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.
Given a finite group G and a subset of G, we say that invariably generatesG if for every . This concept was introduced by Dixon with motivations from computational Galois theory; see  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 graph of G as follows. The vertices are the conjugacy classes of G different from , and two vertices and are adjacent if and only if 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 , and Guralnick–Malle  independently, that finite simple groups are invariably generated by two elements so that, in this case, is nonempty. It is also known that is invariably generated by two elements if and only if (see [15, Proposition 4.10]). Set
The first two results of the paper are the following.
Assume , and let . Then does not have isolated vertices if and only if and n is a prime satisfying and .
The following statements hold.
Let be a sequence of alternating or symmetric groups so that . Assume that, for every i, is not alternating of prime degree. Then the number of isolated vertices of tends to infinity.
Assume n is a prime not contained in . Then the number of isolated vertices of is at most 2.
In Theorem 1.1, we assumed . We will keep this assumption throughout the paper; see Lemma 2.15 for the remaining cases. For the sake of clarity, we mention that and have 5 and 6 isolated vertices, respectively (Lemma 3.2). The only case not addressed in Theorem 1.2 is with prime. In Remark 3.3, we will obtain a partial result, dealing with the case with .
Once Theorems 1.1 and 1.2 are proved, one may ask what happens if the isolated vertices are removed from . With this purpose, we define a graph which is obtained from by removing the isolated vertices. The next result states that, except in case , this graph is connected with bounded diameter (we already recalled that is not invariably generated by two elements, hence is the null graph).
Assume , and let , with . Then is connected with diameter at most 6.
In many cases, we prove better estimates on the diameter.
Assume . If and n is odd, or if and n is even, then . If and is prime, then .
The proofs of Theorems 1.3 and 1.4 rely on some recent results, proved in  and , which classify the primitive subgroups of containing elements having certain cycle types. These results depend on the Classification of Finite Simple Groups.
Let . If n is sufficiently large, then .
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 for all primes , 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.
Let G be a finite simple group. Is the graph connected?
1.2 Some context
The invariably generating graph is the analogue, for invariable generation, of the so-called generating graph 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 .
Many properties of generation of a finite simple group by two elements can be stated in terms of the generating graph. Guralnick and Kantor  proved that if G is a finite simple group, then, for every , there exists such that . Later, Breuer, Guralnick and Kantor  showed that if , then there exists such that . These properties can be stated respectively as follows: 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 .
1.3 An alternative definition
In light of Theorem 1.2, one could ask how the proportion of isolated vertices of behaves as 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 to be the nontrivial conjugacy classes of G. One could define a graph 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 , there is a dependence on the size of the conjugacy classes which does not exist in .
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.
The vertices of our graphs are conjugacy classes of alternating and symmetric groups. We will identify conjugacy classes of with their cycle type, i.e., we will represent conjugacy classes of as partitions of n. We now introduce some terminology about partitions.
Let , and let be a partition of n. We will say that belongs to H, or that is contained in H, if H contains elements with cycle type . If C is the conjugacy class of corresponding to , this is equivalent to the condition . Of course, this condition depends only on the -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 belongs to the -conjugacy class of H.
Let and be two partitions of n. If and both belong to H, we will say that and shareH. When , with , we will say that i is a partial sum in . This is indeed equivalent to the condition that the integer i can be written as the sum of some parts of . In a partition, will mean m parts of length i. Therefore, will mean parts of length , parts of length , …, and parts of length .
Occasionally, given a partition and a positive integer i, we will write to denote “the i-th power of ”, namely the partition obtained by replacing each part of length by d parts of length , where . Note that if has cycle type , then has cycle type .
Finally, we define
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 ; 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].
Let n be a natural number, m a nontrivial divisor of n and . The partition belongs to if and only if either m divides or divides .
If belongs to , the induced permutation on the blocks has at most two cycles. If it has two cycles, then m divides . If it is an -cycle, then divides . The converse implication is proved in the same way. ∎
Let n be a natural number, m a nontrivial divisor of n and . The partition belongs to if and only if one of the following conditions is satisfied:
m divides for every i;
divides for every i;
there exist and such that
Similar to the previous lemma. In case (a), the induced permutation on the blocks has cycle type . In case (b), it is an -cycle. In case (c), it has cycle type , where . ∎
We now move to primitive subgroups. Our main tool is a theorem which classifies the primitive subgroups of 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 different from 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 ().
