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Publicly Available Published by De Gruyter May 7, 2019

Application of character estimates to the number of 𝑇2-systems of the alternating group

Stefan-C. Virchow EMAIL logo
From the journal Journal of Group Theory

Abstract

We use character theory and character estimates to show that the number of T2-systems of the alternating group An is at least

18⁒n⁒3⁒exp⁑(2⁒π6⁒n1/2)⁒(1+o⁒(1)).

Applying this result, we obtain a lower bound for the number of connected components of the product replacement graph Ξ“2⁒(An).

1 Introduction

Let G be a finite group, and let d⁒(G) be the minimal number of generators of G. Fix some integer k⩾d⁒(G), and denote by

𝒩k⁒(G):-{(g1,…,gk)∈Gk:γ€ˆg1,…,gk〉=G}

the set of generating k-tuples of G. We identify 𝒩k⁒(G) with the set of epimorphisms Epi⁑(Fkβ† G), where Fk is the free group on k generators. Now consider the group action

(Aut⁑(Fk)Γ—Aut⁑(G))Γ—Epi⁑(Fkβ† G)β†’Epi⁑(Fkβ† G)
((Ο„,Οƒ),Ο•)β†¦Οƒβˆ˜Ο•βˆ˜Ο„-1.

Systems of transitivity or, in short, Tk-systems are defined to be the orbits of this group action. Denote by Ο„k⁒(G) the number of Tk-systems of G. Neumann and Neumann [25] introduced Tk-systems in 1950. Since then, they have been widely investigated by Dunwoody [10], Evans [12], Gilman [17], Guralnick and Pak [18], Neumann [24] and Pak [27].

There has been renewed interest in Tk-systems in the recent years because they can be applied to the product replacement algorithm (PRA). The PRA is a practical algorithm to generate random elements of finite groups. The algorithm was introduced in 1995 by Celler et al. [4]. Since then, it has been widely studied [2, 14, 21, 27]. The PRA is defined as follows. Let (g1,…,gk)βˆˆπ’©k⁒(G) be a generating k-tuple of G. A move to another generating k-tuple is determined by two steps. First, select uniformly a pair (i,j) with 1β©½iβ‰ jβ©½k. Second, apply one of the subsequent four operations with equal probability:

Ri,jΒ±:(g1,…,gi,…,gk)β†’(g1,…,giβ‹…gjΒ±1,…,gk),
Li,jΒ±:(g1,…,gi,…,gk)β†’(g1,…,gjΒ±1β‹…gi,…,gk).

We apply the above move several times (the choice of the moves must be uniform and independent at each step). Finally, we return a random component of the resulting generating k-tuple. This gives us the desired β€œrandom” element of the group G.

The product replacement graph (PRA graph)Ξ“k⁒(G) is a graph with vertices corresponding to generating k-tuples of G and edges corresponding to the moves Ri,jΒ±, Li,jΒ±. The PRA can be described as a simple random walk on this graph. Denote by Ο°k⁒(G) the number of connected components of Ξ“k⁒(G).

The connection between the PRA and Tk-systems of G is the following. Nielsen showed in [26] that Aut⁑(Fk) is generated by the Nielsen moves

Ri,j:(x1,…,xi,…,xk)β†’(x1,…,xiβ‹…xj,…,xk),
Li,j:(x1,…,xi,…,xk)β†’(x1,…,xjβ‹…xi,…,xk),
Pi,j:(x1,…,xi,…,xj,…,xk)β†’(x1,…,xj,…,xi,…,xk),
Ii:(x1,…,xi,…,xk)β†’(x1,…,xi-1,…,xk)

for 1β©½iβ‰ jβ©½k. Therefore, we can define a graph with vertices corresponding to 𝒩k⁒(G) and edges corresponding to the Nielsen moves and to the automorphisms of G. The connected components of this graph are exactly the Tk-systems of G. Now it follows immediately from the definitions that

(1.1)Ο„k⁒(G)β©½Ο°k⁒(G).

Little is known about the problem to estimate Ο„k⁒(G) as a function of k and G. Of special interest are Tk-systems of a finite simple group G. A conjecture attributed to Wiegold states that Ο„k⁒(G)=1 for kβ©Ύ3 (see [27, Conjecture 2.5.4]). Cooperman and Pak [5], David [7], Evans [12], Garion [15] and Gilman [17] proved this conjecture for some families of simple groups. The case k=2, however, appears to be different. In 2009, Garion and Shalev [16, Theorem 1.8] showed that Ο„2⁒(G)β†’βˆž as |G|β†’βˆž, where G is a finite simple group. Hence they confirmed a conjecture of Guralnick and Pak [18]. In addition, they proved that

Ο„2⁒(An)β©Ύn(12-Ο΅)⁒log⁑n,

where An denotes the alternating group on n letters. In 1985, Evans already established in his Ph.D. Thesis [11] that

Ο„2⁒(A2⁒n+4)β©Ύ#⁒{C:C⁒is conjugacy class of⁒Sn},
Ο„2⁒(A2⁒n+5)β©Ύ#⁒{C:C⁒is conjugacy class of⁒Sn}

for nβ©Ύ2. However, this thesis is hard to find and difficult to access.

Schlage-Puchta [29] applied character theory and character estimates to some statistical problems for the symmetric group. Our aim is to show that these methods and concepts can be adopted to establish a lower bound for the number of T2-systems of An. Our main result is the following theorem.

Theorem 1.1.

Let n be sufficiently large. Then we have

Ο„2⁒(An)β©Ύ18⁒n⁒3⁒exp⁑(2⁒π6⁒n1/2)⁒(1+o⁒(1)).

Taking (1.1) into account, this theorem has an immediate corollary.

Corollary 1.2.

Let n be sufficiently large. Then we have

Ο°2⁒(An)β©Ύ18⁒n⁒3⁒exp⁑(2⁒π6⁒n1/2)⁒(1+o⁒(1)).

Remark 1.3.

