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Almost cyclic elements in cross-characteristic representations of finite groups of Lie type

  • Lino Di Martino EMAIL logo , Marco A. Pellegrini and Alexandre E. Zalesski
Published/Copyright: October 16, 2019

Abstract

This paper is a significant contribution to a general programme aimed to classify all projective irreducible representations of finite simple groups over an algebraically closed field, in which the image of at least one element is represented by an almost cyclic matrix (that is, a square matrix M of size n over a field 𝔽 with the property that there exists α𝔽 such that M is similar to diag(αIdk,M1), where M1 is cyclic and 0kn). While a previous paper dealt with the Weil representations of finite classical groups, which play a key role in the general picture, the present paper provides a conclusive answer for all cross-characteristic projective irreducible representations of the finite quasi-simple groups of Lie type and their automorphism groups.

1 Introduction

Let V be a vector space of finite dimension n over an arbitrary field 𝔽, and let M be a square matrix of size n over 𝔽. Then M is said to be cyclic if the characteristic polynomial and the minimum polynomial of M coincide. Note that a matrix MMatn(𝔽) is cyclic if and only if the 𝔽M-module V is cyclic, that is, is generated by a single element. This is standard terminology in module theory, and the source of the term “cyclic matrix”. Matrices with simple spectrum often arising in applications are cyclic. We consider a generalisation of the notion of cyclic matrix, namely, we define a matrix MMatn(𝔽) to be almost cyclic if there exists α𝔽 such that M is similar to diag(αIdk,M1), where M1 is cyclic and 0kn.

Examples of almost cyclic matrices arise naturally in the study of matrix groups over finite fields. For instance, if an element gGL(V) acts irreducibly on V/V, where V is some eigenspace of g on V, then g is almost cyclic. Transvections and reflections are other examples, as well as unipotent matrices with Jordan form consisting of a single non-trivial block.

The aim of the present paper is to implement a crucial step within a general project, which can be stated as follows: determine all irreducible subgroups G of GL(V) which are generated by almost cyclic matrices, mainly when 𝔽 is algebraically closed. The motivations for such a programme (including the connections with the area of linear group recognition) have been described in full detail in [6], to which paper we also refer for the citations of all relevant past results connected to our subject.

As current applications seem to focus on p-elements, we will limit ourselves to the study of elements gG of any p-power order, p a prime. Furthermore, we will make a systematic use of representation theory and will take for granted the classification theorem of finite simple groups. In view of it, the non-abelian finite simple groups consist of 26 sporadic groups, the alternating groups of degree >4, and the so-called finite groups of Lie type. The latter come together with their defining characteristics, and in this paper, it will be convenient for us to distinguish two groups of Lie type of distinct defining characteristic even if they are isomorphic. For instance, we view PSL3(2) and PSL2(7) as distinct.

The sporadic simple groups and their covering groups have been completely dealt with in [3]. In [5], we started to deal with finite groups of Lie type and determined all the irreducible representations of a quasi-simple group of Lie type G over an algebraically closed field 𝔽 of characteristic coprime to the defining characteristic of G, in which the image of at least one unipotent element g is represented by an almost cyclic matrix. The complementary case, when g is unipotent in G, and the characteristic of 𝔽 is the defining characteristic of G, has been settled for classical groups in [26] and for the exceptional groups of Lie type in [28]. This leaves open the case when g is a semisimple element of prime-power order of G. The case where g2=1 already attracted attention many years ago and was treated in [24, 31, 32, 33, 37, 38].

We should also mention that the occurrence of almost cyclic p-elements in irreducible 𝔽-representation of finite simple groups, where p==char𝔽, has been studied recently in [2].

In [6], the focus was on Weil representations of finite classical groups (there, an ample overview of these representations is given in Section 5.1). The reason to deal first with Weil representations was the strong evidence that most examples of semisimple almost cyclic elements occur in Weil representations, and the actual occurrence of such elements was thoroughly examined in [6].

In the present paper, we will consider the occurrence of almost cyclic elements of prime-power order in projective cross-characteristic irreducible representations ϕ of finite simple classical groups L which are not Weil (Sections 5, 6, 7). Furthermore, we will also consider the group extensions G of groups L by outer automorphisms (note that a separate ad hoc section (Section 4) is required for the case when PSL2(q)GAutPSL2(q)). Finally, we will also deal with the projective irreducible representations of finite simple exceptional groups of Lie type and their automorphism groups (Section 8).

We note explicitly that, for the sake of completeness, we choose to consider as simple groups of Lie type the commutator subgroups of the groups PSp4(2), G2(2), F42(2) and G22(3). The main result of the present paper is summarised in the following.

Main theorem.

Let L be a finite simple group of Lie type, let gAutL be a p-element for some prime p, and let G=L,g. Let ϕ be an irreducible projective representation of G over an algebraically closed field F of characteristic , different from the defining characteristic of L. Suppose that ϕ is non-trivial on L; furthermore, assume that g21, LPSL2(q) and ϕ(g) is almost cyclic. Then either ϕ|L is an irreducible Weil representation of L, where

L{PSLn(q)(n3),PSUn(q)(n3),PSp2n(q)(n2,q𝑜𝑑𝑑)}

and g is semisimple, or L belongs to the following list of groups:

={PSL3(2),PSL3(4),PSL4(2),PSU3(3),PSU3(4),PSU4(2),PSp4(3),PSU5(2),PSp4(2),PSp4(4),PSp6(2),PΩ8+(2),B22(23),G2(2),G2(3),G2(4),G22(3)}.

A more detailed statement, providing full information on the groups L belonging to , is given in Theorem 9.1. The case when g is an involution is fully dealt with in Lemma 2.10 for classical groups, and in Lemma 8.1 for the groups of exceptional type.

