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BY 4.0 license Open Access Published by De Gruyter July 10, 2020

The invariably generating graph of the alternating and symmetric groups

Daniele Garzoni EMAIL logo
From the journal Journal of Group Theory

Abstract

Given a finite group G, the invariably generating graph of G is defined as the undirected graph in which the vertices are the nontrivial conjugacy classes of G, and two classes are adjacent if and only if they invariably generate G. In this paper, we study this object for alternating and symmetric groups. The main result of the paper states that if we remove the isolated vertices from the graph, the resulting graph is connected and has diameter at most 6.

1 Introduction

Given a finite group G and a subset {x1,,xt} of G, we say that {x1,,xt}invariably generatesG if x1g1,,xtgt=G for every g1,,gtG. This concept was introduced by Dixon with motivations from computational Galois theory; see [5] for details. Note that invariable generation can be thought of as a property of conjugacy classes, rather than individual elements.

1.1 The invariably generating graph

Given a finite group G, we define the invariably generating graphΛ(G) of G as follows. The vertices are the conjugacy classes of G different from {1}, and two vertices xG and yG are adjacent if and only if {x,y} invariably generates G. The purpose of this paper is to initiate the study of this object for finite (almost) simple groups, and more precisely, for alternating and symmetric groups. It was proved by Kantor–Lubotzky–Shalev [12], and Guralnick–Malle [9] independently, that finite simple groups are invariably generated by two elements so that, in this case, Λ(G) is nonempty. It is also known that Sn is invariably generated by two elements if and only if n6 (see [15, Proposition 4.10]). Set

𝒫={qd-1q-1:qis a prime power andd2}.

The first two results of the paper are the following.

Theorem 1.1.

Assume n5, and let G{An,Sn}. Then Λ(G) does not have isolated vertices if and only if G=An and n is a prime satisfying nP{11,23} and n-1mod12.

Theorem 1.2.

The following statements hold.

  1. Let (Gi) be a sequence of alternating or symmetric groups so that |Gi|. Assume that, for every i, Gi is not alternating of prime degree. Then the number of isolated vertices of Λ(Gi) tends to infinity.

  2. Assume n is a prime not contained in 𝒫{11,23}. Then the number of isolated vertices of Λ(An) is at most 2.

In Theorem 1.1, we assumed n5. We will keep this assumption throughout the paper; see Lemma 2.15 for the remaining cases. For the sake of clarity, we mention that Λ(A11) and Λ(A23) have 5 and 6 isolated vertices, respectively (Lemma 3.2). The only case not addressed in Theorem 1.2 is G=An with n𝒫 prime. In Remark 3.3, we will obtain a partial result, dealing with the case n=(qd-1)/(q-1) with d.

Once Theorems 1.1 and 1.2 are proved, one may ask what happens if the isolated vertices are removed from Λ(G). With this purpose, we define a graph Ξ(G) which is obtained from Λ(G) by removing the isolated vertices. The next result states that, except in case G=S6, this graph is connected with bounded diameter (we already recalled that S6 is not invariably generated by two elements, hence Ξ(S6) is the null graph).

Theorem 1.3.

Assume n5, and let G{An,Sn}, with GS6. Then Ξ(G) is connected with diameter at most 6.

In many cases, we prove better estimates on the diameter.

Theorem 1.4.

Assume n5. If G=Sn and n is odd, or if G=An and n is even, then d(Ξ(G))4. If G=An and nP is prime, then d(Ξ(G))3.

The proofs of Theorems 1.3 and 1.4 rely on some recent results, proved in [11] and [8], which classify the primitive subgroups of Sn containing elements having certain cycle types. These results depend on the Classification of Finite Simple Groups.

The upper bound in Theorem 1.3 is attained since d(Ξ(S8))=6 (Lemma 3.7). However, we conjecture that this can happen only for finitely many groups.

Conjecture 1.5.

Let G{An,Sn}. If n is sufficiently large, then d(Ξ(G))4.

In Section 4, we obtain a partial result towards a proof of this conjecture. It is interesting to observe that, in this partial result, we use the Prime Number Theorem, but we do not use the CFSG. See also Lemma 4.5, which establishes that d(Ξ(Sn))4 for all primes n7, so the bound in Conjecture 1.5, if true, is attained infinitely often.

Theorem 1.3 suggests a natural question for all finite simple groups.

Question 1.6.

Let G be a finite simple group. Is the graph Ξ(G) connected?

1.2 Some context

The invariably generating graph is the analogue, for invariable generation, of the so-called generating graphΓ(G) of a finite group G. This is defined as follows. The vertices are the nonidentity elements of G, and two vertices x and y are adjacent if and only if x,y=G.

Many properties of generation of a finite simple group by two elements can be stated in terms of the generating graph. Guralnick and Kantor [10] proved that if G is a finite simple group, then, for every 1xG, there exists yG such that x,y=G. Later, Breuer, Guralnick and Kantor [3] showed that if 1x1,x2G, then there exists yG such that x1,y=x2,y=G. These properties can be stated respectively as follows: Γ(G) has no isolated vertices, and it is connected with diameter at most 2.

Theorems 1.1 and 1.2 say that, for alternating and symmetric groups, the invariably generating graph has quite different properties; it usually has isolated vertices, and the number of isolated vertices usually grows as the order of the group grows.

On the other hand, Theorem 1.3 says that if we remove the isolated vertices, we obtain a graph which in some sense shares similarities with the generating graph Γ(G).

1.3 An alternative definition

In light of Theorem 1.2, one could ask how the proportion of isolated vertices of Λ(G) behaves as |G| tends to infinity. The elementary approach used in the proof of Theorem 1.2 is not sufficient to address this problem. Here, however, one comment is in order. We have chosen the vertices of Λ(G) to be the nontrivial conjugacy classes of G. One could define a graph Λe(G) in which the vertices are the nontrivial elements of G, and two vertices are adjacent if and only if they invariably generate G. Of course, the property of connectedness and the value of the diameter are the same in the two graphs. However, when one counts the edges, or the vertices having certain properties, the situation can radically change. Indeed, in Λe(G), there is a dependence on the size of the conjugacy classes which does not exist in Λ(G).

