1 Introduction
Let G be a finite group, and let dβ’(G) be the minimal number of generators of G.
Fix some integer kβ©Ύdβ’(G), and denote by
π©kβ’(G):-{(g1,β¦,gk)βGk:γg1,β¦,gkγ=G}
the set of generating k-tuples of G.
We identify π©kβ’(G) with the set of epimorphisms Epiβ‘(Fkβ G), where Fk is the free group on k generators.
Now consider the group action
(Autβ‘(Fk)ΓAutβ‘(G))ΓEpiβ‘(Fkβ G)βEpiβ‘(Fkβ G)
((Ο,Ο),Ο)β¦ΟβΟβΟ-1.
Systems of transitivity or, in short, Tk-systems are defined to be the orbits of this group action.
Denote by Οkβ’(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,β¦,gk)βπ©kβ’(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,β¦,gi,β¦,gk)β(g1,β¦,giβ
gjΒ±1,β¦,gk),
Li,jΒ±:(g1,β¦,gi,β¦,gk)β(g1,β¦,gjΒ±1β
gi,β¦,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 βrandomβ element of the group G.
The product replacement graph (PRA graph)Ξkβ’(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°kβ’(G) the number of connected components of Ξkβ’(G).
The connection between the PRA and Tk-systems of G is the following.
Nielsen showed in [26] that Autβ‘(Fk) is generated by the Nielsen moves
Ri,j:(x1,β¦,xi,β¦,xk)β(x1,β¦,xiβ
xj,β¦,xk),
Li,j:(x1,β¦,xi,β¦,xk)β(x1,β¦,xjβ
xi,β¦,xk),
Pi,j:(x1,β¦,xi,β¦,xj,β¦,xk)β(x1,β¦,xj,β¦,xi,β¦,xk),
Ii:(x1,β¦,xi,β¦,xk)β(x1,β¦,xi-1,β¦,xk)
for 1β©½iβ jβ©½k.
Therefore, we can define a graph with vertices corresponding to π©kβ’(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)Οkβ’(G)β©½Ο°kβ’(G).
Little is known about the problem to estimate Οkβ’(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 Οkβ’(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 Ο2β’(G)ββ as |G|ββ, where G is a finite simple group.
Hence they confirmed a conjecture of Guralnick and Pak [18].
In addition, they proved that
Ο2β’(An)β©Ύn(12-Ο΅)β’logβ‘n,
where An denotes the alternating group on n letters.
In 1985, Evans already established in his Ph.D. Thesis [11] that
Ο2β’(A2β’n+4)β©Ύ#β’{C:Cβ’is conjugacy class ofβ’Sn},
Ο2β’(A2β’n+5)β©Ύ#β’{C:Cβ’is conjugacy class ofβ’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
Ο2β’(An)β©Ύ18β’nβ’3β’expβ‘(2β’Ο6β’n1/2)β’(1+oβ’(1)).
Taking (1.1) into account, this theorem has an immediate corollary.
Corollary 1.2.
Let n be sufficiently large.
Then we have
Ο°2β’(An)β©Ύ18β’nβ’3β’expβ‘(2β’Ο6β’n1/2)β’(1+oβ’(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-1β’h-1β’gβ’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 dβ’(G)β©½2.
Then, for each T2-system S of G, the set of Autβ‘(G)-conjugates of {[g1,g2]Β±1} is invariant for all (g1,g2)βS.
Higmanβs lemma leads us to the following lemma.
Lemma 2.2.
Let J:-{[Ο,Ο]:(Ο,Ο)βN2β’(An)}.
For all sufficiently large n, we have
Ο2β’(An)β©Ύ#β’{C:Cβ’is a conjugacy class ofβ’Snβ’π€ππ‘ββ’Cβ©Jβ β
}.
Proof.
Note that (cf. [9, Theorem 8.2A])
Autβ‘(An)={Ξ±|An:Ξ±βInnβ‘(Sn)}βforβ’n>6.
Thus we have
β¬:-{K:Kβ’is anβ’Autβ‘(An)β’-conjugacy class withβ’Kβ©Jβ β
}={C:Cβ’is a conjugacy class ofβ’Snβ’withβ’Cβ©Jβ β
}.
Taking into account that dβ’(An)β©½2 (cf. [8]), Lemma 2.1 yields
Ο2β’(An)β©Ύ#β’{KβͺK-1:Kββ¬}=|β¬|
as desired.
β
Definition 2.3.
Let C be a conjugacy class of Sn.
Denote by fβ’(C) the number of fixed points of elements of C.
