The invariably generating graph of the alternating and symmetric groups

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 connected if and only if they invariably generate $G$. In this paper we study this object for alternating and symmetric groups. First we observe that in most cases it has isolated vertices. Then, we prove that if we take them out we obtain a connected graph. Finally, we bound the diameter of this new graph from above and from below --- apart from trivial cases, it is between $3$ and $6$ ---, and in about half of the cases we compute it exactly.


Introduction
Given a group G and a subset S of G, we say that S invariably generates G if s g(s) , s ∈ S = G for every choice of g(s) ∈ G. The motivation for considering this kind of generation comes from computational Galois Theory, as now we breafly explain.
Let f ∈ Z[x] be monic, of degree n 1, separable over Q. If p is a prime number not dividing disc(f ), we can associate to p a partition of n whose parts are the degrees of the distinct factors of f mod pZ [x]. A theorem due to Dedekind states that Gal(f ) contains an element having this cycle type. Actually much more is true: the Frobenius Density Theorem says that, asymptotically, the proportion of primes having a given cycle type is equal to the proportion of elements of Gal(f ) having that cycle type. This allows to gain informations on Gal(f ) factoring f modulo different primes; the point is to understand how many factorisations are necessary. For instance, if one suspects that Gal(f ) = S n , the expected number of factorisations needed to prove this fact coincides with the expected number of elements needed to invariably generate S n . These problems have been considered first in [5]. Then, several results have been obtained, for example in [17], [20], [7], [8], [4]. In [20] it is shown that 4 random permutations invariably generate S n with probability bounded away from zero, whilst in [8] it is shown that, if we replace 4 with 3, the same probability tends to zero as n → ∞.
We forget for a while invariable generation, and we recall that it is possible to associate to every finite group G an undirected graph Γ(G), called the generating graph of G, in which the vertices are the nonidentity elements of G, and two vertices x and y are connected if and only if x, y = G.
If G is a nonabelian finite simple group, it is known that Γ(G) is nonempty. In fact, in [11] and in [3] it is shown that Γ(G) is connected with diameter 2. The same holds for symmetric groups of degree at least 5, as shown in [1] and [2]. For other properties of the generating graph, see for example [14], [15], [16]. Now we put together generating graph and invariable generation. If G is a finite group, we define an undirected graph Λ(G), called the invariably generating graph of G, in which the vertices are the conjugacy classes of G different from {1}, and two vertices Cl(x) and Cl(y) are connected if and only if {x, y} invariably generates G. In [13] it is shown that, if G is a nonabelian finite simple group, Λ(G) is nonempty.
The aim of this paper is to study this object for alternating and symmetric groups, in order to understand whether there are or not similarities with its classical analogue. First of all, it is easy to see that, expect for the case of S 6 , these groups are invariably generated by 2 elements. Moreover, it is easy to observe that in most cases Λ has isolated vertices. However, if we cancel them we obtain a new graph, that we denote with Ξ, which is connected and whose diameter is quite small -although not as small as that of the generating graph. These last two sentences summarise the content of this paper; we state them explicitly in the following theorems. or all the following conditions are satisfied: G n = A n , n is prime, n ≡ −1 mod 12, n = 11, 23, n is different from the cardinality of any projective space.
(2) As n → ∞, with n nonprime if G n = A n , the number of isolated vertices of Λ(G n ) goes to infinity as well.
Theorem 1.2. Let n 3 be a natural number, and let G ∈ {S n , A n }. Then, Ξ(G) is connected unless G = S 6 , in which case Λ(S 6 ) is the empty graph. In Table 1 the diameters for the groups of degree at most 10 are listed. Assume now n 11.
(b) n even: 4 d(Ξ(S n )) 6.  The following is dedicated to the proof of the previous results. The computations for the cases of degree at most 10 are mostly missing. Often, in these cases, it is necessary to draw all the vertices and all the edges and look at the picture. In order to draw the edges, the arguments are the same as those used in the general case: including them thoroughly would be essentially a repetition.

Preliminary observations
The vertices of our graph are conjugacy classes of alternating and symmetric groups. A conjugacy class of S n is faithfully represented by its cycle type. The same holds for a conjugacy class of A n , unless its cycle type is made of distinct odd parts, in which case there are two conjugacy classes with the same cycle type. Except for the latter case, we will consider the vertices of the graph as partitions of n; we fix now some language about this.
Given G a subgroup of S n and given (a 1 , . . . , a t ) a partition of n, we will say that (a 1 , . . . , a t ) belongs to G, or that (a 1 , . . . , a t ) is contained in G, if G contains elements with cycle type (a 1 , . . . , a t ).
When two cycle types (a 1 , . . . , a t ) and (b 1 , . . . , b s ) belong to G, we will say that they share G.
When (a 1 , . . . , a t ) ∈ S i × S n−i , with 1 i n/2, we will say that i is partial sum in (a 1 , . . . , a t ).
In a cycle type, a writing of the form i m will mean m cycles long i (and not one cycle long i m ). Therefore, (a m 1 1 , a m 2 2 , . . . , a mt t ) will mean m 1 cycles long a 1 + m 2 cycles long a 2 + . . . + m t cycles long a t .
We will call nontrivial primitive group of degree n a primitive group of degree n not containing A n (the degree, when clear, will be omitted). For the rest, the terminology we will use is standard.
Cycle types with 1, 2 and 3 orbits will be crucial for us; our arguments will often concern the existence of edges with them. For this reason, it is important to establish which (maximal) subgroups of the symmetric groups such elements can belong to. We state two theorems, copied word-to-word respectively from [12] and [18], that address this problem for what concerns primitive subgroups. Before them, it seems appropriate to place a classical theorem in permutation groups.  (2, p) with n = p + 1 for some prime p 5, or (c) G = M 11 , M 12 or M 24 with n = 12, 12 or 24 respectively. 3. k = 2 and P GL(2, q) G P ΓL(2, q) with n = q + 1 for some prime power q. Theorem 2.3. Let G be a primitive permutation group of degree n which contains an element with exactly two cycles of lengths k and n − k k. Then one of the following holds, where G 1 denotes the stabiliser of a point.

(Affine action) G AGL(m, p)
is an affine permutation group, where n = p m and p is a prime number. Furthermore, one of the following holds.

(Almost simple action) S
G Aut(S) for a simple, nonabelian group S, and one of the following holds.
We mention that, although we do not report here the statements, we will make use of [10], that extends Theorem 2.3 to 3 and 4 orbits. The following remark, instead, deals with transitive imprimitive maximal subgroups.

Remark 2.4.
Let n be a natural number, m be a nontrivial divisor of n and 1 i n/2; the cycle type (i, n − i) belongs to S m ≀ S n/m if and only if either m divides i or n/m divides i. In the first case, the i-cycle permutes all the points of i/m blocks, and the (n − i)-cycle permutes all the points of (n − i)/m blocks. In the second case, the i-cycle permutes i/(n/m) points from each block, and the (n − i)-cycle permutes (n − i)/(n/m) points from each block.
