k$
, we have
$$ \begin{align*}\bigg(\frac{x}{p_i},\frac{x}{p_j}\bigg)=\frac{x}{p_ip_j}\in G_{S}\bigg(\frac{x}{p_i}\bigg)\cap G_{S}\bigg(\frac{x}{p_j}\bigg)\end{align*} $$
and
$$ \begin{align*} \bigg[\frac{x}{p_i},\frac{x}{p_j}\bigg] & =[p_1^{e_1}\cdots p_i^{e_i-1}\cdots p_j^{e_j}\cdots p_k^{e_k},p_1^{e_1}\cdots p_i^{e_i}\cdots p_j^{e_j-1}\cdots p_k^{e_k}] \\ & =p_1^{e_1}\cdots p_i^{e_i}\cdots p_j^{e_j}\cdots p_k^{e_k}=x. \end{align*} $$
Thus,
${x}/{p_i}$
and
${x}/{p_j}$
satisfy the condition
$\mathcal {G}$
, which implies that the element
$x\in S$
satisfies the condition
$\mathcal {G}$
. Therefore, Theorem 3.3 is proved.
For any factor-closed set S and for any integer
$a>1$
, the set
$aS:=\{ax: x\in S\}$
is not factor closed since
$a\in aS$
,
$1\mid a$
and
$1\not \in aS$
. One can easily check that if S is a gcd-closed set satisfying the condition
$\mathcal {G}$
, then so is the set
$aS$
. Bourque and Ligh [Reference Bourque and Ligh4] showed that for any factor-closed set S, one has
$(S)\mid [S]$
in the ring
$M_{|S|}(\mathbb {Z})$
of
$|S|\times |S|$
matrices over the integers. If a and b are positive integers with
$a\mid b$
, then
$(S^a)\mid (S^b), (S^a)\mid [S^b]$
and
$[S^a]\mid [S^b]$
when the set S is a divisor chain (see [Reference Hong21]), or is a gcd-closed set with
$\max _{x\in S}\{|G_S(x)|\}=1$
(see [Reference Zhu29, Reference Zhu and Li31]), or is a factor-closed set (from this paper). We believe that such factorisations are still true when S is a gcd-closed set satisfying the condition
$\mathcal {G}$
.
Conjecture 3.4. Let a and b be positive integers with
$a\mid b$
and let S be a gcd-closed set satisfying the condition
$\mathcal {G}$
. Then,
$(S^a)\mid (S^b), (S^a)\mid [S^b]$
and
$[S^a]\mid [S^b]$
in
$M_{|S|}({\mathbb Z})$
.
Conjecture 3.5. Let a and b be positive integers with
$a\mid b$
. Let
$S=S_1\cup S_2$
with
$S_1$
and
$S_2$
being gcd-closed sets satisfying the condition
$\mathcal {G}$
such that
$\gcd (\mathrm {lcm}(S_1),\mathrm {lcm}(S_2))=1$
and
$\min(S)>1$
. Then,
$(S^a)\mid (S^b),(S^a)\mid [S^b]$
and
$[S^a]\mid [S^b]$
in
$M_{|S|}({\mathbb Z})$
.
Conjecture 3.6. Let
$a\ge 1$
be an integer. Let
$S_1$
and
$S_2$
be gcd-closed sets such that
$\gcd (\mathrm {lcm}(S_1), \mathrm {lcm}(S_2))=1$
. Then, (i)
$(S_1^a)\mid [S_1^a]$
if and only if
$S_1$
satisfies the condition
$\mathcal {G}$
. (ii) If
$S=S_1\cup S_2$
and
$\min(S)>1$
, then
$(S^a)\mid [S^a]$
if and only if
$S_1$
and
$S_2$
satisfy the condition
$\mathcal {G}$
.
A general question is to characterise the couples
$(S,f)$
with S a set of distinct positive integers, and f integer-valued and multiplicative for which the matrix
$(f[S])$
is divisible by the matrix
$(f(S))$
in the ring
$M_{|S|}(\mathbb {Z})$
. If S is factor closed or multiple closed, the answer to this question is known (see, for instance, [Reference Bourque and Ligh7, Reference Hong18]). If S is gcd closed, we give a conjectural answer to this question as the following conjecture.
Conjecture 3.7. Let S be a gcd-closed set satisfying the condition
$\mathcal {G}$
and let f be a multiplicative function such that
$f(d)\in \mathbb {Z}$
whenever
$d\mid \mathrm {lcm}(S)$
and
$f(u)\mid f(v)$
whenever
$u\mid v$
and
$u,v\in S$
. Then,
$(f(S))$
divides
$(f[S])$
in
$M_n(\mathbb {Z})$
.
Conjecture 3.8. Let
$S=S_1\cup S_2$
with
$S_1$
and
$S_2$
being gcd-closed sets satisfying the condition
$\mathcal {G}$
such that
$\gcd({\rm lcm}(S_1), {\rm lcm}(S_2))=1$
and
$\min(S)>1$
. Let f be a multiplicative function such that
$f(d)\in \mathbb {Z}$
whenever
$d\mid \mathrm {lcm}(S)$
, and
$f(u)\mid f(v)$
whenever
$u\mid v$
and
$u,v\in S$
. Then,
$(f(S))$
divides
$(f[S])$
in
$M_n(\mathbb {Z})$
.
Some evidence to support Conjecture 3.8 is given in [Reference Chen, Feng, Hong and Qiu8]. An interesting problem is to give explicit formulas for the determinants of the matrices
$(f(S))$
and
$(f[S])$
when S is a gcd-closed set satisfying the condition
$\mathcal{G}$
. Another general question is to characterise the couples
$(S,f)$
with S gcd closed and f multiplicative such that the matrix
$(f[S])$
is nonsingular. We will explore this interesting question in the near future. Regarding the counterexamples to the Bourque–Ligh conjecture [Reference Bourque and Ligh4] that the LCM matrix
$[S]$
defined on any gcd-closed set S is nonsingular, we surprisingly found that all the known gcd-closed sets S with the LCM matrix
$[S]$
being singular, given in [Reference Haukkanen, Mattila and Mantysalo10, Reference Hong14, Reference Hong15, Reference Hong19, Reference Hong20, Reference Korkee, Mattila and Haukkanen24], do not satisfy the condition
$\mathcal {G}$
. We believe that this phenomenon is true in general. In other words, we propose the following conjectures that are closely relating to the conjecture raised in [Reference Hong16].
Conjecture 3.9. Let
$a\ge 1$
be an integer. Then, the ath power LCM matrix
$[S^a]$
defined on any gcd-closed set S satisfying the condition
$\mathcal {G}$
is nonsingular.
Conjecture 3.10. Let
$a\ge 1$
be an integer. Let
$S=S_1\cup S_2$
with
$S_1$
and
$S_2$
being gcd-closed sets satisfying the condition
$\mathcal {G}$
such that
$\gcd (\mathrm {lcm}(S_1), \mathrm {lcm}(S_2))=1$
and
$\min(S)>1$
. Then, the ath power LCM matrix
$[S^a]$
defined by S is nonsingular.
Acknowledgements
The author thanks Dr. S.Y. Qiang for her help in typing the manuscript. The author would also like to thank the anonymous referee for helpful comments and a careful reading of the manuscript and for correcting some typos as well as errors.