Probabilistic invariable generation has always been considered in terms of elements (cf. [5, 13, 14, 6]). Still, we believe it is worth exploring the problem of counting conjugacy classes – although, in this paper, we do not address any question of this kind.

2 Notation and preliminary results

In this section, we fix some notation, and we gather some preliminary lemmas and observations that we will use throughout the paper.

2.1 Notation

The vertices of our graphs are conjugacy classes of alternating and symmetric groups. We will identify conjugacy classes of Sn with their cycle type, i.e., we will represent conjugacy classes of Sn as partitions of n. We now introduce some terminology about partitions.

Let HSn, and let 𝔭 be a partition of n. We will say that 𝔭 belongs to H, or that 𝔭 is contained in H, if H contains elements with cycle type 𝔭. If C is the conjugacy class of Sn corresponding to 𝔭, this is equivalent to the condition HC. Of course, this condition depends only on the Sn-conjugacy class of the subgroup H. Therefore, in the above terminology, we are allowed, if we wish, to replace H by its conjugacy class, and to say that 𝔭 belongs to the Sn-conjugacy class of H.

Let 𝔭1 and 𝔭2 be two partitions of n. If 𝔭1 and 𝔭2 both belong to H, we will say that 𝔭1 and 𝔭2shareH. When 𝔭Si×Sn-i, with 1i<n, we will say that i is a partial sum in 𝔭. This is indeed equivalent to the condition that the integer i can be written as the sum of some parts of 𝔭. In a partition, im will mean m parts of length i. Therefore, (a1m1,a2m2,,atmt) will mean m1 parts of length a1, m2 parts of length a2, …, and mt parts of length at.

Occasionally, given a partition 𝔭 and a positive integer i, we will write 𝔭i to denote “the i-th power of 𝔭”, namely the partition obtained by replacing each part of length by d parts of length /d, where d=(,i). Note that if xSn has cycle type 𝔭, then xi has cycle type 𝔭i.

Finally, we define

𝒫={qd-1q-1:qis a prime power andd2}.

2.2 Maximal overgroups of certain elements

Most of the arguments will rely heavily on the knowledge of the maximal overgroups of certain elements in Sn; specifically, cycles, or elements having few orbits in the natural action on n points. The intransitive maximal subgroups are easily determined. For convenience, we now isolate some elementary observations regarding transitive imprimitive subgroups, while then moving to the more difficult case of primitive subgroups. We use some of the language introduced in the previous subsection. The following two lemmas are a consequence of [2, Theorem 2.5].

Lemma 2.1.

Let n be a natural number, m a nontrivial divisor of n and 1i<n. The partition (a1,a2) belongs to SmSn/m if and only if either m divides a1 or n/m divides a1.

Proof.

If (a1,a2) belongs to SmSn/m, the induced permutation on the blocks has at most two cycles. If it has two cycles, then m divides a1. If it is an (n/m)-cycle, then n/m divides a1. The converse implication is proved in the same way. ∎

Lemma 2.2.

Let n be a natural number, m a nontrivial divisor of n and 1i<n. The partition (a1,a2,a3) belongs to SmSn/m if and only if one of the following conditions is satisfied:

  1. m divides ai for every i;

  2. n/m divides ai for every i;

  3. there exist 1t<n/m and ij{1,2,3} such that

    ak=tbk𝑓𝑜𝑟k=i,j,𝑤𝑖𝑡ℎbi+bj=m.

Proof.

Similar to the previous lemma. In case (a), the induced permutation on the blocks has cycle type (a1/m,a2/m,a3/m). In case (b), it is an (n/m)-cycle. In case (c), it has cycle type (t,a/m), where {1,2,3}{i,j}. ∎

We now move to primitive subgroups. Our main tool is a theorem which classifies the primitive subgroups of Sn containing a cycle, and which relies on the CFSG. This should be seen as a generalization of a classical theorem of Jordan (see e.g. [16, Theorem 13.9]) stating that there are no proper primitive subgroups of Sn different from An containing a cycle of prime length fixing at least 3 points. Since we will apply this result several times, for convenience, we report here the statement.

Theorem 2.3 ([11]).

Let G be a primitive permutation group of finite degree n, not containing the alternating group An. Suppose that G contains a cycle fixing k points, where 0kn-2. Then one of the following holds:

  1. k=0 and

    1. either CpGAGL1(p) with n=p prime,

    2. or PGLd(q)GPΓLd(q) with n=(qd-1)/(q-1) and d2 for some prime power q,

    3. or G=PSL2(11), M11 or M23 with n=11,11 or 23 , respectively.

  2. k=1 and

    1. either AGLd(q)GAΓLd(q) with n=qd and d1 for some prime power q,

    2. or G=PSL2(p) or PGL2(p) with n=p+1 for some prime p5,

    3. or G=M11, M12 or M24 with n=12,12 or 24 , respectively.

  3. k=2 and PGL2(q)GPΓL2(q) with n=q+1 for some prime power q.

Note that the statement implies that there are no proper primitive subgroups of Sn different from An containing a cycle fixing at least 3 points, generalizing indeed Jordan’s theorem. We note the following immediate consequence.

Corollary 2.4.

Assume xSn is such that a suitable power of x is a nontrivial cycle fixing at least 3 points. Then x does not lie in proper primitive subgroups of Sn different from An.

We also mention that we will make essential use of the main result from [8], which classifies the primitive subgroups of Sn containing an element having at most 4 cycles.

We will shortly apply the previous results to certain elements (or partitions) of particular interest to us.

2.3 Conjugacy classes of An

If a partition of n is made of distinct odd parts, the corresponding Sn-conjugacy class splits into two An-conjugacy classes (and vice versa), giving rise to two vertices of Λ(An). Often, this does not represent a serious change; the following technical lemmas give conditions under which the two vertices may be essentially thought of as a unique vertex.

Lemma 2.5.

Let xAn, and let HSn be such that HAn=HSn. Then H contains elements belonging to xAn if and only if it contains elements belonging to (x)An for every xxSn.

Proof.

Assume xzH with zAn, and assume x=xzg with gSn. Then x=xzgHg=Hh for some hAn by hypothesis, hence (x)h-1H. This concludes the proof. ∎

Notation 2.6.