Let G be a primitive permutation group of finite degree n, not containing the alternating group . Suppose that G contains a cycle fixing k points, where . Then one of the following holds:
either with prime,
or with and for some prime power q,
or , or with or 23 , respectively.
either with and for some prime power q,
or or with for some prime ,
or , or with or 24 , respectively.
and with for some prime power q.
Note that the statement implies that there are no proper primitive subgroups of different from containing a cycle fixing at least 3 points, generalizing indeed Jordan’s theorem. We note the following immediate consequence.
Assume 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 different from .
We also mention that we will make essential use of the main result from , which classifies the primitive subgroups of 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
If a partition of n is made of distinct odd parts, the corresponding -conjugacy class splits into two -conjugacy classes (and vice versa), giving rise to two vertices of . 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.
Let , and let be such that . Then H contains elements belonging to if and only if it contains elements belonging to for every .
Assume with , and assume with . Then for some by hypothesis, hence . This concludes the proof. ∎
Assume are such that is adjacent to in for any and for any . Under these assumptions, we say with slight abuse of notation that is adjacent to in .
This notation will be convenient, as we will represent -conjugacy classes as partitions, and we will be allowed to say that “a partition is adjacent to a partition ”, rather than “any -conjugacy class of elements with cycle type is adjacent to any -conjugacy classes of elements with cycle type ”. We will now see that, in many cases, the assumption of Notation 2.6 is satisfied.
Let , and assume . Let . Then is adjacent in to if and only if and are adjacent (in the terminology of Notation 2.6).
Choose . We show first that if is not adjacent to , then is not adjacent to . By assumption, we can write for some integer i and for some . Again by assumption, for some , and for some proper subgroup H of . Then
whence and is not adjacent to , as required.
Assume now . We show that if is not adjacent to , then is not adjacent to . This will conclude the proof. We may assume and ; otherwise, the statement is easy. Choose . By the previous paragraph, if is not adjacent to , then it is not adjacent to so that for some proper subgroup H of , some and . Then , and since is -conjugate of , the proof is concluded. ∎
We now apply the previous considerations to certain elements and subgroups of .
Let H be a maximal subgroup of which is either intransitive, or transitive and imprimitive. Then .
We may assume . We have if and only if . In our case, since every maximal intransitive or imprimitive subgroup of contains transpositions. ∎
Assume belongs to no proper primitive subgroup of different from . Let . Then and are adjacent in if and only if and are adjacent (in the terminology of Notation 2.6).
By assumption, x lies in no proper primitive subgroups different from . Therefore, is adjacent to if and only if, for every , 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 -conjugacy class. The lemma follows. ∎
Let be coprime with n. Then does not belong to proper primitive subgroups different from , and does not belong to transitive imprimitive subgroups.
Theorem 2.3 can be used to generalize the previous lemma to the case ; one just needs to take care of some specific examples of primitive subgroups. The following easy lemma will be used with this purpose.
Let q be a prime power and d a positive integer.
An element of , in the natural action on points, either is a derangement, or fixes a number of points equal to for some .
Assume fixes at least 3 points in the natural action on points. Then g fixes a number of points having the same parity of . Moreover, if , then .
(1) If fixes some point, we may assume that it fixes 0. Hence, . Now just observe that the set of fixed points of an element of is an -subspace of .
(2) Consider acting (on the right) on the set Ω of 1-dimensional subspaces of . Write with p prime. For , denote by the permutation of Ω induced by the mapping of . Then we may express each element as , where and .
Note that is 3-transitive on Ω. Hence, if fixes at least 3 points, we may assume that it fixes , and . It follows that and . Then g fixes points, where is a divisor of r; in particular, it fixes a number of points having the same parity of . Moreover, if , then . The lemma is proved. ∎
Assume n is either an odd prime, or . Assume is an n-cycle. Then .
If is an odd integer, then an n-cycle is normalized by elements having cycle type . If n is an odd prime, then an n-cycle is normalized by an -cycle. In particular, if x is as in the statement, then , from which . ∎
We deduce another consequence of the previous lemmas.
