1 Introduction
Let G be a finite group. The generating graph Γ(G) of G is the graph on the non-identity elements of G where two vertices are connected by an edge if and only if they generate G.
The generating graph has been quite extensively studied for a variety of groups with respect to various graph properties. For example, it is known that if G is a symmetric or alternating group of degree at least 5, then Γ(G) is connected with diameter at most 2. This is a consequence of a result of Binder [2] for G=Sn and of Brenner and Wiegold [4] for G=An. (Indeed the same result holds for any finite simple group by a result of Breuer, Guralnick and Kantor [6].)
A well-known conjecture of Breuer, Guralnick, Lucchini, Maróti and Nagy [7] states that for a finite group G with |G|>3, the graph Γ(G) has a Hamiltonian cycle if and only if every proper quotient of G is cyclic. They established the conjecture for certain types of groups. In particular, they showed that the graph Γ(G) is Hamiltonian for G an alternating or symmetric group of sufficiently large degree. However, their proof uses asymptotic results and contains no information at all as to how large the degree should be. In this paper we obtain explicit lower bounds on the degrees of these groups. More precisely, we prove the following.
Theorem 1.1.
The graphs Γ(Sn) and Γ(An) are Hamiltonian for n≥107.
In fact, we prove a stronger result. More precisely, we prove, for n≥107, that the third closure of the graph Γ(Sn) satisfies Chvátal’s criterion and the graph Γ(An) satisfies Pósa’s criterion (see Sections 6 and 7 for the definition of the closure of a graph and for Chvátal’s and Pósa’s criteria).
We note that the Hamiltonicity of the generating graphs of the symmetric and alternating groups of degrees n with 5≤n≤13 has already been established by computer calculations (see [5, Section 4.4]).
The proof of Theorem 1.1 essentially relies on results concerning vertex degrees in the generating graphs of Sn and An. Since, by the very definition of the generating graph, this is related to 2-element subsets of the group that generate the whole group, we are mainly interested in the problem of generating symmetric and alternating groups by their 2-element subsets. In 1969, Dixon [10] proved that the probability that a random pair of permutations from the symmetric group Sn generates the alternating group An or Sn tends to 1 as n tends to infinity. As a generalization of Dixon’s result, Babai and Hayes [1, Theorem 1] proved the following. Let G be a permutation group of degree n. Assume that the number of points fixed by every element of G is o(n) (that is, a function f such that f(n)/n tends to 0 as n tends to infinity). Then the probability that G and a random permutation in Sn generate either An or Sn tends to 1 as n→∞. (This extends previous work of Łuczak and Pyber [14]
who proved that the probability that an element of Sn is contained in a proper transitive subgroup other than An tends to 0 as n→∞. There are many asymptotic results in this direction, the most recent of these being [9, 11, 12].) We note that the result of Babai and Hayes is a crucial component to the proof of the above mentioned result of Breuer et al. In this paper we obtain an explicit version of the result of Babai and Hayes in the case G is cyclic (see Theorem 2.3), which is used as a key tool in the proof of Theorem 1.1.
The paper is structured as follows. Put G=Sn and let g∈G be a permutation with at most n fixed points. In Section 2 we introduce the basic setup and establish a close connection between the degree of the vertex g of the graph Γ(G) and the maximal subgroups of G which contain g. In Sections 3, 4 and 5 we investigate three cases where such a maximal subgroup is either primitive, intransitive or transitive imprimitive. Together these provide a strong lower bound on the degree of the vertex g (see Theorem 2.3). Then in Sections 6 and 7 we apply certain graph theoretic criteria in conjunction with this bound to prove Theorem 1.1. We emphasize that the overall strategy of the proof of Theorem 1.1 is the same as in [7, Sections 5 and 6]. Finally, we note that the bound of 107 in Theorem 1.1 is the best possible in view of the methods and results in Sections 3–7.
2 Basic observations
Let G=Sn and let Γ=Γ(G). Let g∈G∖{1}. Denote by d(g,Γ) the degree of the vertex g of Γ. The vertex g is adjacent to another vertex h in the graph Γ if and only if 〈g,h〉=G. Note that this condition is equivalent to h not being contained in any maximal subgroup of G that contains g. Therefore, we have
d(g,Γ)=|{h∈G∖{1}:〈g,h〉=G}|=|G∖⋃M∈ℳ(g)M|,
where ℳ(g) is the set of maximal subgroups of G which contain g.
For the purposes of this paper, we need lower bounds on vertex degrees in Γ(G). Thus we are mainly interested in bounding from above the probability that a random permutation in G is contained in a maximal subgroup that contains g, that is, bounding from above the ratio
(2.1)|⋃M∈ℳ(g)M||G|,
where ℳ(g) is the set of maximal subgroups of G which contain g.