Assume x,yAn are such that (x)An is adjacent to (y)An in Λ(An) for any xxSn and for any yySn. Under these assumptions, we say with slight abuse of notation that xSn is adjacent to ySn in Λ(An).

This notation will be convenient, as we will represent Sn-conjugacy classes as partitions, and we will be allowed to say that “a partition 𝔮 is adjacent to a partition 𝔭”, rather than “any An-conjugacy class of elements with cycle type 𝔮 is adjacent to any An-conjugacy classes of elements with cycle type 𝔭”. We will now see that, in many cases, the assumption of Notation 2.6 is satisfied.

Lemma 2.7.

Let xAn, and assume xAn=xSn. Let yAn. Then xAn is adjacent in Λ(An) to yAn if and only if xSn and ySn are adjacent (in the terminology of Notation 2.6).

Proof.

Choose xxSn. We show first that if yAn is not adjacent to xAn, then yAn is not adjacent to (x)An. By assumption, we can write x=(xi)z for some integer i and for some zAn. Again by assumption, {xg1,yg2}H for some g1,g2An, and for some proper subgroup H of An. Then

(x)z-1g1=(xi)g1=(xg1)iH,

whence {(x)z-1g1,yg2}H and yAn is not adjacent to (x)An, as required.

Assume now yySn. We show that if yAn is not adjacent to xAn, then (y)An is not adjacent to xAn. This will conclude the proof. We may assume yAn(y)An and xAnxSn; otherwise, the statement is easy. Choose xxSnxAn. By the previous paragraph, if yAn is not adjacent to xAn, then it is not adjacent to (x)An so that {xg1,yg2}H for some proper subgroup H of An, some g1SnAn and g2An. Then {x,yg2g1-1}Hg1-1, and since yg2g1-1 is An-conjugate of y, the proof is concluded. ∎

We now apply the previous considerations to certain elements and subgroups of Sn.

Lemma 2.8.

Let H be a maximal subgroup of Sn which is either intransitive, or transitive and imprimitive. Then HAn=HSn.

Proof.

We may assume n3. We have HAn=HSn if and only if NSn(H)⩽̸An. In our case, H⩽̸An since every maximal intransitive or imprimitive subgroup of Sn contains transpositions. ∎

Lemma 2.9.

Assume xAn belongs to no proper primitive subgroup of Sn different from An. Let yAn. Then xAn and yAn are adjacent in Λ(An) if and only if xSn and ySn are adjacent (in the terminology of Notation 2.6).

Proof.

By assumption, x lies in no proper primitive subgroups different from An. Therefore, xAn is adjacent to yAn if and only if, for every g1,g2An, xg1,yg2 is primitive or, equivalently, is not contained in intransitive or imprimitive maximal subgroups. By Lemmas 2.5 and 2.8, this condition depends only on the cycle type of the elements, rather than on their An-conjugacy class. The lemma follows. ∎

Lemma 2.10.

Let 2in/2 be coprime with n. Then (i,n-i) does not belong to proper primitive subgroups different from An, and does not belong to transitive imprimitive subgroups.

Proof.

The fact that (i,n-i) does not lie in imprimitive subgroups follows from Lemma 2.1. Regarding primitive subgroups, (i,n-i)n-i=(i,1n-i)(1n), and n-i3 (the conditions on i imply n5); hence the statement follows from Corollary 2.4. ∎

Theorem 2.3 can be used to generalize the previous lemma to the case i=1; one just needs to take care of some specific examples of primitive subgroups. The following easy lemma will be used with this purpose.

Lemma 2.11.

Let q be a prime power and d a positive integer.

  1. An element of AGLd(q), in the natural action on qd points, either is a derangement, or fixes a number of points equal to qs for some 0sd.

  2. Assume gPΓL2(q) fixes at least 3 points in the natural action on q+1 points. Then g fixes a number of points having the same parity of q+1. Moreover, if gPGL2(q), then g=1.

Proof.

(1) If gAGLd(q) fixes some point, we may assume that it fixes 0. Hence, gGLd(q). Now just observe that the set of fixed points of an element of GLd(q) is an 𝐅q-subspace of 𝐅qd.

(2) Consider PΓL2(q) acting (on the right) on the set Ω of 1-dimensional subspaces of 𝐅q2. Write q=pr with p prime. For ϕGal(𝐅q/𝐅p), denote by fϕ the permutation of Ω induced by the mapping (λ1,λ2)(λ1ϕ,λ2ϕ) of 𝐅q2. Then we may express each element gPΓL2(q) as g=xfϕ, where xPGL2(q) and ϕGal(𝐅q/𝐅p).

Note that PΓL2(q) is 3-transitive on Ω. Hence, if g=xfϕPΓL2(q) fixes at least 3 points, we may assume that it fixes (1,0), (0,1) and (1,1). It follows that x=1 and g=fϕ. Then g fixes p+1 points, where is a divisor of r; in particular, it fixes a number of points having the same parity of q+1. Moreover, if gPGL2(q), then fϕ=1. The lemma is proved. ∎

Lemma 2.12.

Assume n is either an odd prime, or n3mod4. Assume xAn is an n-cycle. Then xAn=xSn.

Proof.

If n3 is an odd integer, then an n-cycle is normalized by elements having cycle type (1,2(n-1)/2). If n is an odd prime, then an n-cycle is normalized by an (n-1)-cycle. In particular, if x is as in the statement, then NSn(x)⩽̸An, from which xAn=xSn. ∎

We deduce another consequence of the previous lemmas.

Lemma 2.13.

Assume n4, and let xAn. Then, for every gSn, xAn and (xg)An are not adjacent in Λ(An).

Proof.

By Lemma 2.8, we may assume that x is not contained in intransitive or imprimitive subgroups. It follows that n is prime and x is an n-cycle. The statement follows then by Lemma 2.12. ∎

This lemma suggests a natural question for all finite simple groups.

Question 2.14.

Let G be a finite simple group. Let xG, and let σAut(G). Is it possible that {x,xσ} invariably generates G?

It is easy to check that the statement is true for G=A6, in which case, S6 has index two in Aut(G). This, together with Lemma 2.13, implies that the question has a negative answer for G=An.