Let now 2 i n/2 be coprime with n. By what just said, (i, n−i) is not contained in imprimitive wreath products. Also, (i, n − i) n−i = (i, 1 n−i ) = (1 n ); since n − i 3, by Theorem 2.1 (i, n − i) is not contained in nontrivial primitive groups. Therefore, if z is a vertex of Λ(S n ), z and (i, n − i) are connected if and only if they share neither A n nor S i × S n−i . Theorem 2.2 can be used to generalise the previous considerations to the case i = 1.
Analogous considerations hold for vertices with 3 orbits. In particular, (a 1 , a 2 , a 3 ) belongs to S m ≀ S n/m if and only if one of the following conditions is satisfied: (a) m divides a i for every i: the a i -cycle permutes all points of a i /m blocks.
(b) n/m divides a i for every i: the a i -cycle permutes a i /(n/m) points from each block.
(c) there exist 1 t < n/m and i = j such that a k = tb k , k = i, j, with b i + b j = m: by what said above, (a i , a j ) belongs to S m ≀ S t ; the remaining cycle permutes the remaining points in such a way that the blocks are preserved.
Instead, if n is even (i 2 , n − 2i) is contained in S 2 ≀ S n/2 (it is of type (c) above).
As already noticed, if a partition of n is made of distinct odd parts then the corresponding S n -conjugacy class splits into two A n -conjugacy classes (and viceversa), giving rise to two vertices of Λ(A n ). The following remark gives conditions under which these two vertices may be essentially thought as a unique vertex. (1) Let x ∈ A n be such that C Sn (x) ⊆ A n and let H S n be such that N Sn (H) A n . Then, H contains elements belonging to Cl An (x) if and only if it contains elements belonging to Cl An (x g ), with g / ∈ A n , if and only if it contains the cycle type of x.
This applies to imprimitive maximal subgroups of S n : these groups themselves are not contained in A n . Therefore, if x is as above and if x does not belong to nontrivial primitive groups, a vertex z of Λ(A n ) is connected to Cl An (x) if and only if it is connected to Cl An (x g ), if and only if the cycle type of z and the cycle type of x do not share imprimitive maximal subgroups of A n . Hence, if x c is the cycle type of x, we are allowed to use the expression "a vertex connected to x c ".
(2) Let x ∈ A n , with C Sn (x) ⊆ A n and N Sn ( x ) A n . If we apply (1) with H = x , we obtain that, if g / ∈ A n , Cl An (x g ) = Cl An (x i ), with i coprime with the order of x. Hence, a vertex z of Λ(A n ) is connected to Cl An (x) if and only if it is connected to Cl An (x g ), if and only if the cycle type of z and the cycle type of x do not share maximal subgroups of A n . Therefore, as above we may use the expression "a vertex connected to x c ", where x c is the cycle type of x.
This applies for instance when x is an n-cycle, where either n is an odd prime or n ≡ 3 mod 4: respectively, (n − 1)-cycles and elements with cycle type (1, 2 (n−1)/2 ) normalise x .
(3) Let x, y ∈ A n be such that C Sn (x) ⊆ A n and C Sn (y) A n . Then, Cl An (y) = Cl Sn (y) is connected to Cl An (x) if and only if it is connected to Cl An (x g ), with g / ∈ A n , if and only if the cycle type of x and the cycle type of y do not share maximal subgroups of A n .
(4) Let n 4, and let x ∈ A n with C Sn (x) ⊆ A n . Then, Cl An (x) is not connected to Cl An (x g ), g / ∈ A n . Indeed, by (1) we may assume that x is not contained in imprimitive groups, hence we may assume that x is an n-cycle and n is prime. By (2), a suitable A n -conjugate of x g lies in x , so the assertion follows.
The previous remark lists cases in which a vertex of Λ(A n ) may be considered as a partition of n, even when the identification is not faithful. This is not always the case: there are examples of x, y ∈ A n , with C Sn (x) ⊆ A n and C Sn (y) ⊆ A n , such that Cl An (x) is connected to Cl An (y) but not to Cl An (y g ), g / ∈ A n . In order to find such an example, by Remark 2.5 the normalisers of x and y must be contained in A n , and x c and y c cannot share imprimitive groups. Hence, we need that x c and y c share a primitive maximal subgroup of A n -let us say exactly one. It is necessary and sufficient that (each copy of) such group contains elements of exactly one A n -class with cycle type x c and exactly one A n -class with cycle type y c (therefore, also the normaliser of the primitive group must be contained in A n ). The example we give is in Λ(A 127 ). (3, 31, 93) and (7,15,105) are contained in P GL (7,2). They are not contained in imprimitive wreath products by Remark 2.4; and they are not contained in other primitive maximal subgroups by [10]. The normalisers of the cyclic groups they generate are indeed even groups; there remains to check the last condition on P GL (7,2). Using GAP, we see that each P GL(7, 2)-conjugacy class with cycle type (3, 31, 93) (resp. (7, 15, 105)) has a representative which is a power of a fixed element (same element for every class). Therefore, the fact that the normalisers of the cyclic groups are even groups is in fact sufficient in order to conclude (note that, given We conclude this section making some easy considerations about the number of fixed points of elements in some groups. We will often be interested in finding edges with (1, n − 1): Remark 2.4 says that, in order to do this, we will not need to worry about imprimitive wreath products; and Theorem 2.3 says that, apart for some exceptional cases, we will just need to exclude the sharing of AGL(m, p) and P GL (2, q), where p and q are primes. Remark 2.6 will be mostly used with this purpose. (2) P ΓL(2, q), with q prime power, is 3-transitive. Hence, if g ∈ P ΓL (2, q) fixes at least 3 points, we may assume that it fixes ∞, 0 and 1, from Such element fixes γ l + 1 points, where l is a divisor of r; in particular, it fixes a number of points having the same parity of q + 1. Moreover, if g ∈ P GL(2, q) then g = 1: only the identity in P GL(2, q) fixes at least 3 points.
Proof. The two mentioned vertices are not connected to (n) (this makes sense; recall Remark 2.5 (2)) because they are contained in AGL(1, n) (and n > 3). Therefore, they might be connected only to something with cycle type (a 1 , . . . , a t ), t 3. In (1, 2 (n−1)/2 ), every 1 i (n − 1)/2 is partial sum, so this vertex is isolated (recall Remark 2.5 (1)). In (1, 3 (n−1)/3 ), the i's which are not partial sum are exactly those ≡ 2 mod 3. If a 1 , a 2 ≡ 2 mod 3 then a 1 + a 2 ≡ 1 mod 3, so one among a 1 , a 2 , a 1 + a 2 is partial sum; (1, 3 (n−1)/3 ) is therefore isolated. Now let Cl(x) be a vertex of Λ(A n ) different from the two above; we want to show it is not isolated. If x / ∈ AGL(1, n), by Theorem 2.2 (and by the fact that there are not transitive imprimitive maximal subgroups), Cl(x) is connected to (n). On the other hand, if x ∈ AGL(1, n) either it is an n-cycle or it has cycle type (1, t (n−1)/t ), t 4. As just remarked, a class of n-cycles is not isolated. If t 5, by Remark 2.4 Cl(x) is connected to (2 2 , n − 4); if t = 4, for the same reason Cl(x) is connected to (3 2 , n − 6).