Assume , and let . Then, for every , and are not adjacent in .
This lemma suggests a natural question for all finite simple groups.
Let G be a finite simple group. Let , and let . Is it possible that invariably generates G?
It is easy to check that the statement is true for , in which case, has index two in . This, together with Lemma 2.13, implies that the question has a negative answer for .
2.4 Small degrees
The main theorems are stated for . For completeness, we address here the cases of degree . We assume in order to avoid trivialities.
Let , and assume . Then has isolated vertices if and only if . Moreover, is connected with diameter at most 2.
This is an easy check. The graphs , , have diameter 1, while has diameter 2; the two classes of 3-cycles are connected by a path of length 2 passing through the class . In , the vertices corresponding to and are isolated. ∎
In this section, we prove the theorems stated in the introduction.
Let be an odd prime. Then and are isolated in (when they make sense and are even permutations). If , then there are no other isolated vertices in .
The two mentioned vertices are not adjacent to (this terminology makes sense by Lemma 2.7, Lemma 2.12 and Notation 2.6) because they are contained in . Therefore, they might be adjacent only to some -conjugacy class with cycle type , with . Every is a partial sum in , so this vertex is isolated (recall Lemmas 2.8 and 2.5), and the values of i that are not partial sums in are exactly those that satisfy . If , then , so one among and is a partial sum. We conclude that is indeed isolated.
Assume now . Let be a vertex of different from the two above; we want to show it is not isolated. If , we deduce from Theorem 2.3 that is adjacent to . On the other hand, if , then either it is an n-cycle, or it has cycle type , with . As just remarked, a class of n-cycles is not isolated. If , by Corollary 2.4, is adjacent to . If , for the same reason, is adjacent to . ∎
For the sake of clarity, we deal with the cases .
The graphs and have 5 and 6 isolated vertices, respectively.
Recall Theorem 2.3. Let . Let . If x does not belong to (for ) and (for ), the same argument as in the previous lemma shows that is adjacent to (note that can be embedded in ). Inspection (using for instance GAP) shows that if and , then belongs to
These are not adjacent to because of . By looking at partial sums, we see that these are all isolated. If and , then either is adjacent to one between and , or belongs to
These are all isolated for the same reason as above. ∎
We begin with Theorem 1.2. Item (2) follows from Lemma 3.1; hence we focus on item (1). Let or , with n nonprime if . For every n, we will define , a subset of the set of isolated vertices of , whose cardinality, as , goes to infinity. This will prove Theorem 1.2. For every m, denote by the set of all partitions of m different from which correspond to even permutation.
We first assume that, whenever n is odd, . Define as the set of all partitions of n of the form , where is whatsoever partition of .
Let . Every is a partial sum, so is not adjacent to classes of elements with at least two cycles. If n is even, then is not adjacent to since is a partial sum in , hence . If n is odd, then , and is not adjacent to since corresponds to even permutations. Therefore, consist of isolated vertices. Clearly, the size of goes to infinity as .
Assume now that and n is odd and not prime. Fix n, and let be the smallest prime divisor of n. Define as the set of all partitions of n of the form , where is any partition of .
If , then ; the first blocks are fixed pointwise. Therefore, is not adjacent to . Moreover, every is a partial sum, so is not adjacent to classes of elements having at least 2 cycles. It follows that is isolated. Note now that ; hence the size of goes to infinity as . 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 . If , by Lemma 3.1, there are isolated vertices in . The cases have been considered in Lemma 3.2. If , let x be any involution lying in a subgroup of conjugate to . The fact that n is odd implies that every is a partial sum in (the cycle type of) x; hence might only be adjacent to a class of n-cycles. However, this does not happen because of the containment in . Therefore, is isolated.
The case with n prime is therefore proved. For the remaining cases, we apply what we proved for Theorem 1.2. We define . This partition belongs to , as defined in this proof; hence it is an isolated vertex of . This concludes the proof. ∎
The unique case not discussed in Theorem 1.2 is the case and prime. We obtain here the partial result that if is prime and , then the number of isolated vertices of tends to infinity. Note that since is prime, must be prime. To establish whether infinitely many such primes (i.e., primes d such that is prime for some prime power q) do actually exist, however, is a hard open problem in number theory; see for instance .
