The invariably generating graph of the alternating and symmetric groups
aa r X i v : . [ m a t h . G R ] J un The invariably generating graph of the alternatingand symmetric groups
Daniele Garzoni
Abstract
Given a finite group G , the invariably generating graph of G isdefined as the undirected graph in which the vertices are the nontrivialconjugacy classes of G , and two classes are connected if and only ifthey invariably generate G . In this paper we study this object foralternating and symmetric groups. First we observe that in most casesit has isolated vertices. Then, we prove that if we take them out weobtain a connected graph. Finally, we bound the diameter of this newgraph from above and from below — apart from trivial cases, it isbetween 3 and 6 —, and in about half of the cases we compute itexactly. Given a group G and a subset S of G , we say that S invariably generates G if h s g ( s ) , s ∈ S i = G for every choice of g ( s ) ∈ G . The motivationfor considering this kind of generation comes from computational GaloisTheory, as now we breafly explain.Let f ∈ Z [ x ] be monic, of degree n >
1, separable over Q . If p is a primenumber not dividing disc( f ), we can associate to p a partition of n whoseparts are the degrees of the distinct factors of f mod p Z [ x ]. A theoremdue to Dedekind states that Gal( f ) contains an element having this cycletype. Actually much more is true: the Frobenius Density Theorem says that,asymptotically, the proportion of primes having a given cycle type is equal tothe proportion of elements of Gal( f ) having that cycle type. This allows togain informations on Gal( f ) factoring f modulo different primes; the pointis to understand how many factorisations are necessary. For instance, if onesuspects that Gal( f ) = S n , the expected number of factorisations needed toprove this fact coincides with the expected number of elements needed toinvariably generate S n .These problems have been considered first in [5]. Then, several resultshave been obtained, for example in [17], [20], [7], [8], [4]. In [20] it isshown that 4 random permutations invariably generate S n with probabilitybounded away from zero, whilst in [8] it is shown that, if we replace 4 with3, the same probability tends to zero as n → ∞ .1e forget for a while invariable generation, and we recall that it is possi-ble 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 h x, y i = 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 diameter2. 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 invariablygenerating graph of G , in which the vertices are the conjugacy classes of G different from { } , and two vertices Cl( x ) and Cl( y ) are connected if andonly if { x, y } invariably generates G . In [13] it is shown that, if G is anonabelian finite simple group, Λ( G ) is nonempty.The aim of this paper is to study this object for alternating and sym-metric groups, in order to understand whether there are or not similaritieswith its classical analogue. First of all, it is easy to see that, expect for thecase of S , 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, whichis connected and whose diameter is quite small — although not as smallas that of the generating graph. These last two sentences summarise thecontent of this paper; we state them explicitly in the following theorems. Theorem 1.1.
Let n > be a natural number and let G n ∈ { S n , A n } .Then,(1) Λ( G n ) does not have isolated vertices if and only if either G n ∈ { S , A , A } or all the following conditions are satisfied: G n = A n , n is prime, n ≡ − mod , n = 11 , , n is different from the cardinality of anyprojective space.(2) As n → ∞ , with n nonprime if G n = A n , the number of isolatedvertices of Λ( G n ) goes to infinity as well. Theorem 1.2.
Let n > be a natural number, and let G ∈ { S n , A n } . Then, Ξ( G ) is connected unless G = S , in which case Λ( S ) is the empty graph. InTable 1 the diameters for the groups of degree at most are listed. Assumenow n > .(1) Symmetric groups:(a) n odd: d (Ξ( S n )) = 4 unless n = 15 , in which case d (Ξ( S )) = 3 .(b) n even: d (Ξ( S n )) . n n d (Ξ( S n )) d (Ξ( A n ))3 1 14 1 25 3 26 Ξ = null graph 27 4 28 6 39 3 310 4 3 (2) Alternating groups:(a) n even: d (Ξ( A n )) = 4 unless n − is not prime and n = 2 p , with p odd prime (both conditions must hold), in which case d (Ξ( A n )) =3 .(b) n odd: d (Ξ( A n )) . Moreover, if n is nonprime d (Ξ( A n )) > , apart possibly in the cases n = 15 , , , ; if none of , and divides n , d (Ξ( A n )) ; if n is a prime different from the car-dinality of any projective space, d (Ξ( A n )) = 3 ; and if n = p ,with p prime, and n is different from the cardinality of any pro-jective space, d (Ξ( A n )) = 4 unless n = 25 , in which case it mightbe d (Ξ( A )) = 3 . The following is dedicated to the proof of the previous results. Thecomputations 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 andlook at the picture. In order to draw the edges, the arguments are thesame as those used in the general case: including them thoroughly would beessentially a repetition.
The vertices of our graph are conjugacy classes of alternating and symmetricgroups. 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 ofdistinct odd parts, in which case there are two conjugacy classes with thesame cycle type. Except for the latter case, we will consider the vertices ofthe graph as partitions of n ; we fix now some language about this.Given G a subgroup of S n and given ( a , . . . , a t ) a partition of n , we willsay that ( a , . . . , a t ) belongs to G , or that ( a , . . . , a t ) is contained in G , if G contains elements with cycle type ( a , . . . , a t ).3hen two cycle types ( a , . . . , a t ) and ( b , . . . , b s ) belong to G , we willsay that they share G .When ( a , . . . , a t ) ∈ S i × S n − i , with 1 i n/
2, we will say that i ispartial sum in ( a , . . . , a t ).In a cycle type, a writing of the form i m will mean m cycles long i (andnot one cycle long i m ). Therefore, ( a m , a m , . . . , a m t t ) will mean m cycleslong a + m cycles long a + . . . + m t cycles long a t .We will call nontrivial primitive group of degree n a primitive group ofdegree n not containing A n (the degree, when clear, will be omitted). Forthe rest, the terminology we will use is standard.Cycle types with 1 , Theorem 2.1.
Let G be a primitive permutation group of degree n . If G contains a cycle fixing at least points, then G > A n . Theorem 2.2.
Let G be a primitive permutation group of finite degree n ,not containing the alternating group A n . Suppose that G contains a cyclefixing k points, where k n − . Then one of the following holds:1. k = 0 and either(a) C p G AGL (1 , p ) with n = p prime, or(b) P GL ( d, q ) G P Γ L ( d, q ) with n = ( q d − / ( q − and d > for some prime power q , or(c) G = L (11) , M or M with n = 11 , or respectively.2. k = 1 and either(a) AGL ( d, q ) G A Γ L ( d, q ) with n = q d and d > for someprime power q , or(b) G = L ( p ) or P GL (2 , p ) with n = p + 1 for some prime p > , or(c) G = M , M or M with n = 12 , or respectively.3. k = 2 and P GL (2 , q ) G P Γ L (2 , q ) with n = q + 1 for some primepower q . Theorem 2.3.
