Primitive permutation IBIS groups
aa r X i v : . [ m a t h . G R ] F e b PRIMITIVE PERMUTATION IBIS GROUPS
ANDREA LUCCHINI, MARTA MORIGI, AND MARIAPIA MOSCATIELLO
Abstract.
Let G be a finite permutation group on Ω. An ordered sequence of elements of Ω,( ω , . . . , ω t ), is an irredundant base for G if the pointwise stabilizer G ( ω ,...,ω t ) is trivial and nopoint is fixed by the stabilizer of its predecessors. If all irredundant bases of G have the same sizewe say that G is an IBIS group. In this paper we show that if a primitive permutation group is IBIS,then it must be almost simple, of affine-type, or of diagonal type. Moreover we prove that a diagonal-type primitive permutation groups is IBIS if and only if it is isomorphic to PSL(2 , f ) × PSL(2 , f )for some f ≥ , in its diagonal action of degree 2 f (2 f − . Introduction
Let G ≤ Sym(Ω) be a finite permutation group. A subset B of Ω is a base for G if the pointwisestabilizer G ( B ) is trivial and we denote by b ( G ) the minimal size of a base for G . An ordered sequenceof elements of Ω, Σ := ( ω , . . . , ω t ), is irredundant for G if no point in Σ is fixed by the stabiliserof its predecessors. Moreover Σ is an irredundant base of G , if it is a base and it is irredundant. Inparticular, such an irredundant base provides the following stationary chain G > G ω > G ( ω ,ω ) > · · · > G ( ω ,...,ω t − ) > G ( ω ,...,ω t ) = 1 , where the inclusion of subgroups are strict.Determining base sizes is a fundamental problem in permutation group theory, with a long historystretching back to the nineteenth century. Since the knowledge of how an element g ∈ G acts on abase B completely determine the action of g on Ω, bases plays an important role in computationalgroup theory. The smaller the base, the less memory is needed to store group elements, so thereare practical reasons for trying to find small bases.Irredundant bases for a permutation group possess some of the features of bases in a vector space.Indeed, B is a basis for a vector space V if and only if it is an irredundant base for the generallinear group GL( V ) in its natural action on V. However, some familiar properties of bases in vectorspaces do not extend to bases for permutation groups: irredundant bases for groups in general arenot preserved by re-ordering, and they can have different sizes.In [2], Cameron and Fon-Der-Flaass showed that all irredundant bases for a permutation group G have the same size if and only if all the irredundant bases for G are preserved by re-ordering.Groups satisfying one of the previous equivalent properties are called Irredundant Bases of InvariantSize groups , IBIS groups for short. Moreover, Cameron and Fon-Der-Flaass [2] also proved that fora permutation group G to be IBIS is a necessary and sufficient for the irredundant bases of G tobe the base of a combinatorial structure known with the name of matroid. If this condition hold,then G acts geometrically on the matroid and when G acts primitively and is not cyclic of primeorder, then the matroid is geometric (see [2] for more details). This brought Cameron to ask fora possible classification of the IBIS groups. As explained by Cameron himself in [1, Section 4-14],there is no hope for a complete classification of IBIS group, when the cardinalities of the bases arelarge. But it might be reasonable to pose this question for primitive groups. Mathematics Subject Classification.
Key words and phrases. primitive group; base size; IBIS group .
It is easy to observe that the symmetric group Sym( k ) in its natural action on [ k ] := { , . . . , k } has base size k − k ] with size k − k ) . Similarlythe base size for the action of the alternating group Alt( k ) acting naturally on [ k ] has base size k − k ] with size k − k ) . There are only finitelymany values of k for which Sym( k ) or Alt( k ) acts as a primitive IBIS group, not in its naturalrepresentation [2, Theorem 3.5]. In particular if the action of Sym( k ) and Alt( k ) on m -sets of [ k ]is IBIS, then k ≤
12 and m ≤
5. In this context, it is worth remembering that Sym(6) and Alt(6)act as primitive IBIS groups on 2-sets (with degree 15) and on partitions into 2-sets of size 3 of [6](with degree 10). There exist other almost simple primitive groups that are IBIS, e.g.
PSL(2 , q )and PGL(2 , q ) on their action of degree q + 1, PSL(2 , f ) acting on the cosets of a dihedral subgroupof order 2(2 f + 1), PSL(3 ,
2) in degree 7, and Alt(7) in degree 15. In another direction, it is notdifficult to observe that Frobenius groups are IBIS group with base size 2. Hence, the affine groupAGL(1 , p ) is a primitive IBIS group with base size 2, for every prime p . More in general, AGL( d, p )is a primitive IBIS group with base size d + 1 for every prime p , and for every natural number d .Therefore we have quite a few examples of almost simple and affine-type primitive groups thatare IBIS. So before launching in a complete classification of primitive IBIS groups, we decided tostart the investigation of primitive permutation groups that can be IBIS in the classes of diagonal-type, product-type and twisted wreath product-type. The main result of this paper is the followingtheorem, that we prove in Section 6. Theorem 1.1.
Let G be a primitive permutation IBIS group. Then one of the following holds: The group G is of affine type. The group G is almost simple. The group G is of diagonal type.Moreover, G is an IBIS primitive group of diagonal type if and only it belongs to the infinitefamily of non-monolithic diagonal type groups { PSL(2 , f ) × PSL(2 , f ) | f ≥ } having degree | PSL(2 , f ) | = 2 f (2 f − . It is not difficult to see that there are no primitive permutation IBIS groups having non-abeliansocle and base size 2 (see Lemma 2.3). This could lead to expect that there are not examples ofIBIS groups with non-abelian socle and small base size. However, in Proposition 3.4, we prove thatPSL(2 , f ) × PSL(2 , f ) in its diagonal action of degree | PSL(2 , f ) | is an IBIS group with base size3. In Propositions 3.2, 3.1 and Theorem 3.20, as already mentioned in Theorem 1.1, we show thatthere are no other diagonal-type primitive permutation groups that are IBIS. Investigating for this,we proved some results regarding non-abelian simple groups (see Subsection 2.1) and about thebase size of primitive diagonal-type permutation groups (Lemma 3.7 and Lemma 3.14). Finally, inSections 4, 5, we prove that there are no IBIS primitive groups that are of product-type nor twistedwreath product groups. 2. Preliminaries
The modern key for analysing a finite primitive permutation group G is to study the socle N of G . The socle of an arbitrary finite group is isomorphic to the non-trivial direct product ofsimple groups; moreover, for finite primitive groups these simple groups are pairwise isomorphic.The O’Nan-Scott theorem describes in details the embedding of N in G and collects some usefulinformation about the action of G . In [10], Liebeck, Praeger and Saxl gave a self-contained proof,precisely five types of primitive groups are defined (depending on the group and action structure ofthe socle), namely the Affine-type , the
Almost Simple , the
Diagonal-type , the
Product-type , and the
Twisted Wreath product , and it is shown that every primitive group belongs to exactly one of thesetypes. We refer the reader to [10] for the precise definition of the classes.