The following result indicates that most permutations have a relatively small number of fixed points.
Proposition 2.1.
Let k be a positive integer. Then the proportion of permutations in Sn with at least k fixed points is at most 1/k!.
The next proposition implies that if g∈Sn has f fixed points, then the probability that a random permutation and g generate Sn is at most 1-f/(2n). This means that if g has a relatively large number of fixed points, then we cannot expect a strong lower bound on this probability and hence on the degree of the vertex g of Γ(Sn).
Proposition 2.2 ([1, Proposition 3]).
Let H≤Sn be a permutation group having f fixed points. Then the probability that the group generated by H and a random permutation has a fixed point is at least f/(2n).
In this paper we will not make an explicit reference to Proposition 2.2, however, we include it for interest and context as well as for a justification for imposing the bound of n on the number of fixed points of g in Sections 3–5.
In the light of Propositions 2.1 and 2.2, in Sections 3–5 we assume that g has at most n fixed points and study the ratio in (2.1) in the cases where M is primitive, intransitive and transitive imprimitive (see Theorems 3.7, 4.5 and 5.4). Together these prove:
Theorem 2.3.
For n≥106, if g∈Sn has at most n fixed points, then the probability that a random permutation and g do not generate An or Sn is at most
1n+16.2n+1n2.
In Sections 6 and 7 the bound in Theorem 2.3 will be used together with a corresponding but weaker bound for elements with more than n fixed points (see Theorems 6.6 and 7.4) to establish the Hamiltonicity of Γ(An) and Γ(Sn) for n≥107.
3 The primitive case
Let G=Sn. In this section we consider the probability that a random permutation in G is contained in a primitive maximal subgroup of G containing g. Equivalently, we consider the ratio in (2.1) in the case M is primitive. For our purposes, we also restrict ourselves to the case where g has at most n fixed points.
First, we obtain an upper bound on the order of the centralizer of g in G. We first consider the case where g is fixed-point-free.
Lemma 3.1.
Let g∈Sn be a fixed-point-free permutation. Then we have
|CSn(g)|≤2⌊n/2⌋⋅⌊n2⌋!
for n≠3.
Proof.
Let g have cycle type (k1m1,k2m2,…,krmr). Write g=g1g2⋯gr where the gi are the products of the mi disjoint ki-cycles in the cycle decomposition of g. Let Ω be the permutation domain and let Ωi⊆Ω be the set of elements that appear in a cycle of gi. Put G=Sn and Gi=Sym(Ωi) for all i=1,2,…,r. Then viewing Gi as a subgroup of G and gi as an element of Gi, we have
CG(g)=∏i=1rCGi(gi).
Note that |Ωi|=miki and n=|Ω|=∑i=1r|Ωi|=∑i=1rmiki. It is sufficient to show that
|CGi(gi)|≤2⌊miki2⌋⋅⌊miki2⌋!
since then it will follow that
|CG(g)|≤∏i=1r(2⌊miki2⌋⋅⌊miki2⌋!)≤2⌊n2⌋⋅⌊n2⌋!.
Thus we may assume that g is a product of b disjoint cycles of length a. Raising g to a suitable power, we may further assume that g is of prime order, i.e., a is a prime. Now we have n=ab and |CG(g)|=abb!. Thus we need to show
(3.1)abb!≤2⌊ab2⌋⋅⌊ab2⌋!.
Note that ab≤2⌊a/2⌋b≤2⌊ab/2⌋ for a≠3,5, and b!≤⌊ab/2⌋!. Therefore (3.1) is satisfied for a≠3,5. We now treat the remaining cases.
a=3: If b=2k for some k≥1, then we have
abb!=32k(2k)!≤23k2k(2k)!≤23k(3k)!=2⌊ab/2⌋⌊ab2⌋!.
If b=2k+1 for some k≥1, then we have
abb!=32k+1(2k+1)!≤24k+1(k+1)k(2k+1)!
=23k+1(2k+2)k(2k+1)!
≤23k+1(3k+1)!
=2⌊ab/2⌋⌊ab2⌋!.
a=5: If b=1, then (3.1) is trivially satisfied. Suppose then that b>1. We have
2⌊ab/2⌋⌊ab2⌋!=2⌊5b/2⌋⌊5b2⌋!
≥22b(2b)!≥22bbbb!=(4b)bb!≥5bb!=abb!.
The proof is complete.
∎
We need the notion of projection of permutations onto a subset of the permutation domain (see [1, Section 4]).
Definition 3.2.