Let G be a primitive permutation group of degree n whichcontains an element with exactly two cycles of lengths k and n − k > k .Then one of the following holds, where G 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 followingholds.(a) n = p m , k = 1 , and GL ( m/t, p t ) G Γ L ( m/t, p t ) for adivisor t of m ;(b) n = p m , k = p , and G = GL ( m, p ) ;(c) n = p , k = p , and G < GL (2 , p ) is the group of monomialmatrices (here p > ;(d) n = 2 m , k = 4 , and G = GL ( m, ;(e) (Sporadic affine cases)i. n = 4 , k = 2 , and G = GL (1 , (so G = A );ii. n = 8 , k = 2 , and G = Γ L (1 , ;iii. n = 9 , k = 3 , and G = Γ L (1 , ;iv. n = 16 , k = 8 , and [Γ L (1 ,
16) : G ] = 3 or G ∈ { Γ L (1 , , ( C × C ) ⋊ C , Σ L (2 , , Γ L (2 , , A , GL (4 , } ;v. n = 16 , k = 4 or , and G ∈ { ( S × S ) ⋊ C , S , S } ;vi. n = 16 , k = 2 or , and G = A < GL (4 , ;vii. n = 25 , k = 5 , and [ GL (2 ,
5) : G ] = 5 ;2. (Product action) One of the following holds.(a) n = r with < r ∈ N , k = ra with gcd ( r, a ) = 1 , G = ( S r × S r ) ⋊ C , and G = ( S r − × S r − ) ⋊ C ;(b) n = ( p + 1) with p > prime, k = p + 1 , G = ( P GL (2 , p ) × P GL (2 , p )) ⋊ C , and G = ( AGL (1 , p ) × AGL (1 , p )) ⋊ C .3. (Almost simple action) S G Aut ( S ) for a simple, nonabeliangroup S , and one of the following holds.(a) n > , A n G S n in natural action;(b) n = 10 , k = 5 , and A G S in the action on the 2-sets of { , , , , } ;(c) n = p + 1 , k = 1 , and P SL (2 , p ) G P GL (2 , p ) for a prime p ;(d) n = ( q m − / ( q − , k = n/ , and P SL ( m, q ) G P Γ L ( m, q ) for an odd prime power q and m > even;(e) n = 10 , k = 2 , and M G P Γ L (2 , ;(f ) n = 21 , k = 7 , and P Σ L (3 , G P Γ L (3 , ;(g) n = 12 , k = 1 or , and G = M in its action on 12 points;(h) n = 12 , k = 1 , , or , and G = M ; i) n = 22 , k = 11 , and M G Aut ( M ) = M ⋊ C ;(j) n = 24 , k = 1 , or , and G = M . We mention that, although we do not report here the statements, we willmake use of [10], that extends Theorem 2.3 to 3 and 4 orbits. The followingremark, 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 ifeither m divides i or n/m divides i . In the first case, the i -cycle permutesall 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 ) pointsfrom each block, and the ( n − i )-cycle permutes ( n − i ) / ( n/m ) points fromeach block.Let now 2 i n/ n . By what just said, ( i, n − i ) is notcontained in imprimitive wreath products. Also, ( i, n − i ) n − i = ( i, n − i ) =(1 n ); since n − i >
3, by Theorem 2.1 ( i, n − i ) is not contained in nontrivialprimitive groups. Therefore, if z is a vertex of Λ( S n ), z and ( i, n − i ) areconnected if and only if they share neither A n nor S i × S n − i . Theorem 2.2can be used to generalise the previous considerations to the case i = 1.Analogous considerations hold for vertices with 3 orbits. In particular,( a , a , a ) belongs to S m ≀ S n/m if and only if one of the following conditionsis 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 fromeach 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 ; theremaining cycle permutes the remaining points in such a way that theblocks are preserved.A typical example is ( i , n − i ). If n is odd and 2 i ( n − / n , ( i , n − i ) is not contained in imprimitive wreath products. If i ( n − /
2, neither is it contained in nontrivial primitive groups, because( i , n − i ) i = (1 i , n − i ) = (1 n ) and 2 i >
4. Therefore, also in this case anedge with ( i , n − i ) concerns only parity and partial sums.Instead, if n is even ( i , n − i ) is contained in S ≀ S n/ (it is of type (c)above).As already noticed, if a partition of n is made of distinct odd parts thenthe 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 remark6ives conditions under which these two vertices may be essentially thoughtas a unique vertex. Remark 2.5.
Recall that, given x ∈ A n and H S n , Cl A n ( x ) = Cl S n ( x ) ifand only if C S n ( x ) * A n ; and Cl A n ( H ) = Cl S n ( H ) if and only if N S n ( H ) (cid:10) A n .(1) Let x ∈ A n be such that C S n ( x ) ⊆ A n and let H S n be such that N S n ( H ) (cid:10) A n . Then, H contains elements belonging to Cl A n ( x ) ifand only if it contains elements belonging to Cl A n ( 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 groupsthemselves 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 A n ( x ) if and only if it is connected to Cl A n ( x g ), ifand only if the cycle type of z and the cycle type of x do not shareimprimitive maximal subgroups of A n . Hence, if x c is the cycle typeof x , we are allowed to use the expression "a vertex connected to x c ".Examples are ( i, n − i ), with n even and 2 i n/ n , and ( i , n − i ), with n odd and 2 i ( n − / n (Remark 2.4).(2) Let x ∈ A n , with C S n ( x ) ⊆ A n and N S n ( h x i ) (cid:10) A n . If we apply(1) with H = h x i , we obtain that, if g / ∈ A n , Cl A n ( x g ) = Cl A n ( x i ),with i coprime with the order of x . Hence, a vertex z of Λ( A n ) isconnected to Cl A n ( x ) if and only if it is connected to Cl A n ( x g ), if andonly if the cycle type of z and the cycle type of x do not share maximalsubgroups of A n . Therefore, as above we may use the expression "avertex 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 anodd prime or n ≡ n − , ( n − / ) normalise h x i .(3) Let x, y ∈ A n be such that C S n ( x ) ⊆ A n and C S n ( y ) * A n . Then,Cl A n ( y ) = Cl S n ( y ) is connected to Cl A n ( x ) if and only if it is connectedto Cl A n ( x g ), with g / ∈ A n , if and only if the cycle type of x and thecycle type of y do not share maximal subgroups of A n .(4) Let n >
4, and let x ∈ A n with C S n ( x ) ⊆ A n . Then, Cl A n ( x ) is notconnected to Cl A n ( 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 h x i , so the assertion follows.7he previous remark lists cases in which a vertex of Λ( A n ) may be con-sidered as a partition of n , even when the identification is not faithful. Thisis not always the case: there are examples of x, y ∈ A n , with C S n ( x ) ⊆ A n and C S n ( y ) ⊆ A n , such that Cl A n ( x ) is connected to Cl A n ( y ) but not toCl A n ( y g ), g / ∈ A n . In order to find such an example, by Remark 2.5 the nor-malisers of h x i and h y i must be contained in A n , and x c and y c cannot shareimprimitive groups. Hence, we need that x c and y c share a primitive maxi-mal subgroup of A n — let us say exactly one. It is necessary and sufficientthat (each copy of) such group contains elements of exactly one A n -classwith 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 ). Theexample we give is in Λ( A ). (3 , ,
93) and (7 , , P GL (7 , P GL (7 , P GL (7 , , ,
93) (resp.(7 , , x ∈ A n , with C S n ( x ) ⊆ A n , we have N S n ( h x i ) A n if and onlyif Cl A n ( x ) ∩ h x i = Cl S n ( x ) ∩ h x i ).We conclude this section making some easy considerations about thenumber of fixed points of elements in some groups. We will often be in-terested in finding edges with (1 , n − AGL ( m, p ) and P GL (2 , q ), where p and q are primes.Remark 2.6 will be mostly used with this purpose. Remark 2.6. (1) Being
AGL ( m, p ) transitive, if g ∈ AGL ( m, p ) fixessome point we may assume that it fixes 0. Hence, g ∈ GL ( m, p ); andthe set of fixed points of an element of GL ( m, p ) is a subspace of F mp :it has cardinality p s , where 0 s m .(2) P Γ L (2 , q ), with q prime power, is 3-transitive. Hence, if g ∈ P Γ L (2 , q )fixes at least 3 points, we may assume that it fixes ∞ , 0 and 1, fromwhich g = t φ ( t ), with φ ∈ Gal ( F γ r / F γ ), q = γ r , γ prime. Suchelement fixes γ l + 1 points, where l is a divisor of r ; in particular, itfixes 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 atleast 3 points. 8 Proof of Theorem 1.1
Lemma 3.1.
