aa r X i v : . [ m a t h . G R ] F e b On the Orbital Diameter of Groups of Diagonal Type
Kamilla Rekv´enyiDepartment of Mathematics, Imperial College London, London, SW7 [email protected] 22, 2021
Abstract
The orbital diameter of a primitive permutation group is the maximal diameter of itsorbital graphs. There has been a lot of interest in bounds for the orbital diameter. In thispaper we provide explicit bounds on the diameters of groups of simple diagonal type. As aconsequence we obtain a classification of simple diagonal groups with orbital diameter lessthan or equal to 4. As part of this, we classify all finite simple groups with covering numberat most 3. We also prove some general bounds on the covering number of groups of Lie type.
Let G be a group acting transitively on a finite set Ω. Then G acts on Ω × Ω componentwise.Define the orbitals to be the orbits of G on Ω × Ω . The diagonal orbital is the orbital of theform ∆ = { ( α, α ) | α ∈ Ω } . The number or orbitals is called the rank of G . Let us denote this by rank ( G, Ω). Let Γ be a non-diagonal orbital. Define the corresponding orbital graph to be theundirected graph with vertex set Ω and edge set { α, β } for ( α, β ) ∈ Γ . Note that G acts transitivelyon the edges and the vertices of Γ so it is an edge-transitive and vertex-transitive graph.By [5, Thm 3.2A] the orbital graphs are all connected if and only if the action of G is primitive. The orbital diameter of a primitive permutation group G is the supremum of thediameters of its orbital graphs, see [21]. Let us denote this by orbdiam ( G ).The O’Nan-Scott theorem classifies the primitive permutation groups to be one of thefollowing five types; affine, almost simple, simple diagonal actions, product actions and twistedwreath actions, see [5]. We call an infinite class C of primitive permutation groups bounded ifthere exists t ∈ N such that orbdiam ( G ) ≤ t for all G ∈ C. The paper [21] describes the O’Nan-Scott classes which are bounded. The description is somewhat qualitative and does not containexplicit diameter bounds. Some explicit bounds were obtained in [26] for some almost simplegroups. In this paper we study the orbital diameters of the class of simple diagonal type primitivepermutation groups and provide bounds for these quantities. Further work along these lines bythe author is under way for the other O’Nan-Scott classes.Let us now describe primitive groups of simple diagonal type, following [22]. Let T bea non-abelian simple group, Γ = { , . . . , k } and W = T wr Γ S k with base group T k . Now let D = { ( a, . . . , a ) | a ∈ T } be a diagonal subgroup and let Ω be the set of right cosets of D in T k .Then T k acts on Ω by right multiplication, S k acts on Ω by permuting the components of the cosetrepresentatives, and α ∈ Aut ( T ) acts on Ω by D ( h , . . . , h k ) α = D ( h α , . . . , h αk ) for h i ∈ T . Thegroups T k , S k and Aut ( T ) generate a group N ∼ = T k . ( Out ( T ) × S k ) and this is the normalizer of T k in Sym (Ω) . We say G ≤ Sym (Ω) is a primitive permutation group of simple diagonal type if T k ≤ G ≤ N and G acts primitively on Ω, see [22]. Note G = T k .X where X ≤ Out ( T ) × S k . D ( k, T ) = N ≤ Sym (Ω) . The result in [21, Lemma 5.1] states that the class of simple diagonal groups G = T k .X ≤ D ( k, T ) is bounded only if k and the rank of T are both bounded. However no explicit bounds areobtained.Before we list our results, we need a few definitions. Notice that if T is simple and t ∈ T \ t T generates T. For a subset S of T and r ∈ N we write S r = { s , . . . , s r | s i ∈ S } . Definition 1.1.
Let T be a non-abelian simple group, = t ∈ T and C = t T .
1. Define the c ( T, t ) to be the smallest number r such that C r = T and define the coveringnumber of T to be c ( T ) = max t ∈ T \ c ( T, t ) .
2. For t ∈ T \ put t ± T = t T ∪ ( t − ) T . Let c i ( T, t ) be the smallest number r such that ( t ± T ) r = T. Define the inverse covering number of T to be c i ( T ) = max t ∈ T \ c i ( T, t ) .
3. Let X be such that T E X ≤ Aut ( T ) and let t ± X = { t ± α | α ∈ X } . Let c X ( T, t ) be the smallestnumber such that ( t ± X ) r = T. Define the X-covering number of T to be c X ( T ) = max t ∈ T \ c X ( T, t ) . When X = Aut ( T ) write c A ( T ) = c X ( T ) . Note that c A ( T ) ≤ c X ( T ) ≤ c i ( T ) ≤ c ( T ) . Covering numbers were introduced in [1]. We shall also need the refinements c i ( T ) and c X ( T ) . We now state our results. Let G = T k .X ≤ D ( k, T ) where X ≤ Out ( T ) × S k . We knowfrom [22] that D ( k, T ) = { ( τ , . . . , τ k ) .π | τ i ∈ Aut ( T ) , π ∈ S m and all τ i lie in the same Inn ( T ) − coset } . Let W = { ( α, . . . , α ) .π | α ∈ Aut ( T ) , π ∈ S k } and put D A = W ∩ G, so D A = D.X.
Since G = D A T k the action of G on Ω is equivalent to the action of G on ( G : D A ) . For a ∈ T write ( a k ) = ( a, . . . , a ) ∈ T k . For t ∈ T \ t = { D A , D A (1 k − , t ) } G . The following theorem gives lower and upper bounds on the diameter of Γ t . In the statement weabuse notation and denote c X ( T ) by c X ( T ) , where X is defined as follows. Let G = T k .X with X ≤ Out ( T ) × S k . Let ρ be the projection of X onto Out ( T ) and let π be the canonical map Aut ( T ) → Out ( T ) . Define X = π − ( ρ ( X )) ≤ Aut ( T ) . Theorem (3.1,3.2) . Let G = T k .X as above. The diameter of Γ t satisfies the bounds
12 ( k − c X ( T ) + 1 ≤ diam (Γ t ) ≤ ( k − c i ( T ) . Theorem (5.1) . Let G be a primitive group of simple diagonal type of the form T k .X ≤ D ( k, T ) .1. If orbdiam ( G ) = 2 , then k = 2 and c A ( T ) = 2 .2. If orbdiam ( G ) = 3 , then k = 2 and c A ( T ) ≤ .3. If orbdiam ( G ) = 4 , then one of the following holds:(a) k = 2 and c A ( T ) ≤ (b) k = 3 and c A ( T ) = 2 . The following result gives a partial converse to parts 1, 2 and 3( a ). Lemma (3.3) .