One of our main tools is Lemma 2.4, which gives an upper bound for dimϕ in terms of the order of gG for an arbitrary finite group G, provided g is almost cyclic. For many pairs (ϕ,g), this upper bound conflicts with the known lower bound for dimϕ, which violates the existence of the pair in question.

In order to apply that lemma to a group G, where LGAutL, and L is a simple group of Lie type, we need an accurate upper bound for ηp(AutL), the exponent of a Sylow p-subgroup of AutL, and for α(g), the minimal number of conjugates of g generating G. Results on ηp(AutL) are collected in Section 3. Rather precise information on α(g) is available in a paper by Guralnick and Saxl [13], which is our main reference for this matter. If gL and CL(g) contains no unipotent elements, then α(g)3 (see Lemma 2.7, deduced from a result of Gow [10]). However, if dimϕ is relatively small in comparison with ηp(G), much more analysis is required. This actually happens either if |L| is small or if G=L and ϕ is a Weil representation. (As already mentioned, the Weil representations have been dealt with in [6]).

In our analysis, we will deal separately with the following cases: (i) |g|=2 (see Lemma 2.10 and Section 8); (ii) L=PSL2(q) (Section 4); (iii) the groups PSLn(q), n>2; PSUn(q), n>2; PSp2n(q), n>1, q odd (Section 5); (iv) some exceptional low-dimensional classical groups (Section 6); (v) the groups Sp2n(q), n>1, q even; Ω2n+1(q), n>2, q odd; PΩ2n±(q), n>3 (Section 7); (vi) the exceptional groups of Lie type (Section 8).

Notation.

The centre of a group G is denoted by Z(G). Let G be a finite group. We write |G| for the order of G. For gG, |g| is the order of g, and o(g) is the order of g modulo Z(G), that is, the minimum integer k such that gkZ(G). If g does not lie in a proper normal subgroup of G, we write α(g) for the minimum number of conjugates of g that generate G. For a prime p, we denote by ηp(G) the exponent of a Sylow p-subgroup of G.

A finite field of order q (where q is a prime-power) is denoted by 𝔽q. Let V be a vector space of finite dimension m>1 over the finite field 𝔽q. The general linear group GL(V), that is, the group of all bijective linear transformations of V, and the special linear group SL(V), that is, the subgroup of GL(V) consisting of the transformations of determinant 1, are denoted by GLm(q) and SLm(q), respectively.

Suppose that the space V is endowed with a non-degenerate orthogonal, symplectic or unitary form. Then I(V) denotes the group of the isometries of V, that is the set of all elements xGL(V) preserving the form. Then we will loosely use the term “finite classical group” for a subgroup G of I(V) containing the commutator subgroup I(V), as well as the projective image of G, that is, the quotient of G modulo its scalars, which will be denoted by PG. In particular, if V is a symplectic space over 𝔽q, I(V) will be denoted by Spm(q); if V is a unitary space over the field 𝔽q2, I(V) will be denoted by GUm(q); and if V is an orthogonal space over 𝔽q, I(V) will be denoted by Om(q). More precisely, if V is an orthogonal space of even dimension m, we will denote I(V) by Om+(q) or Om-(q) if V has maximal or non-maximal Witt index, respectively. Accordingly, the subgroups of the unitary and orthogonal isometry groups consisting of the elements of determinant 1 will be denoted by SUm(q), SOm(q), SOm+(q) and SOm-(q), respectively. We recall that Spm(q)=Spm(q) unless m=2 and q3, or (m,q)=(4,2), whereas GUm(q)=SUm(q) unless m=2 and q3, or (m,q)=(3,2). In the orthogonal case, the groups SOm(q), SOm+(q) and SOm-(q) contain I(V) as a subgroup of index 2. We will denote by Ωm(q) and Ωm±(q) the commutator subgroups of the corresponding orthogonal groups, and accordingly, we will refer to a classical group as a subgroup of I(V) containing the commutator subgroup I(V). Furthermore, we will denote by CSp2n(q) the conformal symplectic group, that is, the group of symplectic similarities. Also, it should be noted that in places the term “classical group” will be meant to include also the groups GLm(q) and SLm(q) (considering V endowed with the identically zero bilinear form). It is well known that, for m>2, or m=2 and q>2, one has GLm(q)=SLm(q).

If G is a finite group of Lie type of defining characteristic r, we assume throughout the paper that is coprime to r. Furthermore, note that (1) the conjugacy classes of elements of G are labelled according to the GAP package [40] and (2) the extensions G of a finite simple group of Lie type L by an outer automorphism will be labelled according to the Atlas notation as given in [1, 19] unless we wish to indicate explicitly that G is the extension of L by the group of field automorphisms, in which case G will be denoted, say, by PΣLm(q) or PΣUm(q) for L=PSLm(q) or L=PSUm(q), respectively.

By a projective 𝔽-representation of an arbitrary group G we mean a homomorphism ϕ:GPGLm(𝔽) for some m. Let G¯ denote the pre-image of ϕ(G) in GLm(𝔽), and let ψ denote the linear representation of G¯ arising in this way from ϕ. If gG and g¯ denotes a pre-image of ϕ(g) in GLm(𝔽), we will say for short that ϕ(g) is almost cyclic if ψ(g¯) in GLm(𝔽) is almost cyclic. Note that, clearly, the property of ϕ(g) of being almost cyclic does not depend on the choice of g¯.

2 Preliminaries

We recall the following basic definition.

Definition 2.1.

Let M be an (n×n)-matrix over an arbitrary field 𝔽. We say that M is almost cyclic if there exists α𝔽 such that M is similar to diag(αIdk,M1), where M1 is cyclic and 0kn.

It is trivial to observe that every (3×3)-matrix over any field 𝔽 is almost cyclic. Another elementary observation, which will be useful throughout the paper, is the following: if MGL(V) is almost cyclic and U is an M-stable subspace of V, then the induced actions of M on U and on V/U yield almost cyclic matrices.