2.4 Small degrees

The main theorems are stated for n5. For completeness, we address here the cases of degree n4. We assume n3 in order to avoid trivialities.

Lemma 2.15.

Let 3n4, and assume G{An,Sn}. Then Λ(G) has isolated vertices if and only if G=S4. Moreover, Ξ(G) is connected with diameter at most 2.

Proof.

This is an easy check. The graphs Ξ(A3), Ξ(S3), Ξ(S4) have diameter 1, while Ξ(A4) has diameter 2; the two classes of 3-cycles are connected by a path of length 2 passing through the class (22). In Λ(S4), the vertices corresponding to (12,2) and (22) are isolated. ∎

3 Proofs

In this section, we prove the theorems stated in the introduction.

3.1 Proof of Theorems 1.1 and 1.2

We begin with a lemma which proves Theorem 1.2 (2) and the “if” part of Theorem 1.1.

Lemma 3.1.

Let n5 be an odd prime. Then (1,2(n-1)/2) and (1,3(n-1)/3) are isolated in Λ(An) (when they make sense and are even permutations). If nP{11,23}, then there are no other isolated vertices in Λ(An).

Proof.

The two mentioned vertices are not adjacent to (n) (this terminology makes sense by Lemma 2.7, Lemma 2.12 and Notation 2.6) because they are contained in AGL1(n). Therefore, they might be adjacent only to some An-conjugacy class with cycle type (a1,,at), with t3. Every 1i(n-1)/2 is a partial sum in (1,2(n-1)/2), so this vertex is isolated (recall Lemmas 2.8 and 2.5), and the values of i that are not partial sums in (1,3(n-1)/3) are exactly those that satisfy i2mod3. If a1,a22mod3, then a1+a21mod3, so one among a1,a2 and a1+a2 is a partial sum. We conclude that (1,3(n-1)/3) is indeed isolated.

Assume now n𝒫{11,23}. Let xAn be a vertex of Λ(An) different from the two above; we want to show it is not isolated. If xAGL1(n), we deduce from Theorem 2.3 that xAn is adjacent to (n). On the other hand, if xAGL1(n), then either it is an n-cycle, or it has cycle type (1,t(n-1)/t), with t4. As just remarked, a class of n-cycles is not isolated. If t5, by Corollary 2.4, xAn is adjacent to (22,n-4). If t=4, for the same reason, xAn is adjacent to (32,n-6). ∎

For the sake of clarity, we deal with the cases n=11,23.

Lemma 3.2.

The graphs Λ(A11) and Λ(A23) have 5 and 6 isolated vertices, respectively.

Proof.

Recall Theorem 2.3. Let n{11,23}. Let 1xAn. If x does not belong to M11 (for n=11) and M23 (for n=23), the same argument as in the previous lemma shows that xAn is adjacent to (n) (note that PSL2(11) can be embedded in M11). Inspection (using for instance GAP) shows that if n=11 and 1xM11, then xAn belongs to

{(13,24),(13,42),(12,33),(2,3,6),(1,2,8)}.

These are not adjacent to (n) because of M11. By looking at partial sums, we see that these are all isolated. If n=23 and 1xM23, then either xAn is adjacent to one between (22,19) and (32,17), or xAn belongs to

{(15,36),(13,54),(17,28),(1,22,32,62),(13,22,44),(1,2,4,82)}.

These are all isolated for the same reason as above. ∎

We are now ready to prove Theorems 1.1 and 1.2.

Proof of Theorems 1.1 and 1.2.

In the proof, we will use Lemmas 2.5, 2.7 and 2.8 with no further mention. Recall also Notation 2.6.

We begin with Theorem 1.2. Item (2) follows from Lemma 3.1; hence we focus on item (1). Let G=An or Sn, with n nonprime if G=An. For every n, we will define In, a subset of the set of isolated vertices of Λ(G), whose cardinality, as n, goes to infinity. This will prove Theorem 1.2. For every m, denote by pmE the set of all partitions of m different from (1m) which correspond to even permutation.

We first assume that, whenever n is odd, G=Sn. Define In as the set of all partitions of n of the form (1n/2,𝔭), where 𝔭 is whatsoever partition of pn/2E.

Let 𝔮In. Every 1in/2 is a partial sum, so 𝔮 is not adjacent to classes of elements with at least two cycles. If n is even, then 𝔮 is not adjacent to (n) since n/2 is a partial sum in 𝔮, hence 𝔮Sn/2S2. If n is odd, then G=Sn, and 𝔮 is not adjacent to (n) since 𝔮 corresponds to even permutations. Therefore, In consist of isolated vertices. Clearly, the size of In goes to infinity as n.

Assume now that G=An and n is odd and not prime. Fix n, and let pn be the smallest prime divisor of n. Define In as the set of all partitions of n of the form (1n(pn-1)/pn,𝔭), where 𝔭 is any partition of pn/pnE.

If 𝔮In, then 𝔮Sn/pnSpn; the first pn-1 blocks are fixed pointwise. Therefore, 𝔮 is not adjacent to (n). Moreover, every 1in/2 is a partial sum, so 𝔮 is not adjacent to classes of elements having at least 2 cycles. It follows that 𝔮 is isolated. Note now that n/pnn; hence the size of In goes to infinity as n. This concludes the proof of Theorem 1.2.

We now move to Theorem 1.1. Note that Lemma 3.1 proves the “if” part. We now prove the “only if” part. Assume first that n is prime and G=An. If n-1mod12, by Lemma 3.1, there are isolated vertices in Λ(An). The cases n=11,23 have been considered in Lemma 3.2. If n=(qd-1)/(q-1), let x be any involution lying in a subgroup of Sn conjugate to PΓLd(q). The fact that n is odd implies that every 1in/2 is a partial sum in (the cycle type of) x; hence xAn might only be adjacent to a class of n-cycles. However, this does not happen because of the containment in PΓLd(q). Therefore, xAn is isolated.

The case G=An with n prime is therefore proved. For the remaining cases, we apply what we proved for Theorem 1.2. We define 𝔮=(1n-3,3). This partition belongs to In, as defined in this proof; hence it is an isolated vertex of Λ(G). This concludes the proof. ∎

Remark 3.3.