For a function a:βββ, let Ξ β’(aβ’(n)) be the set of all primes p in the range 12β’n<pβ©½aβ’(n).
We establish the following basic result that is very useful for the proof of our theorem.
Theorem 2.4.
Let pβΞ β’(35β’n), and let ΟβAn 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 CβAn is a conjugacy class of Sn such that
fβ’(C)β©½Ξ΄β
nβπππβ’Ξ΄β(0,14].
Then, for all sufficiently large n, we have
#β’{ΟβAn:[Ο,Ο]βCβ’πππβ’γΟ,Ογ=An}=|C|β’(1+πͺβ’(Ξ΄)).
As an immediate consequence,
for Ξ΄>0 sufficiently small,
we have
#β’{ΟβAn:[Ο,Ο]βCβ’andβ’γΟ,Ογ=An}>0.
We will give the proof of this theorem in the next sections.
By Lemma 2.2, Ο2β’(An) is bounded below by the number of conjugacy classes of Sn containing the commutator [Ο,Ο] for some generating tuple (Ο,Ο)βπ©2β’(An).
Theorem 2.4 shows that this number is in turn bounded below by the number of conjugacy classes C of Sn satisfying CβAn and fβ’(C)β©½Ξ΄β’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,β¦,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 Ξ»β’n to indicate that Ξ» is a partition of n.
Denote by Pβ’(n) the number of partitions of n.
Suppose Ξ»=(Ξ»1,Ξ»2,β¦,Ξ»l)β’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
Pβ’(n)=14β’nβ’3β’expβ‘(2β’Ο6β’n1/2)β’(1+oβ’(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])
#{Ξ»β’n:number of even parts inΞ»is odd}=#{Ξ»β’n:number of even parts inΞ»is even}-#{Ξ»β’n:Ξ»consists of distinct odd parts}.
Thus we have
#β’{CβAn:Cβ’is a conjugacy class ofβ’Sn}β©Ύ12β
Pβ’(n).
Second, choose Ξ΄>0 sufficiently small.
By Lemma 2.2, Theorem 2.4 and the above inequality, we obtain
Ο2β’(An)β©Ύ#β’{CβAn:Cβ’is a conjugacy class ofβ’Snβ’withβ’fβ’(C)β©½Ξ΄β
n}β©Ύ12β’Pβ’(n)-Pβ’(n-βΞ΄β’nβ).
Applying Lemma 2.6 yields our assertion.
β
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 Irrβ‘(Sn) the set of irreducible characters of Sn.
For a conjugacy class C of Sn and ΟβIrrβ‘(Sn), we write Οβ’(C) to denote Οβ’(Ο) for ΟβC.
Lemma 3.1.
Let C1 and C2 be conjugacy classes of Sn, and let ΟβSn.
Then
#β’{(x,y)βC1ΓC2:xβ’y=Ο}=|C1|β’|C2|n!β’βΟβIrrβ‘(Sn)Οβ’(C1)β’Οβ’(C2)β’Οβ’(Ο-1)Οβ’(1).
Proof.
See [6, Proposition 9.33] or [20, Theorem 6.3.1].
β
Recall that the irreducible characters of Sn are explicitly parametrized by the partitions of n.
Denote by ΟΞ» the irreducible character of Sn corresponding to the partition Ξ» of n.
We shall frequently apply the MurnaghanβNakayama rule.
Definition 3.2.
Let Ξ»β’n be a partition.
A rim hookh is an edgewise connected part of the Ferrers diagram of Ξ», 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 Ξ»\h.
An r-rim hook is a rim hook containing r boxes.
The leg length of a rim hook h is
lβ’lβ’(h):-(the number of rows ofβ’h)-1.
Let ΟβSn be a permutation with cycle type (1Ξ±1,β¦,qΞ±q,β¦,nΞ±n) and Ξ±qβ©Ύ1.
Denote by Ο\qβSn-q a permutation with cycle type
(1Ξ±1,β¦,qΞ±q-1,β¦,(n-q)Ξ±n-q).
Lemma 3.3 (MurnaghanβNakayama rule).
Let Ξ»β’n be a partition.
Suppose that ΟβSn is a permutation which contains a q-cycle.
Then we have
ΟΞ»β’(Ο)=βhqβ’-rim hookππβ’Ξ»(-1)lβ’lβ’(h)β’ΟΞ»\hβ’(Ο\q).
Proof.
See [23, Β§β9] or [28, Theorem 4.10.2].
β
The dimension ΟΞ»β’(1) of the irreducible representation associated with Ξ» can be computed via the hook formula.