Let n be an odd prime different from the cardinality of anyprojective space, n = 11 , . Then, the isolated vertices of Λ( A n ) are (1 , ( n − / ) and (1 , ( n − / ) (when they make sense and are even permuta-tions).Proof. The two mentioned vertices are not connected to ( n ) (this makessense; recall Remark 2.5 (2)) because they are contained in AGL (1 , n ) (and n > a , . . . , a t ), t >
3. In (1 , ( n − / ), every 1 i ( n − / , ( n − / ), the i ’s which are not partial sum are exactly those ≡ a , a ≡ a + a ≡ a , a , a + a is partial sum;(1 , ( n − / ) is therefore isolated.Now let Cl( x ) be a vertex of Λ( A n ) different from the two above; wewant to show it is not isolated. If x / ∈ AGL (1 , n ), by Theorem 2.2 (andby 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 isan n -cycle or it has cycle type (1 , t ( n − /t ), t >
4. As just remarked, a classof n -cycles is not isolated. If t >
5, by Remark 2.4 Cl( x ) is connected to(2 , n − t = 4, for the same reason Cl( x ) is connected to (3 , n − S , A and A can be easily checked, the "if" part ofTheorem 1.1 (1) follows by Lemma 3.1. There remains to show that in theother 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 (resp. M ) issuch that every 1 i ( n − / n is odd): itis not connected to vertices with more than 1 orbit (keep always in mindRemark 2.5 (1)). Moreover, it is not connected to ( n ) because of the Mathieugroup; it is therefore an isolated vertex. If n = ( q d − / ( q − P Γ L ( d, q ) is isolated for the same reason. If n
6≡ − , ( n − / ) and (1 , ( n − / ) 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 − , i n/ z is not connected to vertices with at least2 orbits (both in the alternating and in the symmetric graph); this provesthe assertion if n is even and G n = A n . If n is even and G n = S n , z isnot connected to ( n ) as n/ 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 evenpermutation; 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 > z fixesall the blocks), so it is connected to no class with cycle type ( n ).9 emark 3.2. Let n be an even natural number, and let z be a vertexbelonging either to Λ( S n ) or to Λ( A n ). If n/ z , z isnot connected to vertices having only even parts because of the sharing of S n/ ≀ S . We will use this several times. Corollary 3.3.
Theorem 1.1 (1) is proved.
Although in this section we should not speak about Theorem 1.2, itseems a good idea to place here the proof for alternating groups of primedegree different from the cardinality of any projective space.
Lemma 3.4.
Let n > be a prime different from the cardinality of anyprojective space. Then, Ξ( A n ) is connected and d (Ξ( A n )) = 3 .Proof. By Theorem 2.1, (1 n − ,
3) is connected to ( n ); it is connected tonothing else because everything is partial sum. Moreover, (1 , (( n − / )is not isolated ( n >
7) but not connected to ( n ). Hence, d (Ξ( A n )) > n = 11 ,
23. In the proof of Lemma 3.1 we showed that avertex of Ξ( A n ) is connected to at least one among ( n ), (2 , n − , n − d (Ξ( A n )) n = 11 and n = 23 we have M and M that may change something:a vertex inside these groups is not connected to ( n ). However, we cancheck that these groups contain vertices which are either isolated (because ofpartial sums) or connected to one between (2 , n − , n − d (Ξ( A n )) Remark 3.5. (1) The proof of Lemma 3.4 shows that d (Ξ( A n )) > n >
11 prime.(2) By Lemma 3.4, in Theorem 1.1 (1) the condition that Λ does not haveisolated vertices is equivalent to the condition that Λ is connected.(3) Since we will need this later, we mention that also d (Ξ( A )) = d (Ξ( A )) =3. In the first case, we have P GL (3 , , d (Ξ( A )) P Γ L (2 , ,
11) and (4 , AGL (1 ,
17) is connected to (3 , ,
11) and (4 ,
9) are pairwise connected, we have d (Ξ( A )) 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 Am the setof all partitions of m different from (1 m ) that define an even permutation.10e 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/ , z ), where z is whatsoverpartition of p An/ ; and if n is odd, define I n as the set of all partitions of n of the form (1 ( n − / , z ), where z is whatsover partition of p A ( n +1) / .Let w ∈ I n . Every 1 i n/ w is not connectedto vertices with at least 2 orbits. Moreover, if n is even w is not connectedto ( n ) by Remark 3.2; and if n is odd, the same holds, because w is aneven permutation. Therefore, really I n is made of isolated vertices. Inorder to conclude, it suffices to observe that | I n | = | p An/ | if n is even and | I n | = | p A ( n +1) / | if n is odd.Assume now n odd nonprime and G n = A n . Fix n , and call p n thesmallest nontrivial divisor of n . Define I n as the set of all partitions of n ofthe form (1 n ( p n − /p n , z ), where z is any partition of p An/p n .If w ∈ I n , w ∈ S n/p n ≀ S p n : the first p n − z . Hence, w is not connected to ( n ). Moreover,every 1 i ( n − / w is not connected to verticeswith more than 1 orbit: w is isolated. Note now that n/p n > √ n ; hence | I n | = | p An/p n | > | p A ⌊√ n ⌋ | , which goes to infinity as n → ∞ . Corollary 3.6.
Theorem 1.1 (2) is proved.
The estimates given in the proof of Theorem 1.1 (2) are far from beingrealistic. A more subtle purpose would be to determine whether the pro-portion 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 ofsome projective space. Here, the whole game is played by P Γ L , and againmore subtle considerations might be needed. Ξ and upper bound to the di-ameter We begin with a small remark. We will start the proofs with a sentence ofthe type "pick a vertex of Ξ", without any preliminary consideration showingthat Ξ is not the null graph. However, this will often be clear; and in anycase, 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 aresaying that, although we will not state it explicitly, in the proofs it will beshown that the groups are invariably generated by 2 elements). Moreover, asalready announced, many of the cases of degree at most 10 will be omitted.
Proposition 4.1. (1) Let n > be an odd natural number. Then, Ξ( S n ) is connected and d (Ξ( S n )) .
2) Let n > be an even natural number. Then, Ξ( A n ) is connected and d (Ξ( A n )) .Proof. (1) Let z be a vertex of Ξ( S n ) not connected to ( i, n − i ) for every2 i ( n − / n . Then, by Remark 2.4 every such i ispartial sum in z . Now, being in Ξ, z will be connected to some w in which,necessarily, no 2 i ( n − / 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 asabove, are pairwise connected.(2) Equal to (1). We can use the expression "connected to ( i, n − i )" if i > n — see Remark 2.5 (1).The previous proof fails in alternating groups of odd degree and in sym-metric groups of even degree. In the first case, there are not vertices with 2orbits; in the second, such vertices are even permutations, so an edge withthem concerns not only partial sums but also parity. For these elementaryreasons, 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 ele-ments with at most 4 orbits. Proposition 4.2.