Throughout the paper, unless stated otherwise, we assume that 𝔽 is an algebraically closed field of characteristic . We emphasise that the representations of G we will consider are all over 𝔽.

2.1 Almost cyclicity and dimension bounds

Let G be a finite group and ϕ a projective irreducible 𝔽-representation of G. Let gG, and suppose that ϕ(g) is almost cyclic and non-scalar. Our results will be essentially based on the fact that dimϕ can be bounded in terms of the minimum number α(g) of conjugates of g sufficient to generate G. In this subsection, we explain this in detail.

The following crucial results on generation by conjugates are due to Guralnick and Saxl [13].

Proposition 2.2.

Let L be a simple group of Lie type, and let 1xAutL. Denote by α(x) the minimum number of L-conjugates of x sufficient to generate x,L. Then the following holds.

  1. [13, Theorem 4.2] Let L be a simple classical group, and assume that the natural module for L has dimension n5. Then α(x)n unless L=PSpn(q) with q even, x is a transvection and α(x)=n+1.

  2. [13, Theorem 5.1] Let L be a simple exceptional group of Lie type, of untwisted Lie rank m. Then α(x)m+3, except possibly for the case L=F4(q) with x an involution, where α(x)8.

Proposition 2.3.

Under the same assumptions as in Proposition 2.2 and assuming additionally that x has prime order, the following holds.

  1. [13, Lemma 3.1] Let L=PSL2(q) with q4. Then α(x)3 unless either (i)x is a field automorphism of order 2 and α(x)4, except that α(x)=5 for q=9; or (ii)q=5, x is a diagonal automorphism of order 2 and α(x)=4. Moreover, if x has odd order, then α(x)=2 unless q=9, |x|=3, α(x)=3.

  2. [13, Lemma 3.2] If L=PSL3(q), then α(x)3 unless x is an involutory graph-field automorphism and α(x)4.

  3. [13, Lemma 3.3] If L=PSU3(q), q>2, then α(x)3 unless q=3 and x is an inner involution with α(x)=4.

  4. [13, Theorem 4.1 (c), (d)] If L=PSL4(q), then α(x)4 unless one of the following holds: (i)q>2, x is an involutory graph automorphism and α(x)6; (ii)q=2, x is an involutory graph automorphism and α(x)7.

  5. [13, Lemma 3.4] If L=PSU4(q), then α(x)4 unless one of the following holds: (i)x is an involutory graph automorphism and α(x)6; (ii)q=2 with x a transvection and α(x)5.

  6. [13, Theorem 4.1 (f)] If L=PSp4(q), then α(x)4 unless x is an involution and α(x)5, or q=3 and α(x)6.

The above quoted bounds on α(x) play a role in the following lemma, which is crucial for our purposes.

Lemma 2.4.

Let H be a finite group, let ϕ:HPGLm(F) be an irreducible projective F-representation of H, where F is an arbitrary algebraically closed field, and let H¯ be the pre-image of ϕ(H) in GLm(F). Let M be the module affording the linear representation ψ of H¯ arising from ϕ. For any hH, denote by h¯ a pre-image of ϕ(h) in H¯; choose hH such that H is generated by some conjugates of h. Let d=dim(λId-h¯)M0 for some λF. Then dimϕdα(h). In particular, if ϕ(h) is almost cyclic, then dimϕα(h)(o(h)-1)α(h)(|h|-1).

Proof.

Let m=α(h) and h1,,hm be conjugates of h generating H, and let Mi=(λId-hi)M. Then every h¯i acts scalarly as λ on M/Mi, and hence it acts scalarly as λ on any quotient M/U, where U is a subspace containing Mi, thus in particular on M/(Mi). This is true for every i, so every h¯i acts as λ on M/(Mi). Since h1,,hm=H, the group H¯ acts scalarly on M/(Mi). It follows that Mi is a non-zero H¯-submodule of M, and hence coincides with M. Since

dim(i=1m(λId-h¯i)M)i=1mdim(λId-h¯i)M=mdim(λId-h¯)M=md,

the first part of the statement follows. The second part is proven in [3, Lemma 2.1]. ∎

We remark here that, throughout the paper, we will frequently make use of Lemma 2.4 without explicit reference to it. The following results concerning regular semisimple elements will be useful for our purposes.

Lemma 2.5 ([10]).

Let G be a simple group of Lie type of characteristic , and let 1gG be a regular semisimple element. Then every semisimple element of G can be expressed as a product ab, where a,b are conjugate to g.

As a corollary of Lemma 2.5, we obtain a small refinement of case (1) of Proposition 2.3, when considering 2-elements.

Lemma 2.6.

Let L=PSL2(q), where q is odd and q>9. Let gL be a 2-element such that g21. Then α(g)=2.

Proof.

Note that g is a regular semisimple element of L (as CL(g) contains no unipotent element). So, by Lemma 2.5, there exists xL conjugate to g such that gx=s, where s is a Singer cycle of L. If q>9, it is well known that every proper subgroup of L containing s is either S=s or N=NL(S), a dihedral group of order 2|S|. Therefore, the result is true unless both g,xN. But the latter is not the case. Indeed, suppose that g,xN. As N is dihedral, all the 2-elements of order 4 of N are in S. Therefore, gS and xS. As S is cyclic and |g|=|x|, it follows readily that |gx|<|g|; this obviously contradicts gx=s, as claimed.∎

Lemma 2.7 ([6, Lemma 2.8]).

Let G be a simple group of Lie type. Then the following holds.

  1. G is generated by two semisimple elements.

  2. Let gG be any regular semisimple element. Then α(g)3, that is, G can be generated by three elements conjugate to g.

The following two lemmas are crucial for us in view of the results above. The first one yields the best known lower bounds for the dimension of an irreducible projective cross-characteristic 𝔽-representation of a finite classical group. The second one provides lower bounds for the dimension of any non-Weil irreducible projective cross-characteristic 𝔽-representation of a finite classical group.