Let T⊆Ω be a subset. We define the projection map
prT:Sym(Ω)→Sym(T)
by setting prT(σ)=σT for σ∈Sym(Ω), where σT is defined as follows. For i∈T, let
iσT=iσk where k is the smallest positive integer with
iσk∈T.
We next consider the general case where g has at most n fixed points.
Lemma 3.3.
Let g∈Sn be a permutation with at most n fixed points. Then we have |CSn(g)|≤2⌊n/2⌋⋅⌊n/2⌋! for n≠3.
Proof.
Put G=Sn and let Ω be the permutation domain. Also let Ω1=fixΩ(g), Ω2=Ω∖Ω1, and let G1=Sym(Ω1), G2=Sym(Ω2). Denote by f the number of fixed points of g, that is, f=|Ω1|. Using the notation in Definition 3.2 and viewing the subgroups G1 and G2 as subgroups of G, we have
CG(g)≅CG1(gΩ1)×CG2(gΩ2).
Note that gΩ1 is the identity permutation on Ω1 and so |CG1(gΩ1)|=f!. Note also that gΩ2 is fixed-point-free on Ω2, and so by Lemma 3.1, we have
|CG2(gΩ2)|≤2⌊(n-f)/2⌋⋅⌊n-f2⌋!.
Thus it follows that
|CG(g)|=|CG1(gΩ1)|⋅|CG2(gΩ2)|≤f!⋅2⌊(n-f)/2⌋⋅⌊n-f2⌋!.
Set
cr=r!⋅2⌊(n-r)/2⌋⋅⌊n-r2⌋!.
We claim that cr≥cr+2 for any non-negative r≤n-2. By cancelling common factors, we see that this holds if and only if 2⌊(n-r)/2⌋≥(r+2)(r+1). Clearly, ⌊(n-r)/2⌋≥(n-r-1)/2, and hence it is enough to show that n-r-1≥(r+2)(r+1). This is true for n-2≥r. Therefore the claim is established. Also, it is trivial to see that c0≥c1, and so it follows that c0≥cr for any non-negative integer r≤n. The result now follows.
∎
We need upper bounds on the numbers of conjugacy classes of primitive maximal subgroups of Sn and of An.
Theorem 3.4 ([16, Lemma 4.1]).
Let G be one of Sn or An and let r(G) be the number of conjugacy classes of primitive maximal subgroups of G other than An. Then the following hold:
r(G)≤n3(log2n)2 for n≥1000.
Also, we need an upper bound on the orders of primitive permutation groups.
Theorem 3.5 ([15, Corollary 1.1]).
Let G<Sn be a primitive permutation group not containing An. Then |G|<50nn.
We also need a formula for the fixed point ratio of a group element in a transitive action in terms of the conjugacy class of the group element.
Lemma 3.6.
Let G be a group acting transitively on a finite set Ω. Let H be the stabilizer of a point in Ω. Then for all g∈G we have
|fixΩ(g)||Ω|=|gG∩H||gG|.
We are now in a position to prove our main result in this section.
Theorem 3.7.
Let g∈Sn be a permutation with at most n fixed points. Let p be the probability that a random permutation in Sn is contained in a primitive maximal subgroup of Sn that contains g and does not contain An. Then we have p≤1/n2 for n≥27.
Proof.
Put G=Sn. Let Σ be the set of primitive maximal subgroups of G other than An. Consider the conjugation action of G on Σ. Note that H∈fixΣ(g) if and only if g∈NG(H)=H. Let 𝒪1,…,𝒪k be the orbits of this action. Pick Hi∈𝒪i for each i. We have
p=|⋃H∈fixΣ(g)H||G|≤∑i=1k|⋃H∈fix𝒪i(g)H||G|
≤∑i=1k|fix𝒪i(g)||Hi||G|=∑i=1k|fix𝒪i(g)||𝒪i|,
where the last equality follows since
|𝒪i|=|G:StabG(Hi)|=|G:NG(Hi)|=|G:Hi|.
Thus it is enough to show that
|fix𝒪i(g)||𝒪i|≤1kn2 for all i=1,…,k.
Now fix an arbitrary orbit 𝒪 and fix H∈𝒪. Note that the action of G is transitive on 𝒪, and the subgroup H is a point stabilizer, namely
StabG(H)=NG(H)=H. Thus by Lemma 3.6, we have
|fix𝒪(g)||𝒪|=|gG∩H||gG|.
Note also that |gG|=|G:CG(g)|. Thus we have
|fix𝒪(g)||𝒪|=|gG∩H|⋅|CG(g)||G|≤|H|⋅|CG(g)||G|≤50nn⋅2⌊n/2⌋⋅(⌊n/2⌋)!n!,
where the last inequality follows from Lemma 3.3 and Theorem 3.5. Thus it suffices to show that
50nn⋅2⌊n/2⌋⋅(⌊n/2⌋)!n!≤1kn2
provided that n≥27.