Let n > be an even natural number. Then, Ξ( S n ) isconnected and d (Ξ( S n )) .Proof. Let z and w be two vertices of Ξ( S n ) which are connected. One ofthe two, say z , must be odd. Assume z is not connected to ( i, n − i ) forevery 1 i n/ n . Then, every such i , i = 1, is partialsum 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 on 12 points, M and M because it is odd). Wehave the first group only if n = 2 m , in which case 3 does not divide n , so3 is partial sum in z . Hence, z = (3 , . . . ): z / ∈ AGL ( m,
2) (Remark 2.6(1) applied to z ). Assume now z ∈ P GL (2 , p ). Since every 3 i n/ n is partial sum, z has odd parts; let a be one of them. Byassumption, 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 .Now we divide the cases n ≡ n ≡ n/ − n , hence it is partial sum in z . Write z = ( a , . . . , a t ) and n/ − P hi =1 a i . If a k = 1 for some k ∈ { , . . . , h } , then n/ − z (here we use n > n/ − w is connected in the first caseto y = (1 , n/ − , n/ y = (1 , n/ − , n/ y and y are contained in nonontrivial primitive group: we have therefore our desired edge.If n ≡ n/ − n/ −
1. Thesame considerations lead us to the search of an edge with v = (1 , n/ − , n/ v = (1 , n/ , n/ − v and v arecontained in no nontrivial primitive group (for v , Theorem 2.1 is sufficient).The unique change is that now v ∈ S n/ ≀ S . This is not a problem, because n/ z , hence z ∈ S n/ ≀ S : w cannot belong to the samegroup.Now we deduce the connectedness of Ξ and the bound to the diame-ter. 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/ ≀ S . For (1 , n − n ≡ n >
14 (at the end we will consider thecases that we exclude along the proof). In order to show d (Ξ( S n )) y , y and ( i, n − i ), 1 i n/ n , have pairwise distance at most 2 in Ξ( S n ). This is true, because allthese vertices are connected to (2 , n −
4) (for (1 , n − n ≡ AGL ( m, n ≡ n >
20. Again, it is sufficient to show thatthe vertices v , v and ( i, n − i ), 1 i n/ n , have pairwisedistance at most 2 in Ξ( S n ). The vertices ( i, n − i ), 1 i n/ n , i = 1 , n/ −
1, are connected to v ; v , v and ( i, n − i ), i = 3 , , , n − v , v and ( i, n − i ), i = 1 , , , n/ − n ) are connected to (2 , , n − n = 12 , , ,
20. Since 5 divides 20, for n = 20it is possible to replace (2 , ,
11) with (2 , ,
13) and everything works. For n = 16, let z and w be as above. As in the general case, 1 , , z . If 2 is not partial sum, it is easy to see that 8 is partialsum, and so z is isolated by Remark 3.2. Therefore, 2 is partial sum in z ,from which w is connected to (1 , , , ,
13) and ( i, − i ), i = 1 , , ,
7, are connected to (4 , , / ∈ AGL (4 , n = 14, a similar argumentworks. For n = 12, in Remark 5.12 we will show that d (Ξ( S )) = 5.Although the small degrees are left to the reader, we should point outthat Λ( S ) is the empty graph: S is not invariably generated by 2 elements.This we can see with the usual arguments; recall also Remark 3.2 and notethat (2 ) ∈ P GL (2 , emma 4.3. Let n > be a natural number. Let z and w be two partitionsof n that do not share any intransitive maximal subgroup of S n . Then, thereexist a ∈ { z, w } and i ∈ { , , , } such that i e i are not partial sums in a .Proof. By contradiction. Without loss of generality, we write 2 in z and 4in w . So, z = (1 . . . ) or z = (2 , . . . ); and w = (3 , , . . . ) or w = (4 , . . . ). Wedivide the different cases.(1) z = (1 , . . . ) and w = (4 , . . . ).(a) We write 3 in z . Then, z = (1 . . . ). Moreover, we write 6 in w , from which w = (4 , , . . . ). Then, we write 5 in z : impossibleunless falling in equality of partial sums.(b) We write 3 in w . Then, w = (4 , , . . . ), from which we write 6 in z : no way to do this.(2) z = (2 , . . . ) and w = (1 , , . . . ). We write 6 in z : z = (2 , , . . . ). It isnot possible to write 5 in w ; hence z = (2 , , , . . . ). Then, we write 10in w , but it is easy to see that this way we write 13 both in z and in w .(3) z = (2 , . . . ) and w = (4 , . . . ).(a) We write 3 in z . Then, z = (2 , , . . . ) or z = (2 , , . . . ). Moreover,we write 6 in w , from which w = (4 , , . . . ). Then, we write 5 in z : the unique way (avoiding equality of partial sums) is (2 , , . . . ).There is no way to write 7 in z ; hence w = (4 , , , . . . ). And nowa patient check shows that it is not possible to write 14 in z .(b) We write 3 in w . Then, w = (4 , , . . . ), from which we write6 in z : z = (2 , , . . . ). It is impossible to write 5 in z ; hence w = (4 , , , . . . ). Then, we should write 10 in z , but this again isimpossible (unless falling in equality of partial sums). Proposition 4.4.
Let n > be an odd natural number. Then, Ξ( A n ) isconnected and d (Ξ( A n )) . If none of , and divides n , d (Ξ( A n )) ;if n = p , with p prime, and n is different from the cardinality of anyprojective space, d (Ξ( A n )) .Proof. We start from the particular cases. Assume first none of 3, 5 and 7divides 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 Lemma3.4 and Remark 3.5 (3), we may assume n >
23. The hypotesis of Lemma4.3, applied to x c and y c , is satisfied (Remark 2.5 (1)); hence, there exist14 ∈ { x, y } and k ∈ { , , , } such that k e 2 k are not partial sums in b c .It follows that Cl( b ) is connected to ( k , n − k ) (Remark 2.4). In orderto conclude, it suffices to observe that the ( i , n − i )’s, i ∈ { , , , } , arepairwise connected ( n >
19 prevents a sharing of products of symmetricgroups among these vertices).Assume now n = p , p prime, with n different from the cardinality of anyprojective space. Let Cl( x ) be a vertex of Ξ( A n ); we show it has distanceat 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. ByLemma 3.4, n >
15; we assume first n >
21. I claim that one between Cl( x )and Cl( y ) is connected to at least one among (1 , n − , n − , , n − , , n − , , n − , , n −
5) and (2 , , n −
10) belong to someimprimitive wreath product: they are contained in S ≀ S n/ . For primitivemaximal subgroups, (1 , n −
2) belongs to P Γ L (2 , n −
1) and to nothingelse (by Theorem 2.2); (2 , n − , , n −
10) and (1 , , n −
4) belong tonothing (the first three by Theorem 2.1, the last by [10]); (1 , , n − AGL ( m, , n − S ≀ S n/ and S n/ ≀ S .Moreover, it belongs neither to AGL ( m, p ) nor to P Γ L (2 , n − , n − , n − , , n − , , n −
5) and (2 , , n −
10) are connected to (6 , n − d (Ξ( A n )) Note also that the above considerations say what we need to worry aboutif we want an edge with one of those vertices: in the following, we will notrepeat the same things.So we have Cl( x ) and Cl( y ). Assume there exists b ∈ { x, y } such that1 and 2 are not partial sums in b c ; without loss of generality, b = x . Then,either Cl( x ) is connected to (1 , n −
2) or x ∈ P Γ L (2 , n − x c , Cl( x ) is connected to (2 , n − x c = (4 , . . . ), from which x fixes an even number of points greater than 3, from which x / ∈ P Γ L (2 , n − x c and 2 is partial sum in y c , that is, x c = (1 , . . . ) and y c = (2 , . . . ).Then, 4 is partial sum in x c unless Cl( x ) is connected to (2 , n − x c = (1 , , . . . ) or x c = (1 , , . . . ). In the first case, Cl( y ) is connectedto (1 , , n − y ) is connected to (1 , , n −
5) unless y iscontained in AGL ( m,
5) or in S ≀ S n/ . The first option is excluded because By Remark 2.5 (1), we may say "a vertex connected to (1 , , n − fixes 2 points. If y ∈ S ≀ S n/ , since 5 is partial sum in x c , the uniquepossibility for the 2-cycle is to contribute in exchanging 2 blocks, from which y c = (2 , , . . . ). At this point, since certainly x / ∈ S ≀ S n/ , Cl( x ) is connectedto (2 , , n − n = 15. Let Cl( x ) and Cl( y ) be as above. Weprove that one of the two is connected to at least one between (1 ,
13) and(15) (this makes sense by Remark 2.5 (2)). Since, by Theorem 2.2, suchvertices are connected, we have d (Ξ( A ) x ) and Cl( y ) are not connected to (1 ,
13) then(without loss of generality) x c = (1 , . . . ) and y c = (2 , . . . ). Again by Theorem2.2, a vertex is not connected to (15) if and only if it is contained in someimprimitive wreath product. It is not possible that x ∈ S ≀ S , because the1-cycle should contribute to fix 1 block, from which 2 would be partial sumin x c , false; hence we may assume x ∈ S ≀ S and y ∈ S ≀ S . In x c , the1-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 ispartial sum, and in the second, 4 is partial sum: false. Ξ In this section, almost all the proofs will show that the diameter of Ξ, inthe 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 connectedto one vertex only, say ˜ z and ˜ w respectively, with ˜ z and ˜ w not connectedin Ξ (actually, sometimes w will be connected to more than one vertex, andnone of such vertices will be connected to ˜ z ). This will show d ( z, w ) > z (resp. w ) and and the verticesdifferent from ˜ z (resp. ˜ w ) having at least 2 orbits, we will use intransitivemaximal subgroups. This lemma will be used to prove that the sharing ofsuch groups indeed occurs. Lemma 5.1.
For every natural number m , denote with m the natural num-ber between m/ and ( m − / . Let N be a positive natural number, let S be a subset of { , . . . , N / } , and let v be a partition of N such that every i ∈ { , . . . , N / } \ S is partial sum in v . If T is a natural number suchthat T N − max S − if N + T is even and T N − max S − if N + T is odd, with the convention max S = − if S is empty, then every i ∈ { , . . . , ( N + T ) / } \ S is partial sum in v T , where v T is the partition of N + T obtained adding a part of length T to v .Proof. The condition that every i ∈ { , . . . , N / } \ S is partial sum in v implies that also every 1 i N − max S − i / ∈ S , is partial sum (withthe convention max S = − S is empty). The condition on T is equivalentto ( N + T ) / N − max S −
1, hence the assertion follows.16he 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 evendegree will be similar, essentially because they all will rely on vertices with2 orbits. Of course, this will not be possible in alternating groups of odddegree: that is why we will treat this case at the end. The case that willrequire the most work is that of alternating groups of even degree n : wewill divide the cases n power of 2, n ≡ n = 2 d with d odd nonprime, n = 2 p with p odd prime. We will use the followingarithmetical observation: Lemma 5.2.
