New lower bounds for permutation arrays using contraction

Abstract

A permutation array A is a set of permutations on a finite set $\Omega$, say of size n. Given distinct permutations $\pi , \sigma \in \Omega$, we let $hd(\pi , \sigma ) = |\{ x\in \Omega : \pi (x) \ne \sigma (x) \}|$, called the Hamming distance between $\pi$ and $\sigma$. Now let $hd(A) =$ min$\{ hd(\pi , \sigma ): \pi , \sigma \in A \}$. For positive integers n and d with $d\le n$ we let M(n, d) be the maximum number of permutations in any array A satisfying $hd(A) \ge d$. There is an extensive literature on the function M(n, d), motivated in part by suggested applications to error correcting codes for message transmission over power lines. A basic fact is that if a permutation group G is sharply k-transitive on a set of size $n\ge k$, then $M(n,n-k+1) = |G|$. Motivated by this we consider the permutation groups AGL(1, q) and PGL(2, q) acting sharply 2-transitively on GF(q) and sharply 3-transitively on $GF(q)\cup \{\infty \}$ respectively. Applying a contraction operation to these groups, we obtain the following new lower bounds for prime powers q satisfying $q\equiv 1$ (mod 3).

• 1. $M(q-1,q-3)\ge (q^{2} - 1)/2$ for q odd, $q\ge 7$,

• 2. $M(q-1,q-3)\ge (q-1)(q+2)/3$ for q even, $q\ge 8$,

• 3. $M(q,q-3)\ge Kq^{2}log(q)$ for some constant $K>0$ if q is odd.

These results resolve a case left open in a previous paper (Bereg et al. in Des Codes Cryptogr 86(5):1095–1111, 2018), where it was shown that $M(q-1, q-3) \ge q^{2} - q$ and $M(q,q-3) \ge q^{3} - q$ for all prime powers q such that $q\not \equiv 1$ (mod 3). We also obtain lower bounds for M(n, d) for a finite number of exceptional pairs n, d, by applying this contraction operation to the sharply 4 and 5-transitive Mathieu groups.

Appendix: Some applications of number theory

In this section we prove some facts from number theory that were used in this paper.

We start with some notation. For an odd prime p and integer $r\not \equiv 0 ($mod p), define the Legendre symbol $(\frac{r}{p})$ to be 1 (resp. -1) if r is a quadratic residue (resp. nonresidue); that is a square (resp. nonsquare) mod p. If $r\equiv 0 ($mod p), then define $(\frac{r}{p}) = 0.$ We give some basic facts about this symbol without proof in the Lemma and Theorem which follow, since such proofs may be found in standard number theory books.

Lemma 19

For an odd prime p and integers r and s we have the following.

• (a) $(\frac{-1}{p}) = 1$ if $p\equiv 1$(mod 4), and $(\frac{-1}{p}) = -1$ if $p\equiv 3$(mod 4).

• (b) $(\frac{rs}{p}) = (\frac{r}{p})(\frac{s}{p})$.

Theorem 20

(Gauss Quadratic Reciprocity Law) For odd primes p and q we have

$$\begin{aligned} \Big (\frac{p}{q}\Big )\Big (\frac{q}{p}\Big ) = (-1)^{\Big (\frac{p-1}{2}\Big )\Big (\frac{q-1}{2}\Big )}. \end{aligned}$$

Now let’s apply these facts to determining $(\frac{-3}{p})$ for odd primes p.

Theorem 21

Let $p > 3$ be an odd prime. Then

• (a) If $p\equiv 1$(mod 6), then $-3$ is a quadratic residue mod p.

• (b) If $p\equiv 5$(mod 6), then $-3$ is a quadratic nonresidue mod p.

Proof

By the lemma above we have $(\frac{-3}{p}) = (\frac{-1}{p})(\frac{3}{p})$, while by quadratic reciprocity we have $(\frac{3}{p}) = (\frac{p}{3})(-1)^{\frac{p-1}{2}}$. Thus

$$\begin{aligned} \Big (\frac{-3}{p}\Big ) = (-1)^{\frac{p-1}{2}}\Big (\frac{-1}{p}\Big )\Big (\frac{p}{3}\Big ). \end{aligned}$$

The factors on the right depend on the residue classes of p mod 4 and mod 3. Since p is odd with $p > 3$, we have $p\equiv 1$ or 3(mod 4), and also $p\equiv 1$ or 2(mod 3). Thus there are four possible ordered pairs $(\alpha , \beta )$ where $p\equiv \alpha ($mod 4) and $p\equiv \beta ($mod 3). We calculate $(\frac{-3}{p})$ by these four possibilities, giving details for two of them.

Case 1: $\alpha = 1$ and $\beta = 1$; equivalently $p\equiv 1($mod 12).

Now $p\equiv 1($mod 3) says that $(\frac{p}{3}) = 1$. Also $p\equiv 1($mod 4) implies $(-1)^{\frac{p-1}{2}} = 1$ and by Lemma 19 also implies $(\frac{-1}{p}) = 1$. So by the formula above we have $(\frac{-3}{p}) = 1$, showing that $-3$ is a quadratic residue when $p\equiv 1($mod 12).

Case 2: $\alpha = 1$ and $\beta = 2$; equivalently $p\equiv 5($mod 12).

Now $p\equiv 2($mod 3) says that $(\frac{p}{3}) = -1$. Also $p\equiv 1($mod 4) implies $(-1)^{\frac{p-1}{2}} = 1$ and also Lemma 19 implies $(\frac{-1}{p}) = 1$. So by the formula above we have $(\frac{-3}{p}) = -1$, showing that $-3$ is a quadratic nonresidue when $p\equiv 5($mod 12).

