A $q$-analogue of Wilson's congruence

TL;DR

This paper establishes a $q$-analogue of Wilson's congruence involving permutation cycles, the major index, and cyclotomic polynomials, extending classical number theory results into the realm of $q$-series.

Contribution

It introduces a novel $q$-analogue of Wilson's congruence, connecting permutation statistics with cyclotomic polynomials and the Möbius function.

Findings

Proves a $q$-analogue of Wilson's congruence involving the major index.

Establishes congruence relations modulo cyclotomic polynomials.

Links permutation cycle enumeration with classical number theory concepts.

Abstract

Let ${\mathcal C}_n$ be the set of all permutation cycles of length $n$ over $\{1,2,\ldots,n\}$. Let $${\mathfrak f}_n(q):=\sum_{\sigma\in{\mathcal C}_{n+1}}q^{{\mathrm maj}\,\sigma} $$ be a $q$-analogue of the factorial $n!$, where ${\mathrm maj}$ denotes the major index. We prove a $q$-analogue of Wilson's congruence $$ {\mathfrak f}_{n-1}(q)\equiv\mu(n)\pmod{\Phi_n(q)}, $$ where $\mu$ denotes the M\"obius function and $\Phi_n(q)$ is the $n$-th cyclotomic polynomial.

Equations

$${\mathfrak{f}}_n(q):=\sum _{\sigma \in {\mathcal{C}}_{n+1}}{q}^{\mathrm{maj}\sigma }$$

$${\mathfrak{f}}_{n-1}(q)\equiv \mu (n)(\mathrm{mod}{\Phi }_n(q)),$$

$$[n{]}_q:=\frac{1-q^n}{1-q}.$$

$$\underset{q\to 1}{\lim }[n{]}_q=\underset{q\to 1}{\lim }(1+q+\cdots +q^{n-1})=n.$$

$$\left(\genfrac{}{}{0ex}{}{ap+b}{cp+d}\right)\equiv \left(\genfrac{}{}{0ex}{}{a}{c}\right)\left(\genfrac{}{}{0ex}{}{b}{d}\right)(\mathrm{mod}p)$$

$${\left[\genfrac{}{}{0ex}{}{n}{k}\right]}_q:=\frac{[n{]}_q!}{[k{]}_q![n-k{]}_q!}$$

$$[n{]}_q!:=[1{]}_q[2{]}_q\cdots [n{]}_q$$

$${\left[\genfrac{}{}{0ex}{}{ap+b}{cp+d}\right]}_q\equiv \left(\genfrac{}{}{0ex}{}{a}{c}\right)\cdot {\left[\genfrac{}{}{0ex}{}{b}{d}\right]}_q(\mathrm{mod}[p{]}_q),$$

$${\left[\genfrac{}{}{0ex}{}{an+b}{cn+d}\right]}_q\equiv \left(\genfrac{}{}{0ex}{}{a}{c}\right)\cdot {\left[\genfrac{}{}{0ex}{}{b}{d}\right]}_q(\mathrm{mod}{\Phi }_n(q)),$$

$${\Phi }_n(q):=\prod _{\begin{array}{c}1\le k\le n\\ (k,n)=1\end{array}}(q-e^{2\pi i\cdot \frac{k}{n}})$$

$$a^{p-1}\equiv 1(\mathrm{mod}p)$$

$$(p-1)!\equiv -1(\mathrm{mod}p)$$

$$\prod _{k=1}^{p-1}[a{]}_{q^k}\equiv 1(\mathrm{mod}[p{]}_q).$$

$$\prod _{k=1}^{p-1}[a{]}_{q^k}=\frac{1}{{\prod }_{k=1}^{p-1}(1-q^k)}\prod _{k=1}^{p-1}(1-q^{ak})\equiv \frac{1}{{\prod }_{k=1}^{p-1}(1-q^k)}\prod _{j=1}^{p-1}(1-q^j)=1(\mathrm{mod}[p{]}_q).$$