1. If G = T then orbdiam ( G ) = c i ( T ) .
2. If G = D (2 , T ) then orbdiam ( G ) = c A ( T ) . We have not determined whether there are examples of groups in case 3( b ) of Theorem 5.1with orbital diameter 4. In view of Lemma 3.3 and Theorem 5.1, to classify all diagonal type groupswith orbital diameter 2 and 3 we need to classify all non-abelian simple groups with X -coveringnumbers 2, 3 for various X. First we prove a general result on covering numbers of simple groups of Lie type. Theupper bound in the following result was proved in [7]. By Lie rank we mean the rank of thecorresponding simple algebraic group.
Theorem (4.1) . There is a constant d such that r − ≤ c A ( T ) ≤ c ( T ) ≤ dr for all simple groups T of Lie type of Lie rank r. It was proved in [3] that the only finite simple group with covering number 2 is thesporadic group J . It turns out that J is also the only finite simple group with inverse coveringnumber 2 (Proposition 4.3). However, there are infinitely many simple groups T with c A ( T ) = 2 . Theorem (4.4) . Let T be a finite simple group. Then c A ( T ) = 2 if and only if T ∼ = J or T ∼ = P SL ( q ) with q ≡ or q = 2 m . We also classify simple groups with c ( T ) = 3 or c A ( T ) = 3 . Theorem (4.5) . Let T be a finite simple group.1. c ( T ) = 3 if and only if T is isomorphic to one of the following: • P SL ( q ) with q > • P SL ( q ) • P SU ( q ) with | q + 1 , q > • B ( q ) with q > • G ( q ) with q > G (3 n ) with n ≥ • A , A , A • M , M , M , M , J , J , M cl , Ru , Ly , O ′ N , F i ′ , T h , M. c i ( T ) = 3 if and only if c ( T ) = 3 .
3. If c A ( T ) = 3 then one of the following holds:(a) c ( T ) = 3 (b) T ∼ = F (2 n ) . Remark
For part 3(a), the groups with c ( T ) = 3 and c A ( T ) < P SL ( q ) with q ≡ mod
4) or q = 2 m . For part 3(b) we have not been able to determine whether c A ( F (2 n )) = 3 . The result in [21, Lemma 5.1] on simple diagonal action states that for a class of primitivegroups T k .X with bounded orbital diameter the Lie rank of T and k are bounded. Conversely, if T has bounded Lie rank, k is bounded and a few more criteria are met, then one obtains a boundedclass.The proof of this result is model theoretic and includes no explicit bounds on the orbitaldiameter. The following result provides an explicit upper bound, giving rise to many boundedfamilies of primitive groups of simple diagonal type. For simplicity we restrict to the class ofsimple diagonal groups of the form T k .S k ≤ D ( k, T ) . Theorem (6.1) . Let k ≥ and let T be a simple group. Then orbdiam ( T k .S k ) ≤ k − c i ( T ) . In this section we include some background material that we use in the proof of our theorems.We begin with a well-known a character theoretic result which gives us a method to findcovering numbers.
Lemma 2.1. [2, Lemma 10.1] Let C , . . . , C d be conjugacy classes of a finite group G with rep-resentatives c , . . . , c d . For z ∈ G , the number of solutions ( x , . . . , x d ) ∈ C × · · · × C d to theequation x . . . x d = z is Q | C i || G | X χ ∈ Irr ( G ) χ ( c ) . . . χ ( c d ) χ ( z − ) χ (1) d − . Note the immediate corollary of this result.
Corollary 2.2.
Let C and D be conjugacy classes of G with representatives c, d. If (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X χ ∈ Irr ( G ) \ G χ ( c ) k χ ( d − ) χ (1) k − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) < , then D ⊆ C k . We include another result that we use in our classification of groups with small coveringnumbers, namely the classification of strongly real groups. A group is strongly real if and only ifany of its elements can be expressed as a product of two involutions.4 heorem 2.3 ([8, 10, 12, 13, 17, 23, 25, 27, 28]) . Let G be a non-abelian finite simple group. Then G is strongly real if and only if it is isomorphic to one of • P Sp n ( q ) where q mod and n ≥ • P Ω n +1 ( q ) where q ≡ mod and n ≥ • P Ω ( q ) where q ≡ mod • P Ω +4 n ( q ) where q mod and n ≥ • P Ω − n ( q ) where n ≥ • P Ω +8 ( q ) or D ( q ) • A , A , A , A , J , J . The following result is on covering numbers of alternating groups from [4, Thm 2].
Theorem 2.4. [4] Let n ≥ , l odd and l ≤ n . Then every permutation in the alternating group A n is a product of three l − cycles if and only if either n ≤ l or n = 7 and l = 3 . We conclude by listing some existing results on the covering numbers of some finite simplegroups.
Theorem 2.5 ([3, 6, 20, 19]) .