Since the cases of S 3 , A 3 and A 4 can be easily checked, the "if" part of Theorem 1.1 (1) follows by Lemma 3.1. There remains to show that in the other cases Λ has isolated vertices. Assume first n is prime and G n = A n . If n = 11 (resp. n = 23), any class of involutions of M 11 (resp. M 23 ) is such that every 1 i (n − 1)/2 is partial sum (because n is odd): it is not connected to vertices with more than 1 orbit (keep always in mind Remark 2.5 (1)). Moreover, it is not connected to (n) because of the Mathieu group; it is therefore an isolated vertex. If n = (q d − 1)/(q − 1), any class of involutions of P ΓL(d, q) is isolated for the same reason. If n ≡ −1 mod 12, at least one between (1, 2 (n−1)/2 ) and (1, 3 (n−1)/3 ) is an element of A n (n = 3): and these are isolated vertices as observed in Lemma 3.1.
Now we have to consider the cases of G n ∈ {S n , A n }, with n nonprime if G n = A n . Assume n 5, and consider the cycle type z = (1 n−3 , 3). Every 1 i n/2 is partial sum, so z is not connected to vertices with at least 2 orbits (both in the alternating and in the symmetric graph); this proves the assertion if n is even and G n = A n . If n is even and G n = S n , z is not connected to (n) as n/2 is partial sum in z (see Remark 3.2 below). If n is odd and G n = S n , z is not connected to (n) because it is an even permutation; and if n is odd nonprime and G n = A n , z is contained in S m ≀ S n/m , where m is any nontrivial divisor of n (certainly m 3: z fixes all the blocks), so it is connected to no class with cycle type (n).

Remark 3.2.
Let n be an even natural number, and let z be a vertex belonging either to Λ(S n ) or to Λ(A n ). If n/2 is partial sum in z, z is not connected to vertices having only even parts because of the sharing of S n/2 ≀ S 2 . We will use this several times.
Although in this section we should not speak about Theorem 1.2, it seems a good idea to place here the proof for alternating groups of prime degree different from the cardinality of any projective space. Proof. By Theorem 2.1, (1 n−3 , 3) is connected to (n); it is connected to nothing else because everything is partial sum. Moreover, (1, ((n − 1)/2) 2 ) is not isolated (n > 7) but not connected to (n). Hence, d(Ξ(A n )) 3.
If n = 11 and n = 23 we have M 11 and M 23 that may change something: a vertex inside these groups is not connected to (n). However, we can check that these groups contain vertices which are either isolated (because of partial sums) or connected to one between (2 2 , n−4), (3 2 , n−6): d(Ξ(A n )) 3 as above. (2) By Lemma 3.4, in Theorem 1.1 (1) the condition that Λ does not have isolated vertices is equivalent to the condition that Λ is connected.
We conclude this section with the proof of Theorem 1.1 (2). For every n, we define I n , a subset of the set of isolated vertices of Λ(G n ), whose cardinality, as n → ∞, goes to infinity. For every m, call p A m the set of all partitions of m different from (1 m ) that define an even permutation.
We first assume that if n is odd then G n = S n . If n is even, define I n as the set of all partitions of n of the form (1 n/2 , z), where z is whatsover partition of p A n/2 ; and if n is odd, define I n as the set of all partitions of n of the form (1 (n−1)/2 , z), where z is whatsover partition of p A (n+1)/2 . Let w ∈ I n . Every 1 i n/2 is partial sum, so w is not connected to vertices with at least 2 orbits. Moreover, if n is even w is not connected to (n) by Remark 3.2; and if n is odd, the same holds, because w is an even permutation. Therefore, really I n is made of isolated vertices. In order to conclude, it suffices to observe that |I n | = |p A n/2 | if n is even and |I n | = |p A (n+1)/2 | if n is odd. Assume now n odd nonprime and G n = A n . Fix n, and call p n the smallest nontrivial divisor of n. Define I n as the set of all partitions of n of the form (1 n(pn−1)/pn , z), where z is any partition of p A n/pn . If w ∈ I n , w ∈ S n/pn ≀ S pn : the first p n − 1 blocks are fixed pointwise, and the last is preserved by z. Hence, w is not connected to (n). Moreover, every 1 i (n − 1)/2 is partial sum, so w is not connected to vertices with more than 1 orbit: w is isolated. Note now that n/p n √ n; hence which goes to infinity as n → ∞.
The estimates given in the proof of Theorem 1.1 (2) are far from being realistic. A more subtle purpose would be to determine whether the proportion of the isolated vertices goes to zero as the degree goes to infinity. Moreover, we have said nothing about Λ(A n ), n odd prime. By Lemma 3.1, the unique case that requires analysis is when n equals the cardinality of some projective space. Here, the whole game is played by P ΓL, and again more subtle considerations might be needed.

Connectedness of Ξ and upper bound to the diameter
We begin with a small remark. We will start the proofs with a sentence of the type "pick a vertex of Ξ", without any preliminary consideration showing that Ξ is not the null graph. However, this will often be clear; and in any case, even if it will not be, during the proof suitable edges will be exhibited, so that the initial choice of the vertex 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 2 elements). Moreover, as already announced, many of the cases of degree at most 10 will be omitted.  Proof.
(1) Let z be a vertex of Ξ(S n ) not connected to (i, n − i) for every 2 i (n − 1)/2 coprime with n. Then, by Remark 2.4 every such i is partial sum in z. Now, being in Ξ, z will be connected to some w in which, necessarily, no 2 i (n − 1)/2 coprime with n is partial sum. Hence, again by Remark 2.4 w is connected to (i, n − i) for every such i.
In order to conclude, it is sufficient to note that the (i, n − i)'s, with i as above, are pairwise connected.
