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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
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

18n3exp(2π6n1/2)(1+o(1)).

Applying this result, we obtain a lower bound for the number of connected components of the product replacement graph Γ2(An).

## 1 Introduction

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

𝒩k(G):-{(g1,,gk)Gk:g1,,gk=G}

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

(Aut(Fk)×Aut(G))×Epi(FkG)Epi(FkG)
((τ,σ),ϕ)σϕτ-1.

Systems of transitivity or, in short, Tk-systems are defined to be the orbits of this group action. Denote by τk(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)𝒩k(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 1ijk. Second, apply one of the subsequent four operations with equal probability:

Ri,j±:(g1,,gi,,gk)(g1,,gigj±1,,gk),
Li,j±:(g1,,gi,,gk)(g1,,gj±1gi,,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 “random” element of the group G.

The product replacement graph (PRA graph)Γk(G) is a graph with vertices corresponding to generating k-tuples of G and edges corresponding to the moves Ri,j±, Li,j±. The PRA can be described as a simple random walk on this graph. Denote by ϰk(G) the number of connected components of Γk(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)(x1,,xixj,,xk),
Li,j:(x1,,xi,,xk)(x1,,xjxi,,xk),
Pi,j:(x1,,xi,,xj,,xk)(x1,,xj,,xi,,xk),
Ii:(x1,,xi,,xk)(x1,,xi-1,,xk)

for 1ijk. 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)τk(G)ϰk(G).

Little is known about the problem to estimate τk(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 τk(G)=1 for k3 (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 τ2(G) as |G|, where G is a finite simple group. Hence they confirmed a conjecture of Guralnick and Pak [18]. In addition, they proved that

τ2(An)n(12-ϵ)logn,

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

τ2(A2n+4)#{C:Cis conjugacy class ofSn},
τ2(A2n+5)#{C:Cis conjugacy class ofSn}

for n2. 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

τ2(An)18n3exp(2π6n1/2)(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

ϰ2(An)18n3exp(2π6n1/2)(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-1h-1gh.

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:-{[π,σ]:(π,σ)N2(An)}. For all sufficiently large n, we have

τ2(An)#{C:Cis a conjugacy class ofSn𝑤𝑖𝑡ℎCJ}.

## Proof.

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

Aut(An)={α|An:αInn(Sn)}forn>6.

Thus we have

:-{K:Kis anAut(An)-conjugacy class withKJ}={C:Cis a conjugacy class ofSnwithCJ}.

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

τ2(An)#{KK-1:K}=||

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:, let Π(a(n)) be the set of all primes p in the range 12n<pa(n).

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

## Theorem 2.4.

Let pΠ(35n), and let πAn 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 CAn is a conjugacy class of Sn such that

f(C)δn𝑓𝑜𝑟δ(0,14].

Then, for all sufficiently large n, we have

#{σAn:[π,σ]C𝑎𝑛𝑑π,σ=An}=|C|(1+𝒪(δ)).

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

#{σAn:[π,σ]Candπ,σ=An}>0.

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

By Lemma 2.2, τ2(An) is bounded below by the number of conjugacy classes of Sn containing the commutator [π,σ] for some generating tuple (π,σ)𝒩2(An). Theorem 2.4 shows that this number is in turn bounded below by the number of conjugacy classes C of Sn satisfying CAn and f(C)δn for δ>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 λ=(λ1,λ2,,λl) of positive integers such that λ1λ2λl and λ1+λ2++λl=n. We write λn to indicate that λ is a partition of n. Denote by P(n) the number of partitions of n.

Suppose λ=(λ1,λ2,,λl)n. The Ferrers diagram of λ is an array of n boxes having l left-justified rows with row i containing λi boxes for 1il.

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

## Lemma 2.6.

We have

P(n)=14n3exp(2π6n1/2)(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])

#{λn:number of even parts inλis odd}=#{λn:number of even parts inλis even}-#{λn:λconsists of distinct odd parts}.

Thus we have

#{CAn:Cis a conjugacy class ofSn}12P(n).

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

τ2(An)#{CAn:Cis a conjugacy class ofSnwithf(C)δn}12P(n)-P(n-δn).

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 χIrr(Sn), we write χ(C) to denote χ(π) for πC.