Now, we state and prove same preliminary results regarding bases.
Remark 2.1.
Since a base of G with minimal size b ( G ) is clearly irredundant, then a permutationgroup G is IBIS if and only if all its irredundant bases have size b ( G ).The proof of the following Lemma is straightforward. Lemma 2.2.
Let G be a transitive permutation group on a set Ω . Then B ⊂ Ω is a an irredundantbase if and only if B g ⊂ Ω is an irredundant base for every g ∈ G . Lemma 2.3.
Let G ≤ Sym(Ω) be a primitive group with non-abelian socle and let M be a normalsubgroup of G containing the socle of G. If M is not regular, then M has an irredundant base ofsize at least 3. In particular there are no IBIS primitive permutation groups G with non-abeliansocle and b ( G ) = 2 . Proof.
Let ω ∈ Ω . The point stabilizer G ω normalizes M ω = M ∩ G ω , hence G ω ≤ N G ( M ω ) . Since M is not regular, M ω is not normal in G , so G ω ≤ N G ( M ω ) < G. On the other hand G ω is a maximalsubgroup of G , so G ω = N G ( M ω ) and consequently N M ( M ω ) = M ω . Now assume by contradictionthat M ω ∩ M xω = 1 , for every x ∈ M \ M ω . Then M would be a Frobenius group with kernel K andcomplement M ω (see [11, 8.5.5]). In particular K = soc( M ) would be a nilpotent normal subgroupof G (see [11, 10.5.6]), in contradiction with the fact that soc( G ) is a direct product of isomorphicnon-abelian simple groups. (cid:3) Results on finite non-abelian simple groups.
In this subsection, the finite non-abeliansimple CT-group are characterized.
Definition 2.4.
A finite group G is said to be a CT-group group when commutativity is a transitiverelation on G \ { } . Lemma 2.5.
A group G is a CT-group if and only if, for every x, y ∈ G \{ } , either C G ( x ) = C G ( y ) or C G ( x ) ∩ C G ( y ) = 1 . Proof.
First, let assume that G is a CT-group. Moreover assume that t is a non-identity elementof C G ( x ) ∩ C G ( y ) . Since t commutes with both x and y , then every element commuting with x commutes with t and consequently with y . Similarly every element commuting with y commuteswith t and consequently with x . That is C G ( x ) = C G ( y ) . Conversely assume that for every x, y ∈ G \ { } , either C G ( x ) = C G ( y ) or C G ( x ) ∩ C G ( y ) = 1 . Let a, b, c ∈ G \{ } and assume that a commutes with b and b commutes with c . Then a ∈ C G ( a ) ∩ C G ( b )and b ∈ C G ( b ) ∩ C G ( c ) , so C G ( a ) = C G ( b ) = C G ( c ). (cid:3) Lemma 2.6.
The group T = PSL (2 f ) , with f ≥ is a CT-group.Proof. Let q = 2 f . The subgroups of PSL ( q ) are known by a theorem of Dickson. A complete listof all subgroups of PSL (2 f ) can be found in [8]. We have the following possibilities:(1) Elementary-abelian 2-groups.(2) Cyclic groups of order z , where z divides q ± z , where z divides q ± f ≡ q − ≡ C m ⋊ C t of elementary-abelian groups of order 2 m with cyclic groupsof order t , where t divides 2 m − (2 m ) if m divides f and PGL (2 m ) if 2 m divides f .Take x ∈ T \ { } , then x is a non-trivial central element of C T ( x ) . Hence, the list of all subgroups ofPSL (2 f ) yields that C T ( x ) is an abelian subgroup of PSL (2 f ). Now let a, b, and t be non-trivialelements of T such that a commutes with t and t commutes with b. Since a and b are elements of C T ( t ) that is an abelian group, then a and b commute. (cid:3) ANDREA LUCCHINI, MARTA MORIGI, AND M. MOSCATIELLO
In 1957, Suzuki [12] showed that a finite non-abelian simple CT-groups is isomorphic to somePSL(2 , f ) with f ≥
2. Hence the following result holds true.
Theorem 2.7.
A finite non-abelian simple group is a CT-group if and only if it is isomorphic to
PSL(2 , f ) for some f ≥ . Diagonal type
Let k ≥ T be a non-abelian simple group. We set [ k ] := { , . . . , k } and W ( k, T ) := { ( x , ..., x k ) π ∈ Aut( T ) ≀ Sym( k ) | x ≡ x i mod T for all i ∈ [ k ] } ,D ( k, T ) := { ( x, . . . , x ) π ∈ Aut( T ) ≀ Sym( k ) } , Ω( k, T ) := D ( k, T ) /W ( k, T ) , the set of the right cosets of D ( k, T ) in W ( k, T ) A ( k, T ) := W ( k, T ) ∩ Aut( T ) k . Observe that W ( k, T ) = A ( k, T ) ⋊ Sym( k ) = T k . (Out( T ) × Sym( k )). The group G is of diagonaltype if there exists an integer k ≥ T such that T k ≤ G ≤ W ( k, T )and G acts primitively on Ω( k, T ). In the whole section we will assume that G is a primitivepermutation group of diagonal type. Note that G has socle T k and degree | T | k − . Recall that G isprimitive on Ω( k, T ) when P G := { π ∈ Sym( k ) | ( g , . . . , g k ) π ∈ G for some ( g , . . . , g k ) ∈ A ( k, T ) } ≤ Sym( k )is primitive on [ k ] or k = 2 and P G = 1 . If k = 2 and P G = 1 then G has two minimal normalsubgroups, hence the group is non-monolithic. In all other cases, the primitive diagonal-type groupsare monolithic.Throughout the section, we shall use without further comments the following notation. Given t, s , . . . , s k ∈ T , we set H := { x ∈ Aut( T ) | ( x, . . . , x ) ∈ G } , C ( t ) := C H ( t ) , and [1 , s , . . . , s k ] := D ( k, T )(1 , s , . . . , s k ) . From Remark 2.1, it is clear that the study of primitive permutation IBIS groups is intimatelyrelated to the base sizes. For primitive diagonal groups, the minimal base size has been studied in[4]. In the following proposition, we summarize the results proved in [4] which we shall use in thissection.