The previous proof fails in alternating groups of odd degree and in symmetric groups of even degree. In the first case, there are not vertices with 2 orbits; in the second, such vertices are even permutations, so an edge with them concerns not only partial sums but also parity. For these elementary reasons, we will need more efforts -and we will prove something weaker. Vertices with 3 orbits will acquire importance: that is why we will use [10] which, as already mentioned, classifies the primitive groups containing elements with at most 4 orbits. Proof. Let z and w be two vertices of Ξ(S n ) which are connected. One of the two, say z, must be odd. Assume z is not connected to (i, n − i) for every 1 i n/2 coprime with n. Then, every such i, i = 1, is partial sum in z; we show that the same holds for i = 1. If that is not the case, by Theorem 2.2 z is contained in one between AGL(m, 2) and P GL(2, p) (z cannot belong to M 11 on 12 points, M 12 and M 24 because it is odd). We have the first group only if n = 2 m , in which case 3 does not divide n, so 3 is partial sum in z. Hence, z = (3, . . .): z / ∈ AGL(m, 2) (Remark 2.6 (1) applied to z 3 ). Assume now z ∈ P GL(2, p). Since every 3 i n/2 coprime with n is partial sum, z has odd parts; let a be one of them. By assumption, a 3. Also, z has even parts, because it is an odd permutation. Hence, 1 = z a fixes a number of points greater or equal to 3: impossible (Remark 2.6 (2)). Therefore, 1 is partial sum in z.
Clearly there is not a sharing of products of symmetric groups. Moreover, the two mentioned vertices are odd, and lie in no imprimitive wreath product (see Remark 2.4). Finally, [10] tells us that y 1 and y 2 are contained in no nontrivial primitive group: we have therefore our desired edge.
If n ≡ 0 mod 4, we do the same with n/2 − 2 replaced by n/2 − 1. The same considerations lead us to the search of an edge with v 1 = (1, n/2 − 2, n/2 + 1) or v 2 = (1, n/2, n/2 − 1). Again, [10] says that v 1 and v 2 are contained in no nontrivial primitive group (for v 2 , Theorem 2.1 is sufficient). The unique change is that now v 2 ∈ S n/2 ≀S 2 . This is not a problem, because n/2 is partial sum in z, hence z ∈ S n/2 ≀ S 2 : w cannot belong to the same group.
Now we deduce the connectedness of Ξ and the bound to the diameter. Note that the considerations above imply that an edge with y i and v i concerns only partial sums, apart in one case, in which one has to cancel S n/2 ≀ S 2 . For (1, n − 1), remember Remark 2.6.
Assume first n ≡ 2 mod 4 and n > 14 (at the end we will consider the cases that we exclude along the proof). In order to show d(Ξ(S n )) 6 it is sufficient to show that the vertices y 1 , y 2 and (i, n − i), 1 i n/2 coprime with n, have pairwise distance at most 2 in Ξ(S n ). This is true, because all these vertices are connected to (2 2 , n − 4) (for (1, n − 1), being n ≡ 2 mod 4, we do not have AGL(m, 2)).
There remain the cases n = 12, 14, 16, 20. Since 5 divides 20, for n = 20 it is possible to replace (2, 7, 11) with (2, 5, 13) and everything works. For n = 16, let z and w be as above. As in the general case, 1, 3, 5 and 7 are partial sum in z. If 2 is not partial sum, it is easy to see that 8 is partial sum, and so z is isolated by Remark 3.2. Therefore, 2 is partial sum in z, from which w is connected to (1,2,13). In order to conclude, it is sufficient to observe that (1, 2, 13) and (i, (4, 2)). For n = 14, a similar argument works. For n = 12, in Remark 5.12 we will show that d(Ξ(S 12 )) = 5.
Although the small degrees are left to the reader, we should point out that Λ(S 6 ) is the empty graph: S 6 is not invariably generated by 2 elements. This we can see with the usual arguments; recall also Remark 3.2 and note that (2 3 ) ∈ P GL (2,5). This is the unique case of finite alternating and symmetric groups with this property. Now we move to alternating groups of odd degree. In some cases we are able to improve the bound thanks to elementary observations.    Proof. We start from the particular cases. Assume first none of 3, 5 and 7 divides n. Let Cl(x) and Cl(y) be two vertices of Ξ(A n ) which are connected, and denote with x c and y c the cycle types of x and y respectively. By Lemma 3.4 and Remark 3.5 (3), we may assume n 23. The hypotesis of Lemma 4.3, applied to x c and y c , is satisfied (Remark 2.5 (1)); hence, there exist b ∈ {x, y} and k ∈ {2, 3, 5, 7} such that k e 2k are not partial sums in b c . It follows that Cl(b) is connected to (k 2 , n − 2k) (Remark 2.4). In order to conclude, it suffices to observe that the (i 2 , n − 2i)'s, i ∈ {2, 3, 5, 7}, are pairwise connected (n > 19 prevents a sharing of products of symmetric groups among these vertices).
Assume now n = p 2 , p prime, with n different from the cardinality of any projective space. Let Cl(x) be a vertex of Ξ(A n ); we show it has distance at most 2 from (n) (this makes sense by Theorem 2.2 and Remark 2.5 (1)). Assume Cl(x) is not connected to (n); then, by Theorem 2.2 x ∈ S p ≀ S p . Being in Ξ(A n ), Cl(x) will be connected to some Cl(y) with, necessarily, y / ∈ S p ≀ S p . Hence, Cl(y) is connected to (n). Now we move to the general case. Let Cl(x) and Cl(y) be as above. By : if we prove the initial claim, we show that Ξ is connected and d(Ξ(A n )) 6. 1 Note also that the above considerations say what we need to worry about if we want an edge with one of those vertices: in the following, we will not repeat the same things.
So we have Cl(x) and Cl(y). Assume there exists b ∈ {x, y} such that 1 and 2 are not partial sums in b c ; without loss of generality, b = x. Then, either Cl(x) is connected to (1 2 , n − 2) or x ∈ P ΓL(2, n − 1). If 4 is not partial sum in x c , Cl(x) is connected to (2 2 , n − 4); we assume that this is not the case. Then, the unique possibility is x c = (4, . . .), from which x 4 fixes an even number of points greater than 3, from which x / ∈ P ΓL(2, n − 1) (Remark 2.6 (2)).
There remains the case n = 15. Let Cl(x) and Cl(y) be as above. We prove that one of the two is connected to at least one between (1 2 , 13) and (15) (15) if and only if it is contained in some imprimitive wreath product. It is not possible that x ∈ S 3 ≀ S 5 , because the 1-cycle should contribute to fix 1 block, from which 2 would be partial sum in x c , false; hence we may assume x ∈ S 5 ≀ S 3 and y ∈ S 3 ≀ S 5 . In x c , the 1-cycle must contribute to fix 1 block, from which 4 is partial sum. In y c , the 2-cycle either fixes 1 block or exchanges 2 blocks: in the first case, 1 is partial sum, and in the second, 4 is partial sum: false.

Lower bound to the diameter of Ξ
In this section, almost all the proofs will show that the diameter of Ξ, in the different cases, is at least 4. This we will always do in the same way: we will define two vertices of Ξ, z and w, each of which will be connected to one vertex only, sayz andw respectively, withz andw not connected in Ξ (actually, sometimes w will be connected to more than one vertex, and none of such vertices will be connected toz). This will show d(z, w) 4.