Let p be a prime. If p = ( q d − / ( q − , with q prime powerand d > , then d = 2 .Proof. Being 2 p even, q is odd and d is even. Moreover, if r is a divisor of d , then 2 p = ( q d − / ( q −
1) = (( q d − / ( q r − q r − / ( q − r ∈ { , d } . Therefore, d is prime.In symmetric groups of odd degree, our ˜ z will be ( n ). In symmetricgroups of even degree, although ( n ) would work, we will use a differentapproach; whilst in alternating groups of even degree our ˜ z will be (( n/ ).This latter vertex should be considered as a faithful substitute of ( n ): boththe cycle types are contained in every imprimitive wreath product and,apart from some exceptional cases, they are contained in the same nontrivialprimitive groups (Theorem 2.3).These considerations explain the meaning of the following lemma. Lemma 5.3.
Let n > , n = 18 , be a natural number. Then, in Ξ( S n ) there exists a vertex z connected only to ( n ) . Assume now n is even. Then,in Ξ( A n ) there exists a vertex z connected only to (( n/ ) .Proof. If n >
11 is odd, define z as the odd permutation between (1 ( n − / , ( n +1) /
2) and (1 ( n +1) / , ( n − / n > n = 18 is even, we will define an even vertex z , belonging toboth Ξ( S n ) and Ξ( A n ), which will be contained in no imprimitive product,in which every 1 i < n/ n ) and (( n/ ) any nontrivial primitive group. In order to do this,we will divide some cases.(1) n ≡ z = (1 n/ − , n/ i < n/ z ∈ S t ≀ S n/t . The unique possibility isthat t divides n/ t = 2, false because n/ z is not contained in nontrivial primitive groups.If n ≡ z does not work anymore: it is contained in S ≀ S n/ .172) n = 2 p , p > z = (1 n/ − , , , n/ i < n/ z ∈ S t ≀ S n/t . One possibility is that t divides n/ t = 2,false because z , having exactly one 3-cycle in the cycle type, cannot belongto S ≀ S n/ . The other possibility is that the ( n/ t divides n/ n/ t is odd. Also, t divides n/ −
4, so it divides 8, false. If instead the 2-cycle collaborates, it must collaborate in permuting 2 blocks, and what isleft must fix 1 block, from which n/ n/ − n = 18, false (indeed, in that case, (1 , , , ∈ S ≀ S ).Moreover, z does not share with ( n ) and (( n/ ) nontrivial primitivegroups: by Theorem 2.3 and Lemma 5.2, if n = 22 we just need to careabout P Γ L (2 , n − : z does not belong to this group because z fixes anodd number of points greater or equal to 3 (Remark 2.6 (2)). If n = 22, wecheck directly that z / ∈ Aut( M ).(3) n = 2 d , 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 imprimitivewreath products for the same reason ( n = 18), and now 3 is coprime with 2and with n/ z n/ is a 3-cycle: not contained in nontrivial primitivegroups.If p >
5, define z = (1 n/ − p − , , p, p + 2 , n/ i < n/ z . p is coprime with 2, p + 2 and n/ z is contained in S t ≀ S n/t . The ( n/ 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/ n = 18 as above. Neither is able the p -cycle, because p and n/ t divides n/ p + 3, from which t is odd, from which t divides p + 3. Since t is odd, it must be t < p + 3, fromwhich also t < p ( p = 3). Hence, the p -cycle permutes points from morethan 1 block, but cannot permute some whole blocks: it must be helped bysomething 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/ )) is not connected to (1 , n − if n = q + 1, with q prime power, n is certainly different from the cardinality of anyother projective space. emma 5.4. Let n > be a natural number. Then, in Ξ( S n ) there existsa vertex w connected only to (1 , n − . Assume now n is even. Then, in Ξ( A n ) there exists a vertex w connected only to both classes with cycle type (1 , n − .Proof. If n is odd, define w = (2 ( n − / ,
3) if n ≡ w =(2 ( n − / , ) if n ≡ w is even, so it is not connected to ( n ).Every 2 i ( n − / v = (2 , S = { } and add the other parts oneby one), so w is connected to nothing different from (1 , n − w / ∈ AGL ( m, p ) (Remark 2.6 (1)), by Theorem 2.2 w is indeed connected to(1 , n − n is even, and consider first Ξ( S n ). Define w = (2 ( n − / , )if n ≡ w = (2 ( n − / , , ) if n ≡ i n/ w might be connected only to (1 , n −
1) (recall Remark 3.2).Since w is odd, if n = 12 w / ∈ M , M ; and if n = 24, w / ∈ M . Therefore,by Theorem 2.3 w is connected to (1 , n −
1) provided that it is not containedin
AGL ( m,
2) and in
P GL (2 , p ). This last claim is true (Remark 2.6).Finally, consider Ξ( A n ). Define the same w as above, swapping the cases n ≡ n ≡ w becomes even. Direct checkshows that w / ∈ M (on 12 points), M , M . Hence, w is connected onlyto both classes with cycle type (1 , n − Corollary 5.5.
Theorem 1.2 holds for symmetric groups of odd prime degree n and for symmetric and alternating groups of even degree n if n − is primeand n = 18 .Proof. By Lemmas 5.3 and 5.4, it suffices to show that for symmetric groups( n ) is not connected to (1 , n − n/ ) is notconnected to any class with cycle type (1 , n − n is odd prime, thisis true because of AGL (1 , n ). If n is even and n − 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/ )) 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 tosay something about S √ n ≀ S in its product action. Remark 5.6.
Let n be a natural number, and z = ( a , . . . , a r ) be a partitionof n . Assume there exists 1 i n/ z ; say i is the minimum with respect to such property. Then, if a , . . . , a t are someparts of z necessary to write every 1 k i − a j ’s), we have P tj =1 a j = i − a j > i + 1 for every j > t + 1.Indeed, let P hj =1 a j = i − m , m = 1 , . . . , i −
1, that every a j , j = 1 , . . . , r such that a j = m a , . . . , a h . This implies also the second assertion, thatis, a j > i + 1 for every j > t + 1.For m = 1 this is true, otherwise we would have ( P hj =1 a j ) + 1 = i .Assume now the assertion is true for every l m −
1; we prove it is truealso for m . We may assume there exists j such that a j = m , otherwise it isobvious. Certainly, in the a , . . . , 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. Lemma 5.7.
Let n ∈ N be a square, n > . Let g = (( σ , σ ) , τ ) ∈ S √ n ≀ S (in its product action). If τ = (1 , , then g fixes at most √ n points. Nowcall g c the cycle type of g .(a) If g c has an odd part of length at least √ n + 1 , then τ = 1 .(b) If τ = 1 and every i √ n − is partial sum in g c , then also every i √ n − is partial sum.Proof. Assume τ = (1 , g fixes at most √ n points.Saying that (( σ , σ ) , (1 , γ , γ ) is the same as saying that γ = γ σ and σ σ ∈ G γ . Hence, the maximum number of fixed points is obtainedwhen σ = σ − : in this case, ( γ , γ σ ) is fixed by g for every choice of γ : g fixes precisely √ n points.Assume now g c has an odd part of length at least √ n + 1, call it b .Assume τ = (1 , g b = (( ˜ σ , ˜ σ ) , ˜ τ ) is such that ˜ τ = (1 , g b fixes at most √ n points. On the other hand, being b a partof g c we have that g b fixes at least b > √ n + 1 points: contradiction. Thisproves (a).Assume now τ = 1. Assume there exists √ n i n/ g c ; call k the minimum such i .By Remark 5.6, g c = ( a , . . . , a r ), with P ti =1 a i = k − a j > k + 1for every j > t + 1. Write (1 i , b i +1 , . . . , b l ) and (1 j , c j +1 , . . . , c s ), with i, j > b k , c k = 1, for the cycle types of σ and σ respectively. By the structureof g c , it must be A = ij + j P lk = i +1 b k + i P sk = j +1 c k k −
1. On the otherhand, B = i + j + P lk = i +1 b k + P sk = j +1 c k = 2 √ n . Since A > B −
1, we have k > √ n , hence (b) is proved.One comment. Along many of the following proofs we will make as-sumptions on n stronger than what the statements will claim. When thiswill happen, it will be meant that sooner or later, during the proofs, theexcluded cases will be checked singularly. Lemma 5.8.