Definition 3.4.
Let Ξ»β’n be a partition.
The hook of (i,j)βΞ» is
Hi,jβ’(Ξ»):-{(i,jβ²)βΞ»:jβ²β©Ύj}βͺ{(iβ²,j)βΞ»:iβ²β©Ύi}
Here we write (i,j)βΞ» to indicate that (i,j) is a box in the Ferrers diagram of Ξ».
Lemma 3.5 (hook formula).
Let Ξ»β’n be a partition.
Then
ΟΞ»β’(1)=n!β(i,j)βΞ»|Hi,jβ’(Ξ»)|.
Proof.
See [13, Theorem 1] or [28, Theorem 3.10.2].
β
Finally, we will use the following estimate, due to MΓΌller and Schlage-Puchta [22, Theorem 1].
Lemma 3.6.
Let ΟβIrrβ‘(Sn) be an irreducible character, let n be sufficiently large, and let ΟβSn be an element with k fixed points.
Then we have
|Οβ’(Ο)|β©½Οβ’(1)1-logβ‘(n/k)32β’logβ‘n.
4 Character estimates
Let ΟβAn be fixed.
In this section, we will count permutations ΟβAn satisfying that the commutator [Ο,Ο] 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 pβΞ β’(58β’n), and let ΟβAn 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 CβAn is a conjugacy class of Sn such that
fβ’(C)β©½Ξ΄β
nβπππβ’Ξ΄β(0,1).
Then, for all sufficiently large n, we have
#β’{ΟβAn:[Ο,Ο]βC}=|C|β’(1+πͺβ’(Ξ΄)).
Proof.
Due to the prime number theorem, there exists a prime pβΞ β’(58β’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.β16f.]).
Taking into account that Οβ’(Ο-1)=Οβ’(Ο)Β―=Οβ’(Ο), it follows with Lemma 3.1 that
(4.1)#β’{ΟβAn:[Ο,Ο]βC}=β(x,y)βKΟΓCxβ’y=Ο#β’{ΟβAn:Ο-1β’Οβ’Ο=x}=n!2β’|KΟ|β’#β’{(x,y)βKΟΓC:xβ’y=Ο}=|C|2β’βΟβIrrβ‘(Sn)Ο2β’(Ο)β
Οβ’(C)Οβ’(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)βΟβIrrβ‘(Sn)Οβ’(1)β 1Ο2β’(Ο)β
Οβ’(C)Οβ’(1)=πͺβ’(Ξ΄).
This completes our proof.
β
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 Ξ»β’n be a partition with ΟΞ»β’(Ο)β 0.
Then the Ferrers diagram of Ξ» has at most two boxes which do not belong to H1,1β’(Ξ»)βͺH2,2β’(Ξ»).
Moreover, let h be a p-rim hook of Ξ» with ΟΞ»\hβ’(Ο\p)β 0 or an (n-p-2)-rim hook of Ξ» with ΟΞ»\hβ’(Ο\(n-p-2))β 0.
Then the Ferrers diagram of Ξ»\h has at most two boxes which do not belong to H1,1β’(Ξ»\h).
In particular, we have |H1,1β’(Ξ»)|β©Ύp.
Proof.
The claim follows from the special cycle type of Ο and the MurnaghanβNakayama rule (Lemma 3.3).
β
Lemma 4.3.
Let p and Ο be chosen as in Proposition 4.1, and let Ξ»β’n be a partition.
Then we have |ΟΞ»β’(Ο)|β©½2.
Proof.
If ΟΞ»β’(Ο)=0, there is nothing to show.
So let us assume that ΟΞ»β’(Ο)β 0.
Let
R:-{h:hβ’is aβ’pβ’-rim hook ofβ’Ξ»β’withβ’ΟΞ»\hβ’(Ο\p)β 0}.
We have Rβ β
as a result of the MurnaghanβNakayama rule.
Due to
|H1,1β’(Ξ»)|β©Ύp>12β’n
(see Lemma 4.2), a rim hook hβR contains boxes of the hook H1,1β’(Ξ»).
Hence
|R|β©½2.
Now let h1βR.
Because of Lemma 4.2, there is exactly one (n-p-2)-rim hook h2 of Ξ»\h1.
Obviously, there exists exactly one 2-rim hook of (Ξ»\h1)\h2 and exactly one 1-rim hook of (Ξ»\h1)\h2.