## Lemma 3.1.

Let C1 and C2 be conjugacy classes of Sn, and let τSn. Then

#{(x,y)C1×C2:xy=τ}=|C1||C2|n!χIrr(Sn)χ(C1)χ(C2)χ(τ-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 λ of n.

We shall frequently apply the Murnaghan–Nakayama rule.

## Definition 3.2.

Let λn 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

ll(h):-(the number of rows ofh)-1.

Let πSn be a permutation with cycle type (1α1,,qαq,,nαn) and αq1. Denote by π\qSn-q a permutation with cycle type

(1α1,,qαq-1,,(n-q)αn-q).

## Lemma 3.3 (Murnaghan–Nakayama rule).

Let λn be a partition. Suppose that πSn is a permutation which contains a q-cycle. Then we have

χλ(π)=hq-rim hook𝑜𝑓λ(-1)ll(h)χλ\h(π\q).

## Proof.

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

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

## Definition 3.4.

Let λn be a partition. The hook of (i,j)λ is

Hi,j(λ):-{(i,j)λ:jj}{(i,j)λ:ii}

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 λn be a partition. Then

χλ(1)=n!(i,j)λ|Hi,j(λ)|.

## Proof.

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

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

## Lemma 3.6.

Let χIrr(Sn) be an irreducible character, let n be sufficiently large, and let σSn be an element with k fixed points. Then we have

|χ(σ)|χ(1)1-log(n/k)32logn.

## 4 Character estimates

Let πAn be fixed. In this section, we will count permutations σAn satisfying that the commutator [π,σ] 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 pΠ(58n), and let πAn 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 CAn is a conjugacy class of Sn such that

f(C)δn𝑓𝑜𝑟δ(0,1).

Then, for all sufficiently large n, we have

#{σAn:[π,σ]C}=|C|(1+𝒪(δ)).

## Proof.

Due to the prime number theorem, there exists a prime pΠ(58n) for all sufficiently large n. Denote by Kπ the conjugacy class of π in Sn. Because of the cycle type of π, Kπ remains a single conjugacy class inside An (cf. [31, pp. 16f.]). Taking into account that χ(τ-1)=χ(τ)¯=χ(τ), it follows with Lemma 3.1 that

(4.1)#{σAn:[π,σ]C}=(x,y)Kπ×Cxy=π#{σAn:σ-1πσ=x}=n!2|Kπ|#{(x,y)Kπ×C:xy=π}=|C|2χIrr(Sn)χ2(π)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)χIrr(Sn)χ(1)1χ2(π)χ(C)χ(1)=𝒪(δ).

This completes our proof. ∎

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

## Lemma 4.2.

Let p and π be chosen as in Proposition 4.1. Let λn be a partition with χλ(π)0. Then the Ferrers diagram of λ has at most two boxes which do not belong to H1,1(λ)H2,2(λ). Moreover, let h be a p-rim hook of λ with χλ\h(π\p)0 or an (n-p-2)-rim hook of λ with χλ\h(π\(n-p-2))0. Then the Ferrers diagram of λ\h has at most two boxes which do not belong to H1,1(λ\h). In particular, we have |H1,1(λ)|p.

## Proof.

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

## Lemma 4.3.

Let p and π be chosen as in Proposition 4.1, and let λn be a partition. Then we have |χλ(π)|2.

## Proof.

If χλ(π)=0, there is nothing to show. So let us assume that χλ(π)0. Let

R:-{h:his ap-rim hook ofλwithχλ\h(π\p)0}.

We have R as a result of the Murnaghan–Nakayama rule. Due to

|H1,1(λ)|p>12n

(see Lemma 4.2), a rim hook hR contains boxes of the hook H1,1(λ). Hence

|R|2.

Now let h1R. 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(π) is equal to

{h1p-rim hookofλh2(n-p-2)-rim hookofλ\h1(-1)ll(h1)+ll(h2)ifn0(2),h1p-rim hookofλh2(n-p-2)-rim hookofλ\h1h32-rim hookof(λ\h1)\h2(-1)ll(h1)+ll(h2)+ll(h3)ifn1(2).

Therefore, it follows that |χλ(π)|2. ∎

## Definition 4.4.

Let λn. Define

m1:-(number of boxes in the first column of the Ferrers diagram ofλ)-1,
m1:-(number of boxes in the first row of the Ferrers diagram ofλ)-1,
m2:-(number of boxes in the second column of the Ferrers diagram ofλ)-2,
m2:-(number of boxes in the second row of the Ferrers diagram ofλ)-2.