By similar calculations we find that $(\frac{-3}{p}) = 1$ when $\alpha = 3$ and $\beta = 1$ (equivalently $p\equiv 7($mod 12)), and $(\frac{-3}{p}) = -1$ when $\alpha = 3$ and $\beta = 2$ (equivalently $p\equiv 11($mod 12)).

Putting together these cases, we see that $-3$ is a quadratic residue mod p when $p\equiv 1($mod 6), while $-3$ is a quadratic nonresidue mod p when $p\equiv 5($mod 6), as required. $\square$

Corollary 22

Consider the prime power $q = p^{m}$, where $p > 3$ is an odd prime. If $q\equiv 1($mod 3), then $-3$ is a square in the finite field GF(q).

Proof

Since $p > 3$ is an odd prime we have either $p\equiv 1($mod 6) or $p\equiv 5($mod 6). If $p\equiv 1($mod 6), then $-3$ is already a square in the prime subfield $GF(p)\subseteq GF(q)$ by Theorem 21, so $-3$ is a square in GF(q), as required.

So suppose $p\equiv 5($mod 6). Consider the quadratic extension $GF(p)(\sqrt{-3})$ of GF(p) obtained by adjoining to GF(p) a root of the irreducible (by Theorem 21) polynomial $x^{2} + 3$ over GF(p). Then $GF(p)(\sqrt{-3}) \cong GF(p^{2})$, and $-3$ is a square in $GF(p^{2})$.

Since $q\equiv 1($mod 3), then since $p\equiv 5($mod 6) we have $p\equiv 2($mod 3), so it follows that m must be even. We recall the basic fact from finite fields that $GF(p^{r})\subseteq GF(p^{s})$ if and only if $r\vert s$. It follows that $GF(p^{2})\subseteq GF(q)$. Thus since $-3$ is a square in $GF(p^{2})$, then $-3$ is a square in GF(q). $\square$

Corollary 23

Let $q = p^{m}$ be a prime power, $q\equiv 1($mod 3).

• (a)The equation $x^{2} + x + 1 = 0$ has two distinct solutions in GF(q). If $x_{1}$ is such a root, then $\frac{1}{x_{1}}$ is the other distinct root.

• (b)For q odd and distinct $i,j\in GF(q)$, the equation $x^{2} - (i+j)x + ij + (i-j)^{2} = 0$ has two distinct roots in GF(q).

Proof

Consider (a), and suppose first that p is odd. Since the characteristic of the field is odd, we may find the solutions by the standard quadratic formula. We obtain the solutions $x = \frac{1}{2}[ -1 + \sqrt{-3}\,], \frac{1}{2}[ -1 - \sqrt{-3}\, ]$, where we have used the existence of $\sqrt{-3}$ in GF(q) by Corollary 22. These solutions are distinct since p is odd.

Now suppose $p=2$. Recall the trace function $Tr_{GF(q)/ GF(2)}(x) = \sum _{i=0}^{m-1} x^{2^{i}}$, defined for any $x\in GF(q)$, which we abbreviate by Tr(x). It can be shown (see [27]) that the quadratic equation $ax^{2} + bx + c = 0$, with $a,b,c\in GF(2^{m})$, $a\ne 0$, has two distinct solutions in $GF(2^{m})$ if and only if $b\ne 0$ and $Tr(\frac{ac}{b^{2}}) = 0.$ In our case we have $a = b = c = 1$, so $\frac{ac}{b^{2}} = 1$. Since $p = 2$ and $q\equiv 1($mod 3), m must be even. Thus there are an even number of terms in the sum defining Tr(x), each of them equal to 1. So since the characteristic is 2, we get $Tr(\frac{ac}{b^{2}}) = 0$ in our case. It follows that $x^{2} + x + 1 = 0$ has two distinct solutions when $p = 2$, as required.

Observe that if $x_{1}$ is a root of $x^{2} + x + 1 = 0$, then by direct substitution so is $\frac{1}{x_{1}}$. To show that $x_{1}$ and $\frac{1}{x_{1}}$ are distinct, assume not. Then $x_{1} = 1$ or $-1$. If q is even, then $x_{1}^{2} + x_{1} + 1 = 0$ implies that $1 = 0$ since the characteristic of the field is 2, a contradiction. Assume q is odd. Then if $x_{1} = 1$ we get $1+1+1 = 0$, implying $q\equiv 0($mod 3), a contradiction. If $x_{1} = -1$, then we get $1 = 0$, contradiction. Thus $x_{1}$ and and $\frac{1}{x_{1}}$ are distinct.

Next consider (b). Applying the quadratic formula in this field of odd characteristic, we get the two solutions $x = \frac{1}{2}[\,i + j \pm \sqrt{(i+j)^{2} - 4(ij + (j-i)^{2})}\,] = \frac{1}{2}[\,i + j \pm \sqrt{-3(i^{2} + j^{2}) + 6ij}\,] = \frac{1}{2}[\,i + j \pm \sqrt{-3(i - j)^{2}}\,] = \frac{1}{2}[\,i + j \pm \sqrt{-3}(i - j)\,].$ Now since $-3$ is a square in GF(q) for $q\equiv 1($mod 3) by Corollary 22, it follows that the two solutions for x can be written as $x_{1} = \frac{1}{2}[i(1+\sqrt{-3}) + j(1-\sqrt{-3})]$, and $x_{2} = \frac{1}{2}[i(1-\sqrt{-3}) + j(1+\sqrt{-3})].$ Also these two solutions are distinct since $i\ne j$ and q is odd. $\square$