1. If n ≥ , then c ( P SL n ( q )) = n. Also for q > , c ( P SL ( q )) =3 . c ( A n ) = ⌊ n ⌋ for n ≥ , and c ( A ) = 3 .3. c ( B ( q )) = 3 for q > . c ( G ( q )) = 3 for q > . We begin with some notation. Let T be simple, k ≥ G = T k .X ≤ D ( k, T ) in a primitivesimple diagonal action, where X ≤ Out ( T ) × S k as in Section 1. Let Γ be an orbital graph of G. Define d Γ ( a, b ) to be the distance between two vertices, a and b in Γ . Denote a path of length atmost m between a and b in Γ by a m b. Denote the element ( t, ..., t ) ∈ T k as ( t k ) . Let t ∈ T \ t be the orbital graph { D A , D A (1 k − , t } G . Recall X = π − (( ρ ( X ))where π is the canonical map Aut ( T ) → Out ( T ) and ρ is the projection of X to Out ( T ) . For g ∈ T we define the length of g with respect to t to be l Xt ( g ) = min { a : g = t ± α . . . t ± α a , for some α i ∈ X } . Recall we write an element of G as ( h , . . . , h k ) σ h where h i ∈ Aut ( T ) , σ h ∈ S k , and identify T with Inn ( T ) when convenient.The first result in this section is a lower bound for the diameter.5 heorem 3.1. Let c = c X ( T, t ) . The diameter of Γ t satisfies diam (Γ t ) ≥ M where M = ( ( k − c + 1 k odd kc k evenProof. Note that G = D A T k so every right coset of D A has a coset representative in T k . Claim 1
Every coset at distance m away from D A is of the form D A ( t , . . . , t k )where t i ∈ T are such that P ki =1 l Xt ( t i ) ≤ m. Proof of Claim 1
We prove Claim 1 by induction on m. We start with the base case m = 1 . Suppose D A ( g , . . . , g k ) is a neighbour of D A where g i ∈ T . Then there exists h =( h , . . . , h k ) σ h ∈ G such that { D A , D A (1 k − , t ) } h = { D A , D A ( g , . . . , g k ) } . Hence either h ∈ D A or D A (1 k − , t ) h = D A . If h ∈ D A then h = ( a k ) σ h , with a ∈ Aut ( T ). Now D A ( g , . . . , g k ) = D A (1 k − , t ) h = D A h − (1 k − , t ) h = D A (1 k − , t a ) σ h as required. If D A (1 k − , t ) h = D A , then D A ( g , . . . , g k ) = D A h = D A h − (1 k − , t − ) h = D A (1 k − , t − h k ) σ h . Hence Claim 1 holds for m = 1.Now let m ≥ . Let D A h be a coset at distance m from D A . Then D A h is a neighbour ofa coset at distance m − D A ( x , . . . , x k )where x i ∈ T and P ki =1 l Xt ( x i ) ≤ m − . There is an edge between D A ( x , . . . , x k ) and D A h. Hencethere is f ∈ G such that { D A , D A (1 k − , t ± a ) σ } f = { D A ( x . . . , x k ) , D A h } with f = ( f , . . . , f k ) π where f i ∈ Aut ( T ) and π ∈ S k . Again either D A f = D A ( x , . . . , x k ) or D A (1 k − , t ± a ) σ f = D A ( x , . . . , x k ) . If D A f = D A ( x , . . . , x k ) then f i x − i = f j x − j for all i, j, so D A h = D A (1 k − , t ± a ) σ f = D A ( x , . . . , x k ) f − (1 k − , t ± a ) σ f = D A ( x , . . . , x k )(1 k − , t ± af k ) σπ and Claim 1 follows. When D A (1 k − , t ± a ) σ f = D A ( x , . . . , x k ) we obtain the conclusion in asimilar way. Claim 2
There exist h , . . . , h k ∈ T such that d Γ t ( D A , D A ( h , . . . , h k )) ≥ M, where M is as in the statement of the Theorem. Proof of Claim 2
By Claim 1 it suffices to find h , . . . , h k ∈ T such that min g ∈ T P ki =1 l Xt ( gh i ) ≥ M. Let h, h ′ ∈ T with h = h ′ and l Xt ( h ) = l Xt ( h ′ ) = c. Define( h , . . . , h k ) = ( (1 , h, , h, . . . , , h ) k even( h, , h, , . . . , h, , h ′ ) k odd . Then l Xt ( h − h ) = l Xt ( h − h ) = · · · = l Xt ( h − k − h k ) = c and l Xt ( h − h k ) ≥ . l Xt ( xy ) ≤ l Xt ( x ) + l Xt ( y ) = l Xt ( x − ) + l Xt ( y ) for all x, y ∈ T , so it follows that for all i ≥ g ∈ T c = l Xt ( h − i h i +1 )) ≤ l Xt ( gh i ) + l Xt ( gh i +1 )and 1 ≤ l Xt ( h − h k ) ≤ l Xt ( gh ) + l Xt ( gh k ) . Summing these up gives2 k X i =1 l Xt ( gh i ) ≥ k − X i =1 l Xt ( h − i h i +1 ) + l ( h − h k ) ≥ ( k − c + 1 . The result now follows for k odd, and for k even we have l Xt ( h − h k ) = c, so we get k c as a lowerbound. The following result is an upper bound. Lemma 3.2.
We have diam (Γ t ) ≤ ( k − c i ( T ) . Proof.
Claim 1
There exist α i ∈ Aut ( T ) such that D A ((1 i − , t α i ) ± a , k − i ) is adjacent to D A forall a ∈ T and all 1 ≤ i ≤ k. Proof of Claim 1 This is clear for k = 2 so we can assume k ≥ . We have D A D A (1 k − , t )by definition of Γ . Apply ( a k ) ∈ T k to this to get D A D A (1 k − , t a ) . As G is primitive, X acts transitively on the symbols 1 , . . . , k , so for 1 ≤ i ≤ k there is an element( α i , . . . , α i ) .σ i ∈ D A such that D A (1 k − , t )( α i , . . . , α i ) .σ i = D A (1 i − , t α i , k − i ) . Applying ( a k ) gives D A D A (1 i − , ( t α i ) a , k − i ) . Furthermore, applying (1 i − , ( t α i ) − a , k − i ) to this gives D A (1 i − , ( t α i ) − a , k − i ) 1 D A and as a was arbitrary Claim 1 follows. Claim 2
Let h i ∈ T (1 ≤ i ≤ k ) and let a ∈ T. Then D A ( h , . . . , h k ) is adjacent to D A ( h , . . . , h i − , ( t α i ) ± a h i , h i +1 . . . , h k ) for 1 ≤ i ≤ k. Proof of Claim 2 Apply ( h , . . . , h k ) to D A D A (1 i − , ( t α i ) ± a , k − i ) . Claim 3
Let c = c i ( T ) . For any h i , . . . , h k ∈ T and 1 ≤ i ≤ k, A ( h , . . . , h i − , , h i +1 , . . . , h k ) c D A ( h , . . . , h k ) . Proof of Claim 3 We know by definition of c that h i can be expressed as a product of c conjugates of ( t α i ) ± , so h i = ( t α i ) ± a . . . ( t α i ) ± a c for some a i ∈ T. Hence by repeatedly applying Claim 2 D A ( h , . . . , h i − , , h i +1 , . . . , h k ) c D A ( h , . . . , h i − , ( t α i ) ± a . . . ( t α i ) ± a c , h i +1 , . . . , h k )so Claim 3 follows.Using Claim 3 repeatedly we have the following path D A c D A (1 , h , k − ) c D A (1 , h , h , k − ) . . . c D A (1 , h , . . . , h k ) . As (1 , h , . . . , h k ) represents an arbitrary coset, the result follows.We have an exact result for the orbital diameter for the case when G = T or G = D (2 , T ) . Lemma 3.3.