## Lemma 4.5.

Let p and π be chosen as in Proposition 4.1. Let λn be a partition such that the Ferrers diagram of λ contains the box (2,2) and χλ(π)0. Then we have

χλ(1)124n2(n|H1,1(λ)|)(|H1,1(λ)|-1m1),
χλ(1)124n2(n|H1,1(λ)|)(|H1,1(λ)|-1m1).

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

m1=m2+|H1,1(λ)|-p-i𝑜𝑟m1=m2+|H1,1(λ)|-p-i.

## Proof.

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

χλ(1)124n2(n|H1,1(λ)|)(|H1,1(λ)|-1m1)(n-|H1,1(λ)|-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 λ such that χλ\h(π\(n-p-2))0. By assumption, we have |H1,1(λ)|p+3. Therefore, h contains either boxes from the first column or the first row of the Ferrers diagram of λ – 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(λ\h)|{0,1,2}. If |H2,2(λ\h)|=1, we obtain n-p-2=n-|H1,1(λ)|-1+m1-m2 and hence m1=m2+|H1,1(λ)|-p-1. The other cases are analogous. This proves our claim. ∎

## Lemma 4.6.

Let p and π be chosen as in Proposition 4.1. Let λn, where λ{(n),(1n)}, be a partition such that χλ(π)0 and the Ferrers diagram of λ 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 λ such that there is an (n-p-2)-rim hook of λ\h1. This proves the last statement. ∎

## Lemma 4.7.

Let p and π be chosen as in Proposition 4.1. Let λn be partition such that λ{(n),(1n),(n-1,1),(2,1n-2)} and χλ(π)0. Then, for all sufficiently large n, we obtain

χλ(1)124n22n/8.

## Proof.

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

χλ(1)(n-1p+1)(n-1p+1)p+12n/8.

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

Case 1: |H1,1(λ)|78n. By applying Lemmas 4.5 and 4.2, we get

χλ(1)124n2(n|H1,1(λ)|)124n2(nn-|H1,1(λ)|)n-|H1,1(λ)|124n22n/8.

Case 2: |H1,1(λ)|78n. Without loss of generality, there is an i{0,1,2} such that m1=m2+|H1,1(λ)|-p-i (see Lemma 4.5). Because of

0m2n-|H1,1(λ)|-1,

we realize that

18nm112n.

If 18nm112(|H1,1(λ)|-1), with Lemma 4.5, we obtain

χλ(1)124n2(|H1,1(λ)|-1m1)124n2(|H1,1(λ)|-1m1)m1124n22n/8.

The case 12(|H1,1(λ)|-1)m112n is analogous. Hence our result follows. ∎

## Lemma 4.8.

Let C be a conjugacy class of Sn. We have χ(n-1,1)(C)=f(C)-1 and |χ(n-1,1)(C)|=|χ(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 π, C and δ be chosen as in Proposition 4.1, and let n be sufficiently large. Then

χIrr(Sn)χ(1)1χ2(π)χ(C)χ(1)=𝒪(δ).

## Proof.

Let

P1:-{λn:λ{(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

χIrr(Sn)χ(1)1χ2(π)χ(C)χ(1)λP1χλ(π)0χλ(1)logδ32logn+λP2|χλ(C)|χλ(1).

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

λP1χλ(π)0χλ(1)logδ32lognn3(124n22n/8)logδ32lognexp(logδ400nlogn),

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

λP2|χλ(C)|χλ(1)2|f(C)-1|n-1δ.

This completes the proof. ∎

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

## Proposition 4.10.

Let pΠ(58n), and let πSn-1 be a fixed permutation with cycle type (p,n-p-2,1). In addition, suppose that CAn-1 is a conjugacy class of Sn-1 such that

f(C)δ(n-1)𝑓𝑜𝑟δ(0,1).

Then, for all sufficiently large n, we have

#{σSn-1:[π,σ]C}=|C|(2+𝒪(exp(logδ400nlogn))).

The proof is similar to that of Proposition 4.1.

## Proposition 4.11.

Let pΠ(58n), and let πSn-2 be a fixed permutation with cycle type (p,n-p-2). In addition, suppose that CAn-2 is a conjugacy class of Sn-2 such that

f(C)δ(n-2)𝑓𝑜𝑟δ(0,23].