The lower bound in our theorem is the best result you can achieve with our method. This is because counting T2-systems of An is reduced to counting partitions of n. Upper bounds for the number of T2-systems of An are not known to the author.

2 Proof of the theorem

As usual, let Sn be the symmetric group and An the alternating group on n letters. For any group G, the commutator of g and h in G is [g,h]=g-1⁒h-1⁒g⁒h.

In this section, we will show that the number of T2-systems of An is bounded below by the number of conjugacy classes of Sn which fulfill certain conditions. Then we will count these conjugacy classes.

We start with a lemma due to Higman (cf. [24]).

Lemma 2.1.

Let G be a finite group with d⁒(G)⩽2. Then, for each T2-system S of G, the set of Aut⁑(G)-conjugates of {[g1,g2]±1} is invariant for all (g1,g2)∈S.

Higman’s lemma leads us to the following lemma.

Lemma 2.2.

Let J:-{[Ο€,Οƒ]:(Ο€,Οƒ)∈N2⁒(An)}. For all sufficiently large n, we have

Ο„2⁒(An)β©Ύ#⁒{C:C⁒is a conjugacy class of⁒Snβ’π‘€π‘–π‘‘β„Žβ’C∩Jβ‰ βˆ…}.

Proof.

Note that (cf. [9, Theorem 8.2A])

Aut⁑(An)={Ξ±|An:α∈Inn⁑(Sn)} for⁒n>6.

Thus we have

ℬ:-{K:K⁒is an⁒Aut⁑(An)⁒-conjugacy class with⁒K∩Jβ‰ βˆ…}={C:C⁒is a conjugacy class of⁒Sn⁒with⁒C∩Jβ‰ βˆ…}.

Taking into account that d⁒(An)⩽2 (cf. [8]), Lemma 2.1 yields

Ο„2⁒(An)β©Ύ#⁒{KβˆͺK-1:Kβˆˆβ„¬}=|ℬ|

as desired. ∎

Definition 2.3.

Let C be a conjugacy class of Sn. Denote by f⁒(C) the number of fixed points of elements of C. For a function a:ℕ→ℝ, let Π⁒(a⁒(n)) be the set of all primes p in the range 12⁒n<pβ©½a⁒(n).

We establish the following basic result that is very useful for the proof of our theorem.

Theorem 2.4.

Let p∈Π⁒(35⁒n), and let Ο€βˆˆAn be a fixed permutation with cycle type (p,n-p-2,1,1) if n is even and with cycle type (p,n-p-2,2) if n is odd. In addition, suppose that CβŠ‚An is a conjugacy class of Sn such that

f⁒(C)β©½Ξ΄β‹…nβ€ƒπ‘“π‘œπ‘Ÿβ’Ξ΄βˆˆ(0,14].

Then, for all sufficiently large n, we have

#⁒{ΟƒβˆˆAn:[Ο€,Οƒ]∈Cβ’π‘Žπ‘›π‘‘β’γ€ˆΟ€,σ〉=An}=|C|⁒(1+π’ͺ⁒(Ξ΄)).

As an immediate consequence, for Ξ΄>0 sufficiently small, we have

#⁒{ΟƒβˆˆAn:[Ο€,Οƒ]∈C⁒andβ’γ€ˆΟ€,σ〉=An}>0.

We will give the proof of this theorem in the next sections.

By Lemma 2.2, Ο„2⁒(An) is bounded below by the number of conjugacy classes of Sn containing the commutator [Ο€,Οƒ] for some generating tuple (Ο€,Οƒ)βˆˆπ’©2⁒(An). Theorem 2.4 shows that this number is in turn bounded below by the number of conjugacy classes C of Sn satisfying CβŠ‚An and f⁒(C)⩽δ⁒n for Ξ΄>0 sufficiently small. Thus our problem results in counting conjugacy classes of Sn or partitions of n.

Definition 2.5.

A partition of n is a sequence Ξ»=(Ξ»1,Ξ»2,…,Ξ»l) of positive integers such that Ξ»1β©ΎΞ»2β©Ύβ‹―β©ΎΞ»l and Ξ»1+Ξ»2+β‹―+Ξ»l=n. We write λ⊒n to indicate that Ξ» is a partition of n. Denote by P⁒(n) the number of partitions of n.

Suppose Ξ»=(Ξ»1,Ξ»2,…,Ξ»l)⊒n. The Ferrers diagram of Ξ» is an array of n boxes having l left-justified rows with row i containing Ξ»i boxes for 1β©½iβ©½l.

Hardy and Ramanujan [19] achieved the subsequent asymptotic formula for the number of partitions of n.

Lemma 2.6.

We have

P⁒(n)=14⁒n⁒3⁒exp⁑(2⁒π6⁒n1/2)⁒(1+o⁒(1)).

We conclude the section with the proof of our first theorem.

Proof of Theorem 1.1.

First, we note that (see [1, Exercise 44])

#{λ⊒n:number of even parts inλis odd}=#{λ⊒n:number of even parts inλis even}-#{λ⊒n:λconsists of distinct odd parts}.

Thus we have

#⁒{CβŠ‚An:C⁒is a conjugacy class of⁒Sn}β©Ύ12β‹…P⁒(n).

Second, choose Ξ΄>0 sufficiently small. By Lemma 2.2, Theorem 2.4 and the above inequality, we obtain

Ο„2⁒(An)β©Ύ#⁒{CβŠ‚An:C⁒is a conjugacy class of⁒Sn⁒with⁒f⁒(C)β©½Ξ΄β‹…n}β©Ύ12⁒P⁒(n)-P⁒(n-⌈δ⁒nβŒ‰).

Applying Lemma 2.6 yields our assertion. ∎

3 Some character theory

The rest of this paper is devoted to the proof of Theorem 2.4. In this section, we review some results from character theory which are essential for our proof.

We denote by Irr⁑(Sn) the set of irreducible characters of Sn. For a conjugacy class C of Sn and Ο‡βˆˆIrr⁑(Sn), we write χ⁒(C) to denote χ⁒(Ο€) for Ο€βˆˆC.