1. If G = T then orbdiam ( G ) = c i ( T ) .
2. If G = D (2 , T ) then orbdiam ( G ) = c A ( T ) . Proof.
1. We first notice that in the case of k = 2 all orbital graphs are of the form Γ t . If G = T then X = T , so c X ( T ) = c i ( T ) . Hence the bounds from Theorem 3.1 and Lemma 3.2 coincide,and the result follows.2. Consider G = D (2 , T ) ∼ = T . ( Out ( T ) × S ). In this case X ∼ = Aut ( T ) and also D A ∼ = Aut ( T ) × S . Now Theorem 3.1 gives orbdiam ( G ) ≥ c A ( T ) . We will show the other directionof this inequality. Let t ∈ T \
1. Consider the orbital graph Γ = { D A , D A (1 , t ) } G . Now D A D A (1 , t ± )are edges in the graph. For all a ∈ Aut ( T ) apply ( a, a ) ∈ G to these to get D A D A (1 , t ± a ) . Now we can construct a path between D A and any arbitrary coset D A (1 , h ) where h = t ± a . . . t ± a c with c = c A ( T ) such that D A c D A (1 , h ) . This shows that orbdiam ( G ) ≤ c A ( T ) and the resultnow follows. In this section we prove bounds on the covering numbers c ( T ), c i ( T ) and c A ( T ) for simple groupsas stated in the Introduction. We start with a result on covering numbers for simple groups of Lietype. In the following result, the upper bound is proved in [7].8 heorem 4.1. There is a constant d such that r − ≤ c A ( T ) ≤ c ( T ) ≤ dr for all simple groups T of Lie type of Lie rank r. More precisely, c A ( T ) ≥ C T where C T is as inTable 1. T C T P SL n ( q ) ( n, q ) = (2 , or (2 , nP SU n ( q ) n ≥ nP Sp n ( q ) n ≥
4, ( n, q ) = (4 , nP Sp (2) ′ P Ω ǫn ( q ) n ≥ ⌊ n ⌋ B ( q ) q > G ( q ) q > G (3 n ) 3 G ( q ) 3 ∤ q D ( q ) 4 F (2 n ) 3 F ( q ) 2 ∤ q F ( q ) q > F (2) ′ E ǫ ( q ) 4 E ( q ) 4 E ( q ) 5Table 1: Lower boundsNote that the inequality c A ( T ) ≤ c i ( T ) ≤ c ( T ) is immediate from the definitions. Henceestablishing a lower bound for c A ( T ) immediately gives a lower bound for c i ( T ) and c ( T ) . Let V = V n ( q ) . For x ∈ P GL ( V ) let e x ∈ GL ( V ) be a preimage of x and define ν ( x ) = n − max λ ∈ F ⋆q dimC V ( λ e x ) , the minimal codimension of an F q -eigenspace of e x. For a subset S ∈ P GL ( V ) define ν ( S ) = max s ∈ S ν ( s ) . Proposition 4.2.
Let T ≤ P GL ( V ) ∼ = P GL n ( q ) be a simple group and let X be a group such that InnT ≤ X ≤ AutT.
Let S be a non-empty X -invariant subset of T \ such that ν ( s ) = ν ( s ′ ) forall s, s ′ ∈ S , and let S = S ∪ S − . Then c X ( T ) ≥ ν ( T ) ν ( S ) . Proof.
Note that S is a union of X -conjugacy classes. Choose s ∈ S and t ∈ T such that ν ( s ) = ν ( S ) and ν ( t ) = ν ( T ) . Put C = s X . By hypothesis, ν ( y ) = ν ( S ) for all y ∈ C. Let k = n − ν ( S )and r = n − ν ( T ) so that k = max λ ∈ F ⋆q dimC V ( λ e s ) and r = max λ ∈ F ⋆q dimC V ( λ e t ) . Let s , . . . , s l ∈ C. Usingelementary linear algebra we see thatmax λ i ∈ F ⋆q dimC V ( λ e s , . . . , λ l e s l ) ≥ lk − ( l − n.
9e know that t can be expressed as the product of c X ( T ) elements of C. Hence r ≥ c X ( T ) k − ( c X ( T ) − n. Rearranging gives c X ( T ) ≥ n − rn − k = ν ( T ) ν ( S ) . Now we prove the theorem.
Proof of Theorem 4.1.