Then, for all sufficiently large n, we have

#{σSn-2:[π,σ]C}=|C|(2+𝒪(δ1/33)).

## 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 πSm be a fixed permutation with cycle type

(m),(m-1,1),(m-2,2)𝑜𝑟(m-2,1,1).

In addition, suppose that CAm is a conjugacy class of Sm such that

f(C)δm𝑓𝑜𝑟δ(0,23].

Then, for all sufficiently large m, we have

#{σSm:[π,σ]C}=|C|(2+𝒪(δ1/33)).

## Proof.

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

Because of the cycle type of π, the Murnaghan–Nakayama rule and the hook formula yield for λm with χλ(π)0 that

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

2. |χλ(π)|=1,

3. χλ(1)=(m-1m1).

Analogous to (4.1), we have

#{σSm:[π,σ]C}=|C|χIrr(Sm)χ2(π)χ(C)χ(1)=|C|(2+χIrr(Sm)χ(1)1χ2(π)χ(C)χ(1)).

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

|χIrr(Sm)χ(1)1χ2(π)χ(C)χ(1)|m1=1m-2(m-1m1)logδ32logm1m112(m-1)(m-1m1)logδ32logmm1.

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

1m1m1/34(m-1m1)logδ32logmm1m1=1δm133=δ1/331-δ1/33δ1/33,
m1/34m112(m-1)(m-1m1)logδ32logmm1mexp(logδlog232m1/34logm),

which is far smaller than necessary. ∎

## 5 Completion of the proof

Choose πAn and CAn as in Theorem 2.4. We shall show that counting permutations σAn such that [π,σ]C and π,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 pn-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 πG containing a p-cycle with pΠ(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}, 2|Y1|n-1 and |Y1|=|Yi| for all i=2,,k. Let a be an element contained in the p-cycle of π. Then we have aYl0, π(a)Yl1,,πp-1(a)Ylp-1 for pairwise distinct li{1,,k}. Hence we get kp. This yields the contradiction n=|Y1|k2p>n.

Second, let

d:-lcm{m:πcontains a cycle of lengthm<p}.

Obviously, πdG 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 pΠ(n-3). Suppose that πAn contains a p-cycle, and let CAn be a conjugacy class of Sn. Then we get

#{σAn:[π,σ]C𝑎𝑛𝑑π,σ=An}=#{σAn:[π,σ]C}-#{σAn:[π,σ]C𝑎𝑛𝑑(π,σis 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 π,σis not transitive.

## Definition 5.4.

For a cycle τ=(i1,,im), define

M(τ):-{i1,,im}.

Let π1πk be the cycle decomposition for πSn (including 1-cycles), and set Ω:-{1,2,,n}. Define ΓH:-iHM(πi)Ω when H{1,,k}. For a non-empty set Ω, let Sym(Ω) denote the symmetric group on Ω. When ΓΩ, we identify Sym(Γ) with the subgroup of Sym(Ω) consisting of all

τSym(Ω)withsupp(τ)Γ.

## Lemma 5.5.

Let π1πk be the cycle decomposition for πSn (including 1-cycles), and let σSn. Then π,σ is not transitive if and only if there exists a non-empty proper subset H of K:-{1,,k} and a tuple

(σ,σ′′)Sym(ΓH)×Sym(ΓK\H)

such that σ=σσ′′.

## Proof.

”. This implication is obvious.

”. This implication is the result of the following stronger statement.

## Claim.

Let σ1σl be the cycle decomposition for σ (including 1-cycles). Let s,t{1,,n} such that τ(s)t for all τπ,σ. For all m with 1ml-1, there exist non-empty subsets Hm,HmK and a subset LmL:-{1,,l} such that

1. for each jLm, we have

M(σj)iHmM(πi)𝑜𝑟M(σj)iHmM(πi),
2. |Lm|m+1,

3. HmHm=,

4. LmL implies that, for all iHm, there exists

τπ,σsuch thatτ(s)M(πi),
5. LmL implies that, for all iHm, there exists

τπ,σsuch thatτ(t)M(πi).

We prove this claim by induction. If m=1, we have sM(σj) and tM(σj) for jj. Set L1:-{j,j} and

H1:-{iK:M(σj)M(πi)},
H1:-{iK:M(σj)M(πi)}.

Then (a)–(e) follow.