Assume first with q (and d) odd. The action of on the 1-dimensional subspaces of gives an embedding . For every , let be a diagonal matrix of with 1’s and -1’s on the diagonal, and assume the number of -1’s is r. Let denote the image of x in . Then has fixed points. In particular, any two distinct give rise to elements of which have a different number of fixed points, and which therefore belong to different -conjugacy classes. It is easy to check that the element arising in this way belongs to . The number of possibilities for r in order to obtain such an element is . As remarked in the previous proof, is isolated in ; therefore, the number of isolated vertices of is at least .
Assume now q is even. For every , consider a unipotent element of with Jordan blocks of size 2, and with the other Jordan blocks of size 1. This is an involution of . Denote again by the image of x in . Then has fixed points; hence any two distinct give rise to elements belonging to different -conjugacy classes. Moreover, (unless , but recall we are assuming ), and is isolated in . Hence the number of isolated vertices of is at least .
One brief comment about the terminology we will adopt. The proofs will begin with a sentence of the type “Let be a vertex of ”, without any preliminary consideration showing that is not the null graph. However, along the proof suitable edges will be exhibited in so that the initial choice of 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).
Assume is a prime and . Then is connected and .
Let , and assume is a vertex of . If , in the proof of Lemma 3.1, we showed that is adjacent to one among , and (see Lemma 2.12). These vertices are pairwise adjacent by Corollary 2.4; hence we have indeed . If or 23, we observed in the proof of Lemma 3.2 that a class inside or is either isolated, or adjacent to one between and . Therefore, by Corollary 2.4, also in this case, . ∎
For a later use, we point out that the same estimate holds for the groups and .
Assume . Inspection shows that every class lying in is either isolated or adjacent to ; hence by the same argument as in the proof of the previous theorem. Assume finally that . Here every class lying in is either isolated or adjacent to one between and . It is easy to deduce that .
Let be an integer.
If n is odd, then is connected and .
If n is even, then is connected and .
(1) Let be a vertex of . Assume is not adjacent to for any coprime with n. Then, by Lemma 2.10, every such i is a partial sum in . Since is a vertex of , will be adjacent to some vertex . Necessarily, no coprime with n is a partial sum in . Hence, again by Lemma 2.10, is adjacent to for every such i.
Now note that the ’s, with i as above, are pairwise adjacent. From this, it follows that, for any fixed coprime with n, any vertex of has distance at most 2 from . This concludes the proof.
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 ; 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  which, as already mentioned in the introduction, classifies the primitive permutation groups having elements with at most four cycles.
Let be an even integer. Then is connected, and we have .
We first assume and consider the remaining cases at the end of the proof. Let and be two vertices of joined by an edge. One of the two, say , must correspond to odd permutations. Assume is not adjacent to for any coprime with n. By Lemma 2.10, every such i, with , is a partial sum in . We now show that also is a partial sum in . Assume this is not the case. Since is not adjacent to , we deduce from Lemma 2.1 and Theorem 2.3 that is contained in one between , , , , . The last three are excluded because they are subgroups of . Assume or . Since every coprime with n is a partial sum, must have odd parts; let a be the smallest such part. Since is not a partial sum, we have . On the other hand, by assumption, corresponds to odd permutations; hence it has even parts. It follows that 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, is indeed a partial sum in .
Now we divide the cases and .
Assume . Then is coprime with n; hence it is a partial sum in . Write , and assume . If for some , then is a partial sum in . Otherwise, is a partial sum in . We show that is adjacent in the first case to , and in the second case to .
By the considerations above, and do not share intransitive subgroups. Moreover, and correspond to odd permutations, and belong to no transitive imprimitive subgroups by Lemma 2.2. Finally, and belong to no core-free primitive subgroups by [8, Theorem 1.1]. We have therefore our desired edge between and or .
Assume . We employ the same argument as above, with replaced by . The same reasoning lead us to look for an edge between and or . Again, and do not share intransitive subgroups. It follows from [8, Theorem 1.1] that and are not contained in core-free primitive subgroups (for , we may also use Corollary 2.4). Regarding maximal transitive imprimitive subgroups, we only have that is contained in . However, by construction, we consider only when is a partial sum in so that belongs to . Since and are adjacent, we deduce that is not contained in . Therefore, we have an edge between and or .