Case 1, Classical Groups
Let T be a classical simple group with natural module V = V n ( q ) . Define S to be the setof long root elements in T and let X = Aut ( T ) . Then S is X -invariant, provided T = P Sp (2 a ) . Suppose first that T is P SL n ( q ) , P Sp n ( q ) or P SU n ( q / ) and T = P Sp (2 a ) . Then thelong root elements of T are transvections, which have fixed space on V of dimenstion n − , so ν ( S ) = 1 . We claim that ν ( T ) = n. This can be seen as follows. Provided T = P SU n ( q / ) with n even, by [14] T has a Singer element y (i.e. an element such that h y i is irreducible on V ) and clearly ν ( y ) = n. And if T = P SU n ( q / )with n = 2 d ≥
4, then SU n ( q / ) has a subgroup SL d ( q ), and a Singer element of this also satisfies ν ( y ) = n. Hence by Proposition 4.2, c A ( T ) ≥ ν ( T ) ν ( S ) = n. Next consider T = P Sp (2 a ) with a > . Let e S be the set of all long or short root elementsof T. Then e S in invariant under Aut ( T ) . The generic character table of T is in the computer packageChevie [11]. This also contains a function, called ClassMult, which calculates the sum in Lemma2.1. Using this it can be checked that e S = T. Hence c A ( T ) ≥ , as required.Finally suppose T = P Ω ǫn ( q ) . For n ≤ T is isomorphic to one of the groups we havealready covered, so assume n ≥ . The long root elements of T have fixed point space of dimension n − V, so ν ( S ) = 2 . If n is even, then T has an element y such that ν ( y ) = n : for ǫ = − take y to be a Singer element of Ω − n ( q ) [14]; and for ǫ = +, take y to be a Singer element of a subgroup SL n ( q ) of Ω + n ( q ) . If n is odd, then T has an element y such that ν ( y ) = n − − n − ( q ) . We conclude that ν ( T ) ≥ ( n n even n − n oddNow the conclusion follows from Proposition 4.2. Case 2, Exceptional Groups
There are only two families of exceptional groups whosecovering number is known; the Suzuki groups, B ( q ) and the small Ree groups G ( q ) both havecovering number 3 by Theorem 2.5. By Theorem 2.3 they are not strongly real, so in fact c A ( B ( q )) = c A ( G ( q )) = c i ( B ( q )) = c i ( G ( q )) = c ( B ( q )) = c ( G ( q )) = 3 . Now consider T = E ( q ), E ( q ) or E ǫ ( q ) . Let V = V n ( q ) be the adjoint module for T ,of dimension 248 ,
133 or 78, respectively, and let S be the set of long root elements of T. Then10 is invariant under
Aut ( T ) . From Tables 9, 8 and 6 of [18], we see that ν ( S ) is as in the table: T E ( q ) E ( q ) E ǫ ( q ) ν ( S ) 58 34 22 .Also T has regular unipotent elements y , and these have fixed point spaces of dimension 8 , , respectively. Hence ν ( T ) ≥ ,
126 or 72, and the bound c A ( T ) ≥ ν ( T ) ν ( S ) gives the conclusion ofthe theorem.Next consider T = F ( q ), q odd. Again let S be the set of long root elements, whichis invariant under Aut ( T ), and consider the action on the 26-dimensional module V = V ( q ) , asgiven in [18, Table 3]. We see that ν ( S ) = 6 while regular unipotent elements show that ν ( T ) ≥ . Hence c A ( T ) ≥ . Next we claim that c A ( T ) ≥ T = D ( q ), F ( q ) ( q > F (2) ′ or G ( q ) ( q = 3 a ).The character tables of these are available in Chevie and GAP. Using Lemma 2.1 we can computethat for a root element, r , c A ( T, r ) ≥ . The last groups remaining to consider are T = F (2 a ) and G (3 a ) . These groups are notstrongly real by Theorem 2.3, so c A ( T ) ≥ There has been some interest around classifying groups with a small covering numbers. In [3, Thm2.1] it is proven that the only finite simple group with covering number 2 is J . It turns out thatthe same is true for the inverse covering number. Proposition 4.3. c i ( T ) = 2 if and only if c ( T ) = 2 . Proof. If c ( T ) = 2 then clearly also c i ( T ) = 2 . For the converse, suppose c i ( T ) = 2 . Then T hasa conjugacy class of involutions C such that T = C . Hence all elements are conjugate to theirinverses, and so c i ( T ) = c ( T ) . However, the next result shows that there is an infinite family of finite simple groups,such that the covering number is not 2 but the automorphism covering number is 2.
Theorem 4.4.
Let T be a finite simple group. Then c A ( T ) = 2 if and only if either T ∼ = J or T ∼ = P SL ( q ) with q ≡ or q = 2 m . Proof.
We begin by finding c A ( P SL ( q )) for all q ≥ . We know by Proposition 2.5 that c ( P SL ( q )) =3 . If q ≡ , then Theorem 2.3 implies that the involution class has covering number greaterthan 2, hence c A ( P SL ( q )) = 3 . If q is a power of 2 , then by [3, Thm 4.2 (a)] the only conjugacyclasses with covering number 3 are those denoted by R j in [3]; these have class representatives( b j ) where b is an element of order q + 1 and 1 ≤ j ≤ q . In fact, [3, Thm 4.2 (a)] gives that R j = G \ C where C are the root elements. For q = 2 m +1 , the class R q +13 is fixed by all outerautomorphisms, so c A ( P SL (2 m +1 ) = 3 . For q = 2 m , there is no class of type R j which is fixedby all outer automorphisms. Using Lemma 2.1 and Chevie[11] we can show that C ⊆ R j R l , where l = 2 j if 2 j ≤ q and l = q + 1 − j if 2 j > q , so c A ( P SL (2 m )) = 2 . For
P SL ( q ) with q ≡ P SL ( q ) has an outer automorphism that interchanges these two classes. UsingLemma 2.1 we can show that every element can be expressed as a product of two root elements,hence c A ( P SL ( q )) = 2 . This proves the right to left implication of the theorem.11or the converse, suppose c A ( T ) = 2 . Then any element of T can be expressed as aproduct of two involutions, so T is strongly real, hence is given by Theorem 2.3. If T is a simplegroup of Lie type, then by Theorems 4.1 and 2.3 we have that T ∼ = P SL ( q ) with q . We proved above that c A ( P SL ( q )) = 3 for q = 2 m +1 , so q ≡ q = 2 m . If T is not ofLie type then by Theorem 2.3, T ∼ = A , A , J or J . All automorphisms of A and A fix theclass of 3-cycles, so their automorphism covering number is not 2. Looking at the character tableof J and using Lemma 2.1 we conclude that c ( J ) = c A ( J ) ≥ . Hence T = J . The next result gives a similar classification of groups with covering number 3.
Theorem 4.5.
Let T be a finite simple group.1. c ( T ) = 3 if and only if T is isomorphic to one of the following: • P SL ( q ) with q > • P SL ( q ) • P SU ( q ) with | q + 1 , q > • B ( q ) with q > • G ( q ) with q > • G (3 n ) with n ≥ • A , A , A • M , M , M , M , J , J , M cl , Ru , Ly , O ′ N , F i ′ , T h , M. c i ( T ) = 3 if and only if c ( T ) = 3 .