In order to avoid the edges between z (resp. w) and and the vertices different fromz (resp.w) having at least 2 orbits, we will use intransitive maximal subgroups. This lemma will be used to prove that the sharing of such groups indeed occurs.
Proof. The condition that every i ∈ {1, . . . , N /2} \ S is partial sum in v implies that also every 1 i N − max S − 1, i / ∈ S, is partial sum (with the convention max S = −1 if S is empty). The condition on T is equivalent to (N + T )/2 N − max S − 1, hence the assertion follows.
The previous lemma will be applied repeatedly, starting from some v and adding parts of length small enough (each step, N will be replaced by N + T and v by v T , of course).
The proofs for symmetric groups of odd degree and for groups of even degree will be similar, essentially because they all will rely on vertices with 2 orbits. Of course, this will not be possible in alternating groups of odd degree: that is why we will treat this case at the end. The case that will require the most work is that of alternating groups of even degree n: we will divide the cases n power of 2, n ≡ 0 mod 4 but not power of 2, n = 2d with d odd nonprime, n = 2p with p odd prime. We will use the following arithmetical observation: Proof. Being 2p even, q is odd and d is even. Moreover, if r is a divisor of d, then 2p = (q d − 1)/(q − 1) = ((q d − 1)/(q r − 1))((q r − 1)/(q − 1)). Since 2 cannot be equal to any of the factors, it follows that one of the factors is trivial, namely, that r ∈ {1, d}. Therefore, d is prime.
In symmetric groups of odd degree, ourz will be (n). In symmetric groups of even degree, although (n) would work, we will use a different approach; whilst in alternating groups of even degree ourz will be ((n/2) 2 ). This latter vertex should be considered as a faithful substitute of (n): both the cycle types are contained in every imprimitive wreath product and, apart from some exceptional cases, they are contained in the same nontrivial primitive groups (Theorem 2.3).
If n 12, n = 18 is even, we will define an even vertex z, belonging to both Ξ(S n ) and Ξ(A n ), which will be contained in no imprimitive product, in which every 1 i < n/2 will be partial sum and which will not share with (n) and ((n/2) 2 ) any nontrivial primitive group. In order to do this, we will divide some cases.
Moreover, z does not share with (n) and ((n/2) 2 ) nontrivial primitive groups: by Theorem 2.3 and Lemma 5.2, if n = 22 we just need to care about P ΓL(2, n − 1) 2 : z does not belong to this group because z 2 fixes an odd number of points greater or equal to 3 (Remark 2.6 (2)). If n = 22, we check directly that z / ∈ Aut(M 22 ). (3) n = 2d, d odd nonprime. Call p the smallest nontrivial divisor of d. If p = 3, we define z as in point (2): it is not contained in imprimitive wreath products for the same reason (n = 18), and now 3 is coprime with 2 and with n/2+1, so z n/2+1 is a 3-cycle: not contained in nontrivial primitive groups.
If p 5, define z = (1 n/2−2p−5 , 2, p, p + 2, n/2 + 1). Every 1 i < n/2 is partial sum in z. p is coprime with 2, p + 2 and n/2 + 1, so again we do not need to worry about primitive groups. Assume now z is contained in S t ≀ S n/t . The (n/2 + 1)-cycle cannot permute some whole blocks because then t = 2 and z has exactly one p-cycle in the cycle type. Also, being 2, p and p + 2 pairwise coprime, exactly one among them collaborates with n/2 + 1 to permute some blocks. The 2-cycle cannot do that, because we would obtain n = 18 as above. Neither is able the p-cycle, because p and n/2 + 1 are coprime. Therefore, the unique possibility is that t divides n/2 + p + 3, from which t is odd, from which t divides p + 3. Since t is odd, it must be t < p + 3, from which also t < p (p = 3). Hence, the p-cycle permutes points from more than 1 block, but cannot permute some whole blocks: it must be helped by something else. The only candidate is the 2-cycle, impossible because 2 and p are coprime. Therefore, z / ∈ S t ≀ S n/t . Sometimes, (n) (or its substitute (n/2) 2 )) is not connected to (1, n − 1). This lemma will be used in these cases. Proof. If n is odd, define w = (2 (n−3)/2 , 3) if n ≡ 3 mod 4 and w = (2 (n−9)/2 , 3 3 ) if n ≡ 1 mod 4. w is even, so it is not connected to (n). Every 2 i (n − 1)/2 is partial sum (again, Lemma 5.1 seems not necessary, but if we want, put v = (2 2 , 3), S = {1} and add the other parts one by one), so w is connected to nothing different from (1, n − 1). Finally, since w / ∈ AGL(m, p) (Remark 2.6 (1)), by Theorem 2.2 w is indeed connected to (1, n − 1).
Finally, consider Ξ(A n ). Define the same w as above, swapping the cases n ≡ 0 mod 4 and n ≡ 2 mod 4. This way, w becomes even. Direct check shows that w / ∈ M 11 (on 12 points), M 12 , M 24 . Hence, w is connected only to both classes with cycle type (1, n − 1). Proof. By Lemmas 5.3 and 5.4, it suffices to show that for symmetric groups (n) is not connected to (1, n − 1), and for alternating groups ((n/2) 2 ) is not connected to any class with cycle type (1, n − 1). If n is odd prime, this is true because of AGL(1, n). If n is even and n − 1 is prime, this is true because of P GL(2, n − 1) (recall Remark 2.5 (3)).
Using the fact that, given a natural number n and a nontrivial divisor p of n, (n) (or its substitute ((n/2) 2 )) and (p, n − p) are contained in S p ≀ S n/p , we prove Theorem 1.2 in some other cases. In order to do this, we need to say something about S √ n ≀ S 2 in its product action. Remark 5.6. Let n be a natural number, and z = (a 1 , . . . , a r ) be a partition of n. Assume there exists 1 i n/2 that is not partial sum in z; say i is the minimum with respect to such property. Then, if a 1 , . . . , a t are some parts of z necessary to write every 1 k i − 1 (we have possibly reordered the a j 's), we have t j=1 a j = i − 1 and a j i + 1 for every j t + 1. Indeed, let h j=1 a j = i − 1 (after reordering). We prove, by induction on m, m = 1, . . . , i − 1, that every a j , j = 1, . . . , r such that a j = m  appears among the a 1 , . . . , a h . This implies also the second assertion, that is, a j i + 1 for every j t + 1.
For m = 1 this is true, otherwise we would have ( h j=1 a j ) + 1 = i. Assume now the assertion is true for every l m − 1; we prove it is true also for m. We may assume there exists j such that a j = m, otherwise it is obvious. Certainly, in the a 1 , . . . , a h we write m − 1; hence, if some a j = m was missing, we could remove m − 1, add m and write i: false.
(a) If g c has an odd part of length at least √ n + 1, then τ = 1.