The MurnaghanβNakayama rule yields that ΟΞ»β’(Ο) is equal to
{βh1pβ’-rim hookofβ’Ξ»βh2(n-p-2)β’-rim hookofβ’Ξ»\h1(-1)lβ’lβ’(h1)+lβ’lβ’(h2)ifβ’nβ‘0β’(2),βh1pβ’-rim hookofβ’Ξ»βh2(n-p-2)β’-rim hookofβ’Ξ»\h1βh32β’-rim hookofβ’(Ξ»\h1)\h2(-1)lβ’lβ’(h1)+lβ’lβ’(h2)+lβ’lβ’(h3)ifβ’nβ‘1β’(2).
Therefore, it follows that |ΟΞ»β’(Ο)|β©½2.
β
Definition 4.4.
Let Ξ»β’n.
Define
m1:-(number of boxes in the first column of the Ferrers diagram ofβ’Ξ»)-1,
m1β²:-(number of boxes in the first row of the Ferrers diagram ofβ’Ξ»)-1,
m2:-(number of boxes in the second column of the Ferrers diagram ofβ’Ξ»)-2,
m2β²:-(number of boxes in the second row of the Ferrers diagram ofβ’Ξ»)-2.
Lemma 4.5.
Let p and Ο be chosen as in Proposition 4.1.
Let Ξ»β’n be a partition such that the Ferrers diagram of Ξ» contains the box (2,2) and ΟΞ»β’(Ο)β 0.
Then we have
ΟΞ»β’(1)β©Ύ124β’n2β’(n|H1,1β’(Ξ»)|)β’(|H1,1β’(Ξ»)|-1m1),
ΟΞ»β’(1)β©Ύ124β’n2β’(n|H1,1β’(Ξ»)|)β’(|H1,1β’(Ξ»)|-1m1β²).
Furthermore, if |H1,1β’(Ξ»)|β©Ύp+3, then there exists an iβ{0,1,2} such that
m1=m2+|H1,1β’(Ξ»)|-p-iβππβm1β²=m2β²+|H1,1β’(Ξ»)|-p-i.
Proof.
Due to m1+m1β²=|H1,1β’(Ξ»)|-1 the first two formulas are equivalent.
In view of Lemma 4.2, we have the three cases |H3,3β’(Ξ»)|=0, |H3,3β’(Ξ»)|=1 and |H3,3β’(Ξ»)|=2.
In each case, the hook formula (Lemma 3.5) yields
ΟΞ»β’(1)β©Ύ124β’n2β’(n|H1,1β’(Ξ»)|)β’(|H1,1β’(Ξ»)|-1m1)β’(n-|H1,1β’(Ξ»)|-1m2).
Hence our first result follows.
We now turn to the second statement.
It follows from the MurnaghanβNakayama rule that there is an (n-p-2)-rim hook h of Ξ» such that ΟΞ»\hβ’(Ο\(n-p-2))β 0.
By assumption, we have |H1,1β’(Ξ»)|β©Ύp+3.
Therefore, h contains either boxes from the first column or the first row of the Ferrers diagram of Ξ» β but not both at the same time.
Without loss of generality, we assume that the first occurs.
With regard to Lemma 4.2, |H2,2β’(Ξ»\h)|β{0,1,2}.
If |H2,2β’(Ξ»\h)|=1, we obtain n-p-2=n-|H1,1β’(Ξ»)|-1+m1-m2 and hence m1=m2+|H1,1β’(Ξ»)|-p-1.
The other cases are analogous.
This proves our claim.
β
Lemma 4.6.
Let p and Ο be chosen as in Proposition 4.1.
Let Ξ»β’n, where Ξ»β{(n),(1n)}, be a partition such that ΟΞ»β’(Ο)β 0 and the Ferrers diagram of Ξ» does not contain the box (2,2).
If Ξ»β{(n-1,1),(2,1n-2)}, we obtain ΟΞ»β’(1)=n-1.
If Ξ»β{(p+1,1n-p-1),(n-p,1p)}, we have ΟΞ»β’(1)=(n-1p).
If Ξ»β{(n-p-1,1p+1),(p+2,1n-p-2)}, it follows ΟΞ»β’(1)=(n-1p+1).
The cases specified above are the only ones that can occur.
Proof.
The computations for ΟΞ»β’(1) follow immediately from the hook formula.
Furthermore, the MurnaghanβNakayama 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.
β
Lemma 4.7.
Let p and Ο be chosen as in Proposition 4.1.
Let Ξ»β’n be partition such that Ξ»β{(n),(1n),(n-1,1),(2,1n-2)} and ΟΞ»β’(Ο)β 0.