$$\prod _{k=1}^{n-1}[a{]}_{q^k}\equiv 1(\mathrm{mod}{\Phi }_n(q))$$

$$[6{]}_q!\equiv 3+3q-4q^3-6q^4-4q^5(\mathrm{mod}[7{]}_q).$$

$$\prod _{k=1}^{p-1}[k{]}_{q^k}\equiv -1(\mathrm{mod}[p{]}_q).$$

$$\mathrm{maj}\sigma :=\sum _{\begin{array}{c}1\le i\le n-1\\ \sigma (i)>\sigma (i+1)\end{array}}i$$

$${\rm inv}\,\sigma:=\big{|}\{(i,j):\,1\leq i<j\leq n,\ \sigma(i)>\sigma(j)\}\big{|}.$$

$$[n{]}_q!=\sum _{\sigma \in {\mathcal{S}}_n}{q}^{\mathrm{maj}\sigma }$$

$$[n{]}_q!=\sum _{\sigma \in {\mathcal{S}}_n}{q}^{\mathrm{inv}\sigma }.$$

$${\mathcal{C}}_n:=\{\sigma \in {\mathcal{S}}_n:\sigma \text{ is a cycle of length }n\}.$$

$${\mathfrak{f}}_n(q):=\sum _{\sigma \in {\mathcal{C}}_{n+1}}{q}^{\mathrm{maj}\sigma }.$$

$$\mu (n):=\{\begin{array}{ll}1,& \text{if }n=1,\\ (-1{)}^k,& \text{if }n=p_1\cdots p_k\text{ where }{p}_1,\dots ,p_k\text{ are distinct primes},\\ 0& \text{if }n>1\text{ is not square-free}.\end{array}$$

$${\mathfrak{f}}_{n-1}(q)\equiv \mu (n)(\mathrm{mod}{\Phi }_n(q)).$$

$${\mathfrak{f}}_{p-1}(q)\equiv -1(\mathrm{mod}[p{]}_q).$$

$$\sum _{\begin{array}{c}1\le r\le n\\ (r,n)=1\end{array}}{e}^{2\pi i\cdot \frac{kr}{n}}.$$

$$\tau (a)=a+1$$

$$T\sigma :=\tau \circ \sigma \circ {\tau }^{-1}.$$

$$T\sigma =(a_1+1,a_2+2,\dots ,a_n+1).$$

Full text

[TABLE]

be a $q$-analogue of the factorial $n!$, where ${\rm maj}\,$ denotes the major index. We prove a $q$-analogue of Wilson’s congruence

[TABLE]

where $\mu$ denotes the Möbius function and $\Phi_{n}(q)$ is the $n$-th cyclotomic polynomial.

1. Introduction

For each $n\in\mathbb{N}=\{0,1,2,\ldots\}$, define the $q$-integer

[TABLE]

The $q$-integer evidently is a $q$-analogue of the original integer, since

[TABLE]

Suppose that $p$ is a prime. The well-known Lucas congruence says that

[TABLE]

where $a,b,c,d\in\mathbb{N}$ and $0\leq b,d\leq p-1$. A $q$-analogue of the binomial coefficients is the $q$-binomial coefficient, which is given by

[TABLE]

for $n,k\in\mathbb{N}$ with $n\geq k$, where

[TABLE]

is the $q$-factorial. Then we have the following $q$-analogue of Lucas’ congruence (cf. [1])

[TABLE]

where the above congruence is considered over the polynomial ring $\mathbb{Z}[q]$. In fact, (1.2) can be extended to

[TABLE]

where

[TABLE]

denotes the $n$-th cyclotomic polynomial.

Another two classical congruences in number theory is the Fermat congruence

[TABLE]

for each prime $p$ and $1\leq a\leq p-1$, and the Wilson congruence

[TABLE]

for each prime $p$. Fermat’s congruence also has its $q$-analogue as follows:

[TABLE]

In fact, since $q^{a}-q^{b}=q^{b}(1-q^{a-b})$, for each prime $p$, we have $q^{a}\equiv q^{b}\ ({\rm{mod}}\ [p]_{q})$ if and only if $a\equiv b\ ({\rm{mod}}\ p)$. So, if $p\nmid a$, then

[TABLE]

Similarly, we can get

[TABLE]

provided that $(a,n)=1$.