The unique case not discussed in Theorem 1.2 is the case G=An and n𝒫 prime. We obtain here the partial result that if ni=(qdi-1)/(q-1) is prime and di, then the number of isolated vertices of Λ(Ani) tends to infinity. Note that since ni is prime, di must be prime. To establish whether infinitely many such primes (i.e., primes d such that (qd-1)/(q-1) is prime for some prime power q) do actually exist, however, is a hard open problem in number theory; see for instance [4].

Assume first n=(qd-1)/(q-1) with q (and d) odd. The action of PGLd(q) on the 1-dimensional subspaces of 𝐅qd gives an embedding PGLd(q)<Sn. For every 1r<d/2, let x=x(r) be a diagonal matrix of GLd(q) with 1’s and -1’s on the diagonal, and assume the number of -1’s is r. Let x¯ denote the image of x in PGLd(q)<Sn. Then x¯ has (qr+qd-r-2)/(q-1) fixed points. In particular, any two distinct 1r<d/2 give rise to elements of Sn which have a different number of fixed points, and which therefore belong to different Sn-conjugacy classes. It is easy to check that the element x¯ arising in this way belongs to An. The number of possibilities for r in order to obtain such an element is (d-1)/2. As remarked in the previous proof, (x¯)An is isolated in Λ(An); therefore, the number of isolated vertices of Λ(An) is at least (d-1)/2.

Assume now q is even. For every 1d/2, consider a unipotent element x=x() of SLd(q) with Jordan blocks of size 2, and with the other Jordan blocks of size 1. This is an involution of SLd(q). Denote again by x¯ the image of x in PSLd(q)<Sn. Then x¯ has (qd--1)/(q-1) fixed points; hence any two distinct 1d/2 give rise to elements belonging to different Sn-conjugacy classes. Moreover, x¯An (unless (q,d)=(2,2), but recall we are assuming n5), and (x¯)An is isolated in Λ(An). Hence the number of isolated vertices of Λ(An) is at least (d-1)/2.

3.2 Proof of Theorems 1.3 and 1.4

In this subsection, we prove Theorems 1.3 and 1.4.

One brief comment about the terminology we will adopt. The proofs will begin with a sentence of the type “Let 𝔭 be a vertex of Ξ(G)”, without any preliminary consideration showing that Ξ(G) is not the null graph. However, along the proof suitable edges will be exhibited in Λ(G) so that the initial choice of 𝔭 will be licit. (In other words, we are saying that, although we will not state it explicitly, in the proofs, it will be shown that the groups are invariably generated by two elements.)

We begin by proving Theorem 1.4 (in two separate results).

Theorem 3.4.

Assume n5 is a prime and nP. Then Ξ(An) is connected and d(Ξ(An))3.

Proof.

Let xAn, and assume xAn is a vertex of Ξ(An). If n11,23, in the proof of Lemma 3.1, we showed that xAn is adjacent to one among (n), (22,n-4) and (32,n-6) (see Lemma 2.12). These vertices are pairwise adjacent by Corollary 2.4; hence we have indeed d(Ξ(An))3. If n=11 or 23, we observed in the proof of Lemma 3.2 that a class inside M11 or M23 is either isolated, or adjacent to one between (22,n-4) and (32,n-6). Therefore, by Corollary 2.4, also in this case, d(Ξ(An))3. ∎

For a later use, we point out that the same estimate holds for the groups A13 and A17.

Assume n=13=(33-1)/(3-1). Inspection shows that every class lying in PGL3(3) is either isolated or adjacent to (33,7); hence d(Ξ(A13))3 by the same argument as in the proof of the previous theorem. Assume finally that n=17=16+1. Here every class lying in PΓL2(16) is either isolated or adjacent to one between (32,11) and (42,9). It is easy to deduce that d(Ξ(A17))3.

Theorem 3.5.

Let n5 be an integer.

  1. If n is odd, then Ξ(Sn) is connected and d(Ξ(Sn))4.

  2. If n is even, then Ξ(An) is connected and d(Ξ(An))4.

Proof.

(1) Let 𝔭 be a vertex of Ξ(Sn). Assume 𝔭 is not adjacent to (i,n-i) for any 2in/2 coprime with n. Then, by Lemma 2.10, every such i is a partial sum in 𝔭. Since 𝔭 is a vertex of Ξ(Sn), 𝔭 will be adjacent to some vertex 𝔮. Necessarily, no 2in/2 coprime with n is a partial sum in 𝔮. Hence, again by Lemma 2.10, 𝔮 is adjacent to (i,n-i) for every such i.

Now note that the (i,n-i)’s, with i as above, are pairwise adjacent. From this, it follows that, for any fixed 2in/2 coprime with n, any vertex 𝔭 of Ξ(Sn) has distance at most 2 from (i,n-i). This concludes the proof.

(2) The statement for n=6 can be checked explicitly; hence we assume n8. Then the proof is identical to (1). Recall Lemmas 2.5 to 2.10, and Notation 2.6. ∎

Now we move to the proof of the general case, i.e., Theorem 1.3. We are left with symmetric groups of even degree and alternating groups of odd degree.

In Theorem 3.5, the strategy was to look for edges with conjugacy classes of elements having two cycles. This approach is not available anymore. Indeed, in alternating groups of odd degree, elements with two cycles do not exist, and in symmetric groups of even degree, such elements belong to An; hence one must take care of the parity of elements when dealing with generation. For these elementary reasons, our strategy will be to look for edges with elements having three cycles. This is where we will make use of [8] which, as already mentioned in the introduction, classifies the primitive permutation groups having elements with at most four cycles.

Theorem 3.6.

Let n8 be an even integer. Then Ξ(Sn) is connected, and we have d(Ξ(Sn))6.

Proof.