Let n > be an odd nonprime natural number, and let p bethe smallest nontrivial divisor of n . Then, in Ξ( S n ) there exists a vertex w connected only to ( p, n − p ) . Assume now n = 2 d , with d > odd nonprime,and let p be the smallest nontrivial divisor of n . Then, in Ξ( A n ) there existsa vertex w connected only to both classes with cycle type ( p, n − p ) . roof. Assume first n odd and p = 3. By Bertrand-Chebyshev Theorem,we may choose a prime q contained in the open interval ( n/p, n/p ). Nowdefine w = (1 p − , ( p + 1) k , p + 2 , q, r ), with k > p + 1 r < 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 − , ( p + 1) , 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 < n/p , weare okay if n > (5 p + 3 p ) / ( p − n > p , we want p > p + 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 > (5 p + 3 p ) / ( p −
2) holds also in these cases.Now we have to show that w is connected only to ( p, n − p ). Since w iseven, 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 − / i = p , is partial sum.Put v = (1 p − , p + 1 , p + 2), S = { p } and apply Lemma 5.1 adding oneby one all the ( p + 1)-cycles (at least there is another) and the r -cycle (thefirst step is sharp, as p + 1 = 3 p + 2 − p − q < n/p , wesee that q ( n − q ) − p − p = 5, in which case we need n > q .There remains to show that w is indeed connected to ( p, n − p ). In orderto do this, we must show that w and ( p, n − p ) do not share imprimitivewreath products and nontrivial primitive groups. For nontrivial primitivegroups, by Theorem 2.3 we just need to take care of AGL ( m, p ) and S p ≀ S . w does not belong to the first group as it fixes p − p − 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 groupis such that, in its cycle type, the number of 1-cycles + the lengths of someother cycles (of length p ) is divided by p . Moreover, since q is a primestrictly bigger than n/p , an element having q in the cycle type does notbelong to S n/p ≀ S p .There remains to consider the cases n = 25 , ,
49. We define respec-tively w = (1 , , w = (1 , , , , , w = (1 , , , , , n odd and p = 3. In [19] it is shown that, if r >
25, thenthere exists a prime q contained in the open interval ( r, r/ r = n/
3, hence we require n >
75. Define w = (1 , k , , q, r ), with k > r
7. If the defined element is odd, we unify two 4-cycles into an8-cycle. n >
39 is enough to guarantee that the element makes sense. Putting v = (1 , ,
5) or v = (1 , , ,
8) (depending on whether two 4-cycles havebeen unified or not), S = { } and applying Lemma 5.1 adding one by onethe 4-cycles, the r -cycle, and finally the q -cycle, we get that every 1 i ( n − / i = 3, is partial sum: w is connected to nothing different from(3 , n − w does not share with (3 , n −
3) primitive groups andimprimitive wreath products for the same reasons as above.21e are left with the check of the cases n <
75 (with n divided by 3).If n >
39, we just need a prime contained in the open interval ( n/ , n/ n = 39 , ,
69, a quick check shows its existence. For n = 21, we take w = (1 , , , n = 27, w = (1 , , , n = 33, w = (1 , , , n = 39, w = (1 , , , , n = 57, w = (1 , , , , , n = 69, w = (1 , , , , n = 2 d , d >
15 odd nonprime. We define w exactly asabove, dividing the cases p = 3 and p = 3. We run through the routinchecks again.For p = 3, using the stronger inequality n > p we see that the definitionof 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 thecases n <
75, that is, with our conditions, n = 54 ,
66. Since in both casesthere exists a prime contained in the open interval ( n/ , n/ Corollary 5.9.
Theorem 1.2 holds for symmetric groups of odd nonprimedegree n and for alternating groups of even degree n = 2 d , with d odd non-prime.Proof. For symmetric groups of odd nonprime degree, Lemmas 5.3 and 5.8leave us with the case n = 15. In order to show that d (Ξ( S )) = 3, let z be a vertex of Ξ( S ) not connected to (15), ( i, d − i ), i = 1 , , ,
7. Then(Remark 2.4), 1 , , , z ; hence, at least one among 3 , z would be isolated). The unique way inwhich 1 , , , , S ≀ S . Therefore, 3 is partial sum.Now it is immediate to see that there are no vertices in which 1 , , , i coprime with 15 is partial sum).As (15) and ( i, d − i ), i = 1 , , , d (Ξ( S )) d (Ξ( S )) > ,
3) is connected onlyto (1 ,
14) and (1 ,
8) is connected only to (15).For alternating groups, we are left with the cases n = 18 , ,
42. Thelatter two are contamplated in Corollary 5.5 (they also could have beenincluded in Lemma 5.8, but it would have been useless). For n = 18, a littlechange seems necessary. (1 , ) is contained in S ≀ S : it is not connected to(9 ). Instead, by Theorem 2.3 it is connected to (4 , , , ,
11) is connected to (8 ,
10) and to (9 ); the diameter is 4 because(4 ,
14) is connected to none of these vertices ( S ≀ S ).The reader will have noted that the proof of Lemma 5.8 works perfectlyalso for symmetric groups of even degree (values contemplated there). Thiswill be true for almost all the proofs we will give for alternating groups of22ven degree. Nevertheless, the symmetric case can be treated separately,with a proof which is faster and which points out once again the reason ofthe difference, namely, the fact that vertices with 2 orbits are not connectedbecause of parity. Lemma 5.10.
Let n > be an even natural number, with n − not primeunless n = 18 . Then, in Ξ( S n ) there exists a vertex z connected only to (3 , n − .Proof. If 3 does not divide n and n >
26 we define z = (1 , k , , r ), with k > r n >
23 ensures this possibility). If the defined elementis even, we meld two 4’s in a 8. As in Lemma 5.8 we see that the unique1 i n/ i = 3. Being n/ z isnot connected to ( n ); hence, it might be connected only to (3 , n − n = 24, Theorem 2.3 tells us that we do not need to care about nontrivialprimitive groups. Moreover, (3 , n −
3) is contained in no imprimitive wreathproduct. Therefore, since z is odd, z is indeed connected to (3 , n − n = 16, define z = (1 , , n = 22, z = (1 , , ,
9) (the latter is notconnected to (4 ,
18) because of S ≀ S ).If instead 3 divides n , we define z as in Lemma 5.8 for the case p = 3; wetake here the odd version. We are left with the cases n <
75, that is, withour conditions, n = 18 , ,
66. For n = 66, there exists a prime containedin the open interval ( n/ , n/
5) so, since 66 >
39, this case is done. For n = 18, define z = (1 , , , n = 36, z = (1 , , , , Corollary 5.11.
Theorem 1.2 holds for symmetric groups of even degree n .Proof. By Corollary 5.5, we may assume that n − n = 18. Then, the assertion follows by Lemmas 5.4 and 5.10, noting that(1 , n −
1) and (3 , n − S n )(in Lemma 5.10, probably also all the cases of n − Remark 5.12.