Unfortunately, seemingly there exists no suitable $q$-analogue of Wilson’s congruence for the $q$-factorial $[p-1]_{q}!$. For examples, we have

[TABLE]

Alternatively, in [2], Chapman and Pan gave a partial $q$-analogue of Wilson’s congruence for those prime $p>3$ with $p\equiv 3\ ({\rm{mod}}\ 4)$:

[TABLE]

However, (1.9) is invalid if the prime $p\equiv 1\ ({\rm{mod}}\ 4)$, though Chapman and Pan also determined $\prod_{k=1}^{p-1}[k]_{q^{k}}$ modulo $[p]_{q}$ for those prime $p\equiv 1\ ({\rm{mod}}\ 4)$, with help of the fundamental unit and the class number of the quadratic field $\mathbb{Q}(\sqrt{p})$.

In this short note, we shall try to obtain a unified $q$-analogue of Wilson’s congruence for all primes, from the viewpoint of combinatorics. Let $\mathcal{S}_{n}$ denote the permutation group of order $n$, i.e., the set of all permutations over $\{1,2,\ldots,n\}$. Clearly $|\mathcal{S}_{n}|=n!$. For each $\sigma\in\mathcal{S}_{n}$, define the major index of $\sigma$

[TABLE]

and the inversion number $\sigma$

[TABLE]

It is know (cf. ) that

[TABLE]

and

[TABLE]

Let

[TABLE]

We also have $|\mathcal{C}_{n}|=(n-1)!$. Define

[TABLE]

Clearly ${\mathfrak{f}}_{n}(q)$ is another $q$-analogue of the factorial $n!$, too. In this note, we shall prove a $q$-analogue of Wilson’s congruence for ${\mathfrak{f}}_{n}(q)$. Define the Möbius function

[TABLE]

Theorem 1.1**.**

Suppose that $n\geq 2$. Then

[TABLE]

In particular, if $p$ is prime, then

[TABLE]

In fact, as we shall see later, the Möbius function $\mu$ in (1.13) arises from the Ramanujan sum

[TABLE]

2. Proof of Theorem 1.1

Suppose that $n\geq 2$. Let $\mathbb{Z}_{n}=\mathbb{Z}/n\mathbb{Z}$ be the cyclic group of order $n$. Below we always identify $\mathbb{Z}_{n}$ with $\{1,2,\ldots,n\}$, and view $\mathcal{S}_{n}$ as the permutation group over $\mathbb{Z}_{n}$. In particular, for each $1\leq a,b\leq n$, we say $a<b$ over $\mathbb{Z}_{n}$ if and only if $a<b$ over $\mathbb{Z}$.

Let $\tau\in\mathcal{S}_{n}$ be defined by

[TABLE]

for each $a\in\{1,2,\ldots,n\}$. For each $\sigma\in\mathcal{S}_{n}$, let

[TABLE]

Then we have $T\mathcal{C}_{n}=\mathcal{C}_{n}$. In fact, for each $\sigma=(a_{1},a_{2},\ldots,a_{n})\in\mathcal{C}_{n}$,

[TABLE]

Thus $T$ can be viewed as a group action on $\mathcal{C}_{n}$, since clearly $T^{n}\sigma=\sigma$ for each $\sigma\in\mathcal{C}_{n}$.

For each $\sigma\in\mathcal{C}_{n}$, let

[TABLE]

denote the orbit of $\sigma$, where $T^{k}$ denotes the $k$-th iteration of $T$. Then we may partition $\mathcal{C}_{n}$ into

[TABLE]

where $X\subseteq\mathcal{C}_{n}$ satisfies that $\sigma\not\in U_{\kappa}$ for any distinct $\sigma,\kappa\in X$. Since $T$ is a group action, we must have $|U_{\sigma}|$ divides $n$ for each $\sigma\in X$.