Lemma 3.1.

Let C1 and C2 be conjugacy classes of Sn, and let Ο„βˆˆSn. Then

#⁒{(x,y)∈C1Γ—C2:x⁒y=Ο„}=|C1|⁒|C2|n!β’βˆ‘Ο‡βˆˆIrr⁑(Sn)χ⁒(C1)⁒χ⁒(C2)⁒χ⁒(Ο„-1)χ⁒(1).

Proof.

See [6, Proposition 9.33] or [20, Theorem 6.3.1]. ∎

Recall that the irreducible characters of Sn are explicitly parametrized by the partitions of n. Denote by χλ the irreducible character of Sn corresponding to the partition Ξ» of n.

We shall frequently apply the Murnaghan–Nakayama rule.

Definition 3.2.

Let λ⊒n be a partition. A rim hookh is an edgewise connected part of the Ferrers diagram of λ, obtained by starting from a box at the right end of a row and at each step moving upwards or rightwards only, which can be removed to leave a proper Ferrers diagram denoted by λ\h. An r-rim hook is a rim hook containing r boxes.

The leg length of a rim hook h is

l⁒l⁒(h):-(the number of rows of⁒h)-1.

Let Ο€βˆˆSn be a permutation with cycle type (1Ξ±1,…,qΞ±q,…,nΞ±n) and Ξ±qβ©Ύ1. Denote by Ο€\q∈Sn-q a permutation with cycle type

(1Ξ±1,…,qΞ±q-1,…,(n-q)Ξ±n-q).

Lemma 3.3 (Murnaghan–Nakayama rule).

Let λ⊒n be a partition. Suppose that Ο€βˆˆSn is a permutation which contains a q-cycle. Then we have

χλ⁒(Ο€)=βˆ‘hq⁒-rim hookπ‘œπ‘“β’Ξ»(-1)l⁒l⁒(h)⁒χλ\h⁒(Ο€\q).

Proof.

See [23, § 9] or [28, Theorem 4.10.2]. ∎

The dimension χλ⁒(1) of the irreducible representation associated with Ξ» can be computed via the hook formula.

Definition 3.4.

Let λ⊒n be a partition. The hook of (i,j)∈λ is

Hi,j⁒(Ξ»):-{(i,jβ€²)∈λ:jβ€²β©Ύj}βˆͺ{(iβ€²,j)∈λ:iβ€²β©Ύi}

Here we write (i,j)∈λ to indicate that (i,j) is a box in the Ferrers diagram of λ.

Lemma 3.5 (hook formula).

Let λ⊒n be a partition. Then

χλ⁒(1)=n!∏(i,j)∈λ|Hi,j⁒(Ξ»)|.

Proof.

See [13, Theorem 1] or [28, Theorem 3.10.2]. ∎

Finally, we will use the following estimate, due to MΓΌller and Schlage-Puchta [22, Theorem 1].

Lemma 3.6.

Let Ο‡βˆˆIrr⁑(Sn) be an irreducible character, let n be sufficiently large, and let ΟƒβˆˆSn be an element with k fixed points. Then we have

|χ⁒(Οƒ)|⩽χ⁒(1)1-log⁑(n/k)32⁒log⁑n.

4 Character estimates

Let Ο€βˆˆAn be fixed. In this section, we will count permutations ΟƒβˆˆAn satisfying that the commutator [Ο€,Οƒ] lies in a given conjugacy class C of Sn. The outcomes of this section will form the technical basis for the proof of Theorem 2.4.

Proposition 4.1.

Let p∈Π⁒(58⁒n), and let Ο€βˆˆAn be a fixed permutation with cycle type (p,n-p-2,1,1) if n is even and with cycle type (p,n-p-2,2) if n is odd. In addition, suppose that CβŠ‚An is a conjugacy class of Sn such that

f⁒(C)β©½Ξ΄β‹…nβ€ƒπ‘“π‘œπ‘Ÿβ’Ξ΄βˆˆ(0,1).

Then, for all sufficiently large n, we have

#⁒{ΟƒβˆˆAn:[Ο€,Οƒ]∈C}=|C|⁒(1+π’ͺ⁒(Ξ΄)).

Proof.

Due to the prime number theorem, there exists a prime p∈Π⁒(58⁒n) for all sufficiently large n. Denote by KΟ€ the conjugacy class of Ο€ in Sn. Because of the cycle type of Ο€, KΟ€ remains a single conjugacy class inside An (cf. [31, pp. 16f.]). Taking into account that χ⁒(Ο„-1)=χ⁒(Ο„)Β―=χ⁒(Ο„), it follows with Lemma 3.1 that

(4.1)#⁒{ΟƒβˆˆAn:[Ο€,Οƒ]∈C}=βˆ‘(x,y)∈Kπ×Cx⁒y=Ο€#⁒{ΟƒβˆˆAn:Οƒ-1⁒π⁒σ=x}=n!2⁒|KΟ€|⁒#⁒{(x,y)∈Kπ×C:x⁒y=Ο€}=|C|2β’βˆ‘Ο‡βˆˆIrr⁑(Sn)Ο‡2⁒(Ο€)⋅χ⁒(C)χ⁒(1).

The contribution of the linear characters to (4.1) is |C|. This is the expected main term. We shall show that the remaining characters give terms which can be absorbed into the error term. More precisely, we will establish in Lemma 4.9 that

(4.2)βˆ‘Ο‡βˆˆIrr⁑(Sn)χ⁒(1)β‰ 1Ο‡2⁒(Ο€)⋅χ⁒(C)χ⁒(1)=π’ͺ⁒(Ξ΄).

This completes our proof. ∎

We need a number of technical lemmas for the estimate of sum (4.2).

Lemma 4.2.