3. If c A ( T ) = 3 then one of the following holds:(a) c ( T ) = 3 (b) T ∼ = F (2 n ) . Proof.
Case 1, Alternating Groups
By Theorem 2.5 the only alternating groups with covering number 3 are A , A and A .Also by Theorem 2.4, c A ( A n ) ≥ n ≥ . Finally c i ( A n ) = 3 for n = 5 , c A ( A n ) = 2,2 , Case 2, Groups of Lie type If T is a group of Lie type such that c A ( T ) ≤ T is isomorphic toone of the following: P SL ( q ) , P SL ( q ) , P SU ( q ) , P Ω ( q ) , B ( q ) , G ( q ) , G (3 a ) , F (2 a ) . By Theorem 2.5, c ( T ) = 3 for T = P SL ( q ), P SL ( q ), B ( q ) and G ( q ); by Proposition 4.3 forthese groups c i ( T ) = 3; and by Theorem 4.4, c A ( T ) = 3 apart from P SL ( q ) with q ≡ mod q = 2 m .
12y [24, Cor 1.9] for T = P SU ( q ) ( q > c ( T ) = ( | q + 14 3 ∤ q + 1Hence if 3 | q + 1 then also c i ( T ) = c A ( T ) = 3. For 3 ∤ q + 1 it is shown in [24, Table 2] that for atransvection t ∈ T , c ( T, t ) = 4 and hence c A ( T ) = 4 and c i ( T ) = 4 . Next consider T = P Ω ( q ) , q odd. We claim that c A ( T ) ≥ . To prove this let V be the8-dimensional spin representation of Spin ( q ) . Let S be the set of long root elements of T , whichis invariant under Aut ( T ). Using triality we can show that the long root elements of Spin ( q )act on V as long root elements of Ω +8 ( q ) , hence they fix a 6-dimensional subspace of V pointwise,and so ν ( S ) = 2 . We know that
Spin +6 ( q ) ∼ = SL ( q ) embeds into Spin ( q ) . Put V as the naturalmodule of SL ( q ) . Then by [16, 2.2.8] SL ( q ) acts on V as on V ⊕ V ∗ . Take R to be a Singercycle in SL ( q ) . Now R on V ∗ is also Singer cycle, and hence ν ( R ) = 8 . Now by Proposition 4.2, c A ( T ) ≥ . Consider T = G (3 a ) . In the case of a = 1 , the character table is in GAP, so usingLemma 2.1 we find that c A ( G (3)) = c i ( G (3)) = c ( G (3)) = 4 . For a ≥
2, even though thegeneric character table is available in Chevie, solving this problem is not possible using onlyChevie. This is due to the fact that when running the ClassMult function, Chevie outputs valuesfor the character sum in Lemma 2.1 together with a set of many ”possible exceptions” which giveconditions under which this value might not hold. For some classes it is possible to deal with theseexceptions, and for others it is not, so we used another method of solution. For the cases thatare possible to do with Chevie we used Chevie, for the rest we used Corollary 2.2. To describethis, we use the notation for conjugacy classes and characters of T given in [9]. We partitionthe non-trivial irreducible characters into sets ∆ k , where the degree of the characters in ∆ k is apolynomial in q of degree k. We see from [9] that
Irr ( T ) \ T = ∆ ∪ ∆ ∪ ∆ and we get | ∆ | = 1, | ∆ | = 2 q − | ∆ | = q + 13. For χ ∈ ∆ we have χ (1) ≥ q + q + 1, for χ ∈ ∆ we have χ (1) ≥ q ( q − ( q − q + 1) and for χ ∈ ∆ we have χ (1) ≥ q ( q − q + 1)( q − | χ ( x ) | for all non-identity conjugacy classes x. The first column lists theclass representatives in the notation of [9]. The other three columns give upper bounds for | χ ( x ) | for χ ∈ ∆ , ∆ and ∆ , respectively. We use these bounds to bound the sum in Corollary 2.2 with k = 3 . We find that this sum is less than one for all pairs of conjugacy classes (
C, D ) with thefollowing exceptions;
C A A A A A B A A A B B C ( i ) C ( i ) D ( i ) D ( i ) D D ranges over all conj. classes D is 1 or A or A .For these exceptions we can show D ⊆ C using Chevie and Lemma 2.1, with the exception ofshowing E ( i ) ⊆ B , E ( i ) ⊆ B , 1 ⊆ C ( i ) , 1 ⊆ C ( i ) , ⊆ D ( i ) , and 1 ⊆ D ( i ) . For theseexceptions we obtained more precise bounds for the character values than those in Table 2 and usedCorollary 2.2 again. Hence we proved that c ( T ) = 3 . It follows by Theorem 4.1, c i ( T ) = c A ( T ) = 3as well. Finally we need to consider F (2 n ) . We can use the argument given for F ( q ), q odd inthe proof of Theorem 4.1 to conclude that c i ( T ) ≥ . We have not been able to determine whetherthe automorphism covering number of F (2 n ) is 3. Case 3, Sporadic Groups
Zisser[29] and Karni[15] showed that the only sporadic simple groups with covering num-ber 3 are the ones listed in Theorem 4.5. Using character tables in GAP and Lemma 2.1 wecomputed c i ( T ) and c A ( T ) for all sporadic groups and obtained the same list.13 ∆ ∆ A q + 1 ( q + 1)( q + 1) ( q + 1)( q + q + 1) A q + 1 ( q + 1)( q + 1) ( q + 1)( q + q + 1) A q ( q + 1) 2 q + 1 A q ( q + 1) 2 q + 1 A q A q A q B q q + 3 q + 2 3( q + 1) B q q + 2 3( q + 1) B q q + 2 3( q + 1) B q + 2 3 B q + 2 3 C q q + 2 3( q + 1) C C q q + 2 3( q + 1) C D q q + 2 3( q + 1) D D q q + 2 3( q + 1) D E ( i, j ) 3 4 6 E ( i ) 1 4 4 E ( i ) 1 4 4 E ( i, j ) 3 4 6 E ( i ) 0 4 6 E ( i ) 0 4 6Table 2: Bounds on character values In this section we prove the following result, which classifies the primitive permutation groups ofsimple diagonal type with orbital diameter at most 4 . Adopt the notation of the introduction: T is simple, k ≥ G = T k .X ≤ D ( k, T ) with X ≤ Out ( T ) × S k and G acts primitively onΩ = ( G : D A ). Theorem 5.1.