Proposition 3.1. [4]
The following hold true. (1)
If the top group P G does not contain Alt( k ) , then b ( G ) = 2 . (2) If k = 2 , then b ( G ) = 3 when P G = 1 , and b ( G ) ∈ { , } otherwise. (3) If P G = Alt( k ) , then b ( G ) = 3 when k = 2 , and b ( G ) = 2 when k is or . (4) If P G = Sym( k ) , then b ( G ) ∈ { , } when k = 2 , and b ( G ) ∈ { , } when k is or . Diagonal non-monolithic primitive groups.
In this subsection, k = 2 and P G = 1 . Here we are going to prove that G is IBIS if and only if G ∼ = (PSL(2 , f )) , for some f ≥ , , [1 , t ] , [1 , t ] , . . . , [1 , t r ]) is an irredundantsequence for G if and only if the following hold true: • t , . . . , t r are non trivial elements of T ; • for 1 ≤ i ≤ r −
1, the subgroup ∩ ≤ j ≤ i C ( t j ) properly contains ∩ ≤ j ≤ i +1 C ( t j ) • ∩ ≤ j ≤ r C ( t j ) = 1 . Proposition 3.2. If k = 2 , P G = 1 and G is an IBIS group, then T = (PSL(2 , f )) , for some f ≥ . Proof.
If soc( G ) is not isomorphic to (PSL(2 , f )) , then by Theorem 2.7 we can choose two nontrivial elements of T , s and t , such that C ( t ) ∩ C ( s ) = 1 and C ( t ) = C ( s ) . Then ([1 , , [1 , s ] , [1 , t ])is an irredundant sequence that is not a base. In particular we can complete ([1 , , [1 , s ] , [1 , t ]) toan irredundant base of G , thus there exists an irredundant base of G with size at least 4. Since b ( G ) = 3 by Proposition 3.1, it follows from Remark 2.1 that G is not IBIS. Therefore we concludethat soc( G ) = T = (PSL(2 , f )) . (cid:3) Proposition 3.3. If k = 2 , P G = 1 , T = PSL(2 , f ) and G is IBIS, then G = soc( G ) = T . Proof.
Let ϕ ∈ Aut( T ) be the automorphism induced by the Frobenius automorphism of the field F f , so that Aut( T ) = h ϕ i T . If G = T , then there exists ϕ i ∈ Aut( T ) such that ( ϕ i , ϕ i ) ∈ G, thatis ϕ i ∈ H. Consider the following elements of T : w := (cid:18) (cid:19) and z := (cid:18) (cid:19) . Note that w ∈ C ( w ) \ C ( z ) . Moreover, it is clear that w ϕ i = w and z ϕ i = z, that is ϕ i ∈ C ( w ) ∩ C ( z ).Hence ([1 , , [1 , w ] , [1 , z ]) is an irredundant sequence that is not a base. Hence there exists anirredundant base of G with size at least 4. Since b ( G ) = 3 by Proposition 3.1, it follows fromRemark 2.1 that G is not IBIS. Therefore we conclude that G = T = (PSL(2 , f )) . (cid:3) Proposition 3.4.
Let G = T = (PSL (2 f )) , where f ≥ . Then G is an IBIS group.Proof. Since b ( G ) = 3 by Proposition 3.1, it suffices to show that each irredundant base has size 3.Denoting by B an irredundant base, by Lemma 2.2 we can assume that [1 ,
1] is the first elementof B . Since |B| ≥ , there exist two different non-trivial elements s and t of T , such that [1 , t ] and[1 , s ] are in B . Moreover, since B is irredundant, then C T ( t ) = C T ( s ). Hence from Theorem 2.7, wededuce that C T ( t ) ∩ C T ( s ) = 1 and so B has size 3. (cid:3) Diagonal monolithic primitive groups.
Here we are going to show that the monolithicprimitive groups of diagonal type are not IBIS group. Recall that if a group G of diagonal typewith T k ≤ G ≤ W ( k, T ) is monolithic, then P G is primitive on its action on k points. Throughoutthe subsection, we shall assume that G is monolithic without further mentioning.First we assume k = 2 . In this case T ≤ G ≤ W (2 , T ) and P G = h σ i , where σ := (1 , . Notethat there exists y ∈ Aut( T ) such that ( y, y ) σ ∈ G. Recall that H = { z ∈ Aut( T ) | ( z, z ) ∈ G } andconsider the coset Hy.
Since (( y, y ) σ ) = ( y , y ), we have that y ∈ H. For any x ∈ Aut( T ), weget that ( x, x ) σ ∈ G if and only if Hx = Hy.
Further, notice that Hy = H if and only if σ ∈ G. For any t ∈ T , define I ( t ) := { x ∈ yH | x − tx = t − } . Lemma 3.5.