Let p and Ο€ be chosen as in Proposition 4.1. Let λ⊒n be a partition with χλ⁒(Ο€)β‰ 0. Then the Ferrers diagram of Ξ» has at most two boxes which do not belong to H1,1⁒(Ξ»)βˆͺH2,2⁒(Ξ»). Moreover, let h be a p-rim hook of Ξ» with χλ\h⁒(Ο€\p)β‰ 0 or an (n-p-2)-rim hook of Ξ» with χλ\h⁒(Ο€\(n-p-2))β‰ 0. Then the Ferrers diagram of Ξ»\h has at most two boxes which do not belong to H1,1⁒(Ξ»\h). In particular, we have |H1,1⁒(Ξ»)|β©Ύp.

Proof.

The claim follows from the special cycle type of Ο€ and the Murnaghan–Nakayama rule (Lemma 3.3). ∎

Lemma 4.3.

Let p and Ο€ be chosen as in Proposition 4.1, and let λ⊒n be a partition. Then we have |χλ⁒(Ο€)|β©½2.

Proof.

If χλ⁒(Ο€)=0, there is nothing to show. So let us assume that χλ⁒(Ο€)β‰ 0. Let

R:-{h:h⁒is a⁒p⁒-rim hook of⁒λ⁒with⁒χλ\h⁒(Ο€\p)β‰ 0}.

We have Rβ‰ βˆ… as a result of the Murnaghan–Nakayama rule. Due to

|H1,1⁒(λ)|⩾p>12⁒n

(see Lemma 4.2), a rim hook h∈R contains boxes of the hook H1,1⁒(λ). Hence

|R|β©½2.

Now let h1∈R. Because of Lemma 4.2, there is exactly one (n-p-2)-rim hook h2 of λ\h1. Obviously, there exists exactly one 2-rim hook of (λ\h1)\h2 and exactly one 1-rim hook of (λ\h1)\h2.

The Murnaghan–Nakayama rule yields that χλ⁒(Ο€) is equal to

{βˆ‘h1p⁒-rim hookofβ’Ξ»βˆ‘h2(n-p-2)⁒-rim hookof⁒λ\h1(-1)l⁒l⁒(h1)+l⁒l⁒(h2)if⁒n≑0⁒(2),βˆ‘h1p⁒-rim hookofβ’Ξ»βˆ‘h2(n-p-2)⁒-rim hookof⁒λ\h1βˆ‘h32⁒-rim hookof⁒(Ξ»\h1)\h2(-1)l⁒l⁒(h1)+l⁒l⁒(h2)+l⁒l⁒(h3)if⁒n≑1⁒(2).

Therefore, it follows that |χλ⁒(Ο€)|β©½2. ∎

Definition 4.4.

Let λ⊒n. Define

m1:-(number of boxes in the first column of the Ferrers diagram of⁒λ)-1,
m1β€²:-(number of boxes in the first row of the Ferrers diagram of⁒λ)-1,
m2:-(number of boxes in the second column of the Ferrers diagram of⁒λ)-2,
m2β€²:-(number of boxes in the second row of the Ferrers diagram of⁒λ)-2.

Lemma 4.5.

Let p and Ο€ be chosen as in Proposition 4.1. Let λ⊒n be a partition such that the Ferrers diagram of Ξ» contains the box (2,2) and χλ⁒(Ο€)β‰ 0. Then we have

χλ⁒(1)β©Ύ124⁒n2⁒(n|H1,1⁒(Ξ»)|)⁒(|H1,1⁒(Ξ»)|-1m1),
χλ⁒(1)β©Ύ124⁒n2⁒(n|H1,1⁒(Ξ»)|)⁒(|H1,1⁒(Ξ»)|-1m1β€²).

Furthermore, if |H1,1⁒(λ)|⩾p+3, then there exists an i∈{0,1,2} such that

m1=m2+|H1,1⁒(Ξ»)|-p-iβ€ƒπ‘œπ‘Ÿβ€ƒm1β€²=m2β€²+|H1,1⁒(Ξ»)|-p-i.

Proof.

Due to m1+m1β€²=|H1,1⁒(Ξ»)|-1 the first two formulas are equivalent. In view of Lemma 4.2, we have the three cases |H3,3⁒(Ξ»)|=0, |H3,3⁒(Ξ»)|=1 and |H3,3⁒(Ξ»)|=2. In each case, the hook formula (Lemma 3.5) yields

χλ⁒(1)β©Ύ124⁒n2⁒(n|H1,1⁒(Ξ»)|)⁒(|H1,1⁒(Ξ»)|-1m1)⁒(n-|H1,1⁒(Ξ»)|-1m2).

Hence our first result follows. We now turn to the second statement. It follows from the Murnaghan–Nakayama rule that there is an (n-p-2)-rim hook h of Ξ» such that χλ\h⁒(Ο€\(n-p-2))β‰ 0. By assumption, we have |H1,1⁒(Ξ»)|β©Ύp+3. Therefore, h contains either boxes from the first column or the first row of the Ferrers diagram of Ξ» – but not both at the same time. Without loss of generality, we assume that the first occurs. With regard to Lemma 4.2, |H2,2⁒(Ξ»\h)|∈{0,1,2}. If |H2,2⁒(Ξ»\h)|=1, we obtain n-p-2=n-|H1,1⁒(Ξ»)|-1+m1-m2 and hence m1=m2+|H1,1⁒(Ξ»)|-p-1. The other cases are analogous. This proves our claim. ∎

Lemma 4.6.

Let p and Ο€ be chosen as in Proposition 4.1. Let λ⊒n, where Ξ»βˆ‰{(n),(1n)}, be a partition such that χλ⁒(Ο€)β‰ 0 and the Ferrers diagram of Ξ» does not contain the box (2,2).

  1. If λ∈{(n-1,1),(2,1n-2)}, we obtain χλ⁒(1)=n-1.

  2. If λ∈{(p+1,1n-p-1),(n-p,1p)}, we have χλ⁒(1)=(n-1p).

  3. If λ∈{(n-p-1,1p+1),(p+2,1n-p-2)}, it follows χλ⁒(1)=(n-1p+1).

The cases specified above are the only ones that can occur.

Proof.