Lemma 2.8 ([18, Table 1]).

Let L be a simple classical group, and let ϕ be an irreducible projective cross-characteristic F-representation of L with dimϕ>1. Then the following holds.

  1. If L=PSL2(q), then dimϕq-1(2,q-1) unless q=4,9; in the latter exceptional cases, dimϕ2,3, respectively.

  2. If L=PSLn(q), n>2, then dimϕqn-1q-1-2 unless (n,q)=(3,2),(3,4),(4,2),(4,3); in the latter exceptional cases, dimϕ2,4,7,26, respectively.

  3. If L=PSUn(q), n>2, (n,q)(4,2),(4,3), then dimϕqn-qq+1 if n is odd, while dimϕqn-1q+1 if n is even; if (n,q)=(4,2),(4,3), then dimϕ4,6, respectively.

  4. If L=PSp2n(q), n>1 and q is odd, then dimϕqn-12.

  5. If L=PSp2n(q), where n>1, q is even, and (n,q)(4,2), then

    dimϕq(qn-1)(qn-1-1)2(q+1).

    If L=PSp4(2), then dimϕ2.

  6. Let L=PΩ2n+1(q), where n>2 and q is odd. If q=3 and n3, then dimϕ(3n-1)(3n-3)32-1; if q>3, then dimϕq2n-1q2-1-2; if (n,q)=(3,3), then dimϕ27.

  7. Let L=PΩ2n+(q), n>3. If q<4 and (n,q)(4,2), then

    dimϕ(qn-1)(qn-1-1)q2-1;

    if q4, then dimϕ(qn-1)(qn-1+q)q2-1-2; if L=PΩ8+(2), then dimϕ8.

  8. Let L=PΩ2n-(q), n>3, where (n,q)(4,2),(4,4),(5,2),(5,3). Then

    dimϕq(qn+1)(qn-2-1)q2-1-1;

    for the exceptional cases, dimϕ32,1026,151,2376, respectively.

From [12, Theorems 2.1, 2.7], [14, Theorem 1.1], [15] and [17, Theorem 1.6], we obtain the following.

Lemma 2.9.

Let L{PSLn(q),n3;PSUn(q),n3;PSp2n(q),n2,q𝑜𝑑𝑑}, and let ϕ be a non-trivial cross-characteristic projective irreducible representation of L. Then either ϕ is a Weil representation, or one of the following occurs.

  1. If L=PSLn(q) and (n,q)(3,2),(3,4),(4,2),(4,3),(6,2),(6,3), then dimϕϑn+(q), where

    ϑn+(q)={(q-1)(q2-1)gcd(3,q-1)𝑖𝑓n=3,(q-1)(q3-1)gcd(2,q-1)𝑖𝑓n=4,(qn-1-1)(qn-2-qq-1-κn-2)𝑖𝑓n5.

    Here κn=1 if divides qn-1q-1 and 0 otherwise. Further, if (n,q)=(4,3),(6,2),(6,3), then dimϕϑn+(q), where ϑn+(q)=26,61,362, respectively.

    1. If L=PSUn(q) with n5 and (n,q)(6,2), then dimϕϑn-(q), where

      ϑn-(q)={qn-2(q-1)(qn-2-q)q+1𝑖𝑓nis odd,qn-2(q-1)(qn-2-1)q+1𝑖𝑓nis even.
    2. If L=PSU4(q) and q>3, then dimϕϑ4-(q), where

      ϑ4-(q)={(q2+1)(q2-q+1)-22𝑖𝑓qis odd,(q2+1)(q2-q+1)-1𝑖𝑓qis even.
    3. If L=PSU3(q) and q5, then dimϕϑ3-(q), where

      ϑ3-(q)={(q-1)(q2+3q+2)6𝑖𝑓gcd(3,q+1)=3,2q3-q2+2q-33𝑖𝑓gcd(3,q+1)=1.

  2. If L=PSp2n(q) and (2n,q)(6,2), then dimϕσn(q), where

    σn(q)=(qn-1)(qn-q)2(q+1).

2.2 The case |g|=2

In this subsection, we consider the case |g|=2, which yields a significant part of the examples in which almost cyclic elements occur.

Lemma 2.10.

Let L be a finite simple classical group, and let G=L,g, where gAutL is an involution. Let ϕ be a cross-characteristic projective irreducible F-representation of G which is non-trivial on L. Then ϕ(g) is almost cyclic if and only if one of the following holds:

    1. G=PSL2(4), dimϕ=2,3 and 2,

    2. G=PΣL2(4) and either dimϕ=4, 2,5, or dimϕ=2,3, =5,

    1. G=PSL2(5), dimϕ=2,3 and 5,

    2. G=PGL2(5), dimϕ=4 and 5,

  1. G=PSL2(7), dimϕ=3 and 7,

    1. G=PSL2(9), dimϕ=3 and 2,3,

    2. G=PΣL2(9) and either dimϕ=5, 2,3, or dimϕ=4, =2,

  2. G=PSL3(2) and either dimϕ=3 and 2 or dimϕ=2 and =7,

  3. G=PΣL3(4), dimϕ=4 and =3,

  4. G=PSL4(2).2, dimϕ=7, 2 and g yields a graph automorphism,

    1. G=PSU4(2), dimϕ=4,5 and 2 (class 2a),

    2. G=PΣU4(2) and either dimϕ=6, 2,3, or dimϕ=5, =3,

  5. G=PSU4(3).22, dimϕ=6 and 3,

    1. G=PSp4(3) and either dimϕ=4 and 3 (class 2a) or dimϕ=5 and 2,3 (class 2a),

    2. G=PCSp4(3), dimϕ=6 and 2,3,

  6. G=PSp6(2), dimϕ=7 and 2 (class 2a),

  7. G=PSO8+(2), dimϕ=8 and 2.

Proof.