Let t , . . . , t r ∈ T and B = ([1 , , [1 , t ] , . . . , [1 , t r ]) . Then B is a base for G if andonly if ∩ ≤ i ≤ r C ( t i ) = 1 and ∩ ≤ i ≤ r I ( t i ) = ∅ . Moreover, B is an irredundant sequence if and only if t , . . . , t r are non-trivial elements of T and,for ≤ j ≤ r − , either the set ∩ ≤ i ≤ j C ( t i ) properly contains ∩ ≤ i ≤ j +1 C ( t i ) or the set ∩ ≤ i ≤ j I ( t i ) properly contains ∩ ≤ i ≤ j +1 I ( t i ) .Proof. We have G [1 , = { ( h, h ) , ( x, x ) σ | h ∈ H, x ∈ yH } . To conclude it is suffices to notice that( h, h ) ∈ G [1 ,t i ] if and only if h ∈ C ( t i ) , while ( x, x ) σ ∈ G [1 ,t i ] if and only if x ∈ I ( t i ) . (cid:3) The following result will play an important role in investigating the case k = 2. ANDREA LUCCHINI, MARTA MORIGI, AND M. MOSCATIELLO
Theorem 3.6. [9, Theorem 1.1]
Let T be a non-abelian finite simple group, that is not Alt(7) , PSL(2 , q ) , PSL(3 , q ) nor PSU(3 , q ) . Then there exist x, t ∈ T such that the following hold: (i) T = h x, t i ;(ii) t is an involution; (iii) there is no involution α ∈ Aut( T ) such that x α = x − , t α = t. Lemma 3.7.
Assume that T ≤ G ≤ W (2 , T ) , with T as in Theorem 3.6. Then b ( G ) = 3 . Proof.
Let T = h x, t i , with x, t as in Theorem 3.6 and let B := ([1 , , [1 , x ] , [1 , t ]). Note that C Aut( T ) ( x ) ∩ C Aut( T ) ( t ) = 1 and I ( x ) ∩ I ( t ) = ∅ , hence B is a base for G , that is b ( G ) ≤ . ByProposition 3.1, b ( G ) ≥ , so the result follows. (cid:3) Proposition 3.8.
Assume that T ≤ G ≤ W (2 , T ) , with T = PSL(2 , f ) for every f ≥ . If b ( G ) ≤ , then G is not an IBIS group.Proof. By Theorem 2.7, there exist two nonidentity elements s and t in T such that C T ( t ) ∩ C T ( s ) = 1and C T ( t ) = C T ( s ) . Then ([1 , , [1 , s ] , [1 , t ]) is an irredundant sequence that is not a base. Inparticular we can complete ([1 , , [1 , s ] , [1 , t ]) to an irredundant base of G , thus there exists anirredundant base of G with size at least 4. Since we are assuming b ( G ) ≤ , the result follows fromRemark 2.1. (cid:3) Combining Lemma 3.7 and Proposition 3.8 we get the following:
Corollary 3.9.
Let T ≤ G ≤ W (2 , T ) , with T as in Theorem 3.6. Then G is not an IBIS group. In what follows, we are going to show that also when T ∈ { Alt(7) , PSL(2 , q ) , PSL(3 , q ) , PSU(3 , q ) } ,the group T ≤ G ≤ W (2 , T ) is not IBIS. Proposition 3.10. If (Alt(7)) ≤ G ≤ W (2 , Alt(7)) , then G is not an IBIS group.Proof. If (1 , ∈ yH then ((1 , , (1 , σ ∈ G. In this case, consider s = (1 ,
2) (3 , , s =(1 ,
2) (3 ,
5) and s = (1 ,
2) (3 , . Since (1 , ∈ I ( s ) ∩ I ( s ) ∩ I ( s ) and C Alt(7) ( h s , s , s i ) In this case, consider t = (1 , , ,t = (1 , , , and t = (1 , , . Since (1 , , ∈ I ( t ) ∩ I ( t ) ∩ I ( t ) and C Alt(7) ( h t , t , t i ) Let T = Alt(5) and let T ≤ G ≤ W ( T, . Then G is not an IBIS group.Proof. We prove the result by considering different cases and, in each of them, we set a := (1 , , . First, let yH = H. Under this assumption we have that H = T = Alt(5) and so, without loss ofgenerality, we can assume that y = (1 , . Let a := (1 , , , , 5) and let B := ([1 , , [1 , a ] , [1 , a ]) . Note that a ∈ C ( a ) \ C ( a ) , C ( a ) ∩ C ( a ) = 1 , and I ( a ) ∩ I ( a ) = ∅ . Hence B is an irredundantbase of size 3. Now, let b := (1 , , a ∈ C ( a ) \ C ( b ) and (1 , ∈ I ( a ) ∩ I ( b ). Hence([1 , , [1 , a ] , [1 , b ]) is an irredundant sequence that is not a base. Completing this to an irredundantbases of G , we produce an irredundant base of size at least 4. By Remark 2.1, G is not IBIS in thiscase. Let yH = H = T , let c := (1 , , S := ([1 , , [1 , a ] , [1 , c ]). Since C ( a ) ∩ C ( c ) = 1and I ( a ) ∩ I ( c ) = ∅ , then S is an irredundant base of size 3. Let d := (1 , , . Then a ∈ C ( a ) \ C ( d ) and d ∈ I ( a ) ∩ I ( d ). Consequently ([1 , , [1 , a ] , [1 , d ]) is an irredudant sequencethat is not a base and we can conclude as in the previous paragraph.Finally, we need to consider the case yH = H = Sym(5). Here, let t := (3 , , s = (1 , 2) (3 , , and s := (1 , 2) (4 , . Since t ∈ C ( t ) \ C ( s ) , s ∈ ( I ( t ) ∩ I ( s )) \ I ( s ) , and (1 , ∈ C ( t ) ∩ C ( s ) ∩ C ( s ), then ([1 , , [1 , t ] , [1 , s ] , [1 , s ]) is an irredundant sequence of G that is not a base. Thuscompleting the irredundant sequences to a base of G , we find an irredundant base of G with size atleast 5. By Proposition 3.1 b ( G ) ≤ , hence the result follows from Remark 2.1. (cid:3) Proposition 3.12. Let T = PSL(2 , q ) . If q > is odd and T ≤ G ≤ W (2 , T ) , then G is not anIBIS group.Proof. The conjugacy classes of subgroups of T = P SL (2 , q ) are well-known (see for instance [8]).Among them there are a conjugacy class of dihedral groups of order q − C ) anda conjugacy class of dihedral groups of order q + 1 (in Aschbacher class C ). We can choose ǫ ∈ {± } such that a = q + ǫ is odd. Let b = q − ǫ . In T there is just one