The computations for χλ⁒(1) follow immediately from the hook formula. Furthermore, the Murnaghan–Nakayama rule yields that there exists a p-rim hook h1 of Ξ» such that there is an (n-p-2)-rim hook of Ξ»\h1. This proves the last statement. ∎

Lemma 4.7.

Let p and Ο€ be chosen as in Proposition 4.1. Let λ⊒n be partition such that Ξ»βˆ‰{(n),(1n),(n-1,1),(2,1n-2)} and χλ⁒(Ο€)β‰ 0. Then, for all sufficiently large n, we obtain

χλ⁒(1)β©Ύ124⁒n2β‹…2n/8.

Proof.

If the Ferrers diagram of Ξ» does not contain the box (2,2), then it follows from Lemma 4.6 that

χλ⁒(1)β©Ύ(n-1p+1)β©Ύ(n-1p+1)p+1β©Ύ2n/8.

Now let us assume that the Ferrers diagram of Ξ» contains the box (2,2).

Case 1: |H1,1⁒(Ξ»)|β©½78⁒n. By applying Lemmas 4.5 and 4.2, we get

χλ⁒(1)β©Ύ124⁒n2⁒(n|H1,1⁒(Ξ»)|)β©Ύ124⁒n2⁒(nn-|H1,1⁒(Ξ»)|)n-|H1,1⁒(Ξ»)|β©Ύ124⁒n2β‹…2n/8.

Case 2: |H1,1⁒(Ξ»)|β©Ύ78⁒n. Without loss of generality, there is an i∈{0,1,2} such that m1=m2+|H1,1⁒(Ξ»)|-p-i (see Lemma 4.5). Because of

0⩽m2⩽n-|H1,1⁒(λ)|-1,

we realize that

18⁒n⩽m1⩽12⁒n.

If 18⁒n⩽m1⩽12⁒(|H1,1⁒(λ)|-1), with Lemma 4.5, we obtain

χλ⁒(1)β©Ύ124⁒n2⁒(|H1,1⁒(Ξ»)|-1m1)β©Ύ124⁒n2⁒(|H1,1⁒(Ξ»)|-1m1)m1β©Ύ124⁒n2β‹…2n/8.

The case 12⁒(|H1,1⁒(λ)|-1)⩽m1⩽12⁒n is analogous. Hence our result follows. ∎

Lemma 4.8.

Let C be a conjugacy class of Sn. We have Ο‡(n-1,1)⁒(C)=f⁒(C)-1 and |Ο‡(n-1,1)⁒(C)|=|Ο‡(2,1n-2)⁒(C)|.

Proof.

See [28, Example 2.3.8] and [3, Lemma 3.6.10]. ∎

We return to our original problem.

Lemma 4.9.

Let Ο€, C and Ξ΄ be chosen as in Proposition 4.1, and let n be sufficiently large. Then

βˆ‘Ο‡βˆˆIrr⁑(Sn)χ⁒(1)β‰ 1Ο‡2⁒(Ο€)⋅χ⁒(C)χ⁒(1)=π’ͺ⁒(Ξ΄).

Proof.

Let

P1:-{λ⊒n:Ξ»βˆ‰{(n),(1n),(n-1,1),(2,1n-2)}},
P2:-{(n-1,1),(2,1n-2)}.

By applying Lemmas 3.6 and 4.3, we get

βˆ‘Ο‡βˆˆIrr⁑(Sn)χ⁒(1)β‰ 1Ο‡2⁒(Ο€)⋅χ⁒(C)χ⁒(1)β‰ͺβˆ‘Ξ»βˆˆP1χλ⁒(Ο€)β‰ 0χλ⁒(1)log⁑δ32⁒log⁑n+βˆ‘Ξ»βˆˆP2|χλ⁒(C)|χλ⁒(1).

We consider separately the two sums. Lemmas 4.2, 4.6 and 4.7 yield

βˆ‘Ξ»βˆˆP1χλ⁒(Ο€)β‰ 0χλ⁒(1)log⁑δ32⁒log⁑nβ‰ͺn3β‹…(124⁒n2β‹…2n/8)log⁑δ32⁒log⁑nβ‰ͺexp⁑(log⁑δ400⁒nlog⁑n),

which is far smaller than necessary. Due to Lemmas 4.6 and 4.8, we have

βˆ‘Ξ»βˆˆP2|χλ⁒(C)|χλ⁒(1)β©½2⁒|f⁒(C)-1|n-1β‰ͺΞ΄.

This completes the proof. ∎

Below, we will state results similar to Proposition 4.1 for different cycle types of Ο€.

Proposition 4.10.

Let p∈Π⁒(58⁒n), and let Ο€βˆˆSn-1 be a fixed permutation with cycle type (p,n-p-2,1). In addition, suppose that CβŠ‚An-1 is a conjugacy class of Sn-1 such that

f⁒(C)β©½Ξ΄β‹…(n-1)β€ƒπ‘“π‘œπ‘Ÿβ’Ξ΄βˆˆ(0,1).

Then, for all sufficiently large n, we have

#⁒{ΟƒβˆˆSn-1:[Ο€,Οƒ]∈C}=|C|⁒(2+π’ͺ⁒(exp⁑(log⁑δ400⁒nlog⁑n))).

The proof is similar to that of Proposition 4.1.

Proposition 4.11.

Let p∈Π⁒(58⁒n), and let Ο€βˆˆSn-2 be a fixed permutation with cycle type (p,n-p-2). In addition, suppose that CβŠ‚An-2 is a conjugacy class of Sn-2 such that

f⁒(C)β©½Ξ΄β‹…(n-2)β€ƒπ‘“π‘œπ‘Ÿβ’Ξ΄βˆˆ(0,23].

Then, for all sufficiently large n, we have

#⁒{ΟƒβˆˆSn-2:[Ο€,Οƒ]∈C}=|C|⁒(2+π’ͺ⁒(Ξ΄1/33)).

Proof.

Apart from one estimate, the arguments are analogous to those in the proof of Proposition 4.1. We will discuss the missing detail below. ∎

Proposition 4.12.