(b) If τ = 1 and every 1 i √ n − 1 is partial sum in g c , then also every 1 i 2 √ n − 1 is partial sum.
Assume now g c has an odd part of length at least √ n + 1, call it b.
Assume τ = (1, 2). g b = ((σ 1 ,σ 2 ),τ ) is such thatτ = (1, 2); hence, by what proved above, g b fixes at most √ n points. On the other hand, being b a part of g c we have that g b fixes at least b √ n + 1 points: contradiction. This proves (a). Assume now τ = 1. Assume there exists √ n i n/2 that is not partial sum in g c ; call k the minimum such i.
By Remark 5.6, g c = (a 1 , . . . , a r ), with t i=1 a i = k − 1 and a j k + 1 for every j t + 1. Write (1 i , b i+1 , . . . , b l ) and (1 j , c j+1 , . . . , c s ), with i, j 1 and b k , c k = 1, for the cycle types of σ 1 and σ 2 respectively. By the structure of g c , it must be A = ij + j l k=i+1 b k + i s k=j+1 c k k − 1. On the other hand, One comment. Along many of the following proofs we will make assumptions on n stronger than what the statements will claim. When this will happen, it will be meant that sooner or later, during the proofs, the excluded cases will be checked singularly. Proof. Assume first n odd and p = 3. By Bertrand-Chebyshev Theorem, we may choose a prime q contained in the open interval (n/p, 2n/p). Now define w = (1 p−1 , (p + 1) k , p + 2, q, r), with k 2 and p + 1 r < 2(p + 1). If the defined element is odd, we meld two 1's into a 2.
In order the element to make sense, it is necessary and sufficient to write (1 p−1 , (p + 1) 2 , p + 2, q . . .) in such a way that what is left is not smaller than p + 1: we ask n − ((p − 1) + 2(p + 1) + (p + 2) + q) p + 1. Since q < 2n/p, we are okay if n (5p 2 + 3p)/(p − 2). Using that n p 2 , we want p 2 7p + 3. If p 11, this is true; if p = 5, we assume n 55, and if p = 7, we assume n 77, so that n (5p 2 + 3p)/(p − 2) holds also in these cases. Now we have to show that w is connected only to (p, n − p). Since w is even, in order to show that it is connected to nothing different from (p, n−p) it is sufficient to show that every 1 i (n − 1)/2, i = p, is partial sum.
Put v = (1 p−1 , p + 1, p + 2), S = {p} and apply Lemma 5.1 adding one by one all the (p + 1)-cycles (at least there is another) and the r-cycle (the first step is sharp, as p + 1 = 3p + 2 − 2p − 1). Then, using q < 2n/p, we see that q (n − q) − 2p − 1 (unless p = 5, in which case we need n 55, already required above), so that at the end we can add also q.
There remains to show that w is indeed connected to (p, n − p). In order to do this, we must show that w and (p, n − p) do not share imprimitive wreath products and nontrivial primitive groups. For nontrivial primitive groups, by Theorem 2.3 we just need to take care of AGL(m, p) and S p ≀ S 2 . w does not belong to the first group as it fixes p − 1 or p − 3 points, and it does not belong to the second because of Lemma 5.7.
For imprimitive wreath products, (p, n − p) belongs to S p ≀ S n/p and to S n/p ≀ S p . w does not belong to the former, because an element of that group is such that, in its cycle type, the number of 1-cycles + the lengths of some other cycles (of length p) is divided by p. Moreover, since q is a prime strictly bigger than n/p, an element having q in the cycle type does not belong to S n/p ≀ S p .
Assume now n odd and p = 3. In [19] it is shown that, if r 25, then there exists a prime q contained in the open interval (r, 6r/5); we apply this to r = n/3, hence we require n 75. Define w = (1 2 , 4 k , 5, q, r), with k 3 and 4 r 7. If the defined element is odd, we unify two 4-cycles into an 8-cycle. n 39 is enough to guarantee that the element makes sense. Putting v = (1 2 , 4, 5) or v = (1 2 , 4, 5, 8) (depending on whether two 4-cycles have been unified or not), S = {3} and applying Lemma 5.1 adding one by one the 4-cycles, the r-cycle, and finally the q-cycle, we get that every 1 i (n − 1)/2, i = 3, is partial sum: w is connected to nothing different from (3, n − 3). Finally, w does not share with (3, n − 3) primitive groups and imprimitive wreath products for the same reasons as above.
Assume now n = 2d, d 15 odd nonprime. We define w exactly as above, dividing the cases p = 3 and p = 3. We run through the routin checks again.
For p = 3, using the stronger inequality n 2p 2 we see that the definition of w makes sense for every n. w is connected to both classes with cycle type (p, n − p) for the same reasons as those above.
For p = 3, everything works in the same way; we just have to check the cases n < 75, that is, with our conditions, n = 54, 66. Since in both cases there exists a prime contained in the open interval (n/3, 2n/5), we are done. Corollary 5.9. Theorem 1.2 holds for symmetric groups of odd nonprime degree n and for alternating groups of even degree n = 2d, with d odd nonprime.
Proof. For symmetric groups of odd nonprime degree, Lemmas 5.3 and 5.8 leave us with the case n = 15. In order to show that d(Ξ(S 15 )) = 3, let z be a vertex of Ξ(S 15 ) not connected to (15), (i, d − i), i = 1, 2, 4, 7. Then (Remark 2.4), 1, 2, 4, 7 are partial sums in z; hence, at least one among 3, 5 and 6 is not partial sum (otherwise z would be isolated). The unique way in which 1, 2, 4 are partial sums but 3 is not partial sum is (1 2 , 4 2 , 5); but this vertex is isolated, as it is contained in S 5 ≀ S 3 . Therefore, 3 is partial sum. Now it is immediate to see that there are no vertices in which 1, 2, 3, 4 are partial sums but (at least) one between 5 and 6 is not partial sum (in such a way that every i coprime with 15 is partial sum).
For alternating groups, we are left with the cases n = 18, 30, 42. The latter two are contamplated in Corollary 5.5 (they also could have been included in Lemma 5.8, but it would have been useless). For n = 18, a little change seems necessary. (1 3 , 5 3 ) is contained in S 6 ≀S 3 : it is not connected to (9 2 ). Instead, by Theorem 2.3 it is connected to (4,14). On the other hand, (1,2,4,11) is connected to (8,10) and to (9 2 ); the diameter is 4 because (4,14) is connected to none of these vertices (S 2 ≀ S 9 ).