conjugacy class of involutions and thecentralizer of an involution is a dihedral group D b of order 2 b . Note that | T | = q ( q − 1) = 2 qab. Thus the total number of involutions is qa . There are qb subgroups isomorphic to a dihedral group D a of order 2 a and each of them contains a involutions. Hence every involution is contained in b different subgroups of type D a . In particular, there exist two different subgroups M and M , both isomorphic to D a , such that M ∩ M contains an involution x .First, assume q > . In this case the dihedral groups of order 2 a are maximal in G . Let y i ∈ M i be an element of order a such that M i = h y i , x i and y xi = y − i , for i = 1 , . Since there is no maximalsubgroup of T containing both y and y , we have that T = h y , y i . The previous argument doesnot work if q ∈ { , , } but it can be easily checked that, also in these cases, T contains twoelements y , y of order a and an involution x such that y x = y − , y x = y − and h y , y i = T. Note that if z ∈ Aut( T ) is such that y zi = y − i , for i = 1 , , then zx − ∈ C Aut( T ) ( h y , y i ) = 1 , hence z = x. Assume that x yH. Then we have that I ( y ) ∩ I ( y ) = ∅ and C ( y ) ∩ C ( y ) = 1, thus B = ([1 , , [1 , y ] , [1 , y ]) is an irredundant base for G of size 3. Hence, by Proposition 3.8, G is notIBIS.Now, assume that x ∈ yH, so that y ∈ Hx = H . Note that there exists a subgroup of type D b containing x and r such that r has order b and r x = r − . From q = 5, it follows b > r = r − . Moreover we have that T = h x, y , r i , x ∈ C ( x ) \ C ( h x, r i ), r b ∈ C ( h x, r i ) \ C ( h x, r, y i ) , and x ∈ I ( x ) ∩ I ( r ) ∩ I ( y ) . Hence ([1 , , [1 , x ] , [1 , r ] , [1 , y ]) is an irredundant sequence that is not a base,so G has an irredundant base of size at least 5. As usual, the result follows from Proposition 3.1and Remark 2.1. (cid:3) Lemma 3.13. Let T = PSL(2 , f ) with f ≥ . Then there exists two maximal subgroups K , K of T such that K ∼ = K ∼ = D q − and | K ∩ K | = 1 . Proof. Let q = 2 f . In T there is just one conjugacy class of involutions and the centralizer of aninvolution is a Sylow 2-subgroup of order q . Note that | T | = q ( q − . Thus the total number ofinvolutions is q − 1. There are q ( q + 1) / D q − (all conjugatedin T ). Denote by Ω the set of these sugroups. Any element of Ω contains q − q/ . Now, fix D ∈ Ω . If K ∈ Ω and K = D, then | D ∩ K | ≤ . Let Ω D = { K ∈ Ω | | D ∩ K | = 2 } . For any involution z ∈ D , there are exactly q/ − K ∈ Ω D with D ∩ K = h z i , hence | Ω D | = ( q − q/ − < | Ω |− q ( q +1) / − K = D with K ∈ Ω \ Ω D . (cid:3) Lemma 3.14. Assume that T ≤ G ≤ W (2 , T ) , with T = PSL(2 , f ) and f ≥ . If either yH = H or yH = H = T , then b ( G ) = 3 . ANDREA LUCCHINI, MARTA MORIGI, AND M. MOSCATIELLO Proof. First, we assume that yH = H . Since q = 2 f , we have that PSL(2 , q ) is isomorphic toSL(2 , q ), so we may deal with T = SL(2 , q ) . Let a be an element of multiplicative order q − F q . Consider the following elements of T : x := (cid:18) a a − (cid:19) , w := (cid:18) (cid:19) , and z := wx = (cid:18) a − a (cid:19) . Then N := h w, z i is a dihedral group of order 2( q − C T ( x ) = h x i , it follows that C T ( N ) = 1.We are going to show that C Aut( S ) ( N ) = 1 . Denote by ϕ ∈ Aut( T ) the automorphism inducedby the Frobenius automorphism of the field F q . Let s ∈ T , let 1 ≤ i ≤ f, and let assume that ϕ i s centralizes N . In particular, ϕ i s centralizes x , and so x ϕ i = x s − and x are conjugated. Therefore { a, a − } = { a i , a − i } .If a = a − i , then a i +1 = 1 and so q − f − i + 1, contradicting the assumption f ≥ 3. Hence a = a i , that is ϕ i is the identity. Consequently s ∈ C T ( N ) = 1 . This proves that C Aut( S ) ( N ) = 1 and in particular that C ( z ) ∩ C ( w ) = 1.If I ( w ) ∩ I ( z ) = ∅ , then there exist i ∈ N and s ∈ T such that ϕ i s ∈ I ( w ) ∩ I ( z ). Since w and z are involutions, then ϕ i s ∈ C Aut( S ) ( N ) = 1. Consequently 1 ∈ yH , that contradicts yH = H .Hence I ( w ) ∩ I ( z ) = ∅ and then ([1 , , [1 , z ] , [1 , w ]) is a base for G . Combining this and Propo-sition 3.1, we deduce that b ( G ) = 3.Now, assume that yH = H = T. By Lemma 3.13, G contains two subgroups H and H isomor-phic to a dihedral group of order 2( q − , with H ∩ H = 1 . Take x ∈ H , x ∈ H , both of order q − . There is no maximal subgroup containing both x and x , so T = h x , x i . In particular, C ( x ) ∩ C ( x ) = 1. Moreover, if z is an involution inverting x , then z ∈ H . Hence, there are noinvolutions in T inverting both x and x . Therefore ([1 , , [1 , x ] , [1 , x ]) is an irredundant base ofsize 3. Combining this and Proposition 3.1, we deduce that b ( G ) = 3. (cid:3) Proposition 3.15. Let T = PSL(2 , f ) , with f ≥ . If T ≤ G ≤ W (2 , T ) then G is not an IBISgroup.Proof. As above, we work in T = SL(2 , f ). First, assume that yH = H . Without loss