By Stirling’s bound, we have
2πn⋅(ne)n≤n! and ⌊n2⌋!≤e⋅n/2⋅(n2e)n/2.
Thus it is enough to show that
50nn⋅2n/2⋅e⋅n/2⋅(n/(2e))n/22πn⋅(n/e)n≤1kn2,
or equivalently,
(3.2)eπ⋅25nn+2⋅k≤(ne)n/2.
By Theorem 3.4 we have k≤36 for n<1000, and k≤n3(log2n)2 for n≥1000. Using these, we see that (3.2) holds for n≥27. This completes the proof.
∎
4 The intransitive case
Let G=Sn. In this section we consider the probability that a random permutation in G is contained in an intransitive maximal subgroup of G containing g. Equivalently, we consider the ratio in (2.1) in the case M is intransitive. For our purposes, we also restrict ourselves to the case where g has at most n fixed points.
In fact, in this section, we consider the more general situation where g is replaced by an arbitrary subgroup H with at most n fixed points. We obtain our desired result by setting H=〈g〉.
We first obtain a result in the case where H is fixed-point-free.
Lemma 4.1.
Let H≤Sn be a fixed-point-free permutation group. Let p be the probability that H and a random permutation generate an intransitive group. Then we have p≤11.9n for n≥83.
Proof.
Put G=Sn and let Ω be the permutation domain. Clearly we may assume that H is intransitive. Let 𝒜 be the set of H-invariant subsets of Ω of size at most n/2. Let g∈G. Note that the group 〈g,H〉 is intransitive if and only if g∈StabG({T}) for some T∈𝒜. We may therefore assume that each H-orbit has size 2 or 3. Suppose then that there are k orbits of size 2 and l orbits of size 3 of H so that n=2k+3l. Now,
p=|⋃T∈𝒜StabG({T})||G|
≤∑T∈𝒜|StabG({T})||G|=∑T∈𝒜|T|!⋅(n-|T|)!n!
(4.1)=∑T∈𝒜1(n|T|)=∑t=2⌊n/2⌋∑0≤r≤k0≤s≤lt=2r+3s(kr)(ls)(nt).
Note that k+l≤n/2, and we have r+s≤t/2 and t≤n/2 in (4.1). Thus, we have
(kr)(ls)≤(k+lr+s)≤(⌊n/2⌋⌊t/2⌋)=(⌊n/2⌋⌊t/2⌋)2≤(nt).
Also, it is easy to see that the number of pairs (r,s) of non-negative integers with t=2r+3s is at most t/6+1≤n/12+1. We now have
p≤∑t=2⌊n/2⌋∑0≤r≤k0≤s≤lt=2r+3s1(nt)
≤∑t=251(nt)+∑t=6⌊n/2⌋(n12+1)⋅1(nt)
≤∑t=251(nt)t+(n2-5)⋅(n12+1)⋅1(n6)6
=2n+(3n)3/2+(4n)2+(5n)5/2+9(n2+2n-120)n3≤11.9n,
where the last inequality holds since n≥83.
∎
As in Section 3, we use projections on to subsets in order to deal with the general case. We require the following two observations from [1].
Lemma 4.2 ([1, Observation 11]).
Let T⊆Ω. Then the projection map
prT is uniform, that is,
|prT-1(g)|=|Ω|!/|T|! for all g∈Sym(T).
Lemma 4.3 ([1, Observation 12]).
Let g∈Sym(Ω) and let T⊆Ω. Let G≤Sym(T) where Sym(T) is considered as a subgroup of Sym(Ω). Then the orbits of the
subgroup of Sym(T) generated by G and gT are exactly the intersection
of T with those orbits of the subgroup of Sym(Ω) generated by G and g
which have non-empty intersection with T.
We are now ready to handle the general case.
Lemma 4.4.
Let H≤Sn be a permutation group with at most n fixed points. Let p be the probability that H and a random permutation generate an intransitive group.
Then we have p≤1n+15.2n for n≥104.
Proof.
Put G=Sn and let Ω be the permutation domain. Also, set F=fixΩ(H) and R=Ω∖F. Let g∈G. Note that the group 〈g,H〉 is intransitive if and only if not all elements of R are in the same 〈g,H〉-orbit or there is a g-invariant subset of F, i.e., g∈StabG({A}) for some A⊆F. Let p1 be the probability that there is a g-invariant subset of F, and let p2 be the probability that not all elements of R are in the same 〈g,H〉-orbit. By the union bound, we have p≤p1+p2. Let f=|F| so that f≤n. Now,
p1=|⋃A⊆FStabG({A})||G|≤∑A⊆F|StabG({A})||G|
=∑i=1f∑A⊆F|A|=i|StabG({A})||G|=∑i=1f(fi)⋅i!(n-i)!n!