Let G be a primitive group of simple diagonal type of the form T k .X ≤ D ( k, T ) .1. If orbdiam ( G ) = 2 , then k = 2 and c A ( T ) = 2 .2. If orbdiam ( G ) = 3 , then k = 2 and c A ( T ) ≤ .3. If orbdiam ( G ) = 4 , then one of the following holds:(a) k = 2 and c A ( T ) ≤ (b) k = 3 and c A ( T ) = 2 . Remark
Note that Lemma 3.3 is a partial converse of this result, as it shows that there are somefamilies of groups of simple diagonal type with k = 2 such that orbdiam ( G ) = c A ( T ), namely G = D (2 , T ).For the proof of this theorem need some preliminary lemmas.14ote that it follows from Theorem 3.1 that k is a lower bound for the orbital diameter.In fact, for c A ( T ) = 2 it is a strict lower bound; Lemma 5.2.
Let c A ( T ) = 2 , k ≥ and G = T k .X ≤ D ( k, T ) . Then orbdiam ( G ) ≥ k + 1 . Proof.
Recall our definition of the length of an element of T with respect to t ∈ T \
1: for g ∈ T,l At ( g ) = min { a : g = t ± α . . . t ± α a , α i ∈ Aut ( T ) } . Since c A ( T ) = 2, we have l At ( g ) ≤ l At ( g ) = 1 if and only if g ∈ t ± Aut ( T ) . Also by Proposition4.4, T = J or P SL ( q ) with q ≡ mod
4) or q = 2 m . Let t be an involution in T and define Γ = { D A , D A (1 k − , t ) } G . Claim 1
We can choose x, y, z ∈ T such that l At ( x ) = l At ( y ) = l At ( x − y ) = l At ( t a x ) = 2 ∀ a ∈ T and l At ( z ) = l At ( x − z ) = l At ( y − z ) = 2 . Proof of Claim 1 For T = J and P SL (4) this can be verified in GAP. Now let T = P SL ( q ), with q ≥ q ≡ mod
4) or q = 2 m . Let Z = Z ( SL ( q )) . Define x, y, z ∈ T to be (cid:18) ω ω − (cid:19) Z , (cid:18) ω ω − (cid:19) Z and (cid:18) ω (cid:19) Z, respectively, where ω ∈ F ⋆q has order q −
1. To seethat these elements satisfy Claim 1 consider the product of an involution in
P SL ( q ) with x or y. An involution in
P SL ( q ) is of the form (cid:18) a bc − a (cid:19) Z, and (cid:18) a bc − a (cid:19) (cid:18) ω i ω − i (cid:19) Z = (cid:18) aω i bc − aω − i (cid:19) Z, which is not an involution. The other assertions in Claim 1 are easily verified. Claim 2
There is a coset D A ( m , . . . , m k ) such that d Γ ( D A , D A ( m , . . . , m k )) ≥ k + 1 . Proof of Claim 2
Assume first that k is odd and let and let x, y, z ∈ T be as in Claim 1. Set( m , . . . , m k ) = ( y, , x, , x, , . . . , , x ) . Suppose that d Γ ( D A , D A ( m , . . . , m k )) ≤ k. Then by Claim 1 in the proof of Theorem 3.1, thereexists g ∈ T such that k X i =1 l At ( gm i ) ≤ k. (1)By Claim 1, 2 k = k − X i =1 l At ( m − i m i +1 ) + l At ( m − m k )and recall that l At ( m − i m i +1 ) ≤ l At ( gm i +1 ) + l At ( gm i ) . Putting these together we get2 k = k − X i =1 l At ( m − i m i +1 )+ l At ( m − m k ) ≤ k X i =1 l At ( gm i ) = 2 (cid:18) l At ( gy ) + k − l At ( gx ) + k − l At ( g ) (cid:19) ≤ k, P ki =1 l At ( gm i ) = k. If l At ( gm i ) = 1for all i , then l At ( g ) = l At ( gx ) = l At ( gy ) = 1 which is a contradiction by Claim 1. Hence thereexist j such that l At ( gm j ) = 0 and so g = 1 , x − or y − . If g = x − then k − l At ( g ) = k − l At ( x − y ) = 2, so P ki =1 l At ( gm i ) ≥ k +1 . If g = y − then k − l At ( g ) = k − k − l At ( y − x ) = k − P ki =1 l At ( gm i ) ≥ k − ≥ k + 1 . And if g = 1 then k − l At ( x ) = k − l At ( y ) = 2, so P ki =1 l At ( gm i ) ≥ k + 1 . These contradictions prove Claim 2 for the case where k is odd.Now assume k is even. In this case let ( m , . . . , m k ) = ( y, z, x, , x, , . . . , d Γ ( D A , D A ( m , . . . , m k )) ≤ k . As before, we deduce that there exists g ∈ T such that P ki =1 l Xt ( gm i ) = k and we reach a contradiction in the same way. These contradictions establishClaim 2 and so the orbital diameter of G is bounded below by k + 1 . Now we include another result regarding simple groups with covering number 3.
Lemma 5.3.
Let T be a simple group and let X be a group such that InnT ≤ X ≤ Aut ( T ) and c X ( T ) = 3 . Let g ∈ T be such that c X ( T, g ) = 3 . Then there exist x, y ∈ T such that l Xg ( x ) = l Xg ( y ) = l Xg ( x − y ) = 3 . Proof.
Let t ∈ T be such that l Xg ( t ) = 3 and let D = t ± X . We prove the statement of the lemmaby contradiction.Assume D ⊆ C ∪ C ∪ { } = E . Since c X ( T ) = 3 ,T = D ⊆ ED ⊆ T and so ED = G = { g ± a g ± b t ± c , g ± b t ± c , t ± c | a, b, c ∈ X } . However since l Xg ( t ) = 3 we have 1 ED, a contradiction. Hence D C ∪ C ∪ { } and so thereexist x, y ∈ D satisfying the statement of the lemma. Lemma 5.4.
Let G = T .X ≤ D (3 , T ) with c X ( T ) = 3 . Choose g ∈ T such that c X ( T, g ) = 3 andlet Γ = { D A , D A (1 , , g ) } G . Then diam (Γ ) > . Proof.