We pause for a moment and we show that d (Ξ( S )) = 5.Let z be a vertex of Λ( S ); we prove that it is either isolated or connectedto one among (1 , , , , , , , d (Ξ( S ))
5. Thethird, the fourth and the fifth vertices are connected to the sixth. Forthe first and the second, this is not true, because we have, respectively,
P GL (2 ,
11) and S ≀ S . However, (5 , , ,
7) and (1 ,
11) are connectedto (3 , ,
6) ((3 , , / ∈ P GL (2 , M , M ); (12) and (1 ,
11) are connectedto (2 , , ,
7) and (3 ,
9) are connected to (1 ,
10) ((3 , / ∈ P GL (2 , ,
9) and (1 ,
11) are connected to (2 , ); (1 , ,
7) and (3 ,
9) are connected.This is sufficient. 23n many occasions, along the following lines, z will be claimed to beisolated: in order to check so, keep in mind Remark 3.2 and the fact thatvertices with 2 orbits are even.Assume first that 1 and 5 are partial sums in z . If 2 is not partial sumand 4 is partial sum, z = (1 , ,
7) or z = (1 , , ) or z = (1 , , S ≀ S ). If 2 and 4 are notpartial sums, z = (1 , , , z = (1 , ,
6) or z = (1 , ):isolated. If 3 is partial sum and 4 is not partial sum, there is no way towrite 5, false; if 4 is partial sum, z is isolated unless 6 is not partial sum, inwhich case it is connected to (12) (by Remark 5.6, it is some ( a , . . . , a t , P a i = 5; 7 is certainly coprime with the a i ’s).Therefore, we may assume that one between 1 and 5 is not partial sum.By Theorem 2.3, this implies that z is connected to (1 ,
11) or to (5 ,
7) unlessit is even or only 1 is not partial sum and z is contained in one among P GL (2 , M , M . M , M are contained in A ; and direct checkshows that the odd elements of P GL (2 ,
11) have cycle type (1 , ), (1 , ): the first is isolated and the others are connected to (5 , z is even. If 1 is not partial sum and 5 is partial sum,the possibilities are z = (2 , , z = (2 , , z = (5 , z = (1 ,
9) or z = (1 , , ) the group S ≀ S )). The last possibility is that 1 and 5 arenot partial sums in z . Then, if z is not connected to (1 , , z = (2 ), z = (2 , ), z = (2 , z = (4 , , , d (Ξ( S ) >
5, note that (4 ,
8) is connected only to(2 , , , , , ), (3 , , , ,
6) suitableimprimitive wreath products. On the other hand, as already noticed (1 , , , , ,
7) is connected only to (12), (4 ) and toeven vertices, hence d ((2 , , , (3 , >
3, hence d ((4 , , (1 , , > n ≡ n > p is too weak in order to define that w . Hence the approach isdifferent; we use the fact that (( n/ ) is not connected to vertices havingonly even parts. This feature will be used also in the case n power of 2. Lemma 5.13.
Let j > be a natural number, and d > be an odd naturalnumber; set n = 2 j d . Then, in Ξ( A n ) there exists a vertex w connected to (2 j − , n − j − ) , and maybe only to some other ( a , . . . , a m ) , with a k evenfor every k and ( a , . . . , a m ) = (( n/ ) .Proof. Define w = (1 j − − , j − + 1 , (2 j ) ( d − / , n/ − j − ).24t might be odd; in order to change sign, if j > j = 2, w = (1 , , ( d − / , n/ − d − / >
3, that is, d > d = 5, w = (1 , , , , , ); and if d = 3, w = (1 , , ), already even.From now on we will use the original definition of w ; at the end we willanalyse whether the changes will have affected something or not.Being n/ − j − ≡ j , w belongs to S j ≀ S n/ j : it is not connectedto (( n/ ). Moreover, it is easy to see that only some even 1 i n/ j − k , k odd); therefore, w might be connected only to some ( a , . . . , 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 − , n − j − ). If n = 12, Theorem 2.3 tells us that we should not worryabout primitive groups (clearly 2 j − = k √ n ). If n = 12, w = (1 , , ) doesnot belong to nontrivial primitive groups by Theorem 2.1.Therefore, there remains to show that w and (2 j − , n − j − ) do not shareimprimitive wreath products. (2 j − , n − j − ) belongs only to S t ≀ S n/ t andto S n/ t ≀ S t for every t j − w does not belong to theformer (see proof of Lemma 5.8).For the latter, assume first the ( n/ − j − )-cycle permute points froma single block. Then, n/ − j − n/ t , that is, 2 j − ( d − j − t d , fromwhich t = 1. But w does not belong to S n/ ≀ S , because w has fixed pointsand n/ n/ − j − )-cycle must permutepoints taken from different blocks.Note that n/ t = 2 j − t d and n/ − j − = 2 j − ( d − n/ − j − )-cycle cannot permute some wholeblocks. The 1-cycles cannot contribute to finish such blocks, from which n/ − j − + k (2 j − + 1) + 2 j m = a ( n/ t ), for some k, m, a , with k ∈ { , } , m > a >
1. If k = 1, the first member is odd whilst the second iseven; so k = 0. n/ − j − + 2 j m = 2 j (( d − /
2) + 2 j m = 2 j (( d − / m ),and a ( n/ 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 usedall cycles of w yet.Now we should make sure that the variations of w do not affect ourconsiderations. It is easy to observe that the unique modification we shouldtake care of is the melding of two 1’s into a 2: a priori it could changesomething about imprimitive wreath products. In particular, the 2-cyclecould help in permuting 2 blocks, from which n/ − j − +2 j m +2 = 2( n/ 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 ≡ mod but not power of . Lemma 5.15.
Let m > , m = 7 be a natural number. Then, in Ξ( A m ) there exists a vertex w connected only to ( i, m − i ) for some even i , i = 2 m − . roof. We divide the cases m even and m odd. In any case, in order to showthat w is not connected to ((2 m − ) ) we will show that w is contained insome imprimitive wreath product.If m is even, 2 m − − m − − t . Define w = (1 , , , , m − − , m − − t , m − −
2) (if m = 6, there are no 8-cycles). w belongs to S m − ≀ S : the (2 m − − t from each block; the 3 cycles, in number 2 m − − t , finishsuch blocks; what is left fixes the remaining block (not pointwise, of course).Applying suitably Lemma 5.1 it is easy to see that every 1 i < m − , i = 2, is partial sum; w is then connected to nothing different from (2 , m − m − is not partial sum (otherwise also 2 would be, and this is not thecase), and w has odd parts, so w / ∈ S m − ≀ S ; and clearly w / ∈ S ≀ S m − .Moreover, w / ∈ AGL ( m,
2) because w fixes 5 points. Hence, by Theorem2.3 we can conclude that w is indeed connected to (2 , m − m is odd, 9 divides exactly one between 2 m − −
4, 2 m − −
10, 2 m − − m − − t , and define w = (1 , , , , ∗ m − − , m − − t , m − − w belongs to S m − ≀ S : the (2 m − − t from each block, the 9-cycles finish these blocks,and what is left preserves each of the remaining 7 blocks. As above, every1 i < m − , i = 4, is partial sum; there remains to show that w isconnected to (4 , m − AGL ( m,
2) because it fixes3 points; by Theorem 2.3 (and Remark 2.4), there remains to exclude that w belongs to S ≀ S m − , S m − ≀ S , S ≀ S m − , S m − ≀ S . For the first twogroups is as above, and for the third, usual consideration (proof of Lemma5.8). For the fourth, the (2 m − − m − − > m − ; and it cannot permute points from 2 blocksbecause 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 3 t fromeach block. The unique other cycles whose length is divisible by 3 are the9-cycles, which permute 3 points from each block. But this would implythat 2 m − is divisible by 3, false.In the second case (9 divides 2 m − − m − −
10 = 9 t , and define w = (1 , , , , ∗ m − − , (2 m − − t ) / , m − − w ∈ S m − ≀ S and that we have an edge only with (10 , m − m − − m − − m − −
34 = 9 t , and define w = (1 , , , ∗ m − − , , (2 m − − − t ) / , m − −
34) (note that m > w ∈ S m − ≀ S and wehave an edge only with (34 , m − m − − t > m − − t ≡ Corollary 5.16.
Theorem 1.2 holds for alternating groups of degree a powerof .Proof. Lemmas 5.3 and 5.15 leave us with the cases of degree 16 , , ,
8) is contained in
AGL (4 , n = 2 p . For this, we need some more considerations about P Γ L (2 , q ). Lemma 5.17.