Lemma 2.1**.**

Suppose that $\sigma\in\mathcal{C}_{n}$. If $T\sigma=\sigma$, then there exists $1\leq r\leq n-1$ such that

[TABLE]

for each $a\in\mathbb{Z}_{n}$.

Proof.

According to the definition of $T$, we have

[TABLE]

for each $a\in\mathbb{Z}_{n}$. Since $T\sigma=\sigma$,

[TABLE]

for each $a\in\mathbb{Z}_{n}$. ∎

Let $\lambda_{r}\in\mathcal{S}_{n}$ be the permutation such that $\lambda_{r}(a)=a+r$ for each $a\in\mathbb{Z}_{n}$. It is not difficult to verify that $\lambda_{r}\in\mathcal{C}_{n}$ if and only if $(r,n)=1$. According to Lemma 2.1, for each $\sigma\in X$, $|U_{\sigma}|=1$ if and only if $\sigma=\lambda_{r}$ for some $r$ prime to $n$. That is,

[TABLE]

Recall that

[TABLE]

Define

[TABLE]

Notice that

[TABLE]

So we always have

[TABLE]

Lemma 2.2**.**

Suppose that $\sigma\in\mathcal{C}_{n}$. Then

[TABLE]

where

[TABLE]

Furthermore,

[TABLE]

Proof.

If $\sigma(i-1),\sigma(i)\neq n$, then

[TABLE]

Assume that $\sigma(i_{0})=n$. Clearly

[TABLE]

as well as $T\sigma(i_{0})>T\sigma(i_{0}+1)$. Hence

[TABLE]

where we identify $\sigma(0)$ with $\sigma(n)$. Apparently

[TABLE]

It follows that

[TABLE]

Finally, since $\sigma(i_{0})=n$ is greater than $\sigma(i_{0}-1)$ and $\sigma(i_{0}+1)$, we have

[TABLE]

and

[TABLE]

(2.2) is concluded.

Similarly, we also have

[TABLE]

∎

Lemma 2.3**.**

For each $\sigma\in\mathcal{C}_{n}$, $\overline{{\rm des}\,}\sigma=1$ if and only if $\sigma=\lambda_{r}$ for some $r$ prime to $n$.

Proof.

Assume that $\sigma(i_{0})=n$. Clearly $i_{0}\neq n$, otherwise $\sigma\not\in\mathcal{C}_{n}$. Since $\overline{{\rm des}\,}\sigma=1$ and $\sigma(i_{0})=n>\sigma(i_{0}+1)$, we must have

[TABLE]

Since $\sigma(i_{0}+1)\geq 1$ and $\sigma(i_{0}-1)\leq b=n-1$, we obtain that $\sigma(i_{0}+j)=j$ for each $1\leq j\leq n-i_{0}$ and $\sigma(k)=n-i_{0}+k$ for each $1\leq k\leq i_{0}-1$, i.e., $\sigma=\lambda_{n-i_{0}}$. ∎

Now we are ready to prove Theorem 1.1. In view of (2.1),

[TABLE]

Suppose that $\sigma\in X$ and $|U_{\sigma}|\geq 2$. By Lemma 2.3, we have $\overline{{\rm des}\,}\sigma\geq 2$. Let $h=|U_{\sigma}|$. According to Lemma 2.2,

[TABLE]

Since $1\leq\overline{{\rm des}\,}\sigma-1\leq n-1$, $1-q^{\overline{{\rm des}\,}\sigma-1}$ is not divisible by $\Phi_{n}(q)$. On the other hand, $T^{h}\sigma=\sigma$ since $|U_{\sigma}|=h$. So, by Lemma 2.2, we must have

[TABLE]

i.e.,

[TABLE]

Thus for each $\sigma\in X$ with $|U_{\sigma}|>1$, we have

[TABLE]

It follows that

[TABLE]

Finally, it suffices to show that

[TABLE]

Let $\zeta$ be a $n$-th primitive root of unity. Then

[TABLE]

So we only need to prove that for each $1\leq k\leq n$ with $(k,n)=1$,

[TABLE]

which immediately follows from our knowledge of Ramanujan’s sum. All are done.∎