Let x, y ∈ T be as in Lemma 5.3. The for all t ∈ T we have l Xg ( tx ) + l Xg ( t ) + l Xg ( ty ) ≥ l Xg ( x ) + l Xg ( y ) + l Xg ( x − y )2 = 92 > . Hence d Γ ( D A , D A ( x, , y )) > . The conclusion follows.
Proof of Theorem 5.1.
Let T be simple and G = T k .X ≤ D ( k, T ) .
1. If orbdiam ( G ) = 2 then by Theorem 3.1 it follows that k = 2 and c A ( T ) = 2 .
2. Suppose orbdiam ( G ) = 3. Then k ≤ k = 2 then Theorem 3.1 shows c A ( T ) ≤ k = 3, then by Theorem 3.1, c A ( T ) = 2, but then Lemma 5.2implies that orbdiam ( G ) >
3, a contradiction.3. Assume that orbdiam ( G ) = 4 . By Theorem 3.1 one of the following occurs:16 k = 2 and c A ( T ) ≤ k = 3 and c A ( T ) = 2III k = 3 and c A ( T ) = 3IV k = 4 and c A ( T ) = 2 . In Cases I and II, conclusions ( a ) and ( b ) of Theorem 5.1 hold.In Cases III and IV, Lemmas 5.2 and 5.4 imply that orbdiam ( G ) ≥
5, which is a contradiction.
In this section we obtain a general upper bound for the orbital diameter of a primitive simplediagonal group. To keep things as simple as possible we restrict attention to groups of the form G = T k .S k ≤ D ( k, T ) . Theorem 6.1.
Let T be a simple group and let G = T k .S k ≤ D ( k, T ) be a primitive simplediagonal group. Then orbdiam ( G ) ≤ k − c i ( T ) . We first need a preliminary lemma.
Lemma 6.2.
Let T be an simple group and u ∈ T an involution. Then there exists x ∈ T suchthat uu x = [ u, x ] has order greater than 2.Proof. Suppose that uu x has order less than or equal to 2 for all x ∈ T . Then u commutes withall of its conjugates, hence it commutes with h u T i = T , as T is simple, and so u ∈ Z ( T ). This is acontradiction. Proof of Theorem 6.1.
For k = 2 conclusion follows from Lemma 3.3 so assume k ≥ . Let Γ bean orbital graph of G. As we already covered the case of Γ t in Lemma 3.2, we may assume that Γ = { D A , D A (1 i , t i +1 , . . . , t k ) } G where i ≤ k − t j ∈ T \ j ≥ i + 1.Suppose there is a path of length at most m from D A to D A ( m , . . . , m k ) where m i ∈ T. Recall our notation for this: D A m D A ( m , . . . , m k ) (2)Applying ( m − , . . . , m − k ) we get D A m D A ( m − , . . . , m − k ) . (3)If we apply ( a k ) to (2) or (3) for any a ∈ T, we get D A m D A ( m ± a , . . . , m ± ak ) . (4)Claim 1 There exists t ∈ T \ D A D A (1 k − , t − , t ) . l, j ≥ i + 1 such that t l = t j . Apply the permutation ( l j )and (1 i , t i +1 , . . . , t k ) to the edge D A D A (1 i , t − i +1 , . . . , t − k ) to get a path D A D A (1 i , t i +1 , . . . , t k ) D A (1 l − , t − j t l , j − l − , t − l t j , k − j ) . Putting u = t − j t l and applying a suitable permutation yields Claim 1 in this case. Now assume t i +1 = · · · = t k = t. Apply (1 k ) and (1 i , t k − i ) to the edge D A D A (1 i , ( t − ) k − i ) to get D A D A (1 i , t k − i ) D A ( t − , k − , t ) . Claim 1 again follows.Claim 2 There exists t ′ ∈ T \ D A c D A (1 k − , t ′ )where c = c i ( T ) . Proof of Claim 2 We know from Claim 1 that for some t ∈ T \ D A D A (1 k − , t − , t ) . Let u = t ± a . . . t ± a c be an involution in T where a i ∈ T. We can construct a path in a similar wayas in the proof of Claim 3 of Lemma 3.2 to get that D A c D A (1 k − , t ± a . . . t ± a c , t ∓ a . . . t ∓ a c ) = D A (1 k − , u, u ′ ) . If there exist b , b ∈ T such that u b u b = 1 and u ′ n u ′ b = 1 then D A c D A (1 k − , u ′ b u ′ b ) = D A . If no such b , b exist then for b ∈ T , uu b = 1 implies u ′ u ′ b = 1 . Hence u ′ is also an invo-lution and C G ( u ) = C G ( u ′ ) . So assume now that this is the case. Applying (1 k − , u ′ , u ) to D A c D A (1 k − , u, u ′ ) , we see that here is a path D A c D A (1 k − , w, w ) , where w = 1 and either w = u = u ′ or w = uu ′ . Note that if k = 3, then the claim follows, soassume k ≥ w is an involution, as u and u ′ commute. By Lemma 6.2 there is x ∈ T such that ww x has order strictly greater than 2. Now D A c D A (1 k − , w x , w x ) . Apply (1 k − , w, w ) to get D A c D A (1 k − , w, w ) 4 c D A (1 k − , w x w, w x w ) . Similarly, D A c D A (1 k − , ww x , ww x ) . h = ww x . Then h − = w x w , so we have D A c D A (1 k − , h − , h − ) . Apply (1 k − , h, h,
1) to get D A (1 k − , h, h,
1) 8 c D A (1 k − , h, , h − ) , so D A c D A (1 k − , h, , h − ) . Apply (1 k − , h − , , h − ) to get D A (1 k − , h − , , h − ) 16 c D A (1 k − , , , h − ) , so D A c D A (1 k − , , , h − ) , where h − = 1. Hence Claim 2 follows.At this point we can apply Lemma 3.2 to deduce that the diameter of Γ is bounded aboveby 24( k − c , completing the proof. Acknowledgements
This paper forms part of work towards a PhD degree under the supervision of Professor MartinLiebeck at Imperial College London, and the author wishes to thank him for his direction andsupport throughout. The author is also very thankful to EPSRC for their financial support.
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