=∑i=1ff(f-1)⋯(f-i+1)n(n-1)⋯(n-i+1)≤∑i=1f(fn)i
≤fn+(fn)2+(f-2)(fn)3
≤1n+1n+n-2n3/2≤1n+2n.
Now set HR={hR : h∈H}, where hR is as in Definition 3.2. Then,
p2=|{g∈G:not all elements of R are in the same 〈g,H〉-orbit}||G|
=|{g∈G:〈gR,HR〉 is intransitive on R}||G|
=(|Ω|!/|R|!)⋅|{g∈Sym(R):〈g,HR〉 is intransitive on R}||Ω|!
=|{g∈Sym(R):〈g,HR〉 is intransitive on R}||R|!
≤11.9n-f≤11.9n-n,
where the second equality follows from Lemma 4.3, the third equality follows from Lemma 4.2, and the first inequality follows from Lemma 4.1 since HR is fixed-point-free on R and |R|=n-f≥n-n≥83. Thus we have
p≤p1+p2≤1n+2n+11.9n-n≤1n+15.2n,
where the last inequality holds since n≥104.
∎
We now state and prove the main results.
Theorem 4.5.
Let g∈Sn be a permutation with at most n fixed points. Let p be the probability that a random permutation is contained in an intransitive maximal subgroup of Sn containing g. Then we have p≤1n+15.2n for n≥104.
Proof.
If a permutation h∈Sn is contained in an intransitive maximal subgroup of Sn containing g, then the group generated by g and h is intransitive. The result now follows from Lemma 4.4 by setting H=〈g〉.
∎
5 The transitive imprimitive case
Let G=Sn. In this section we consider the probability that a random permutation in G is contained in a transitive imprimitive maximal subgroup of G containing g. Equivalently, we consider the ratio in (2.1) in the case M is transitive imprimitive. For our purposes, we also restrict ourselves to the case where g has at most n fixed points.
Note that the transitive imprimitive maximal subgroups of Sn containing g correspond to g-invariant partitions of the permutation domain into subsets of equal size greater than 1. Thus we are really interested in bounding the proportion of such g-invariant partitions.
We first state and prove our result in the case where the blocks have size at least 3 and then consider the remaining case in another lemma.
Lemma 5.1.
Let g∈Sn be a permutation with at most n fixed points. Let S be the set of partitions of the permutation domain into b blocks of size a, where a>2 and b>1. Let p be the proportion of g-invariant partitions in S.
Then we have p≤1/(2n3/2) for n≥57.
Proof.
Let Ω be the permutation domain and let c1,…,cr be a subset of Ω containing exactly one element from each cycle of g. Denote by f the number of fixed points of g so that f≤n. Consider an arbitrary but fixed partition in 𝒮 and label its blocks from 1 to b. We call such a partition a labelled partition. It is easy to see that a g-invariant labelled partition is uniquely determined by the placement of elements c1,…,cr in its blocks and the induced action of g on the set of its blocks. The number of ways of distributing the elements c1,…,cr into the blocks is at most br. Therefore, the number of g-invariant labelled partitions is at most brb!. Since there are b! ways of labelling the blocks of a given partition in 𝒮, it follows that the number of g-invariant partitions in 𝒮 is at most br. Note also that |𝒮|=n!/(a!bb!). Using Stirling’s bound, we have
p≤brn!/((a!)b⋅b!)≤br⋅(a!)b⋅b!n!
≤br⋅(ea(a/e)a)b⋅eb(b/e)b2πn(n/e)n
=e⋅ab/22πn⋅bn-r-b-1/2.
Thus it suffices to show that
(5.1)e⋅ab/22πn⋅bn-r-b-1/2≤12n3/2
for n≥57. Note that n=ab and so (5.1) holds if and only if
(5.2)e2π⋅nb/2+1≤bn-r-b/2-1/2.
Taking logarithms of both sides of (5.2),
we see that (5.1) holds if and only if
1+ln(2π)+(b2+1)ln(n)+b2ln(b)≤(n-r-12)ln(b).
Note that r, the number of cycles of g, is maximal when all cycles of g are of lengths 1 or 2. Thus it follows that r≤f+(n-f)/2≤(n+n)/2. Therefore, it is enough to show that
(5.3)1+ln(2/π)ln(b)+(b2+1)⋅ln(n)ln(b)+b2≤n-n-12.