We first assume n12 and consider the remaining cases at the end of the proof. Let 𝔭 and 𝔮 be two vertices of Ξ(Sn) joined by an edge. One of the two, say 𝔭, must correspond to odd permutations. Assume 𝔭 is not adjacent to (i,n-i) for any 1in/2 coprime with n. By Lemma 2.10, every such i, with i1, is a partial sum in 𝔭. We now show that also i=1 is a partial sum in 𝔭. Assume this is not the case. Since 𝔭 is not adjacent to (1,n-1), we deduce from Lemma 2.1 and Theorem 2.3 that 𝔭 is contained in one between AGLd(2), PGL2(p), M11, M12, M24. The last three are excluded because they are subgroups of An. Assume 𝔭AGLd(2) or PGL2(p). Since every 3in/2 coprime with n is a partial sum, 𝔭 must have odd parts; let a be the smallest such part. Since i=1 is not a partial sum, we have a3. On the other hand, by assumption, 𝔭 corresponds to odd permutations; hence it has even parts. It follows that (1n)𝔭a fixes a number of points which is greater or equal to 3, and which is a multiple of a. This contradicts Lemma 2.11. Therefore, i=1 is indeed a partial sum in 𝔭.

Now we divide the cases n2mod4 and n0mod4.

Assume n2mod4. Then n/2-2 is coprime with n; hence it is a partial sum in 𝔭. Write 𝔭=(a1,,at), and assume n/2-2=i=1hai. If ak=1 for some k{1,,h}, then n/2-3 is a partial sum in 𝔭. Otherwise, n/2-1 is a partial sum in 𝔭. We show that 𝔮 is adjacent in the first case to 𝔞1=(1,n/2-3,n/2+2), and in the second case to 𝔞2=(1,n/2-2,n/2+1).

By the considerations above, 𝔮 and 𝔞i do not share intransitive subgroups. Moreover, 𝔞1 and 𝔞2 correspond to odd permutations, and belong to no transitive imprimitive subgroups by Lemma 2.2. Finally, 𝔞1 and 𝔞2 belong to no core-free primitive subgroups by [8, Theorem 1.1]. We have therefore our desired edge between 𝔮 and 𝔞1 or 𝔞2.

Assume n0mod4. We employ the same argument as above, with n/2-2 replaced by n/2-1. The same reasoning lead us to look for an edge between 𝔮 and 𝔟1=(1,n/2-2,n/2+1) or 𝔟2=(1,n/2,n/2-1). Again, 𝔮 and 𝔟i do not share intransitive subgroups. It follows from [8, Theorem 1.1] that 𝔟1 and 𝔟2 are not contained in core-free primitive subgroups (for 𝔟2, we may also use Corollary 2.4). Regarding maximal transitive imprimitive subgroups, we only have that 𝔟2 is contained in Sn/2S2. However, by construction, we consider 𝔟2 only when n/2 is a partial sum in 𝔭 so that 𝔭 belongs to Sn/2S2. Since 𝔭 and 𝔮 are adjacent, we deduce that 𝔮 is not contained in Sn/2S2. Therefore, we have an edge between 𝔮 and 𝔟1 or 𝔟2.

Now we deduce the connectedness of Ξ(Sn) and the bound to the diameter. The considerations above imply that an edge with 𝔞i and 𝔟i concerns only intransitive subgroups (i.e., partial sums), except for 𝔟2, where one has to deal also with Sn/2S2.

Assume first n2mod4. The argument given above shows that every vertex of Ξ(Sn) has distance at most 2 from one among 𝔞1,𝔞2 and (i,n-i) for some 1in/2 coprime with n. Hence, in order to conclude, it is sufficient to show that these vertices have pairwise distance at most 2. For n=14, this can be checked directly. Assume then n>14. We show that all these vertices are adjacent to (22,n-4), which clearly concludes the proof. For all the vertices except (1,n-1), this follows from Lemma 2.10 and from the considerations of the previous paragraph. For (1,n-1), by Theorem 2.3, we need to exclude the sharing of AGLd(p), PGL2(p), M24. The last is contained in An, while (22,n-4) is not. Moreover, n2mod4; hence n is not a power of 2 and we do not have affine subgroups. Finally, (14,n-4)=(22,n-4)2 fixes at least 4 points; hence it does not belong to PGL2(p) by Lemma 2.11(2). This concludes the proof in case n2mod4.

Assume now n0mod4. We assume first n>12. As in case n2mod4, in order to conclude, it is sufficient to prove that the vertices 𝔟1,𝔟2 and (i,n-i) with 1in/2 coprime with n have pairwise distance at most 2. The vertices 𝔟1,𝔟2 and (i,n-i) with 1in/2 coprime with n and i3,5 are adjacent to (2,3,n-5). The vertices (i,n-i) with 1in/2 coprime with n and i1,n/2-1 are adjacent to both 𝔟1 and 𝔟2. The vertices (i,n-i) with 2in/2 coprime with n are adjacent to (22,n-4). By [8, Theorem 1.1] (which in the affine case relies on [7, Theorem 1.5]), we deduce that (22,n-4) is not contained in affine subgroups; hence also (1,n-1) is adjacent to (22,n-4). These considerations imply that the vertices have pairwise distance at most 2.

Consider now the case n=12. The argument of the previous paragraph does not work, and we need more detailed inspection. Let 𝔭 and 𝔮 be as at the beginning of the proof, with 𝔭 corresponding to odd permutations, and such that 1 and 5 are partial sums in 𝔭. If 2 is not a partial sum and 4 is a partial sum, then it is easy to deduce 𝔭=(1,4,7). If 2 and 4 are not partial sums, then 𝔭=(1,5,6). Assume now 2 is a partial sum. If 3 is not a partial sum, then it is easy to check that 𝔭 must have four cycles, false. If 3 is a partial sum and 4 is not a partial sum, then 5 cannot be a partial sum, false. If 4 is a partial sum, then 𝔭 is isolated unless 6 is not a partial sum and 𝔭 is adjacent to (12). With this more detailed information, it is not difficult to deduce d(Ξ(S12))6. The proof of the theorem for n12 is now concluded. In the next lemma, we consider the case n=8. The case n=10 can be dealt with similarly, and we omit the details. ∎

We compute the exact diameter of Ξ(S8); this shows that the upper bound in Theorem 1.3 can be attained.

Lemma 3.7.

The graph Ξ(S8) is connected with diameter 6.

Proof.