Let Ο€βˆˆSm be a fixed permutation with cycle type

(m),(m-1,1),(m-2,2)β€ƒπ‘œπ‘Ÿβ€ƒ(m-2,1,1).

In addition, suppose that CβŠ‚Am is a conjugacy class of Sm such that

f⁒(C)β©½Ξ΄β‹…mβ€ƒπ‘“π‘œπ‘Ÿβ’Ξ΄βˆˆ(0,23].

Then, for all sufficiently large m, we have

#⁒{ΟƒβˆˆSm:[Ο€,Οƒ]∈C}=|C|⁒(2+π’ͺ⁒(Ξ΄1/33)).

Proof.

We sketch the argument for the case that our permutation Ο€βˆˆSm has cycle type (m). The other cases are similar.

Because of the cycle type of Ο€, the Murnaghan–Nakayama rule and the hook formula yield for λ⊒m with χλ⁒(Ο€)β‰ 0 that

  1. the Ferrers diagram of Ξ» does not contain the box (2,2),

  2. |χλ⁒(Ο€)|=1,

  3. χλ⁒(1)=(m-1m1).

Analogous to (4.1), we have

#⁒{ΟƒβˆˆSm:[Ο€,Οƒ]∈C}=|C|β’βˆ‘Ο‡βˆˆIrr⁑(Sm)Ο‡2⁒(Ο€)⋅χ⁒(C)χ⁒(1)=|C|⁒(2+βˆ‘Ο‡βˆˆIrr⁑(Sm)χ⁒(1)β‰ 1Ο‡2⁒(Ο€)⋅χ⁒(C)χ⁒(1)).

We now use Lemma 3.6 and (i)–(iii) to estimate the error term.

|βˆ‘Ο‡βˆˆIrr⁑(Sm)χ⁒(1)β‰ 1Ο‡2⁒(Ο€)⋅χ⁒(C)χ⁒(1)|β©½βˆ‘m1=1m-2(m-1m1)log⁑δ32⁒log⁑mβ‰ͺβˆ‘1β©½m1β©½12⁒(m-1)(m-1m1)log⁑δ32⁒log⁑m⁒m1.

We split the summation over m1 in two ranges and consider separately the subsums. We obtain

βˆ‘1β©½m1β©½m1/34(m-1m1)log⁑δ32⁒log⁑m⁒m1β©½βˆ‘m1=1∞δm133=Ξ΄1/331-Ξ΄1/33β‰ͺΞ΄1/33,
βˆ‘m1/34β©½m1β©½12⁒(m-1)(m-1m1)log⁑δ32⁒log⁑m⁒m1β©½mβ‹…exp⁑(log⁑δ⋅log⁑232⁒m1/34log⁑m),

which is far smaller than necessary. ∎

5 Completion of the proof

Choose Ο€βˆˆAn and CβŠ‚An as in Theorem 2.4. We shall show that counting permutations ΟƒβˆˆAn such that [Ο€,Οƒ]∈C and γ€ˆΟ€,σ〉=An can be reduced to the problems dealt with in the last section.

In this context, the following result on permutation groups due to Jordan (cf. [9, Theorem 3.3E] or [30, Theorem 13.9]) is very useful.

Lemma 5.1.

Let G be a primitive subgroup of Sn. Suppose that G contains at least one permutation which is a p-cycle for a prime pβ©½n-3. Then either G=Sn or G=An.

We now show how this lemma can be applied to our problem.

Lemma 5.2.

Let G be a transitive subgroup of Sn. Suppose that there exists a permutation Ο€βˆˆG containing a p-cycle with p∈Π⁒(n-3). Then either G=Sn or G=An.

Proof.

First, we observe that G is primitive. You can see this as follows: Let us assume G is imprimitive. Then there exist blocks Y1,…,Yk for G such that Y1,…,Yk form a partition of {1,…,n}, 2β©½|Y1|β©½n-1 and |Y1|=|Yi| for all i=2,…,k. Let a be an element contained in the p-cycle of Ο€. Then we have a∈Yl0, π⁒(a)∈Yl1,…,Ο€p-1⁒(a)∈Ylp-1 for pairwise distinct li∈{1,…,k}. Hence we get kβ©Ύp. This yields the contradiction n=|Y1|⁒kβ©Ύ2⁒p>n.

Second, let

d:-lcm⁑{m:π⁒contains a cycle of length⁒m<p}.

Obviously, Ο€d∈G is a p-cycle. Since G is primitive, we can conclude with Lemma 5.1 that G=Sn or G=An. ∎

The last lemma immediately yields the following corollary.

Corollary 5.3.

Let p∈Π⁒(n-3). Suppose that Ο€βˆˆAn contains a p-cycle, and let CβŠ‚An be a conjugacy class of Sn. Then we get

#⁒{ΟƒβˆˆAn:[Ο€,Οƒ]∈Cβ’π‘Žπ‘›π‘‘β’γ€ˆΟ€,σ〉=An}=#⁒{ΟƒβˆˆAn:[Ο€,Οƒ]∈C}-#⁒{ΟƒβˆˆAn:[Ο€,Οƒ]∈Cβ’π‘Žπ‘›π‘‘β’(γ€ˆΟ€,σ〉⁒is not transitive)}.

In the above equation, we are able to compute the first term on the right-hand side (see Proposition 4.1). In order to calculate the second one, we have to analyze the condition that γ€ˆΟ€,σ〉is not transitive.

Definition 5.4.

For a cycle Ο„=(i1,…,im), define

M⁒(Ο„):-{i1,…,im}.

Let Ο€1⋅…⋅πk be the cycle decomposition for Ο€βˆˆSn (including 1-cycles), and set Ξ©:-{1,2,…,n}. Define Ξ“H:-⋃i∈HM⁒(Ο€i)βŠ‚Ξ© when HβŠ‚{1,…,k}. For a non-empty set Ξ©, let Sym⁒(Ξ©) denote the symmetric group on Ξ©. When Ξ“βŠ‚Ξ©, we identify Sym⁒(Ξ“) with the subgroup of Sym⁒(Ξ©) consisting of all

Ο„βˆˆSym⁒(Ξ©) with⁒supp⁑(Ο„)βŠ‚Ξ“.