of generality,denoting again by ϕ the automorphism induced by the Frobenius automorphism of the field F f , wecan say that y = ϕ t , for some t ∈ N . Since y ∈ H , then f = 2 r is even. Moreover, we can assumethat t divides r. Since 2 t + 1 divides 2 f − , there exists a nonzero element c ∈ F f of multiplicativeorder 2 t + 1. Consider the following elements x := (cid:18) c c − (cid:19) and w := (cid:18) (cid:19) . Note that x ∈ C ( x ) \ C ( h x, w i ). Further, x y = x ϕ t = x − and w y = w ϕ t = w = w − , so that y ∈ I ( x ) ∩ I ( w ) = ∅ . Hence ([1 , , [1 , x ] , [1 , w ]) is an irredundant sequence that is not a base. FromLemma 3.14, we have that b ( G ) = 3, hence the result follows from Remark 2.1.Therefore we can assume that yH = H , so that σ ∈ G .Let assume that there exists 1 = ϕ i ∈ Aut( T ) such that ( ϕ i , ϕ i ) ∈ G, that is H > T. Considerthe following elements of T : x := (cid:18) (cid:19) , w := (cid:18) (cid:19) and z := (cid:18) (cid:19) . Note that N := h x, w i is a dihedral group of order 2 | x | and w inverts x . In particular x ∈ C ( x ) \ C ( w ) ,w ∈ I ( x ) ∩ I ( w ) , and w / ∈ I ( z ). Moreover, it is clear that x ϕ i = x , w ϕ i = w, and z ϕ i = z , that is ϕ i ∈ C ( x ) ∩ C ( w ) ∩ C ( z ). Hence ([1 , , [1 , x ] , [1 , w ] , [1 , z ]) is an irredundant sequence that is not abase. This produces a base of G with size at least 5. Since b ( G ) ≤ Finally, assume Hy = H = T. Notice that T contains two involutions z , z generating a dihedralgroup of order 2(2 f + 1) . Since C ( z ) ∩ C ( z ) = I ( z ) ∩ I ( z ) = 1 , for every x ∈ T with | x | > , ([1 , , [1 , z ] , [1 , z ] , [1 , x ]) is an irredundant base of size 4. Now, the result follows combining 3.14and Remark 2.1. (cid:3) Proposition 3.16. If T = PSL(3 , q ) and T ≤ G ≤ W (2 , T ) , then G is not an IBIS group.Proof. Let q = p f , let N := x y z (cid:12)(cid:12) x, y, z ∈ F p ⊆ SL(3 , q ) , and consider the following elements of N : a := , b := , c := . Since N intersects the center of SL(3 , q ) trivially, we may identify N with a subgroup of T. Note that N is non-abelian of order p , Z ( N ) = h c i , and N is isomorphic to a dihedral group of order 8 when p = 2, and it has exponent p when p is odd. In particular a ∈ C T ( c ) \ ( C T ( c ) ∩ C T ( b )), b ∈ ( C T ( c ) ∩ C T ( b )) \ ( C T ( c ) ∩ C T ( b ) ∩ C T ( a )) and c ∈ ( C T ( c ) ∩ C T ( b ) ∩ C T ( a )). Thus ([1 , , [1 , c ] , [1 , b ] , [1 , a ]) isan irredundant sequence that is not a base. By Proposition 3.1 we have that b ( G ) ≤ 4, hence byRemark 2.1 G is not IBIS. (cid:3) Proposition 3.17. Assume that S ≤ G ≤ W (2 , S ) , with T = PSU(3 , q ) . Then G is not an IBISgroup.Proof. Let q = p f , with p prime, and let α be the automorphism of order 2 of the field F q with q elements defined by α ( x ) = x q . All α -Hermitian forms in a vector space of dimension 3 over F q are equivalent, so we may assume that GU(3 , q ) is the isometry group of the form B such that B (( x , x , x ) , ( y , y , y )) = x y q + x y q + x y q associated to the matrix . Let M := − α q β α (cid:12)(cid:12) α, β ∈ F q , β + β q + αα q = 0 . It is not difficult to prove that M is a subgroup of the special unitary group SU(3 , q ), and since M intersects the center of SU(3 , q ) trivially, it can be identified with a subgroup of T = PSU(3 , q ) . Weshall use without further comment the fact that the trace map from F q to F q given by tr( β ) = β + β q is surjective. Taking a non-zero element γ ∈ F q such that γ + γ q = 0 , then c := γ is a non-trivial central element of M. Denote by ω a primitive element of F q , then ω q +1 ∈ F q andconsequently there exists δ ∈ F q such that δ + δ q = − ωω q . Hence we can consider the followingelements of M : a := − and b := − ω q δ ω . Since ω q = ω, then a and b do not commute. Hence, we have that a ∈ C ( c ) \ ( C ( c ) ∩ C ( b )), b ∈ ( C ( c ) ∩ C ( b )) \ ( C ( c ) ∩ C ( b ) ∩ C ( a )), and c ∈ ( C ( c ) ∩ C ( b ) ∩ C ( a )). So ([1 , , [1 , c ] , [1 , b ] , [1 , a ]) isan irredundant sequence of size 4 that is not a base. Now, the result follows combining Propoition 3.1and Remark 2.1. (cid:3) So far, we proved that when G is monolithic primitive, then T ≤ G ≤ W (2 , T ) is a non IBISgroup. Hence we need to consider the case k ≥ Proposition 3.18. If T ≤ G ≤ W (3 , T ) then G is not IBIS.Proof. Assume by contradiction that G is an IBIS group. By Proposition 3.1 and Lemma 2.3 wecan assume P G = Sym(3). In particular σ := (1 , ∈ P G , so there exist y ∈ Aut( T ), s , s ∈ T suchthat ( y, s y, s y ) σ ∈ G. That is ( y, y, y ) σ ∈ G for some y ∈ Aut( T ) . It follows from the classificationof the finite simple groups that there exists a non-trivial element t in C T ( y ) (see for example ([6,1.48]). So we have that ( y, y, y ) σ ∈ G ([1 , , , [1 , ,t ]) , while ( y, y, y ) σ / ∈ G [1 ,t,t ] . If s / ∈ C T ( t ) , then( s, s, s ) ∈ G [1 , , \ G [1 , ,t ] . Moreover ( t, t, t ) ∈ G ([1 , , , [1 , ,t ] , [1 ,t,t ]) . Hence ([1 , , , [1 , , t ] , [1 , t, t ]) isan irredundant sequence that is not a base. In particular G has an irredundant base of size at least4. Now, the result follows combining 3.1 and Remark 2.1. (cid:3) Proposition 3.19. Let k ≥ and let T k ≤ G ≤ W ( k, T ) . Then G is not an IBIS group.Proof. If P G does not contain Alt( k ), then b ( G ) = 2 by Proposition 3.1, hence the result followsfrom Lemma 2.3. So we can assume that Alt( k ) ≤ P G . Let ( u , . . . , u k ) ∈ T k ≤ G. Recall that g = ( y, . . . , y ) σ stabilises [ u , . . . , u k ] if and only if g ( u ,u ,u ...,u k ) − = ( u yu − σ , . . . , u k yu − kσ ) σ ∈ D ( k, T ) , i.e. if and only if(3.1) u yu − σ = · · · = u k yu − kσ . Consider in particular ( u , . . . , u k ) = ( t , t , , . . . , 1) where t and t are chosen so that they arenot conjugated in Aut( T ) and T = h t , t i (choose for instance an involution t , then the existenceof t follows from the results in [7]). Let τ := [ u , . . . , u k ] . We are going to deduce from (3.1) that g = ( y, . . . , y ) σ stabilises τ if and only if 1 and 2 are fixed by σ and y = 1 . Let w i := u i yu − iσ − with u i = t i if i ≤ , u i = 1 otherwise. Assume 1 σ − = j > . Then w j = yt − , and w i ∈ { y, yt − } , when i ≥ i = j . This would imply w i = w j in contradiction with (3.1). So we must have 1 σ − ∈{ , } . A similar argument shows that 2 σ − ∈ { , } . If follows that w = w = w = · · · = w k = y. If 1 σ = 2, then t yt − = y, i.e. t = t y in contradiction with the assumption that t and t arenot conjugated in Aut( T ) . So we must have 1 σ = 1 and 2 σ = 2 . This implies t yt − = t yt − = y, hence y ∈ C Aut( T ) ( h t , t i ) = C Aut( T ) ( T ) = 1.Let ι := [(1 , . . . , G ( ι,τ ) = { σ ∈ G | σ = 1 , σ = 2 } . Since Alt( k ) ≤ P G , there exists a ∈ Aut( T ) such that z = ( a, . . . , a )(1 , , ∈ G. Let 1 = s ∈ C T ( a ) and consider η := [ s, s, , . . . , . Clearly G ( ι,τ ) ≤ G ( ι,η ) and z ∈ G ( ι,η ) \ G ( ι,τ ) . Further, there exists b ∈ Aut( T )such that v := ( b, . . . , b )(1 , , ∈ G. If follows from (3.1) and the fact that s = 1 , that v ∈ G ι \ G η .So we have G ι > G ( ι,η ) > G ( ι,η,τ ) = G ( ι,τ ) and therefore G is not IBIS. (cid:3) Combining Propositions 3.9, 3.10, 3.11 , 3.12, 3.15, 3.16, 3.17,3.18, 3.19, we finally deduce thefollowing result. Theorem 3.20. Every monolithic primitive group of diagonal type is a non IBIS group. Product type Let H ≤ Sym(Γ) be a primitive group of almost simple type or of diagonal type with socle T .For k ≥ , we consider W = H ≀ Sym( k ) acting on Ω = Γ k , with its natural product action. Then G is of product type if T k ≤ G ≤ W and the group P of the elements σ ∈ Sym( k ) such that( h , . . . , h k ) σ ∈ G for some h i ∈ H is transitive. Write Ω = Γ × · · · × Γ k where Γ i = Γ for each i .Without loss of generality, we may assume that G induces H on each of the k factors Γ i of Ω. Inthe whole section we will assume that G is a primitive permutation group of product type. Theorem 4.1. The group G is not an IBIS group.Proof. To show this result, let assume by contradiction that G is an IBIS group. We are going toconstruct two irredundant bases of G and we will compare their lengths.Let S := { γ , γ , . . . , γ r } be an irredundant base of T ≤ Sym(Ω). By Lemma 2.3 we deduce that(4.1) r ≥ . Let α i = ( γ i , . . . , γ i ) with 0 ≤ i ≤ r and let β i,j be the k -tuple having the i -th component equal to γ j and the others equal to γ , for 1 ≤ i ≤ k and 1 ≤ j ≤ r . The following sequences B s = { α , α , . . . , α r } and B l = { α , β , , . . . , β ,r , β , , . . . , β ,r , . . . , β k, , . . . , β k,r } are irredundant sequences of G , and B s has size r + 1 , while B l has size 1 + kr . Indeed, fromthe choice of S , for every 1 ≤ a ≤ r − 1, there exists u a ∈ T that stabilizes γ , . . . , γ a but doesnot stabilize γ a +1 , and consequently ( u a , u a . . . , u a ) ∈ G stabilises α , α , . . . , α a but not α a +1 .Moreover ( u a , , . . . , ∈ G stabilises α , β , , . . . , β ,a but not β ,a +1 , (1 , u a , . . . , ∈ G stabilises α , β , , . . . , β ,r , β , , . . . , β ,a but not β ,a +1 and proceeding with this argument we finally get that(1 , , . . . , u a ) ∈ G stabilises α , β , , . . . , β ,r , β , , . . . , β ,r , . . . , β k, , . . . , β k,a , but not β k,a +1 .Let X := G ∩ H k . It is not difficult to observe that X ( B s ) = X ( B l ) = { ( h , . . . , h k ) ∈ X | h ∈ H ( S ) , . . . , h k ∈ H ( S ) } . Now, let A = { δ , . . . , δ b } be a base of minimal size of X ( B s ) = X ( B l ) . Hence B s ∪ A and B l ∪ A arestill irredundant sequences for G .Here, let δ , . . . , δ c , and ν , . . . , ν c ′ be elements of G such that the sequences B = B s ∪ A ∪{ δ , . . . , δ c } and B = B l ∪ A ∪ { ν , . . . , ν c ′ } are irredundant bases of G . Since G is IBIS, then |B | = r + 1 + b + c = 1 + rk + b + c ′ = |B | . Hence