In Figure 1, we have drawn the graph Ξ(S8). The group S8 has 21 nontrivial conjugacy classes; one can compute explicitly the neighborhood of each of them in Λ(S8). We can save some computations in view of the following observations. Whenever 4 is a partial sum in a partition 𝔭, 𝔭 is not adjacent to partitions having only parts of even length because of the sharing of S4S2. It follows that if, in a partition 𝔭, the integers 1, 3 and 4 are partial sums, then 𝔭 is isolated. This implies that the set of vertices of Ξ(S8) is a subset of

A:={(13,5),(12,6),(1,2,5),(1,7),(3,5),(2,32),(42),(2,6),(24),(22,4),(8)}.

Note also that partitions which, for every odd integer , have an even number (possibly zero) of parts of length are not adjacent to partitions having only parts of even length because of S2S4. We observe finally that the only core-free maximal primitive subgroups of S8 are AGL3(2) and PGL2(7) (up to conjugation). Among the partitions in A, AGL3(2) contains (1,7), (42), (2,6), (24), and PGL2(7) contains (12,6), (1,7), (42), (24), (8). It is now easy to draw the graph. ∎

Figure 1 The graph Ξ⁢(S8){\Xi(S_{8})}.
Figure 1

The graph Ξ(S8).

The last case to consider is alternating groups of odd degree.

Theorem 3.8.

Let n5 be an odd integer. Then Ξ(An) is connected, and we have d(Ξ(An))6.

Proof.

The cases n=5,7,9 can be checked explicitly; we omit the details and assume n11. The cases n=11,13,17,19 have been considered in Theorem 3.4 and in the comments following it. Therefore, we need to consider the cases n=15 and n21. We first assume n21, and deal with the case n=15 at the end of the proof. Throughout the proof, recall Lemmas 2.8 and 2.5.

Let x,yAn, and assume xAn is adjacent to yAn in Ξ(An). We will show that xAn or yAn is adjacent to at least one among (12,n-2), (22,n-4), every class with cycle type (1,3,n-4), (1,4,n-5) and (2,8,n-10). This will show that every vertex of Ξ(An) has distance at most 2 from one of these vertices. The argument given in the next two paragraphs shows that all these vertices are adjacent to (62,n-12). It will follow that every vertex of Ξ(An) has distance at most 3 from (62,n-12), which clearly will conclude the proof.

Let us analyze the classes (12,n-2), (22,n-4), (1,3,n-4), (1,4,n-5) and (2,8,n-10). By Lemma 2.2, only (1,4,n-5) and (2,8,n-10) belong to some maximal transitive imprimitive subgroup; they are contained in S5Sn/5. Regarding maximal core-free primitive subgroups, by Theorem 2.3, (12,n-2) belongs only to PΓL2(q) with n=q+1. Moreover, (22,n-4), (2,8,n-10) and (1,3,n-4) belong to no core-free primitive subgroups, the first two by Corollary 2.4, the last by [8, Theorem 1.1]. Again by [8, Theorem 1.1], (1,4,n-5) belongs possibly only to AGLd(5).

Now consider the class (62,n-12). The maximal transitive imprimitive subgroups it is contained in are S3Sn/3 and Sn/3S3. Moreover, by Lemma 2.11, it belongs neither to AGLm(p) nor to PΓL2(q). Therefore, as claimed (12,n-2), (22,n-4), every class with cycle type (1,3,n-4), (1,4,n-5) and (2,8,n-10) are adjacent to (62,n-12).

Hence, in order to conclude the proof, it is sufficient to prove the initial claim, that is, to prove that xAn or yAn is adjacent to at least one among (12,n-2), (22,n-4), every class with cycle type (1,3,n-4), (1,4,n-5) and (2,8,n-10). In a previous paragraph, we determined the maximal overgroups of these classes. In the following, we will freely use this information with no further mention.

Denote by c(x) and c(y) the cycle types of x and y, respectively. Assume there exists z{x,y} such that 1 and 2 are not partial sums in c(z); without loss of generality, z=x. Then, by Theorem 2.3, either xAn is adjacent to (12,n-2), or xPΓL2(q). If moreover 4 is not a partial sum in c(x), then xAn is adjacent to (22,n-4). Assume then that 4 is a partial sum in c(x). The unique possibility is c(x)=(4,), from which 1x4 fixes an even number of points greater than 3, from which xPΓL2(q) by Lemma 2.11 (2).

Therefore, we assume (without loss of generality) that 1 is a partial sum in c(x) and 2 is a partial sum in c(y), that is, c(x)=(1,) and c(y)=(2,). Then either xAn is adjacent to (22,n-4), or 4 is a partial sum in c(x). In the latter case, we have c(x)=(1,3,) or c(x)=(1,4,). If c(x)=(1,3,), then yAn is adjacent to (1,3,n-4). If c(x)=(1,4,), then yAn is adjacent to (1,4,n-5) unless y is contained in AGLd(5) or S5Sn/5. The option yAGLd(5) is excluded by Lemma 2.11 (1) because y2 fixes 2 points. If yS5Sn/5, since 5 is a partial sum in c(x), the unique possibility for the 2-cycle is to act as a 2-cycle on the blocks, from which c(y)=(2,8,). Now, since certainly xS5Sn/5, xAn is adjacent to (2,8,n-10). This concludes the proof in case n21.

There remains the case n=15. Let x,yAn, and assume xAn is adjacent to yAn in Ξ(An). We prove that xAn or yAn is adjacent to at least one between (12,13), (1,72) and (15) (this makes sense by Lemmas 2.7 and 2.12). By Theorem 2.3 and Lemma 2.2, (12,13) and (15) are adjacent. Moreover, (1,72) and (15) are both adjacent to (22,11), and (1,72) and (12,13) are both adjacent to (32,9). It follows that (12,13), (1,72) and (15) have pairwise distance at most 2, from which indeed d(Ξ(A15)6.