Lemma 5.5.

Let Ο€1⋅…⋅πk be the cycle decomposition for Ο€βˆˆSn (including 1-cycles), and let ΟƒβˆˆSn. Then γ€ˆΟ€,σ〉 is not transitive if and only if there exists a non-empty proper subset H of K:-{1,…,k} and a tuple

(Οƒβ€²,Οƒβ€²β€²)∈Sym⁒(Ξ“H)Γ—Sym⁒(Ξ“K\H)

such that Οƒ=σ′⁒σ′′.

Proof.

β€œβ‡β€. This implication is obvious.

β€œβ‡’β€. This implication is the result of the following stronger statement.

Claim.

Let Οƒ1⋅…⋅σl be the cycle decomposition for Οƒ (including 1-cycles). Let s,t∈{1,…,n} such that τ⁒(s)β‰ t for all Ο„βˆˆγ€ˆΟ€,σ〉. For all m with 1β©½mβ©½l-1, there exist non-empty subsets Hm,Hmβ€²βŠ‚K and a subset LmβŠ‚L:-{1,…,l} such that

  1. for each j∈Lm, we have

    M⁒(Οƒj)βŠ‚β‹ƒi∈HmM⁒(Ο€i)β€ƒπ‘œπ‘Ÿβ€ƒM⁒(Οƒj)βŠ‚β‹ƒi∈Hmβ€²M⁒(Ο€i),
  2. |Lm|β©Ύm+1,

  3. Hm∩Hmβ€²=βˆ…,

  4. Lmβ‰ L implies that, for all i∈Hm, there exists

    Ο„βˆˆγ€ˆΟ€,σ〉 such that τ⁒(s)∈M⁒(Ο€i),
  5. Lmβ‰ L implies that, for all i∈Hmβ€², there exists

    Ο„βˆˆγ€ˆΟ€,σ〉 such that τ⁒(t)∈M⁒(Ο€i).

We prove this claim by induction. If m=1, we have s∈M⁒(Οƒj) and t∈M⁒(Οƒjβ€²) for jβ‰ jβ€². Set L1:-{j,jβ€²} and

H1:-{i∈K:M⁒(Οƒj)∩M⁒(Ο€i)β‰ βˆ…},
H1β€²:-{i∈K:M⁒(Οƒjβ€²)∩M⁒(Ο€i)β‰ βˆ…}.

Then (a)–(e) follow.

Now let 1⩽m⩽l-2, and suppose that the statement is true for m. The case Lm=L is trivial. Hence let us assume that Lm⊊L.

Case 1: there exists j0∈Lβˆ–Lm such that M⁒(Οƒj0)βˆ©β‹ƒi∈HmM⁒(Ο€i)β‰ βˆ… or M⁒(Οƒj0)βˆ©β‹ƒi∈Hmβ€²M⁒(Ο€i)β‰ βˆ…. Without loss of generality, assume that

M⁒(Οƒj0)βˆ©β‹ƒi∈HmM⁒(Ο€i)β‰ βˆ….

Define Lm+1:-Lmβˆͺ{j0}, Hm+1:-Hmβˆͺ{i∈K:M⁒(Οƒj0)∩M⁒(Ο€i)}β‰ βˆ… and Hm+1β€²:-Hmβ€². Then (a)–(e) result from the induction hypothesis and some easy computations.

Case 2: M⁒(Οƒj)βˆ©β‹ƒi∈HmM⁒(Ο€i)=βˆ… and M⁒(Οƒj)βˆ©β‹ƒi∈Hmβ€²M⁒(Ο€i)=βˆ… for all j∈Lβˆ–Lm. Set Lm+1:-L, Hm+1:-Kβˆ–Hmβ€² and Hm+1:-Hmβ€². Obviously, (a)–(e) are fulfilled. ∎

Definition 5.6.

Denote by [Sn] the set of conjugacy classes of Sn.

Let C∈[Sn], C1∈[Sk] and C2∈[Sn-k]. Denote by ml, sl and tl the number of l-cycles in C, C1 and C2, respectively. We say that C=C1⁒C2 if and only if ml=sl+tl for all l=1,…,n.

Finally, we draw our attention to βˆ‘|C1|⁒|C2|, where the sum runs over all (C1,C2)∈[Sk]Γ—[Sn-k] satisfying C1⁒C2=C for a fixed conjugacy class C of Sn. In this context, the following lemma is very useful.

Lemma 5.7.

Let C be a conjugacy class of Sn with cycle type (1m1,2m2,…,nmn), where ml is the number of l-cycles in C. Then we obtain

|C|=n!1m1⁒m1!β‹…2m2⁒m2!⋅…⋅nmn⁒mn!.

Proof.

See [28, Proposition 1.1.1]. ∎

We now observe the following.

Lemma 5.8.

Suppose that C is a conjugacy class of Sn such that f⁒(C)β©½Ξ΄β‹…n for δ∈(0,14], and let n be sufficiently large.

  1. Then, for 12⁒n<k⩽58⁒n, we have

    βˆ‘C1∈[Sk]C2∈[Sn-k]C=C1⁒C2|C1|⁒|C2|β©½exp⁑(-350⁒n)⁒|C|.
  2. Furthermore, we obtain

    βˆ‘C1∈[Sn-1]C2∈[S1]C=C1⁒C2|C1|⩽δ⁒|C|β€ƒπ‘Žπ‘›π‘‘β€ƒβˆ‘C1∈[Sn-2]C2∈[S2]C=C1⁒C2|C1|β©½2⁒δ2⁒|C|.

Proof.

Denote by (1m1,2m2,…,nmn) the cycle type of C.

We begin with the first claim. If C contains an l-cycle with l>k, then the sum is equal to 0. Therefore, we assume that C has cycle type

(1m1,2m2,…,kmk,(k+1)0,…,n0).