Let n be an even natural number, with n − prime power,and let z be a partition of n . If z belongs to P Γ L (2 , n − and has an evennumber of 1-cycles, then, for every i n odd, z has an even number of i -cycles as well. In particular, z ∈ S ≀ S n/ .Proof. The proof is by induction on i , i = 1 , . . . , n , i odd. The base caseholds by assumption. Assume now the induction holds for every k i − k odd). In order to prove the assertion for i , just observe that, if in z therewas an odd number of i -cycles, then, by the inductive hypotesis, z i wouldfix 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 Γ L (2 , n − , z is even.If there are not 1-cycles, we are done. If there are at least two, eitherthere are no others, in which case we are happy, or there are others, in whichcase they must be in even number by Remark 2.6 (2), happy again. Nowassume by contradiction there is exactly one 1-cycle. Then, there cannotbe 2-cycles or 4-cycles, for otherwise z or z would fix an odd number ofpoints greater or equal to 3. Then, the unique way to write 1 , z = (1 , , , . . . ). z will fix 3 k + 1 points, k >
1, from which 3 k + 1 = γ l + 1( n − γ ), from which γ = 3. The same consideration, with 3replaced by 5, implies γ = 5, contradiction. Lemma 5.19.
Thorem 1.2 holds for alternating groups of even degree n ,with n = 2 p , p > odd prime.Proof. If n − n − d (Ξ( A n )) = 3. In order to do this, we27how that a vertex z of Ξ( A n ) is connected to ( p ) or to ( i, n − i ) for some1 i n/ n (this makes sense by Remark 2.5 (1)). Sincethe mentioned vertices are pairwise connected (important: (1 , n −
1) and( p ) are connected), this implies d (Ξ( A n ))
3. Then, d (Ξ( A n )) > z be as above; assume it is not connected to ( p ). Then, Theorem2.3 tells us that either z is contained in some imprimitive wreath product,that is, in S p ≀ S or in S ≀ S p , or z ∈ P Γ L (2 , n − n = 22 and z ∈ Aut( M ). Assume also z is connected to no class with cycle type( i, n − i ) for every 1 i n/ n . Then, such i ’s are partialsums in z (Remark 2.4 for i >
3; for i = 1, see Theorems 2.2 or 2.3); wewant to get to a contradiction.An inspection of the cycle types of Aut( M ) says that such group doesnot contain z ’s with the required properties unless they are isolated. Assumenow z ∈ S p ≀ S or z ∈ S ≀ S p . z might be connected only to some ( a , . . . , a t ), a i even for every i ; but this does not happen, as such elements belong toboth the imprimitive wreath products.The last possibility is that z does not belong to any imprimitive wreathproduct, and belongs instead to P Γ L (2 , n − , z : Remark 5.18 implies z ∈ S ≀ S p .In order to conclude the proof of Theorem 1.2 we are left with alternatinggroups of odd nonprime degree. Here, our attention moves to vertices with3 orbits. Lemma 5.20.
Let n > be an odd nonprime natural number. Then,in Ξ( A n ) there exists a vertex z connected only to (1 , n − . If moreover n = 33 , in Ξ( A n ) there exists a vertex w connected only to (2 , n − , exceptfor n = 35 , in which case w is connected also to (2 , , .Proof. Let p be the smallest nontrivial divisor of n . Both z and w will becontained in S n/p ≀ S p , so that they will not be connected to any class of n -cycles.We start with z . Assume first p = 3. We start with z = (3 , , , . . . ),where two 3-cycles permute points from 1 block, the 4-cycle and another3-cycle from another block and the 5-cycle from the third block. Then, inorder to finish each block we use 3-cycles and finally an m i -cycle, 3 m i i = 1 , ,
3. If the defined element is odd, in order to change sign we unifytwo 3-cycles into a 6-cycle (the element still belongs to S n/ ≀ S because two3-cycles permute points from the same block). In order the element to makesense we need n/ − >
3, true.If p >
5, we start in the same way, but here one 3-cycle, instead of work-ing together with the 4-cycle, permutes points from another block. Then,each block is finished as above (also the blocks not touched yet are preserved28y 3-cycles and with a final cycle). Again, in order to change sign we pos-sibly unify two 3-cycles into a 6-cycle. Here we need n/p − >
3, hence werequire n = 35 , i ( n − / z are i = 1 , z might be connected only to(1 , n − z ∈ P Γ L (2 , n − z fixes an even number of pointsgreater than 3 (Remark 2.6 (2)). For n = 35, define z = (3 , , , , n = 49, z = (3 , , , , , S n/ ≀ S . Instead, it is not true that only 1 and 2 are not partial sum; it istrue, however, that these vertices share a product of symmetric groups withevery vertex with at least 3 orbits different from (1 , n − w . If p = 3, we start this way: w = (1 , , , , . . . ).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 completethe 2 moved blocks with some 6-cycles, and at the end with an m , -cycle,with m , ∈ { , , } . We complete the fixed block with 5-cycles, and atthe end with an m -cycle, with 5 m
9. For this to make sense, wemust have n/ − > n/ − >
3: we require n = 39 ,
45. If thedefined 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 two6-cycles permute points from 2 blocks; the 1-cycle and one 5-cycle permutepoints from 1 block; the other 5-cycle from another block, and the samefor the remaining 6-cycle. We complete the blocks as above (if p >
5, theblocks that have not been touched yet are preserved with 5-cycles and witha final cycle). We change sign as above. Here we need n/p − > n/p − >
5: we ask n = 35 , , , , i ( n − / , ,
4. Therefore, w is connected only to (2 , d −
4) (Remark2.4). For n = 39, define w = (1 , , , , n = 45, w = (1 , , , , , n = 35, w = (1 , , , , n = 55, w = (1 , , , , n = 77, w = (1 , , , , ); for n = 121, w = (1 , , , , ); for n = 49, w = (1 , , , ,
10) (same comment as for the sporadic cases of z , with anexception: for n = 35, w is connected also to (2 , , Corollary 5.21.
Theorem 1.2 holds for alternating groups of odd nonprimedegree n .Proof. Since (1 , n −
2) is not connected to (2 , n −
4) and to (2 , n − n = 33. In this case, we define w = (1 , ,
31) is not connected to any class of 33-cycles because of P Γ L (2 , z defined in Lemma 5.3, with sign changed, this argument works in factfor every n >
33 such that n − As remarked at the end of Section 3, one would like to say something moreprecise about the number of isolated vertices of Λ in the different cases.Moreover, it might be interesting to give suitable descriptions of the isolatedvertices, in order to determine, for instance, when nontrivial primitive groupsinfluence their list, and when instead they depend uniquely on imprimitivegroups and (possibly) on parity. However, this is not strictly connected tothe content of this paper.What is strictly connected is the fact that Theorem 1.2 should be im-proved: there remains to compute the exact diameter for alternating groupsof 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 equalto the cardinality of some projective space. As observed in Remark 3.5 (3), d (Ξ( A )) = d (Ξ( A )) = 3. Essentially the same arguments and a littlebit of patience show that also d (Ξ( A )) = d (Ξ( A )) = 3. Hence one askswhether 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 knowwhether the given lower bound holds also in the cases n = 15 , , , d (Ξ( S )) = 6 and d (Ξ( S )) = 5. Ξ( S ) has 9 vertices and only 8 edges. Ingeneral, symmetric groups of even degree might have more isolated verticesand less edges than the other cases, the reason being in Remark 3.2 andin the fact that vertices with 2 orbits are even. Indeed, for instance, thesefeatures imply that even vertices in which every even i is partial sum areisolated; this is the main reason for which Λ( S ) is the empty graph.However, it might be synonymous of laziness being content with the ex-amples of S and S : one should discover whether these are exceptions, dueto 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 con-cerns nontrivial primitive groups and imprimitive wreath products — haveprobably 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 some ˜ z ; call ˜ z c the cycle type. Itseems reasonable that, summing some of the parts of ˜ z c in a suitable way, stillthere is not a sharing of nontrivial primitive groups and imprimitive wreathproducts. This way, caring also about parity, we get that z is connectedto a vertex with very few cycles, say at most 3 or 4 (by [10], it is rare30hat 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 cyclesshould have pairwise distance at most 2 in Ξ, we speculate that indeed d (Ξ) Acknowledgments
I would like to thank Professor Andrea Lucchini for introducing me to theproblem studied in this paper, for patiently reading various drafts of thesame paper and for providing many helpful comments aimed to improve it.
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Università degli Studi di Padova, Dipartimento di Matem-atica "Tullio Levi-Civita", Via Trieste 63, 35121 Padova, Italy
E-mail address : [email protected]@gmail.com