Inequality (5.3) holds for b=2, n≥31, and also for b=3, n≥26. Also holding n fixed, the left-hand side of (5.3) is an increasing function of b on [4,∞]. Note that b≤n/3 by assumption, and hence for b≥4 inequality (5.3) is implied by
1+ln(2/π)ln(n/3)+(n6+1)⋅ln(n)ln(n/3)+n6≤n-n-12,
which holds if and only if n≥57. This completes the proof.
∎
To prove the general case, we need the following lemma.
Lemma 5.2.
Let g∈Sym(Σ) be a fixed-point-free permutation all of whose cycles are of the same size. Denote by n(g,Σ) the number of g-invariant partitions of Σ into blocks of size 2. Then n(g,Σ) is maximal when the cycles of g are of size 2.
Proof.
Let x∈Sym(Σ) be a fixed-point-free involution and let h∈Sym(Σ) be a fixed-point-free permutation all of whose cycles are of the same size. We need to show that n(h,Σ)≤n(x,Σ). We may assume by raising h to a suitable power that h is fixed-point-free of order a prime p. If p=2, there is nothing to prove. If p>2, then we have |Σ|=2pk for some integer k (note that we may assume Σ has even size since otherwise there is nothing to prove). This means that there are pk cycles of x, each of length 2. Suppose first that k is even. The number of x-invariant partitions of Σ into blocks of size 2 where no block is fixed is equal to
(pk)!(2!)pk/2⋅(pk/2)!⋅2pk/2=(pk)!(pk/2)!.
This follows since each such partition can be obtained by breaking the pk cycles into groups each consisting of two cycles and for each group forming a partition that is left invariant by these cycles. In particular, we have
n(x,Σ)≥(pk)!(pk/2)!.
On the other hand, h is a permutation having 2k cycles, each of length p>2, and so an h-invariant partition can have no block fixed. Arguing as above and noting that there are p distinct ways of forming a partition out of two cycles of length p that is also left invariant by these cycles (distribute the elements in one cycle into p empty blocks and then there are p distinct ways of distributing the elements in the other cycle into the blocks), we see that
(5.4)n(h,Σ)=(2k)!(2!)k⋅k!⋅pk.
Assuming pk≥8, that is, (p,k)≠(3,2), we have
n(x,Σ)≥(pk)!(pk/2)!≥(pk2)pk/2≥(pk2)3k/2≥(pk)k≥n(h,Σ).
Also for (p,k)=(3,2), we have n(x,Σ)≥6!/3!=120≥27=n(h,Σ). Thus, the result holds for even k. Suppose now that k is odd. Arguing as before, the number of x-invariant partitions with exactly one block fixed is equal to
(pk1)⋅(pk-1)!(2!)(pk-1)/2⋅((pk-1)/2)!⋅2(pk-1)/2=(pk)!((pk-1)/2)!.
In particular, we have
n(x,Σ)≥(pk)!((pk-1)/2)!.
Moreover, (5.4) holds.
Assuming pk≥8, that is, (p,k)∉{(3,1),(5,1),(7,1)}, we have
n(x,Σ)≥(pk)!((pk-1)/2)!≥(pk+12)(pk+1)/2
≥(pk2)3k/2≥(pk)k≥n(h,Σ).
Note that n(h,Σ)≤n(x,Σ) trivially holds for (p,k)∈{(3,1),(5,1),(7,1)}. Thus, the result also holds for odd k. The proof is complete.
∎
We now state and prove our result in the general case.
Lemma 5.3.
Let g∈Sn be a permutation with at most n fixed points. Let S be the set of partitions of the permutation domain into b blocks of size a, where a>1 and b>1. Let p be the proportion of g-invariant partitions in S.
Then we have p≤1/(2n3/2) for n≥106.
Proof.
By Lemma 5.1 we may assume that a=2. Note that, in this case, a set of blocks cyclically permuted by g is permuted either by one cycle of even length or by two cycles of equal length. Write g=gi1gi2⋯gik where 1≤i1<i2<⋯<ik≤n and gij is the product of cycles of length ij in the cycle decomposition of g. Let Ωj be the subset of Ω consisting of the elements in the cycles of gij so that Ω=⊔j=1kΩj. Note that if g is to fix a partition, the number of cycles of length ij must be even for ij odd. Thus we may assume that |Ωj| is even for all j. Now, viewing gij as an element of Sym(Ωj), a g-invariant partition of Ω can be regarded as a union of gij-invariant partitions of Ωj, for j=1,…,k. We now set some notation. For a set Σ and a permutation h∈Sym(Σ), denote by n(h,Σ) the number of h-invariant partitions of Σ into blocks of size 2. Thus we have
n(g,Ω)=∏j=1kn(gij,Ωj).
Fix some ij≥2, and set h=gij, Σ=Ωj. Let x∈Sym(Σ) be a fixed-point-free involution. By Lemma 5.2 we have n(h,Σ)≤n(x,Σ).