Now we deduce the connectedness of and the bound to the diameter. The considerations above imply that an edge with and concerns only intransitive subgroups (i.e., partial sums), except for , where one has to deal also with .
Assume first . The argument given above shows that every vertex of has distance at most 2 from one among and for some coprime with n. Hence, in order to conclude, it is sufficient to show that these vertices have pairwise distance at most 2. For , this can be checked directly. Assume then . We show that all these vertices are adjacent to , which clearly concludes the proof. For all the vertices except , this follows from Lemma 2.10 and from the considerations of the previous paragraph. For , by Theorem 2.3, we need to exclude the sharing of , , . The last is contained in , while is not. Moreover, ; hence n is not a power of 2 and we do not have affine subgroups. Finally, fixes at least 4 points; hence it does not belong to by Lemma 2.11(2). This concludes the proof in case .
Assume now . We assume first . As in case , in order to conclude, it is sufficient to prove that the vertices and with coprime with n have pairwise distance at most 2. The vertices and with coprime with n and are adjacent to . The vertices with coprime with n and are adjacent to both and . The vertices with coprime with n are adjacent to . By [8, Theorem 1.1] (which in the affine case relies on [7, Theorem 1.5]), we deduce that is not contained in affine subgroups; hence also is adjacent to . These considerations imply that the vertices have pairwise distance at most 2.
Consider now the case . The argument of the previous paragraph does not work, and we need more detailed inspection. Let and be as at the beginning of the proof, with corresponding to odd permutations, and such that 1 and 5 are partial sums in . If 2 is not a partial sum and 4 is a partial sum, then it is easy to deduce . If 2 and 4 are not partial sums, then . Assume now 2 is a partial sum. If 3 is not a partial sum, then it is easy to check that 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 is isolated unless 6 is not a partial sum and is adjacent to . With this more detailed information, it is not difficult to deduce . The proof of the theorem for is now concluded. In the next lemma, we consider the case . The case can be dealt with similarly, and we omit the details. ∎
We compute the exact diameter of ; this shows that the upper bound in Theorem 1.3 can be attained.
The graph is connected with diameter 6.
In Figure 1, we have drawn the graph . The group has 21 nontrivial conjugacy classes; one can compute explicitly the neighborhood of each of them in . We can save some computations in view of the following observations. Whenever 4 is a partial sum in a partition , is not adjacent to partitions having only parts of even length because of the sharing of . It follows that if, in a partition , the integers 1, 3 and 4 are partial sums, then is isolated. This implies that the set of vertices of is a subset of
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 . We observe finally that the only core-free maximal primitive subgroups of are and (up to conjugation). Among the partitions in A, contains , , , , and contains , , , , . It is now easy to draw the graph. ∎
The last case to consider is alternating groups of odd degree.
Let be an odd integer. Then is connected, and we have .
The cases can be checked explicitly; we omit the details and assume . The cases have been considered in Theorem 3.4 and in the comments following it. Therefore, we need to consider the cases and . We first assume , and deal with the case at the end of the proof. Throughout the proof, recall Lemmas 2.8 and 2.5.
Let , and assume is adjacent to in . We will show that or is adjacent to at least one among , , every class with cycle type , and . This will show that every vertex of 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 . It will follow that every vertex of has distance at most 3 from , which clearly will conclude the proof.
Let us analyze the classes , , , and . By Lemma 2.2, only and belong to some maximal transitive imprimitive subgroup; they are contained in . Regarding maximal core-free primitive subgroups, by Theorem 2.3, belongs only to with . Moreover, , and 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], belongs possibly only to .
Now consider the class . The maximal transitive imprimitive subgroups it is contained in are and . Moreover, by Lemma 2.11, it belongs neither to nor to . Therefore, as claimed , , every class with cycle type , and are adjacent to .
Hence, in order to conclude the proof, it is sufficient to prove the initial claim, that is, to prove that or is adjacent to at least one among , , every class with cycle type , and . 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 and the cycle types of x and y, respectively. Assume there exists such that 1 and 2 are not partial sums in ; without loss of generality, . Then, by Theorem 2.3, either is adjacent to , or . If moreover 4 is not a partial sum in , then is adjacent to . Assume then that 4 is a partial sum in . The unique possibility is , from which fixes an even number of points greater than 3, from which by Lemma 2.11 (2).