Then, for all sufficiently large n, we obtain
ΟΞ»β’(1)β©Ύ124β’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
ΟΞ»β’(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: |H1,1β’(Ξ»)|β©½78β’n.βBy applying Lemmas 4.5 and 4.2, we get
ΟΞ»β’(1)β©Ύ124β’n2β’(n|H1,1β’(Ξ»)|)β©Ύ124β’n2β’(nn-|H1,1β’(Ξ»)|)n-|H1,1β’(Ξ»)|β©Ύ124β’n2β
2n/8.
Case 2: |H1,1β’(Ξ»)|β©Ύ78β’n.βWithout loss of generality, there is an iβ{0,1,2} such that m1=m2+|H1,1β’(Ξ»)|-p-i (see Lemma 4.5).
Because of
0β©½m2β©½n-|H1,1β’(Ξ»)|-1,
we realize that
18β’nβ©½m1β©½12β’n.
If 18β’nβ©½m1β©½12β’(|H1,1β’(Ξ»)|-1), with Lemma 4.5, we obtain
ΟΞ»β’(1)β©Ύ124β’n2β’(|H1,1β’(Ξ»)|-1m1)β©Ύ124β’n2β’(|H1,1β’(Ξ»)|-1m1)m1β©Ύ124β’n2β
2n/8.
The case 12β’(|H1,1β’(Ξ»)|-1)β©½m1β©½12β’n is analogous.
Hence our result follows.
β
Lemma 4.8.
Let C be a conjugacy class of Sn.
We have Ο(n-1,1)β’(C)=fβ’(C)-1 and |Ο(n-1,1)β’(C)|=|Ο(2,1n-2)β’(C)|.
Proof.
See [28, Example 2.3.8] and [3, Lemma 3.6.10].
β
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
βΟβIrrβ‘(Sn)Οβ’(1)β 1Ο2β’(Ο)β
Οβ’(C)Οβ’(1)=πͺβ’(Ξ΄).
Proof.
Let
P1:-{Ξ»β’n:Ξ»β{(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
βΟβIrrβ‘(Sn)Οβ’(1)β 1Ο2β’(Ο)β
Οβ’(C)Οβ’(1)βͺβΞ»βP1ΟΞ»β’(Ο)β 0ΟΞ»β’(1)logβ‘Ξ΄32β’logβ‘n+βΞ»βP2|ΟΞ»β’(C)|ΟΞ»β’(1).
We consider separately the two sums.
Lemmas 4.2, 4.6 and 4.7 yield
βΞ»βP1ΟΞ»β’(Ο)β 0ΟΞ»β’(1)logβ‘Ξ΄32β’logβ‘nβͺn3β
(124β’n2β
2n/8)logβ‘Ξ΄32β’logβ‘nβͺexpβ‘(logβ‘Ξ΄400β’nlogβ‘n),
which is far smaller than necessary.
Due to Lemmas 4.6 and 4.8, we have
βΞ»βP2|ΟΞ»β’(C)|ΟΞ»β’(1)β©½2β’|fβ’(C)-1|n-1βͺΞ΄.
This completes the proof.
β
Below, we will state results similar to Proposition 4.1 for different cycle types of Ο.
Proposition 4.10.
Let pβΞ β’(58β’n), and let ΟβSn-1 be a fixed permutation with cycle type (p,n-p-2,1).
In addition, suppose that CβAn-1 is a conjugacy class of Sn-1 such that
fβ’(C)β©½Ξ΄β
(n-1)βπππβ’Ξ΄β(0,1).
Then, for all sufficiently large n, we have
#β’{ΟβSn-1:[Ο,Ο]βC}=|C|β’(2+πͺβ’(expβ‘(logβ‘Ξ΄400β’nlogβ‘n))).
The proof is similar to that of Proposition 4.1.
Proposition 4.11.
Let pβΞ β’(58β’n), and let ΟβSn-2 be a fixed permutation with cycle type (p,n-p-2).
In addition, suppose that CβAn-2 is a conjugacy class of Sn-2 such that
fβ’(C)β©½Ξ΄β
(n-2)βπππβ’Ξ΄β(0,23].
Then, for all sufficiently large n, we have
#β’{ΟβSn-2:[Ο,Ο]βC}=|C|β’(2+πͺβ’(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.
β
Proposition 4.12.
Let ΟβSm be a fixed permutation with cycle type
(m),(m-1,1),(m-2,2)βππβ(m-2,1,1).
In addition, suppose that CβAm is a conjugacy class of Sm such that
fβ’(C)β©½Ξ΄β
mβπππβ’Ξ΄β(0,23].