(I) Suppose that ϕ(g) is almost cyclic. By Lemma 2.4, this implies that

(2.1)dimϕα(g)(|g|-1)=α(g).

An upper bound for the value of α is given by Propositions 2.2 and 2.3, while a lower bound for the value of dimϕ is given in Lemma 2.8.

For the sake of brevity, we give full details of the computations yielding the statement of the lemma only in the case where L=PSLn(q). The other classical groups can be dealt with in a similar way.

Let L=PSL2(q) with 7q9. Then (2.1) yields q-1(q-1,2)4, which holds only if q=7. If L{PSL2(4),PSL2(5)}, then dimϕ2 and α(g)=3 if gL, α(g)4 otherwise. If L=PSL2(7), then dimϕ3, α(g)=3. If L=PSL2(9), then dimϕ3 and α(g)=3 unless gL, in which case α(g)5.

Let L=PSL3(q) with q2,4. In this case, (2.1) yields q2+q-14, which is false. If L=PSL3(2), then dimϕ2 and α(g)=3. If L=PSL3(4), then dimϕ4 and α(g)=3 unless gL, in which case α(g)=4.

Let L=PSL4(q) with q4. Then (2.1) yields q3+q2+q-16, which is clearly impossible. If L=PSL4(2), then dimϕ7 and α(g)6 unless gL, in which case α(g)7. If L=PSL4(3), then dimϕ26, whence a contradiction.

Finally, let L=PSLn(q) with n5. Then (2.1) yields qn-1q-1-2n, a contradiction. By the computations above, we readily get items (1) to (7) of the statement.

(II) Conversely, in all the occurrences listed in the statement of the lemma (items (1) to (12)), there actually exists at least one representation ϕ of G and a conjugacy class of involutions g such that ϕ(g) is almost cyclic. These ϕ’s and these g can be readily determined using [1, 19] and the GAP package. ∎

Remark 2.11.

The outer automorphism group of the group L=PSL2(9) is elementary abelian of order 4. Let g be an outer automorphism of L of order 2. Then g is either a field automorphism, or a diagonal automorphism, or the product of the two. In the first case, G=L,g=PΣL2(9), in the second case, G=L,g=PGL2(9), in the third case, G=L,g is isomorphic to M10, a maximal subgroup of the Mathieu group M11, a sporadic simple group. It is also well known that PSL2(9)Sp4(2)Ω4-(3) and Sp4(2)PΣL2(9). If we view L as Sp4(2) and let G=L,g, where gAutL is an involution, then the above lemma misses the 𝔽-representations of G for =3. So, let ϕ be a cross-characteristic projective irreducible 𝔽-representation of G which is non-trivial on L. Then direct inspection shows that ϕ(g) is almost cyclic if and only if one of the following occurs:

  1. dimϕ=3, G=Sp4(2) and 2,

  2. dimϕ=3, G=PGL2(9) and =3,

  3. dimϕ=4, G=Sp4(2), and =3,

  4. dimϕ=5, G=Sp4(2), and 2,3.

3 Maximum order of a p-element in AutL

Let L be one of the following groups:

PSLn(q),wheren>1and(n,q)(2,2),(2,3),
PSUn(q),wheren>2and(n,q)(3,2),
PSp2n(q),wheren>1and(n,q)(2,2).

So, L is simple. Let r be a prime and q=rα.

Recall that every automorphism of L is a product of inner, diagonal, field and graph automorphisms. Let Ad be the subgroup of AutL generated by all inner and diagonal automorphisms. It is well known that Ad is normal in AutL. Moreover, AdPGLn(q) or PGUn(q) if L=PSLn(q) or PSUn(q), respectively. Thus, |Ad/L|=(q-1,n) if L=PSLn(q), and (q+1,n) if L=PSUn(q). If L=PSp2n(q), then AdPCSp2n(q). Thus, we have |Ad:L|=2 for q odd, and Ad=L for q even. The group AutL/Ad is abelian. Set Φ=Gal(𝔽q/𝔽r) and Φ2=Gal(𝔽q2/𝔽r). If L=PSLn(q), then AutL/AdC2×Φ, where C2 is the cyclic group of order 2. If L=PSUn(q), then AutL/AdGal(𝔽q2/𝔽r)=Φ2. If L=PSp2n(q), n>2, or n=2 for q odd, then AutL/AdΦ. The case L=PSp4(q) with q even is more complex. In [25, Theorem 28], an automorphism ϕ of Sp4(q), q even, is constructed such that ϕ2 is a generator of Φ. It follows that AutSp4(q)/Sp4(q) is a cyclic group of order 2|Φ|Φ2. So, if gAutL is a 2-element and |Φ|2=2m, then g2m+1L. See [22, Table 5.1.A] for details.

Let p be a prime. For any positive integer m, we denote by |m|p the p-part of m, that is, the highest power of p dividing m. Furthermore, for any positive integer m, if p does not divide m, then e=ep(m) is defined to be the minimum integer i>0 such that p divides mi-1 if p>2. If p=2, then e2(m) is defined to be 1 if 4(m-1) and 2 if 4(m+1). Note that e1, and if e>1, then e<me-1m-1.

Lemma 3.1.

Let p be a prime.

  1. ep(q)=ep(qpk). In particular, if |𝔽qpk×|p>1, then |𝔽q×|p>1.

  2. Suppose that p(q-1) if p>2, and 4(q-1) otherwise. Let k>0 be an integer. Then |𝔽qpk×|p=pk|𝔽q×|p, equivalently, |qpk-1|p=pk|q-1|p.

Proof.