The reader will have noted that the proof of Lemma 5.8 works perfectly also for symmetric groups of even degree (values contemplated there). This will be true for almost all the proofs we will give for alternating groups of even degree. Nevertheless, the symmetric case can be treated separately, with a proof which is faster and which points out once again the reason of the difference, namely, the fact that vertices with 2 orbits are not connected because of parity. Proof. If 3 does not divide n and n 26 we define z = (1 2 , 4 k , 5, r), with k 3 and 4 r 7 (n 23 ensures this possibility). If the defined element is even, we meld two 4's in a 8. As in Lemma 5.8 we see that the unique 1 i n/2 which is not partial sum is i = 3. Being n/2 partial sum, z is not connected to (n); hence, it might be connected only to (3, n − 3). Since n = 24, Theorem 2.3 tells us that we do not need to care about nontrivial primitive groups. Moreover, (3, n − 3) is contained in no imprimitive wreath product. Therefore, since z is odd, z is indeed connected to (3, n − 3). For n = 16, define z = (1 2 , 4 2 , 6); for n = 22, z = (1 2 , 5, 6, 9) (the latter is not connected to (4, 18) because of S 11 ≀ S 2 ).
If instead 3 divides n, we define z as in Lemma 5.8 for the case p = 3; we take here the odd version. We are left with the cases n < 75, that is, with our conditions, n = 18, 36, 66.  Proof. By Corollary 5.5, we may assume that n − 1 is not prime unless n = 18. Then, the assertion follows by Lemmas 5.4 and 5.10, noting that (1, n−1) and (3, n−3), being even permutations, are not connected in Ξ(S n ) (in Lemma 5.10, probably also all the cases of n − 1 prime could have been included: there were only some cases to check singularly. However, there was no reason for doing that).
In many occasions, along the following lines, z will be claimed to be isolated: in order to check so, keep in mind Remark 3.2 and the fact that vertices with 2 orbits are even.
Being n/2−2 j−1 ≡ 0 mod 2 j , w belongs to S 2 j ≀S n/2 j : it is not connected to ((n/2) 2 ). Moreover, it is easy to see that only some even 1 i n/2 are not partial sum (to be precise, they are the 2 j−1 k, k odd); therefore, w might be connected only to some (a 1 , . . . , a m ), with a k even for every k. In order to conclude, there remains to show that w is indeed connected to (2 j−1 , n − 2 j−1 ). If n = 12, Theorem 2.3 tells us that we should not worry about primitive groups (clearly 2 j−1 = k √ n). If n = 12, w = (1, 3, 4 2 ) does not belong to nontrivial primitive groups by Theorem 2.1. Therefore, there remains to show that w and (2 j−1 , n−2 j−1 ) do not share imprimitive wreath products. (2 j−1 , n − 2 j−1 ) belongs only to S 2 t ≀ S n/2 t and to S n/2 t ≀ S 2 t for every t j − 1 (Remark 2.4). w does not belong to the former (see proof of Lemma 5.8).
For the latter, assume first the (n/2 − 2 j−1 )-cycle permute points from a single block. Then, n/2 − 2 j−1 n/2 t , that is, 2 j−1 (d − 1) 2 j−t d, from which t = 1. But w does not belong to S n/2 ≀ S 2 , because w has fixed points and n/2 is not partial sum. Therefore, the (n/2 − 2 j−1 )-cycle must permute points taken from different blocks.
Note that n/2 t = 2 j−t d and n/2 − 2 j−1 = 2 j−1 (d − 1): the first does not divide the second; so the (n/2 − 2 j−1 )-cycle cannot permute some whole blocks. The 1-cycles cannot contribute to finish such blocks, from which n/2 − 2 j−1 + k(2 j−1 + 1) + 2 j m = a(n/2 t ), for some k, m, a, with k ∈ {0, 1}, m 0 and a 1. If k = 1, the first member is odd whilst the second is even; so k = 0. n/2 − 2 j−1 + 2 j m = 2 j ((d − 1)/2) + 2 j m = 2 j ((d − 1)/2 + m), and a(n/2 t ) = 2 j−t ad; equality implies that 2 t divides a, from which 2 t = a, false, because otherwise all points would be permuted, and we have not used all cycles of w yet. Now we should make sure that the variations of w do not affect our considerations. It is easy to observe that the unique modification we should take care of is the melding of two 1's into a 2: a priori it could change something about imprimitive wreath products. In particular, the 2-cycle could help in permuting 2 blocks, from which n/2−2 j−1 +2 j m+2 = 2(n/2 t ). But this equation, seen mod 4, gives a contradiction.
Corollary 5.14. Theorem 1.2 holds for alternating groups of even degree n, with n ≡ 0 mod 4 but not power of 2. Proof. We divide the cases m even and m odd. In any case, in order to show that w is not connected to ((2 m−1 ) 2 ) we will show that w is contained in some imprimitive wreath product.
Applying suitably Lemma 5.1 it is easy to see that every 1 i < 2 m−1 , i = 2, is partial sum; w is then connected to nothing different from (2, 2 m −2). Now, 2 m−1 is not partial sum (otherwise also 2 would be, and this is not the case), and w has odd parts, so w / ∈ S 2 m−1 ≀ S 2 ; and clearly w / ∈ S 2 ≀ S 2 m−1 . Moreover, w / ∈ AGL(m, 2) because w 4 fixes 5 points. Hence, by Theorem 2.3 we can conclude that w is indeed connected to (2, 2 m − 2).
If m is odd, 9 divides exactly one between 2 m−1 −4, 2 m−1 −10, 2 m−1 −34. In the first case, write 2 m−1 −4 = 9t, and define w = (1 3 , 5, 7, 9, w belongs to S 2 m−4 ≀ S 16 : the (2 m−1 − 4)-cycle permutes points taken from 9 blocks in number t from each block, the 9-cycles finish these blocks, and what is left preserves each of the remaining 7 blocks. As above, every 1 i < 2 m−1 , i = 4, is partial sum; there remains to show that w is connected to (4, 2 m − 4). It is not contained in AGL(m, 2) because it fixes 3 points; by Theorem 2.3 (and Remark 2.4), there remains to exclude that w belongs to S 2 ≀ S 2 m−1 , S 2 m−1 ≀ S 2 , S 4 ≀ S 2 m−2 , S 2 m−2 ≀ S 4 . For the first two groups is as above, and for the third, usual consideration (proof of Lemma 5.8). For the fourth, the (2 m−1 − 4)-cycle cannot permute points from 1 block because 2 m−1 − 4 > 2 m−2 ; and it cannot permute points from 2 blocks because one would need two 2-cycles or one 4-cycle to finish these blocks. Therefore, it must permute points taken from 3 blocks, in number 3t from each block. The unique other cycles whose length is divisible by 3 are the 9-cycles, which permute 3 points from each block. But this would imply that 2 m−2 is divisible by 3, false.
Essentially, the same arguments as above show that w ∈ S 2 m−4 ≀ S 16 and that we have an edge only with (10, 2 m − 10). Again the (2 m−1 − 10)cycle permutes points from 9 blocks, but here, to finish such blocks, we use 18-cycles, which permute 2 points from each block.