Then, for all sufficiently large m, we have
#β’{ΟβSm:[Ο,Ο]βC}=|C|β’(2+πͺβ’(Ξ΄1/33)).
Proof.
We sketch the argument for the case that our permutation ΟβSm has cycle type (m).
The other cases are similar.
Because of the cycle type of Ο, the MurnaghanβNakayama rule and the hook formula yield for Ξ»β’m with ΟΞ»β’(Ο)β 0 that
the Ferrers diagram of Ξ» does not contain the box (2,2),
Analogous to (4.1), we have
#β’{ΟβSm:[Ο,Ο]βC}=|C|β’βΟβIrrβ‘(Sm)Ο2β’(Ο)β
Οβ’(C)Οβ’(1)=|C|β’(2+βΟβIrrβ‘(Sm)Οβ’(1)β 1Ο2β’(Ο)β
Οβ’(C)Οβ’(1)).
We now use Lemma 3.6 and (i)β(iii) to estimate the error term.
|βΟβIrrβ‘(Sm)Οβ’(1)β 1Ο2β’(Ο)β
Οβ’(C)Οβ’(1)|β©½βm1=1m-2(m-1m1)logβ‘Ξ΄32β’logβ‘mβͺβ1β©½m1β©½12β’(m-1)(m-1m1)logβ‘Ξ΄32β’logβ‘mβ’m1.
We split the summation over m1 in two ranges and consider separately the subsums.
We obtain
β1β©½m1β©½m1/34(m-1m1)logβ‘Ξ΄32β’logβ‘mβ’m1β©½βm1=1βΞ΄m133=Ξ΄1/331-Ξ΄1/33βͺΞ΄1/33,
βm1/34β©½m1β©½12β’(m-1)(m-1m1)logβ‘Ξ΄32β’logβ‘mβ’m1β©½mβ
expβ‘(logβ‘Ξ΄β
logβ‘232β’m1/34logβ‘m),
which is far smaller than necessary.
β
5 Completion of the proof
Choose ΟβAn and CβAn as in Theorem 2.4.
We shall show that counting permutations ΟβAn such that [Ο,Ο]βC and γΟ,Ογ=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 ΟβG containing a p-cycle with pβΞ β’(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,β¦,Yk for G such that Y1,β¦,Yk form a partition of {1,β¦,n}, 2β©½|Y1|β©½n-1 and |Y1|=|Yi| for all i=2,β¦,k.
Let a be an element contained in the p-cycle of Ο.
Then we have aβYl0, Οβ’(a)βYl1,β¦,Οp-1β’(a)βYlp-1 for pairwise distinct liβ{1,β¦,k}.
Hence we get kβ©Ύp.
This yields the contradiction n=|Y1|β’kβ©Ύ2β’p>n.
Second, let
d:-lcmβ‘{m:Οβ’contains a cycle of lengthβ’m<p}.
Obviously, ΟdβG is a p-cycle.
Since G is primitive, we can conclude with Lemma 5.1 that G=Sn or G=An.
β
The last lemma immediately yields the following corollary.
Corollary 5.3.
Let pβΞ β’(n-3).
Suppose that ΟβAn contains a p-cycle, and let CβAn be a conjugacy class of Sn.
Then we get
#β’{ΟβAn:[Ο,Ο]βCβ’πππβ’γΟ,Ογ=An}=#β’{ΟβAn:[Ο,Ο]βC}-#β’{ΟβAn:[Ο,Ο]βCβ’πππβ’(γΟ,Ογβ’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 γΟ,Ογis not transitive.
Definition 5.4.
For a cycle Ο=(i1,β¦,im), define
Mβ’(Ο):-{i1,β¦,im}.
Let Ο1β
β¦β
Οk be the cycle decomposition for ΟβSn (including 1-cycles), and set Ξ©:-{1,2,β¦,n}.
Define ΞH:-βiβHMβ’(Οi)βΞ© when Hβ{1,β¦,k}.
For a non-empty set Ξ©, let Symβ’(Ξ©) denote the symmetric group on Ξ©.
When ΞβΞ©, we identify Symβ’(Ξ) with the subgroup of Symβ’(Ξ©) consisting of all
ΟβSymβ’(Ξ©)βwithβ’suppβ‘(Ο)βΞ.
Lemma 5.5.
Let Ο1β
β¦β
Οk be the cycle decomposition for ΟβSn (including 1-cycles), and let ΟβSn.
