Chapter VII: Symmetric groups and linear groups
7.1 The natural action of a symmetric group
We introduce the following notations aboutm-schemes:
Definition 7.3. Forn∈N+, letm(n)(resp. m0(n)) be the smallest positive integer such that any non-discrete antisymmetric m(n)-scheme (resp. m0(n)-scheme) on [n]has a matching (resp. is not strongly antisymmetric).
It is easy to see that m(n) and m0(n) are nondecreasing in n. We also have dSym(n)≤m0(n)≤m(n)by Lemma 2.7 and Lemma 2.10.
It was proven [Gua09] and independently in [Aro13] thatm(n)≤
2 log 12
logn+ O(1). We review the proof of this bound, starting from the following lemma:
Lemma 7.1. LetΠ = {P1, . . . , Pm}be an antisymmetricm-scheme on a finite set S where m ≥ 3. SupposeB ∈ P1 satisfies|B| ≥ 3. Letxbe an element of B so thatΠ|x ={P10, . . . , Pm−10 }is an(m−1)-scheme onS− {x}(see Definition 6.3).
Then at least one of the two conditions is satisfied.
1. There existsB0 ∈P10 contained inBsatisfying|B0| ≤(|B| −1)/4.
2. There exist distinct elementsy, z ∈B− {x}such that for the(m−2)-scheme Π|x,y = {P100, . . . , Pm−200 } on S − {x, y}, the block B00 of P100 containing z satisfies|B00| ≤ (|B|+ 1)/12. Furthermore, (x, y), (y, z), and(z, x)are in the same block ofP2.
Proof. By replacingΠ withΠkB, we may assumeΠis homogeneous andS =B. By antisymmetry, we know |P2| is even. If |P2| ≥ 4, there exists B1 ∈ P2 of cardinality at most|B|(|B| −1)/4. LetB0 :={y ∈B : (x, y)∈B1}. ThenB0 is a block ofP10by definition, and its cardinality is|B1|/|B| ≤(|B| −1)/4by regularity ofΠ. And the first condition is met.
So assume|P2|= 2. ThenP2contains two blocksB1andB2of the same cardinality
|B|(|B| −1)/2. Choose y ∈ B − {x} such that (x, y) ∈ B1. Such an element y exists by regularity and homogeneity of Π. By Lemma 2.11 and Lemma 2.12, we have an antisymmetric association scheme P(Π) = P2 ∪ {1B} that has three blocks. By Lemma 2.20, the number of elements z ∈ B − {x, y} satisfying
(y, z),(z, x) ∈ B1 is precisely (|B|+ 1)/4 > 0. The cardinality of the set T :=
{(a, b, c) : (a, b),(b, c),(c, a)∈B1}is then|B1|·(|B|+1)/4. Choosez ∈B−{x, y}
such that(x, y, z)∈T. LetB10,B20, andB30 be the blocks ofP3containing(x, y, z), (y, z, x) and (z, x, y) respectively, which are all subsets of T. They have the same cardinality by invariance of Π, and are distinct by antisymmetry of Π. So
|B10| ≤ |T|/3 = |B1| ·(|B|+ 1)/12. By regularity ofΠ, the cardinality of the set {u ∈ S − {x, y} : (x, y, u) ∈ B10} is |B10|/|B1| ≤ (|B|+ 1)/12, and this set is exactly the blockB00 ofP100 containingz by definition. So the second condition is satisfied.
Lemma 7.1 implies the following recursive relation:
Lemma 7.2. Forn≥3, m(n)≤max
m
n−1 4
+ 1, m
n+ 1 12
+ 2
.
The inequality also holds form0(·)in replaced ofm(·).
Proof. Let Π = {P1, . . . , Pm} be a non-discrete antisymmetric m-scheme on a finite setSof cardinalityn, wherem ≥3. Also assume
m ≥max
m
n−1 4
+ 1, m
n+ 1 12
+ 2
. We want to show thatΠhas a matching.
Choose B ∈ P1 such that |B| > 1. Let x be an element of B and suppose Π|x ={P10, . . . , Pm−10 }. ThenΠ|xis an antisymmetric(m−1)-scheme onS− {x}. Note thatΠkB is a homogeneous antisymmetricm-scheme onB by Lemma 6.14, which implies|B| ≥3. Then either of the two conditions in Lemma 7.1 is satisfied.
If the first condition is satisfied, there exists B0 ∈ P10 contained in B satisfying
|B0| ≤ (|B| −1)/4 ≤ (n−1)/4. If |B0| > 1, we see (Π|x)kB0 is a non-discrete antisymmetric(m−1)-scheme onB0. It has a matching since m−1 ≥ m((n− 1)/4) ≥ m(|B0|). SoΠ also has a matching by Lemma 6.3 and Lemma 6.14. On the other hand, if|B0|= 1, we letybe the unique element inB0 and letB1 be the block ofP2containing(x, y). Note that|B0|=|B1|/|B|, which implies|B1|=|B|. Asx, y ∈B, we haveπ21(B1) =π22(B1) =B. ThenB1 is a matching ofΠ.