Therefore, we assume (without loss of generality) that 1 is a partial sum in and 2 is a partial sum in , that is, and . Then either is adjacent to , or 4 is a partial sum in . In the latter case, we have or . If , then is adjacent to . If , then is adjacent to unless y is contained in or . The option is excluded by Lemma 2.11 (1) because fixes 2 points. If , since 5 is a partial sum in , the unique possibility for the 2-cycle is to act as a 2-cycle on the blocks, from which . Now, since certainly , is adjacent to . This concludes the proof in case .
There remains the case . Let , and assume is adjacent to in . We prove that or is adjacent to at least one between , and (this makes sense by Lemmas 2.7 and 2.12). By Theorem 2.3 and Lemma 2.2, and are adjacent. Moreover, and are both adjacent to , and and are both adjacent to . It follows that , and have pairwise distance at most 2, from which indeed .
By Theorem 2.3, if and are not adjacent to , then (without loss of generality) and . Assume now and are not adjacent to . We consider the various possibilities. Notice that because otherwise 2 would be a partial sum in . Direct inspection (using for instance GAP) shows that if , then it must be ; hence we may assume this is not the case. Therefore, by Theorem 2.3, it must be , and or . Since , we have ; hence 4 is not a partial sum in . Inspection immediately implies , from which . In , 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 , and in the second case, 4 is a partial sum in . In both cases, we get a contradiction, and the proof is finished. ∎
4 Some comments on Conjecture 1.5
Let . There exists an absolute constant such that if is a vertex of , then is adjacent to a class which has at most c cycles.
A way to think about this is that, since is a vertex of , by definition, is adjacent to some other class . It seems conceivable that, summing the parts of the cycle type of y in a suitable way, one obtains that is indeed adjacent to some , 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 , only a “few” core-free primitive subgroups contain elements having at most 4 cycles.
We now record a consequence of the Prime Number Theorem.
Fix . Denote by the number of primes less or equal to n. Then is asymptotic to .
The Prime Number Theorem states that is asymptotic to ; hence the statement follows from an easy computation. ∎
It is sufficient to show that, for n large, vertices which have at most c cycles have pairwise distance at most 2 in . Let and be two such vertices, and denote by and 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
and are not partial sums in and in ,
and do not divide n.
Let us prove the claim. By Theorem 4.2, the number of primes contained in the interval is asymptotic . The number of divisors of n is much smaller; it is known that for every fixed (cf. [1, Theorem 13.12, or Exercise 13, p. 303]).
Notice now that there are at most integers i such that i is a partial sum in at least one between and .
The claim now follows because the number of primes in the interval 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 and n is even, or and n is odd, then both and are adjacent to . Indeed, by (c) is not contained in transitive imprimitive subgroups, and a power of 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 with n odd, or with n even. Then we claim that both and are adjacent to every class with cycle type . It is easy to deduce from Lemma 2.2 and from (c) above that does not belong to transitive imprimitive subgroups of . Moreover, since , we have . It follows that a power of is a p-cycle, hence does not belong to core-free primitive subgroups by Jordan’s theorem.
We have shown that and have distance at most 2 in , and the proof is concluded. ∎
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 of is adjacent to a vertex such that the number of integers i which are partial sums in (the cycle type of) y is at most 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 in some cases.
Assume is prime. Then .
The strategy is to define two partitions and such that is adjacent in only to , and is adjacent in only to . Since and are not adjacent because of the sharing of , this will prove indeed that .
Note that corresponds to odd permutations. Moreover, every is a partial sum in ; hence is not adjacent to partitions having at least 2 parts. Finally, is adjacent to by Theorem 2.3.
Note that corresponds to even permutations, so it is not adjacent to . It is easy to check that every is a partial sum in ; hence is adjacent to nothing different from . Finally, since , we deduce by Theorem 2.3 that is indeed adjacent to . This concludes the proof of the lemma. ∎
Similar methods should suffice to give sensible lower bounds to 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.
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