By Theorem 2.3, if xAn and yAn are not adjacent to (12,13), then (without loss of generality) c(x)=(1,) and c(y)=(2,). Assume now xAn and yAn are not adjacent to (15). We consider the various possibilities. Notice that xS3S5 because otherwise 2 would be a partial sum in c(x). Direct inspection (using for instance GAP) shows that if xPGL4(2), then it must be c(x)=(1,72); hence we may assume this is not the case. Therefore, by Theorem 2.3, it must be xS5S3, and yS3S5 or yPGL4(2). Since xS5S3, we have c(x)=(1,4,); hence 4 is not a partial sum in c(y). Inspection immediately implies yPGL4(2), from which yS3S5. In c(y), the 2-cycle either acts trivially on the blocks, or acts as a 2-cycle on the blocks. In the first case, 1 is a partial sum in c(y), and in the second case, 4 is a partial sum in c(y). In both cases, we get a contradiction, and the proof is finished. ∎

Now the proof of Theorem 1.3 and Theorem 1.4 follows immediately from Theorems 3.4, 3.5, 3.6 and 3.8.

4 Some comments on Conjecture 1.5

Conjecture 1.5 states that if G{An,Sn}, then, up to finitely many exceptions, one has d(Ξ(G))4. Here we reduce this conjecture to the following one (and in fact to something much weaker, see Remark 4.4).

Conjecture 4.1.

Let G{An,Sn}. There exists an absolute constant c>0 such that if xG is a vertex of Ξ(G), then xG is adjacent to a class which has at most c cycles.

A way to think about this is that, since xG is a vertex of Ξ(G), by definition, xG is adjacent to some other class yG. It seems conceivable that, summing the parts of the cycle type of y in a suitable way, one obtains that xG is indeed adjacent to some zG, where z has a bounded number of cycles. In fact, we believe that the value of c should be rather small, say at most 4; by [8], only a “few” core-free primitive subgroups contain elements having at most 4 cycles.

We now record a consequence of the Prime Number Theorem.

Theorem 4.2.

Fix ξ>0. Denote by π(n) the number of primes less or equal to n. Then π((1+ξ)n)-π(n) is asymptotic to ξn/lnn.

Proof.

The Prime Number Theorem states that π(n) is asymptotic to n/lnn; hence the statement follows from an easy computation. ∎

Theorem 4.3.

Conjecture 4.1 implies Conjecture 1.5.

Proof.

It is sufficient to show that, for n large, vertices which have at most c cycles have pairwise distance at most 2 in Ξ(G). Let xG and yG be two such vertices, and denote by c(x) and c(y) the cycle type of x and y, respectively.

We first claim that if n is sufficiently large, then there exist distinct prime numbers p and r such that

  1. n/3<p,rn/2,

  2. p,r and p+r are not partial sums in c(x) and in c(y),

  3. p,r and p+r do not divide n.

Let us prove the claim. By Theorem 4.2, the number of primes contained in the interval (n/3,n/2] is asymptotic n/6lnn. The number of divisors d(n) of n is much smaller; it is known that d(n)=o(nϵ) for every fixed ϵ>0 (cf. [1, Theorem 13.12, or Exercise 13, p. 303]).

Notice now that there are at most 2c+1=O(1) integers i such that i is a partial sum in at least one between c(x) and c(y).

The claim now follows because the number of primes in the interval (n/3,n/2] is much larger than all the other quantities considered above; hence, among all possibilities for p and r, we certainly find one satisfying (b) and (c).

At this point, we conclude the proof. If G=An and n is even, or G=Sn and n is odd, then both xG and yG are adjacent to (p,n-p). Indeed, by (c) (p,n-p) is not contained in transitive imprimitive subgroups, and a power of (p,n-p) is a p-cycle, hence does not lie in core-free primitive subgroups by a classical theorem of Jordan [16, Theorem 13.9]. Assume now G=An with n odd, or G=Sn with n even. Then we claim that both xG and yG are adjacent to every class with cycle type (p,r,n-p-r). It is easy to deduce from Lemma 2.2 and from (c) above that (p,r,n-p-r) does not belong to transitive imprimitive subgroups of Sn. Moreover, since min{p,r}>n/3, we have n-p-r<min{p,r}. It follows that a power of (p,r,n-p-r) is a p-cycle, hence does not belong to core-free primitive subgroups by Jordan’s theorem.

We have shown that xG and yG have distance at most 2 in Ξ(G), and the proof is concluded. ∎

Remark 4.4.

In the proof of the previous theorem, we have used a much weaker hypothesis than the validity of Conjecture 4.1. Indeed, the same argument works provided each vertex xG of Ξ(G) is adjacent to a vertex yG such that the number of integers i which are partial sums in (the cycle type of) y is at most δn/logn for some explicit fixed constant δ. This statement seems more suitable for a combinatorial proof. Any proof of this sort would very likely avoid the use of the CFSG.

We conclude with a lemma providing a lower bound to d(Ξ(G)) in some cases.

Lemma 4.5.

Assume n7 is prime. Then d(Ξ(Sn))4.

Proof.

The strategy is to define two partitions 𝔭 and 𝔮 such that 𝔭 is adjacent in Ξ(Sn) only to (n), and 𝔮 is adjacent in Ξ(Sn) only to (1,n-1). Since (n) and (1,n-1) are not adjacent because of the sharing of AGL1(n), this will prove indeed that d(Ξ(Sn))4.

Define

𝔭={(1(n-1)/2,(n+1)/2)ifn3mod4,(1(n+1)/2,(n-1)/2)ifn1mod4.

Note that 𝔭 corresponds to odd permutations. Moreover, every 1in/2 is a partial sum in 𝔭; hence 𝔭 is not adjacent to partitions having at least 2 parts. Finally, 𝔭 is adjacent to (n) by Theorem 2.3.

Now define

𝔮={(2(n-3)/2,3)ifn3mod4,(2(n-9)/2,33)ifn1mod4.

Note that 𝔮 corresponds to even permutations, so it is not adjacent to (n). It is easy to check that every 2in/2 is a partial sum in 𝔮; hence 𝔮 is adjacent to nothing different from (1,n-1). Finally, since 𝔮AGL1(n), we deduce by Theorem 2.3 that 𝔮 is indeed adjacent to (1,n-1). This concludes the proof of the lemma. ∎

Similar methods should suffice to give sensible lower bounds to d(Ξ(G)) in all the various cases. However, the details would become slightly technical, as one would need to specialize the argument depending on the arithmetic of n.


Communicated by Timothy C. Burness


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Received: 2019-12-05
Revised: 2020-06-20
Published Online: 2020-07-10
Published in Print: 2020-11-01

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