Again, the argument is the same as that above; w ∈ S 2 m−4 ≀ S 16 and we have an edge only with (34, 2 m − 34). Here, the 54-cycle permutes 6 points from each of the 9 blocks, and the 36-cycles permute 4 points from each block (this is possible as 2 m−4 − t 6 and 2 m−4 − t ≡ 2 mod 4). In order to conclude (for alternating groups of even degree) there remains the case n = 2p. For this, we need some more considerations about P ΓL(2, q). Lemma 5.17. Let n be an even natural number, with n − 1 prime power, and let z be a partition of n. If z belongs to P ΓL(2, n − 1) and has an even number of 1-cycles, then, for every 1 i n odd, z has an even number of i-cycles as well. In particular, z ∈ S 2 ≀ S n/2 .
Proof. The proof is by induction on i, i = 1, . . . , n, i odd. The base case holds by assumption. Assume now the induction holds for every k i − 1 (k odd). In order to prove the assertion for i, just observe that, if in z there was an odd number of i-cycles, then, by the inductive hypotesis, z i would fix an odd number of points greater or equal to 3, impossible by Remark 2.6 (2).
Remark 5.18. The previous lemma applies for instance if z ∈ P ΓL(2, n−1) is such that 1, 3 and 5 are partial sums: we show that this condition implies that the number of 1-cycles in z is even.
If there are not 1-cycles, we are done. If there are at least two, either there are no others, in which case we are happy, or there are others, in which case they must be in even number by Remark 2.6 (2), happy again. Now assume by contradiction there is exactly one 1-cycle. Then, there cannot be 2-cycles or 4-cycles, for otherwise z 2 or z 4 would fix an odd number of points greater or equal to 3. Then, the unique way to write 1, 3 and 5 is z = (1, 3, 5, . . .). z 3 will fix 3k + 1 points, k 1, from which 3k + 1 = γ l + 1 (n − 1 power of γ), from which γ = 3. The same consideration, with 3 replaced by 5, implies γ = 5, contradiction. Proof. If n − 1 is prime, the result follows by Corollary 5.5 (in even degrees, this is the first time we make serious use of Corollary 5.5). Then, by Lemma 5.2 either we have not projective groups or n − 1 is prime power but not prime. We prove that, in these cases, d(Ξ(A n )) = 3. In order to do this, we by 3-cycles and with a final cycle). Again, in order to change sign we possibly unify two 3-cycles into a 6-cycle. Here we need n/p − 6 3, hence we require n = 35, 49.
Applying suitably Lemma 5.1 we see that the unique 1 i (n − 1)/2 which are not partial sum in z are i = 1, 2: z might be connected only to (1 2 , n − 2). By Theorem 2.2, in order this to happen we just have to exclude that z ∈ P ΓL(2, n − 1). This is true, as z 4 fixes an even number of points greater than 3 (Remark 2.6 (2)). For n = 35, define z = (3,4,6,8,14); for n = 49, z = (3, 4, 7 2 , 6,8,14). It is easy to see that these vertices belong to S n/3 ≀ S 3 . Instead, it is not true that only 1 and 2 are not partial sum; it is true, however, that these vertices share a product of symmetric groups with every vertex with at least 3 orbits different from (1 2 , n − 2). Now we move to w. If p = 3, we start this way: w = (1, 5 2 , 6 3 , 8, . . .). The 8-cycle and the three 6-cycles permute points from the same 2 blocks; the 1-cycle and the two 5-cycles permute points from 1 block. We complete the 2 moved blocks with some 6-cycles, and at the end with an m 2,3 -cycle, with m 2,3 ∈ {6, 8, 10}. We complete the fixed block with 5-cycles, and at the end with an m 1 -cycle, with 5 m 1 9. For this to make sense, we must have n/3 − 11 5 and n/3 − 13 3: we require n = 39, 45. If the defined element is odd, we meld two 6-cycles and we get a 12-cycle.
If p 5, we start w in the same way, but here, the 8-cycle and two 6-cycles permute points from 2 blocks; the 1-cycle and one 5-cycle permute points from 1 block; the other 5-cycle from another block, and the same for the remaining 6-cycle. We complete the blocks as above (if p > 5, the blocks that have not been touched yet are preserved with 5-cycles and with a final cycle). We change sign as above. Here we need n/p − 10 3 and n/p − 6 5: we ask n = 35, 49, 55, 77, 121.

Questions and comments
As remarked at the end of Section 3, one would like to say something more precise about the number of isolated vertices of Λ in the different cases. Moreover, it might be interesting to give suitable descriptions of the isolated vertices, in order to determine, for instance, when nontrivial primitive groups influence their list, and when instead they depend uniquely on imprimitive groups and (possibly) on parity. However, this is not strictly connected to the content of this paper.
What is strictly connected is the fact that Theorem 1.2 should be improved: there remains to compute the exact diameter for alternating groups of odd degree and for symmetric groups of even degree. In the first case, call n the degree, if n is prime there remains the case in which n is equal to the cardinality of some projective space. As observed in Remark 3.5 (3), d(Ξ(A 13 )) = d(Ξ(A 17 )) = 3. Essentially the same arguments and a little bit of patience show that also d(Ξ(A 31 )) = d(Ξ(A 73 )) = 3. Hence one asks whether there are cases in which the diameter is at least 4.
If n is odd nonprime, for the sake of completeness one may want to know whether the given lower bound holds also in the cases n = 15, 21, 25, 27. Then, one asks for the existence of cases in which the diameter is at least 5.
This already has an answer for symmetric groups of even degree, being d(Ξ(S 8 )) = 6 and d(Ξ(S 12 )) = 5. Ξ(S 8 ) has 9 vertices and only 8 edges. In general, symmetric groups of even degree might have more isolated vertices and less edges than the other cases, the reason being in Remark 3.2 and in the fact that vertices with 2 orbits are even. Indeed, for instance, these features imply that even vertices in which every even i is partial sum are isolated; this is the main reason for which Λ(S 6 ) is the empty graph.
However, it might be synonymous of laziness being content with the examples of S 8 and S 12 : one should discover whether these are exceptions, due to the degree being too small, or there are other cases with this property. If the degree is not too small, it is unlikely that sporadic examples exist: groups whose degrees have similar arithmetical properties -for what concerns nontrivial primitive groups and imprimitive wreath products -have probably the same diameter.
The perception is that the diameter should always be at most 4. Indeed, pick a vertex z of Ξ. z will be connected to somez; callz c the cycle type. It seems reasonable that, summing some of the parts ofz c in a suitable way, still there is not a sharing of nontrivial primitive groups and imprimitive wreath products. This way, caring also about parity, we get that z is connected to a vertex with very few cycles, say at most 3 or 4 (by [10], it is rare that elements with at most 4 orbits belong to nontrivial primitive groups). Observing that, if the degree is not too small, vertices with at most 4 cycles should have pairwise distance at most 2 in Ξ, we speculate that indeed d(Ξ) 4.