Fermat’s little theorem states that p divides np-n for any integer n, and hence p(n-1) if and only if p(np-1). So, p(qi-1) if and only if p((qpk)i-1), whence (1). For (2), see [7, Lemma 7.5]. ∎

Recall that, for a group X, we denote by ηp(X) the exponent of a Sylow p-subgroup of X. The following lemma is well known: for instance, see [23].

Lemma 3.2.

Let G=GLn(q), p a prime dividing |G|, (p,q)=1. If e=ep(q), then ηp(G)=pl|qe-1|p, where l is such that plen<pl+1e.

Corollary 3.3.

Let q=q02 be odd. Then η2(GLn(q))=2η2(GLn(q0)).

Proof.

Suppose first that n=2t, t>0. By assumption, 4 divides q-1, and by Lemma 3.2, η2(GLn(q))=2t|q-1|2. Then

η2(GLn(q0))={2t|q0-1|2if 4dividesq0-1,2t|q0+1|2if 4dividesq0+1.

As |q02-1|2=2|q0-1|2 in the former case and |q02-1|2=2|q0+1|2 in the latter case, the result follows. If 2t+1>n>2t, then

η2(GLn(q))=η2(GL2t(q))=2η2(GL2t(q0))=2η2(GLn(q0)),

whence the statement. ∎

Lemma 3.4 ([36, Lemma 3.2]).

Let p be an odd prime and n>1. Let G=SLn(q) with p(q-1), or G=SUn(q) with p(q+1). Then G has no irreducible p-element.

Lemma 3.5.

Let q=q0pk, k>0, and

G=G(q){GLn(q),SLn(q),GUn(q),SUn(q),Sp2n(q),Ω2n+1(q),Ω2n±(q)}.

Let Ψ be the group of field automorphisms of G of order pk, p>2, and set H=GΨ. Then the following holds.

  1. ηp(H)=max{p(k-i)ηp(G(q0pi)):0ik}.

  2. If pq, then ηp(H)=pkηp(G).

  3. If pq, then ηp(H)=ηp(G).

  4. If pq, then ηp(L)=ηp(LΨ), where L is the simple non-abelian composition factor of G.

Proof.

(1) The statement follows from [39, Corollary 14] and its immediate generalisation to any G(q) as remarked in [30], taking into account the well-known fact that, as p is odd, G(q0)G(q0p) (see [22, Table 4.5.A]).

(2) The statement follows from (1) and the known fact that

ηp(G(q0pi))=ηp(G(q0))fori=1,,k.

For an even stronger statement, see, e.g., [29, Lemma 2.32].

(3) The statement follows from [30, Lemma 3.10].

(4) It is well known that L contains a subgroup isomorphic to a Sylow p-subgroup of G unless n is a p-power, L=PSLn(q) and p divides (n,q-1) or L=PSUn(q) and p divides (n,q+1). Let us consider the exceptional cases. By Lemma 3.4, every p-element of SLn(q) and SUn(q) is reducible. It follows that

ηp(SLn(q))ηp(GLn/p(q))andηp(SUn(q))ηp(GUn/p(q)).

Moreover, there is an embedding j:GLn/p(q)SLn(q) such that j(GLn/p(q)) contains no scalar matrix (except the identity), which in turn yields an embedding j:GLn/p(q)PSLn(q). Thus, ηp(PSLn(q))=ηp(GLn/p(q))=ηp(SLn(q)). Similarly, we have ηp(PSUn(q))=ηp(GUn/p(q))=ηp(SUn(q)). By (3),

ηp(LΨ)=ηp(SLn(q)Ψ)=ηp(SLn(q))=ηp(L)forL=PSLn(q),

and similarly for L=PSUn(q). ∎

Lemma 3.6.

Suppose q=q02k, where k>0 and q is odd. Let

G=G(q){GLn(q),SLn(q),PSLn(q)},

and let Ψ be the group of field automorphisms of G of order 2k. Set H=GΨ. Then η2(H)=η2(G).

Proof.

By [39, Corollary 14], η2(H)=max{2(k-i)η2(G(q02i)):0ik}. For G=GLn(q), the statement then follows by iterated application of Corollary 3.3. Otherwise, the argument used to prove item (4) of Lemma 3.5 remains valid for p=2. (As k>0, Lemma 3.4 remains valid for n=p=2.) So,

η2(PSLn(q))=η2(GLn/2(q))=η2(SLn(q)).

Again by Corollary 3.3, we get η2(GLn/2(q))=2k-iη2(GLn/2(q02i)) for every i. It follows that η2(H)=η2(GLn/2(q))=η2(G), as claimed. ∎

Lemma 3.7.

Let G=GLn(q), q odd, and suppose that 2tn<2t+1. Then η2(G)=2t|q-1|2 if 4(q-1) and 2t|q+1|2 if 4(q+1).

Proof.

By Lemma 3.2, η2(G)=|qe-1|22l, where 2len<2l+1e. If e=1, then t=l, and the statement follows. Let e=2. Then

|q2-1|2=|q-1|2|q+1|2=2|q+1|2and2l+1n<2l+2.

So, l=t-1, whence the statement. ∎

4 The groups PSL2(q)

In this section, we deal with the groups L=PSL2(q). Set G=L,g, where 1gAutL is a p-element for some prime p. Note that here we may assume g21 in view of Lemma 2.10 and g(GPGL2(q)) as the case gPGL2(q) is settled in [6]. Furthermore, recall that L=PSL2(q) has no graph automorphism, and therefore AutL/PGL2(q)Gal(𝔽q/𝔽r). As above, let Φ=Gal(𝔽q/𝔽r) and pm=|Φ|p.

Theorem 4.1.

Let L=PSL2(q), q4, and let G=L,g, where 1gAutL is a p-element for some prime p and gPGL2(q). Let ϕ be a cross-characteristic irreducible projective F-representation of G, which is non-trivial on L. Suppose that g21. Then ϕ(g) is not almost cyclic unless one of the following holds.