Then γΟ,Ογ is not transitive if and only if there exists a non-empty proper subset H of K:-{1,β¦,k} and a tuple
(Οβ²,Οβ²β²)βSymβ’(ΞH)ΓSymβ’(ΞK\H)
such that Ο=Οβ²β’Οβ²β².
Proof.
βββ. This implication is obvious.
βββ.
This implication is the result of the following stronger statement.
Claim.
Let Ο1β
β¦β
Οl be the cycle decomposition for Ο (including 1-cycles).
Let s,tβ{1,β¦,n} such that Οβ’(s)β t for all ΟβγΟ,Ογ.
For all m with 1β©½mβ©½l-1, there exist non-empty subsets Hm,Hmβ²βK and a subset LmβL:-{1,β¦,l} such that
for each jβLm, we have
Mβ’(Οj)ββiβHmMβ’(Οi)βππβMβ’(Οj)ββiβHmβ²Mβ’(Οi),
Lmβ L implies that, for all iβHm, there exists
ΟβγΟ,Ογβsuch thatβΟβ’(s)βMβ’(Οi),
Lmβ L implies that, for all iβHmβ², there exists
ΟβγΟ,Ογβsuch thatβΟβ’(t)βMβ’(Οi).
We prove this claim by induction.
If m=1, we have sβMβ’(Οj) and tβMβ’(Οjβ²) for jβ jβ².
Set L1:-{j,jβ²} and
H1:-{iβK:Mβ’(Οj)β©Mβ’(Οi)β β
},
H1β²:-{iβK:Mβ’(Οjβ²)β©Mβ’(Οi)β β
}.
Then (a)β(e) follow.
Now let 1β©½mβ©½l-2, and suppose that the statement is true for m.
The case Lm=L is trivial.
Hence let us assume that LmβL.
Case 1: there exists j0βLβLm such that Mβ’(Οj0)β©βiβHmMβ’(Οi)β β
or Mβ’(Οj0)β©βiβHmβ²Mβ’(Οi)β β
.βWithout loss of generality, assume that
Mβ’(Οj0)β©βiβHmMβ’(Οi)β β
.
Define Lm+1:-Lmβͺ{j0},
Hm+1:-Hmβͺ{iβK:Mβ’(Οj0)β©Mβ’(Οi)}β β
and
Hm+1β²:-Hmβ².
Then (a)β(e) result from the induction hypothesis and some easy computations.
Case 2: Mβ’(Οj)β©βiβHmMβ’(Οi)=β
and Mβ’(Οj)β©βiβHmβ²Mβ’(Οi)=β
for all jβLβLm.βSet Lm+1:-L, Hm+1:-KβHmβ² and Hm+1:-Hmβ².
Obviously, (a)β(e) are fulfilled.
β
Definition 5.6.
Denote by [Sn] the set of conjugacy classes of Sn.
Let Cβ[Sn], C1β[Sk] and C2β[Sn-k].
Denote by ml, sl and tl the number of l-cycles in C, C1 and C2, respectively.
We say that C=C1β’C2 if and only if ml=sl+tl for all l=1,β¦,n.
Finally, we draw our attention to β|C1|β’|C2|, where the sum runs over all (C1,C2)β[Sk]Γ[Sn-k] satisfying C1β’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,β¦,nmn), where ml is the number of l-cycles in C.
Then we obtain
|C|=n!1m1β’m1!β
2m2β’m2!β
β¦β
nmnβ’mn!.
Proof.
See [28, Proposition 1.1.1].
β
We now observe the following.
Lemma 5.8.
Suppose that C is a conjugacy class of Sn such that fβ’(C)β©½Ξ΄β
n for Ξ΄β(0,14], and let n be sufficiently large.
Then, for 12β’n<kβ©½58β’n, we have
βC1β[Sk]C2β[Sn-k]C=C1β’C2|C1|β’|C2|β©½expβ‘(-350β’n)β’|C|.
Furthermore, we obtain
βC1β[Sn-1]C2β[S1]C=C1β’C2|C1|β©½Ξ΄β’|C|βπππββC1β[Sn-2]C2β[S2]C=C1β’C2|C1|β©½2β’Ξ΄2β’|C|.
Proof.
Denote by (1m1,2m2,β¦,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,β¦,kmk,(k+1)0,β¦,n0).
By computing the cardinality of C and of the conjugacy classes C1β[Sk] and C2β[Sn-k] with C=C1β’C2 via Lemma 5.7, we obtain
βC1β[Sk]C2β[Sn-k]C=C1β’C2|C1|β’|C2|β©½|C|β’(nk)-1β’β(i1,β¦,in-k)0β©½ijβ©½mjβl=1n-k(mlil)=|C|β’(nk)-1β’2(m1+β―+mn-k).