r ( k − 1) = c − c ′ ≤ c. (4.2)Note that G ( B s ∪A ) ∩ X = 1 , then G ( B s ∪A ) ∼ = G ( B s ∪A ) / ( G ( B s ∪A ) ∩ X ) ∼ = G ( B s ∪A ) X/X ≤ Sym( k ). So,denoting by ℓ (Sym( k )) the maximal length of a chain of subgroups of Sym( k ) , we can estimate c with ℓ (Sym( k )) . From (4.2) and (4.1) we deduce that(4.3) 2( k − ≤ ℓ (Sym( k )) . By [3] we have that ℓ (Sym( k )) < k/ , hence (4.3) yields k < 4. Since ℓ (Sym(2)) = 1 and ℓ (Sym(3)) = 2, (4.3) allows us to exclude also k = 2 and k = 3. This contradicts the assumption k ≥ (cid:3) Twisted wreath product type Let T be a non-abelian simple group, and let k be an integer that is at least 2 . A group G oftwisted wreath product type with socle T k acts primitively on a set | Ω | with degree | T k | and is alsoa twisted wreath product, which is a split extension of T k by a transitive subgroup P of Sym( k ). Inthe whole section we will assume that G is a primitive permutation group of twisted wreath producttype. Theorem 5.1. The group G is not an IBIS group. Proof. We may identify G with a subgroup of the wreath product W = Aut( S ) ≀ P , with theproperty that soc( G ) = T k and T k has a maximal complement, say H , in G , which is isomorphic to P. In particular for every σ ∈ P there exists a unique element ( a , . . . , a k ) ∈ (Aut( T )) k such that( a , . . . , a k ) σ ∈ H. By [5, Theorem 5.0.1], if P is a primitive subgroup of Sym( k ) , then b ( G ) = 2. In this case, sincesoc( G ) is non-abelian, the result follows from Lemma 2.3. So we may assume that k is not a prime.Since P is transitive, it contains a fixed-point free permutation σ. Either σ or one of its powers (inthe case when σ is a k -cycle) admits at least two orbits of size at least 2. So it is not restrictive toassume that P contains an element ρ = (1 , . . . , r )( r + 1 , r + 2 , . . . , r + s ) τ (5.1)for some r, s ≥ τ ∈ Sym( k ) which pointwise fixes the set { , . . . , r + s } . There exists aunique element ( a , . . . , a k ) ∈ (Aut( T )) k such that h := ( a , . . . , a k ) ρ ∈ H. It follows from theclassification of the finite simple groups that C T ( a · · · a r ) = 1 ([6, 1.48]). In particular there exists t ∈ C T ( a · · · a r ) of prime order, say p. Choose u ∈ T of prime order q , with q = p, and considerthe following elements of T k : y = ( t, t a , t a a , . . . , t a a ··· a r − , , . . . , , y = (1 , . . . , , u, , . . . , , where u is the entry in position r + 1 , and y = y y . We have that h ∈ C H ( y ) = C H ( y q ) but h / ∈ C H ( y ). Hence H ∩ H y q ∩ H y = H ∩ H y = C H ( y ) < C H ( y q ) = H ∩ H y q < H and therefore G isnot IBIS. (cid:3) Proof of Theorem 1.1 We work through the non-affine and non-almost simple classes of the O’Nan-Scott Theorem. If G is of diagonal type, as showed in Propositions 3.4, 3.2, 3.1 and Theorem 3.20, there is a uniqueinfinite family of IBIS group. If G is of product action type or twisted wreath product, then theresult follows immediately from Section 4 and Sections 5 respectively. Hence the proof is complete.ACKNOWLEDGMENTS. All authors are members of GNSAGA. References [1] P. J. Cameron, Permutation Groups, London Mathematical Society Student Texts. Cambridge University Press,1999.[2] P. J. Cameron and D. G. Fon-Der-Flaas, Bases for Permutation Groups and Matroids, Europ. J. Combin. (1995), 537–544.[3] P. J. Cameron, R. Solomon and A. Turull, Chains of Subgroups in Symmetric Groups, J. Algebra (1989),340–352.[4] J. B. Fawcett, The base size of a primitive diagonal group, J. Algebra (2013), 302–321.[5] J. B. Fawcett, Bases of primitive permutation groups, Phd dissertation.[6] D. Gorenstein, Finite simple groups. An introduction to their classification. University Series in Mathematics.Plenum Publishing Corp., New York, 1982.[7] R. M. Guralnick and W. M. Kantor, Probabilistic generation of finite simple groups. Special issue in honor ofHelmut Wielandt. J. Algebra 234 (2000), 743–792.[8] B. Huppert, Endliche Gruppen I, Springer, Berlin, 1967.[9] D. Leemans and M. W. Liebeck, Chiral polyhedra and finite simple groups, Bull. Lond. Math. Soc. (2017),no. 4, 581–592.[10] M. W. Liebeck, C. E. Praeger and J. Saxl, On the O’Nan-Scott theorem for finite primitive permutation groups. J. Australian Math. Soc. (A) (1988), 389–396.[11] D. J. S. Robinson, A course in the theory of groups, Springer Science & Business Media, 1996. [12] M. Suzuki, The nonexistence of a certain type of simple groups of odd order, Proc. Amer. Math. Soc. (1957),686–695. Andrea Lucchini, Dipartimento di Matematica “Tullio Levi-Civita”, Universit`a degli studi diPadova, Via Trieste 63, 35121, Padova, Italy Email address : [email protected] Marta Morigi, Universit`a di Bologna, Piazza di Porta San Donato 5, 40126 Bologna, Italy Email address : [email protected] Mariapia Moscatiello, Dipartimento di Matematica, Universit`a di Bologna, Piazza di Porta SanDonato 5, 40126 Bologna, Italy Email address ::