  1. p=3, q=8, |g|=9 and dimϕ=7,8 if 3, or dimϕ=7 if =3.

  2. p=2, q=4, |g|=4 and dimϕ=5 if 3,5, or dimϕ=4 if =3, or dimϕ=2,3,5 if =5.

  3. p=2, q=9 and one of the following holds:

    1. if 2,5, then

      1. G=L.21=PΣL2(9), |g|=4 and dimϕ=4,5,

      2. G=L.22=PGL2(9), |g|=8 and dimϕ=6,8,9,

      3. G=L.23=M10, |g|=8 and dimϕ=6,8,9;

    2. if =2, then

      1. G=L.21=PΣL2(9), |g|=4 and dimϕ=4,

      2. G=L.22=PGL2(9), |g|=8 and dimϕ=8,

      3. G=L.23=M10, |g|=8 and dimϕ=8;

    3. if =5, then

      1. G=L.21=PΣL2(9), |g|=4 and dimϕ=4,5,

      2. G=L.22=PGL2(9), |g|=8 and dimϕ=6,8,

      3. G=L.23=M10, |g|=8 and dimϕ=8.

Proof.

Let q=q0pm. By Lemma 2.8 (1), unless q=4,9,

dimϕq-1(q-1,2)=q0pm-1(q0-1,2).

Note that if g is semisimple, then, by Lemma 3.1 (2),

|g|pm|L|p=pm|(q02)pm-1|p=pm|q02-1|p.

Suppose that p>2 (and hence q4,9). If g is unipotent, the result follows from [5, Lemma 4.13]. So, assume that g is semisimple. Then |g|pm(q0+1) and, by Lemma 2.3, α(g)=2. If ϕ(g) is almost cyclic, the inequality

q0pm-12(2,q0-1)(pm(q0+1)-1)

must hold by Lemma 2.4. However, this happens only for (q,p)=(8,3), which gives item (1). Indeed (see [1, 19]), the group G=AutPSL2(8) has irreducible representations of degree 7,8 over the complex numbers, in which the elements of order 9 are almost cyclic. This implies that the elements of order 9 are almost cyclic in every irreducible constituent of degree >1 of a Brauer reduction of any of these representations modulo any prime. If =3, we only get constituents of degree 7, whereas if =7, these are of degree 7 and 8. Note that the case of =2 is irrelevant here, but will appear for G=G22(3)SL2(8)3.

Now, suppose that p=2, and consider first the case when q>9 is odd. By the above, |g|2m+1(q0+1). Suppose that ϕ(g) is almost cyclic. Then, by Lemma 2.4, q-12dimϕα(g)(|g|-1). Thus, as g21, by Lemma 2.6, we must have

q02m-122(2m+1(q0+1)-1).

This inequality holds only if either m=1 and q0=3,5,7,9,11,13, or m=2 and q0=3. However, η2(PSL2(81))=16, contradicting Lemma 2.4. Similarly, we can rule out q=121,169 since η2(PSL2(121))16 and η2(PSL2(169))16. So, we are left with the cases q=25,49.

Next, suppose that q>4 is even. Then dimϕq-1 and clearly |g|2m+1. If ϕ(g) is almost cyclic, then q-1dimϕα(g)(|g|-1). By Lemma 2.3 (1), α(g)4. Thus, since g21, q02m-14(2m+1-1), which holds if and only if q0=2 and m=1,2, i.e., if and only if q=4,16.

In conclusion, we are left with the cases where q{4,9,16,25,49}.

(1) Suppose q=4. Then |g|=4, α(g)=2, and using the packages GAP and MAGMA, we get item (2) of the statement.

(2) Suppose q=9. Then |g|=4,8 and α(g)=2. If 2, using the packages GAP and MAGMA, we get the results recorded in the statement, item (3). So, suppose =2. Assume first |g|=4. Then ϕ(g) almost cyclic implies dimϕ6, which in turn implies dimϕ=4,6. If G=PΣL2(9), then G has a 2-modular representation of degree 4 in which ϕ(g) is actually cyclic, whereas 3.G has a faithful 2-modular representation of degree 6 in which ϕ(g) is almost cyclic. There are no other occurrences for |g|=4. Next, assume that |g|=8. Then ϕ(g) almost cyclic implies dimϕ14. Then either G=PGL2(9) or G=M10. In the first case, there is a 2-modular representation of G of degree 8 in which ϕ(g) is almost cyclic, and a faithful 2-modular representation of 3.G of degree 6 in which ϕ(g) is almost cyclic. In the latter case, there is a 2-modular representation of G of degree 8 in which ϕ(g) is almost cyclic, and two faithful 2-modular representations of 3.G of degree 6 and 9, respectively, in which ϕ(g) is almost cyclic.

(3) Suppose q=16. Then |g|=4,8 and α(g)=2. Since dimϕ15, ϕ(g) cannot be almost cyclic by Lemma 2.4. So, this case is ruled out.

(4) Suppose q=25. Then |g|=4,8 and α(g)=2. Since dimϕ12, we may assume |g|=8 by Lemma 2.4. However, L.22=PΣL2(25) does not contain elements of order 8. So, we are left to consider the projective representations of L.21=PGL2(25) and of L.23. Making use of GAP, we see that if 2, then ϕ(g) is not almost cyclic. On the other hand, if =2, then dimϕ24, and hence ϕ(g) is not almost cyclic. So, this case is ruled out.