The Stirling formula yields the inequality (nk)β©Ύ(4125)n for 12β’n<kβ©½58β’n.
In addition, we have
βi=1nmiβ©½fβ’(C)+n-fβ’(C)2β©½12β’nβ’(1+Ξ΄)β©½58β’n.
Hence our first estimate follows.
We now show the second statement.
Let iβ{1,2}.
We again apply Lemma 5.7.
By calculating the cardinality of C and of conjugacy classes C1β[Sn-i] such that there exists C2β[Si] with C=C1β’C2, we get
βC1β[Sn-1]C2β[S1]C=C1β’C2|C1|β©½|C|β’m1nβ©½Ξ΄β’|C|,
βC1β[Sn-2]C2β[S2]C=C1β’C2|C1|β©½|C|β’1nβ’(n-1)β’(m1β’(m1-1)+2β’m2)β©½|C|β’(Ξ΄2+1n-1)β©½2β’Ξ΄2β’|C|.
This proves our claim.
β
We are now in the position to give the proof of our second theorem.
Proof of Theorem 2.4.
Let Ο1β
β¦β
Οk be the cycle decomposition of Ο (including 1-cycles), and set K:-{1,β¦,k}.
Then Lemma 5.5 yields
B:-#β’{ΟβAn:[Ο,Ο]βCβ’andβ’(γΟ,Ογβ’is not transitive)}β©½β{H,K\H}HβK,Hβ β
#β’{(Οβ²,Οβ²β²)βSymβ’(ΞH)ΓSymβ’(ΞK\H):[Ο,Οβ²β’Οβ²β²]βC}.
Let H be a proper non-empty subset of K.
Define
Οβ²:-βiβHΟiβandβΟβ²β²:-βiβK\HΟi.
If (Οβ²,Οβ²β²)βSymβ’(ΞH)ΓSymβ’(ΞK\H), we get
[Ο,Οβ²β’Οβ²β²]=[Οβ²,Οβ²]β
[Οβ²β²,Οβ²β²].
Obviously, we have [Οβ²,Οβ²]βSymβ’(ΞH) and [Οβ²β²,Οβ²β²]βSymβ’(ΞK\H).
Therefore, [Ο,Οβ²β’Οβ²β²]βC holds if and only if there exist conjugacy classes C1βSymβ’(ΞH) and C2βSymβ’(ΞK\H) with [Οβ²,Οβ²]βC1, [Οβ²β²,Οβ²β²]βC2 and C=C1β’C2.
Case 1: nβ‘1β’(2).βBy assumption,
k=3 and K={1,2,3}.
For Ξ»β’m, let ΟΞ»βSm be a fixed permutation with cycle type Ξ».
The above considerations yield
Bβ©½βC1β[Sn-2]C2β[S2]C=C1β’C2#β’{ΟβSn-2:[Ο(p,n-p-2),Ο]βC1}
β
#β’{ΟβS2:[Ο(2),Ο]βC2}
ββ+βC1β[Sp]C2β[Sn-p]C=C1β’C2#β’{ΟβSp:[Ο(p),Ο]βC1}
β
#β’{ΟβSn-p:[Ο(n-p-2,2),Ο]βC2}
ββ+βC1β[Sp+2]C2β[Sn-p-2]C=C1β’C2#β’{ΟβSp+2:[Ο(p,2),Ο]βC1}
β
#β’{ΟβSn-p-2:[Ο(n-p-2),Ο]βC2}
=πͺβ’(βC1β[Sn-2]C2β[S2]C=C1β’C2|C1|+βC1β[Sp]C2β[Sn-p]C=C1β’C2|C1|β’|C2|+βC1β[Sp+2]C2β[Sn-p-2]C=C1β’C2|C1|β’|C2|)
=πͺβ’(Ξ΄2β’|C|).
For iβ{0,1,2}, we have 12β’n<p+iβ©½58β’n.
In addition, fβ’(C1)β©½fβ’(C) is valid for C=C1β’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: nβ‘0β’(2).βWith an argument similar to the first case, we obtain
B=πͺβ’(Ξ΄β’|C|).
Hence it follows from Corollary 5.3 and Proposition 4.1 that
#β’{ΟβAn:[Ο,Ο]βCβ’andβ’γΟ,Ογ=An}=|C|β’(1+πͺβ’(Ξ΄)).
So we are done.
β