We may therefore assume that g is an involution. Let c be the number of 2-cycles of g so that n=f+2c. Arguing as above, we have n(g,Ω)=n(g1,Ω1)⋅n(g2,Ω2). Clearly,
n(g1,Ω1)=f!(2!)f/2⋅(f/2)!≤(f2)f/2.
The number of g2-invariant partitions of Ω2 into blocks of size 2 with exactly i blocks fixed is
ci={(ci)⋅(c-i)!(2!)(c-i)/2⋅((c-i)/2)!⋅2(c-i)/2if i-c≡0(mod2),0if i-c≡1(mod2).
Note that ci≤c! for each i. Thus,
n(g2,Ω2)=∑i=0cci≤c⋅c!.
Note also that
|𝒮|=n!(2!)n/2⋅(n/2)!≥(n4)n/2.
Using Stirling’s bound, we have
p=n(g,Ω)|𝒮|=n(g1,Ω1)⋅n(g2,Ω2)|𝒮|≤(f/2)f/2⋅c⋅c!(n/4)n/2
=c⋅c!⋅(2fn)f/2⋅(4n)c
≤c⋅ec⋅(2fn)f/2⋅(4cen)c.
Note that f≤n and (n-n)/2≤c≤n/2. Thus we have
p≤e⋅(n2)3/2⋅(2e)n-n2≤12n3/2,
where the last inequality holds for n≥106.
The proof is complete.
∎
We now state and prove the main results of this section.
Theorem 5.4.
Let g∈Sn be a permutation with at most n fixed points. Let p be the probability that a random permutation is contained in a transitive imprimitive maximal subgroup of Sn containing g. Then we have p≤1/n for n≥106.
Proof.
Let Σ be the collection of transitive imprimitive maximal subgroups of Sn containing g. For a,b>1 with ab=n, let Σa,b⊆Σ be the set of members of Σ that leave invariant a (unique) partition of the permutation domain into b blocks of size a so that
Σ=⊔a,b>1n=abΣa,b.
By Theorem 5.3, we have
|Σa,b|≤12n3/2⋅n!a!bb!.
Also, for M∈Σa,b, we have |M|=a!bb!. Thus,
p=|⋃M∈ΣM||Sn|≤∑M∈Σ|M||Sn|=∑a,b>1n=ab∑M∈Σa,b|M||Sn|
=∑a,b>1n=ab|Σa,b|⋅a!bb!|Sn|≤1n,
where the last inequality holds because the number of positive divisors of n is clearly at most 2n.
∎
6 Hamiltonicity of Γ(Sn)
In this section we prove Theorem 1.1 for Sn. We first need a result that provides a sufficient condition for a graph to be Hamiltonian and a characterization of Hamiltonian graphs.
Definition 6.1.
Let Γ be a graph with n≥3 vertices and vertex degrees d1≤d2≤⋯≤dn. Then Γ satisfies Chvátal’s criterion if dn-i≥n-i whenever di≤i<n/2.
Theorem 6.2 (Chvátal [8]).
A graph is Hamiltonian if it satisfies Chvátal’s criterion.
The best vertex degree characterization of Hamiltonian graphs is due to Bondy and Chvátal.
Definition 6.3.
Let Γ be a graph with n vertices. The closurecl(Γ) of Γ is the graph (with the same vertex set) constructed from Γ by adding for all non-adjacent pairs of vertices u and v with d(Γ,u)+d(Γ,v)≥n the new edge uv.
Theorem 6.4 (Bondy, Chvátal [3]).
A graph is Hamiltonian if and only if its closure is Hamiltonian.
To apply Theorems 6.2 and 6.4 we need lower bounds on vertex degrees in Γ(Sn).
Theorem 6.5.
Let g∈Sn be a permutation with at most n fixed points. Let p be the probability that g and a random permutation generate Sn. Then for n≥106, we have
p≥{12-(1n+16.2n+1n2)if g∈An,1-(1n+16.2n+1n2)if g∈Sn∖An.
Proof.
Note that
p=|{x∈Sn:〈g,x〉=Sn}||Sn|.
If g∈Sn∖An, then we have
p=|{x∈Sn:〈g,x〉≥An}||Sn|.
If g∈An, then we have
p≥|{x∈Sn:〈g,x〉≥An}|-n!/2|Sn|=|{x∈Sn:〈g,x〉≥An}||Sn|-12.
The result now follows from Theorem 2.3.
∎
We also need a result on minimal vertex degrees in a certain subgraph of Γ(Sn).
Theorem 6.6 ([7, Theorem 6.1]).