Next assume the second condition is satisfied. So there exist distinct elementsy, z ∈ B−{x}such that for the(m−2)-schemeΠ|x,y ={P100, . . . , Pm−200 }onS−{x, y}, the
cardinality of the blockB ofP1 containingzis at most(|B|+ 1)/12≤(n+ 1)/12. Furthermore,(x, y), (y, z), and(z, x)are in the same blockB0 ofP2. If|B00| >1, we see(Π|x,y)kB00 is a non-discrete antisymmetric(m−2)-scheme onB00. It has a matching sincem−2≥ m((n+ 1)/12) ≥m(|B00|). SoΠalso has a matching by Lemma 6.3 and Lemma 6.14. On the other hand, if |B00| = 1, we let B00 be the block of P3 containing (x, y, z). We have π13(B00) = π33(B00) = B0 since (x, y),(y, z) ∈ B0. Also note that|B00| = |B00|/|B0|, which implies |B0| = |B00|. SoB00 is a matching ofΠ.
This proves the inequality for m(·). The proof form0(·)is similar, and we leave it to the reader.
Theorem 7.1([Gua09; Aro13]). For alln ∈N+, m(n)≤
2 log 12
logn+O(1).
More generally, an antisymmetric m-scheme Π = {P1, . . . , Pm} on a finite setS always has a matching ifP1 has a blockB of cardinalityk >1andm≥m(k). In particular it holds for sufficiently largem =
2 log 12
logk+O(1).
Proof. Notem(1) = 1andm(2) = 2. The first claim then follows from Lemma 7.2 and a simple induction. The second claim follows by consideringΠkBand applying Lemma 6.14.
Theorem 7.1 implies a bound for dSym(n), and also a bound ford(G) by Corol- lary 6.3, whereGis an arbitrary permutation group on a set of cardinalityn: Corollary 7.1. LetGbe a permutation group on a set of cardinalityn ∈N+. Then d(G)≤dSym(n)≤
2 log 12
logn+O(1).
We conclude this section with the following technical lemma, which is used later in the proof of Theorem 7.5.
Lemma 7.3. Let G be a permutation group on a finite set S, and let P be the corresponding system of stabilizers of depthmwhere1≤m ≤ |S|. LetC ={CH : H ∈ P}be a strongly antisymmetricP-scheme. SupposeC is non-discrete onGx for some x ∈ S. Then there exists (x1, . . . , xm) ∈ S(m) such thatCGx
1,...,xm has a block of cardinality at least2(log 124 )m2−O(m).
Proof. Let P be the system of stabilizers of depth m with respect to the natural action of G0 := Sym(S) on S. Let C0 = {CH0 : H ∈ P} be the induction of C to P0 (see Definition 6.4), which is strongly antisymmetric by Lemma 6.1 and is non-discrete onG0xforx∈S in the lemma sinceC is non-discrete onGx. Assume the lemma holds forSym(S), P0, and an m-tuple(y1, . . . , ym) ∈ S(m), i.e., there existsB0 ∈CG0 0
y1,...,ym of cardinality at least2(log 124 )m2−O(m). By Definition 6.4, we know B0 is of the form φG0y1,...,ym,g(B), where g ∈ G0, φG0y1,...,ym,g is an injection from(G∩gG0y1,...,ymg−1)\GtoG0y1,...,ym\G0, andB is a block ofCG∩gG0y
1,...,ymg−1. Letxi =gyifori∈[m]. ThenG∩gG0y1,...,ymg−1 =Gx1,...,xm. So(x1, . . . , xm)and B ∈CGx1,...,xm satisfy the condition in the lemma.
Thus we may assumeG = Sym(S)and it acts naturally onS. By Lemma 2.12, it suffices to show that for any non-discrete strongly antisymmetric m-scheme Π = {P1, . . . , Pm}onS, the partitionPmhas a block of cardinality at least2(log 124 )m2−cm, where c = O(1). We prove this claim by induction on m. The case m = 1 is trivial. For m > 1, assume the claim for m0 < m. Let B0 be a block of P1 of cardinality k > 1. By Theorem 7.1, we have m ≤
2 log 12
logk +c0 for some c0 = O(1), or equivalently k ≥ 2log 122 (m−c0). Choose x ∈ B0 and consider the (m−1)-schemeΠ-schemeΠ0 := Π|x ={P10, . . . , Pm−10 }onS− {x}. It is strongly antisymmetric by Lemma 6.3. Let B1 be a block of P10 contained in B0, which exists by compatibility of Π and the fact k > 1. If |B1| = 1, we have seen in the proof of Lemma 7.2 that Π has matching, contradicting the assumption that Π is strongly antisymmetric. So |B1| > 1. By Lemma 6.14, the homogeneous (m−1)-schemeΠ0kB1 ={P100, . . . , Pm−100 }onB1is strongly antisymmetric. By the induction hypothesis, the partitionPm−100 has a blockB0 ⊆B1(m−1) of cardinality at least2(log 124 )(m−1)2−c(m−1). AndB0 is also a block ofPm−10 ∈ Π0 by definition and compatibility ofΠ0. ThenPm ∈ Πhas a blockB containing(x, x1, . . . , xm−1)for all(x1, . . . , xm−1)∈B0. By regularity ofΠ, we have
|B|=|B0||B0| ≥2log 122 (m−c0)·2(log 124 )(m−1)2−c(m−1) ≥2(log 124 )m2−cm for sufficiently largec=O(1).