By computing the cardinality of C and of the conjugacy classes C1∈[Sk] and C2∈[Sn-k] with C=C1⁒C2 via Lemma 5.7, we obtain

βˆ‘C1∈[Sk]C2∈[Sn-k]C=C1⁒C2|C1|⁒|C2|β©½|C|⁒(nk)-1β’βˆ‘(i1,…,in-k)0β©½ijβ©½mj∏l=1n-k(mlil)=|C|⁒(nk)-1⁒2(m1+β‹―+mn-k).

The Stirling formula yields the inequality (nk)⩾(4125)n for 12⁒n<k⩽58⁒n. In addition, we have

βˆ‘i=1nmiβ©½f⁒(C)+n-f⁒(C)2β©½12⁒n⁒(1+Ξ΄)β©½58⁒n.

Hence our first estimate follows.

We now show the second statement. Let i∈{1,2}. We again apply Lemma 5.7. By calculating the cardinality of C and of conjugacy classes C1∈[Sn-i] such that there exists C2∈[Si] with C=C1⁒C2, we get

βˆ‘C1∈[Sn-1]C2∈[S1]C=C1⁒C2|C1|β©½|C|⁒m1n⩽δ⁒|C|,
βˆ‘C1∈[Sn-2]C2∈[S2]C=C1⁒C2|C1|β©½|C|⁒1n⁒(n-1)⁒(m1⁒(m1-1)+2⁒m2)β©½|C|⁒(Ξ΄2+1n-1)β©½2⁒δ2⁒|C|.

This proves our claim. ∎

We are now in the position to give the proof of our second theorem.

Proof of Theorem 2.4.

Let Ο€1⋅…⋅πk be the cycle decomposition of Ο€ (including 1-cycles), and set K:-{1,…,k}. Then Lemma 5.5 yields

B:-#⁒{ΟƒβˆˆAn:[Ο€,Οƒ]∈C⁒and⁒(γ€ˆΟ€,σ〉⁒is not transitive)}β©½βˆ‘{H,K\H}H⊊K,Hβ‰ βˆ…#⁒{(Οƒβ€²,Οƒβ€²β€²)∈Sym⁒(Ξ“H)Γ—Sym⁒(Ξ“K\H):[Ο€,σ′⁒σ′′]∈C}.

Let H be a proper non-empty subset of K. Define

Ο€β€²:-∏i∈HΟ€i and π′′:-∏i∈K\HΟ€i.

If (Οƒβ€²,Οƒβ€²β€²)∈Sym⁒(Ξ“H)Γ—Sym⁒(Ξ“K\H), we get

[Ο€,σ′⁒σ′′]=[Ο€β€²,Οƒβ€²]β‹…[Ο€β€²β€²,Οƒβ€²β€²].

Obviously, we have [Ο€β€²,Οƒβ€²]∈Sym⁒(Ξ“H) and [Ο€β€²β€²,Οƒβ€²β€²]∈Sym⁒(Ξ“K\H). Therefore, [Ο€,σ′⁒σ′′]∈C holds if and only if there exist conjugacy classes C1βŠ‚Sym⁒(Ξ“H) and C2βŠ‚Sym⁒(Ξ“K\H) with [Ο€β€²,Οƒβ€²]∈C1, [Ο€β€²β€²,Οƒβ€²β€²]∈C2 and C=C1⁒C2.

Case 1: n≑1⁒(2). By assumption, k=3 and K={1,2,3}. For λ⊒m, let Ο€Ξ»βˆˆSm be a fixed permutation with cycle type Ξ». The above considerations yield

Bβ©½βˆ‘C1∈[Sn-2]C2∈[S2]C=C1⁒C2#⁒{ΟƒβˆˆSn-2:[Ο€(p,n-p-2),Οƒ]∈C1}
β‹…#⁒{ΟƒβˆˆS2:[Ο€(2),Οƒ]∈C2}
  +βˆ‘C1∈[Sp]C2∈[Sn-p]C=C1⁒C2#⁒{ΟƒβˆˆSp:[Ο€(p),Οƒ]∈C1}
β‹…#⁒{ΟƒβˆˆSn-p:[Ο€(n-p-2,2),Οƒ]∈C2}
  +βˆ‘C1∈[Sp+2]C2∈[Sn-p-2]C=C1⁒C2#⁒{ΟƒβˆˆSp+2:[Ο€(p,2),Οƒ]∈C1}
β‹…#⁒{ΟƒβˆˆSn-p-2:[Ο€(n-p-2),Οƒ]∈C2}
=π’ͺ⁒(βˆ‘C1∈[Sn-2]C2∈[S2]C=C1⁒C2|C1|+βˆ‘C1∈[Sp]C2∈[Sn-p]C=C1⁒C2|C1|⁒|C2|+βˆ‘C1∈[Sp+2]C2∈[Sn-p-2]C=C1⁒C2|C1|⁒|C2|)
=π’ͺ⁒(Ξ΄2⁒|C|).

For i∈{0,1,2}, we have 12⁒n<p+i⩽58⁒n. In addition, f⁒(C1)⩽f⁒(C) is valid for C=C1⁒C2. Therefore, the last but one equality follows from Proposition 4.11 and Proposition 4.12. The last step results from Lemma 5.8.

Case 2: n≑0⁒(2). With an argument similar to the first case, we obtain

B=π’ͺ⁒(δ⁒|C|).

Hence it follows from Corollary 5.3 and Proposition 4.1 that

#⁒{ΟƒβˆˆAn:[Ο€,Οƒ]∈C⁒andβ’γ€ˆΟ€,σ〉=An}=|C|⁒(1+π’ͺ⁒(Ξ΄)).

So we are done. ∎


Communicated by MiklΓ³s AbΓ©rt


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Received: 2016-01-01
Revised: 2019-04-02
Published Online: 2019-05-07
Published in Print: 2019-09-01

Β© 2020 Walter de Gruyter GmbH, Berlin/Boston

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