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

Case 1: there exists j0LLm such that M(σj0)iHmM(πi) or M(σj0)iHmM(πi). Without loss of generality, assume that

M(σj0)iHmM(πi).

Define Lm+1:-Lm{j0}, Hm+1:-Hm{iK:M(σj0)M(πi)} and Hm+1:-Hm. Then (a)–(e) result from the induction hypothesis and some easy computations.

Case 2: M(σj)iHmM(πi)= and M(σj)iHmM(πi)= for all jLLm. Set Lm+1:-L, Hm+1:-KHm and Hm+1:-Hm. 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=C1C2 if and only if ml=sl+tl for all l=1,,n.

Finally, we draw our attention to |C1||C2|, where the sum runs over all (C1,C2)[Sk]×[Sn-k] satisfying C1C2=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!1m1m1!2m2m2!nmnmn!.

## 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)δn for δ(0,14], and let n be sufficiently large.

1. Then, for 12n<k58n, we have

C1[Sk]C2[Sn-k]C=C1C2|C1||C2|exp(-350n)|C|.
2. Furthermore, we obtain

C1[Sn-1]C2[S1]C=C1C2|C1|δ|C|𝑎𝑛𝑑C1[Sn-2]C2[S2]C=C1C2|C1|2δ2|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=C1C2 via Lemma 5.7, we obtain

C1[Sk]C2[Sn-k]C=C1C2|C1||C2||C|(nk)-1(i1,,in-k)0ijmjl=1n-k(mlil)=|C|(nk)-12(m1++mn-k).

The Stirling formula yields the inequality (nk)(4125)n for 12n<k58n. In addition, we have

i=1nmif(C)+n-f(C)212n(1+δ)58n.

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=C1C2, we get

C1[Sn-1]C2[S1]C=C1C2|C1||C|m1nδ|C|,
C1[Sn-2]C2[S2]C=C1C2|C1||C|1n(n-1)(m1(m1-1)+2m2)|C|(δ2+1n-1)2δ2|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 π1πk be the cycle decomposition of π (including 1-cycles), and set K:-{1,,k}. Then Lemma 5.5 yields

B:-#{σAn:[π,σ]Cand(π,σis not transitive)}{H,K\H}HK,H#{(σ,σ′′)Sym(ΓH)×Sym(ΓK\H):[π,σσ′′]C}.

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

π:-iHπiandπ′′:-iK\Hπi.

If (σ,σ′′)Sym(ΓH)×Sym(ΓK\H), we get

[π,σσ′′]=[π,σ][π′′,σ′′].

Obviously, we have [π,σ]Sym(ΓH) and [π′′,σ′′]Sym(ΓK\H). Therefore, [π,σσ′′]C holds if and only if there exist conjugacy classes C1Sym(ΓH) and C2Sym(ΓK\H) with [π,σ]C1, [π′′,σ′′]C2 and C=C1C2.

Case 1: n1(2). By assumption, k=3 and K={1,2,3}. For λm, let πλSm be a fixed permutation with cycle type λ. The above considerations yield

BC1[Sn-2]C2[S2]C=C1C2#{σSn-2:[π(p,n-p-2),σ]C1}
#{σS2:[π(2),σ]C2}
+C1[Sp]C2[Sn-p]C=C1C2#{σSp:[π(p),σ]C1}
#{σSn-p:[π(n-p-2,2),σ]C2}
+C1[Sp+2]C2[Sn-p-2]C=C1C2#{σSp+2:[π(p,2),σ]C1}
#{σSn-p-2:[π(n-p-2),σ]C2}
=𝒪(C1[Sn-2]C2[S2]C=C1C2|C1|+C1[Sp]C2[Sn-p]C=C1C2|C1||C2|+C1[Sp+2]C2[Sn-p-2]C=C1C2|C1||C2|)
=𝒪(δ2|C|).

For i{0,1,2}, we have 12n<p+i58n. In addition, f(C1)f(C) is valid for C=C1C2. 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: n0(2). With an argument similar to the first case, we obtain

B=𝒪(δ|C|).

Hence it follows from Corollary 5.3 and Proposition 4.1 that

#{σAn:[π,σ]Candπ,σ=An}=|C|(1+𝒪(δ)).

So we are done. ∎

Communicated by Miklós Abért

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

© 2020 Walter de Gruyter GmbH, Berlin/Boston

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