Let Γb(Sn) be the bipartite subgraph of Γ(Sn) obtained by removing edges between elements in Sn∖An. Then for n>15, the degree of every vertex in Γb(Sn) is at least n!/n3.
Notation 6.7.
Let A1(n) and A2(n) be the sets of permutations in Sn∖An and An, respectively, with less than n fixed points. Set B1(n)=(Sn∖An)∖A1(n) and B2(n)=An∖(A2(n)∪{1}).
Lemma 6.8.
For i=1,2, we have
|Bi(n)|≤n!2(⌈n⌉)!.
Proof.
This follows from Proposition 2.1.
∎
Definition 6.9.
For a graph Γ, we set cl(1)(Γ)=cl(Γ) and inductively set cl(i)(Γ)=cl(cl(i-1)(Γ)) for every positive integer i≥2. The graph cl(i)(Γ) is called the i-th closure of the graph Γ.
In the next lemma we investigate adjacency in the graph cl(3)(Γ(Sn)).
Lemma 6.10.
Let n≥107 be an integer. The set Sn∖An induces a complete subgraph in the graph cl(3)(Γ(Sn)). Furthermore, every vertex in A1(n) is adjacent to every other vertex and every vertex in B1(n) is adjacent to at least (n!/2)-1+(n!/n3) other vertices in the graph cl(3)(Γ(Sn)).
Proof.
Set Γ0=Γ(Sn). Since 1/n+16.2/n+1/n2≤1/4 for n≥107, by Theorem 6.5, for u∈A1(n) we have
d(Γ0,u)≥(1-(1n+16.2n+1n2))⋅n!≥34n!,
and for v∈A2(n) we have
d(Γ0,v)≥14n!.
The rest of the proof follows from exactly the same arguments as in the proof of [7, Lemma 6.4] with Ai(n) and Bi(n), i=1,2, defined as in Notation 6.7.
∎
We are now ready to prove Theorem 1.1 in the case of symmetric groups.
Proof of Theorem 1.1 in the case of symmetric groups.
The proof exactly follows the proof of [7, Lemma 6.5] with Ai(n) and Bi(n), i=1,2, defined as in Notation 6.7, and δn replaced by ⌈n⌉.
∎
7 Hamiltonicity of Γ(An)
In this section we prove Theorem 1.1 for An. We first need a result of Pósa that provides a sufficient condition for a graph to be Hamiltonian.
Definition 7.1.
Let Γ be a graph with n vertices and vertex degrees d1≤d2≤⋯≤dn. Then Γ satisfies Pósa’s criterion if dk≥k+1 for all positive integers k with k<n/2.
Theorem 7.2 (Pósa [17]).
A graph is Hamiltonian if it satisfies Pósa’s criterion.
To apply Theorem 7.2 we need lower bounds on vertex degrees in Γ(An).
Theorem 7.3.
Let g∈An be a permutation with at most n fixed points. Let p be the probability that g and a random permutation generate An. Then we have
p≥1-2⋅(1n+16.2n+1n2) for n≥106.
Proof.
Note that
p=|{x∈An:〈g,x〉=An}||An|≥|{x∈Sn:〈g,x〉≥An}|-n!/2|An|
=2⋅|{x∈Sn:〈g,x〉≥An}||Sn|-1.
The result now follows from Theorem 2.3.
∎
Also, we need a result on minimal vertex degrees in Γ(An) which is a consequence of the proof of [13, Proposition 7.1].
Theorem 7.4 ([7, Theorem 5.3]).
Let n≥8. Then the degree of every vertex in Γ(An) is at least n!/(10n3).
We are now ready to prove Theorem 1.1 in the case of alternating groups.
Proof of Theorem 1.1 in the case of alternating groups.
Let A(n) be the set of permutations in An with less than n fixed points and let B(n)=An∖A(n). Let d1≤d2≤⋯≤dn!/2-1 be the vertex degrees in the graph Γ(An). Let k<(n!/2-1)/2. We need to show dk≥k+1. If there exists i≤k such that di is equal to the vertex degree of a permutation in A(n), then by Theorem 7.3 we have
dk≥di≥(1-2⋅(1n+16.2n+1n2))⋅n!2≥n!4≥k+1,
where the third inequality holds since 1/n+16.2/n+1/n2≤1/4.
Thus we may assume that di is the vertex degree of a permutation in B(n) for each i≤k. But then k≤|B(n)| and hence by Theorem 7.4 we have
dk≥n!10n3≥n!2(⌈n⌉)!+1≥|B(n)|+1≥k+1,
where the second inequality holds since n≥101.
This shows that Γ(An) satisfies Pósa’s criterion. In particular, by Theorem 7.2, Γ(An) is Hamiltonian.
∎