(5) Suppose q=49. Then |g|=4,8,16 and α(g)=2. Since dimϕ24, we may assume |g|=16 again by Lemma 2.4. However, L.22=PΣL2(49) does not contain elements of order 16. So, we are left to consider the projective representations of L.21=PGL2(49) and L.23. Making use of GAP, we see that if 2,3, then ϕ(g) is not almost cyclic. On the other hand, if =2, then dimϕ48, and hence ϕ(g) is not almost cyclic. So, also this last case is ruled out. ∎

5 The groups PSLn(q), n>2; PSUn(q), n>2; PSp2n(q), n>1, q odd

In this section, we consider the simple groups L=PSLn(q), n>2; PSUn(q), n>2 with (n,q)(3,2); PSp2n(q), q odd.

Let 1gAutL be a p-element, set G=L,g, and let ϕ be a cross-characteristic projective irreducible representation of G that is non-trivial on L. Observe that if the restriction of ϕ to L has a constituent which is not a Weil representation of L, a lower bound of dimϕ is provided by Lemma 2.9. On the other hand, the occurrence of almost cyclic elements in the Weil representations of the groups L listed above is thoroughly dealt with in our earlier paper [6]. There, it is also shown that they extend to the subgroup Ad of AutL generated by all inner and diagonal automorphisms. Now, suppose that gAdL and ϕ restricted to L decomposes into the sum of, say, t constituents which are all Weil representations of L. By Clifford’s theorem, these constituents are all g-conjugates. However, they extend to Ad. Hence, t=1; in other words, ϕ itself is a Weil representation of Ad and is disposed of in [6]. In view of this, this section will be mainly concerned with answering the following cases: (A) the case when the restriction of ϕ to L has a constituent which is not a Weil representation of L; (B) the case when gAd and all the constituents of the restriction of ϕ to L are Weil representations of L.

5.1 Case (A)

Apart from very few groups, case (A) is answered by the following.

Lemma 5.1.

Let L be one of the following simple groups:

  1. PSLn(q), where n3 and (n,q){(3,2),(3,4),(4,2)},

  2. PSUn(q), where n3 and (n,q){(3,3),(3,4),(4,2),(4,3),(5,2)},

  3. PSp2n(q), where n2, q is odd and (n,q)(2,3).

Let 1gAutL be a p-element for some prime p, and set G=L,g. Let ϕ be a cross-characteristic projective irreducible representation of G that is non-trivial on L. If the restriction of ϕ to L has a constituent which is not a Weil representation of L, then ϕ(g) is not almost cyclic.

Proof.

Suppose that ϕ(g) is almost cyclic. By Lemma 2.4 the following bound must be met:

(5.1)dimϕα(g)(|g|-1).

Since the restriction of ϕ to L has a constituent which is not a Weil representation of L, lower bounds for dimϕ are provided by Lemma 2.9. Upper bounds for α(g) are given in Propositions 2.2 and 2.3. Finally, by [11, Theorem 2.16], we have |g|μ(L), where

μ(L)={qn-1q-1ifL=PSLn(q),qn-1-1ifL=PSUn(q)withnodd andqnot prime,qn-1+qifL=PSUn(q)withnodd andqprime,qn-1+1ifL=PSUn(q)withneven andq>2,4(2n-3+1)ifL=PSUn(q)withneven andq=2,qn+1q-1ifL=PSp2n(q).

Suppose first that L=PSLn(q). If n=3 and q5, then α(g)4. The bound ϑ3+(q)4(q2+q) only holds for q=5,7,13. Next, assume that n=4 and q4. In this case, α(g)6; the bound ϑ4+(q)6(q3+q2+q) only holds for q=4,5,7,9,11,13. Now, assume that n5 and (n,q)(6,2),(6,3). In this case, the inequality ϑn+(q)nqn-qq-1 only holds for (n,q)=(5,2).

If L=PSL3(3), then |g|{2,3,4,8,13} and dimϕ16. If |g|3, then α(g)3, whereas if |g|4, then α(g)=2. Thus, we readily get a contradiction for |g|13. Likewise, for |g|=13, using the GAP package, we see that ϕ(g) is not almost cyclic.

If L{PSL4(3),PSL5(2),PSL6(2),PSL6(3)}, we use the following data to get a contradiction with the bound (5.1).

  1. L=PSL4(3): |g|{2,3,4,5,8,9,13} and dimϕ26. If |g|5, then α(g)=2; if |g|4 then α(g)6.

  2. L=PSL5(2): |g|{2,3,4,5,7,8,16,31} and dimϕ75. If |g|5, then α(g)=2; if |g|4 then α(g)5.

  3. L=PSL6(2): |g|{2,3,4,5,7,8,9,16,31} and dimϕ61. If |g|5, then α(g)=2, whereas if |g|4, then α(g)6.

  4. L=PSL6(3): |g|{2,3,4,5,7,8,9,11,13,16,121} and dimϕ362. If |g|9 then α(g)=2; if |g|8 then α(g)6.

Next, suppose that L=PSUn(q). If n=3 and q5, the bound

ϑ3-(q)3(μ(PSU3(q))-1)

only holds for q=5,8,11,17. If n=4 and q4, the bound

ϑ4-(q)6(μ(PSU4(q))-1)

only allows the possibilities q=4,5,7,9,11. If n5 and (n,q)(6,2), the bound

ϑn-(q)n(μ(PSUn(q))-1)

only holds when (n,q){(5,2),(5,3),(7,2)}.

If L=PSU6(2), then |g|{2,3,4,5,7,8,9,11,16}, dimϕ21. If |g|5, then α(g)=2, while if |g|4, then α(g)6. Thus, if |g|16, by Lemma 2.4, ϕ(g) is not almost cyclic. If |g|=16 (gL), we need to examine the representations ϕ such that 21dimϕ30. Using the GAP package, we find that ϕ(g) is never almost cyclic.

Finally, suppose that L=PSp2n(q). If n=2 and q5, then the bound

σ2(q)5q3-q+1q-1

only holds for q=5,7,9,11. If n3, the bound

σn(q)2nqn+1-q+1q-1

only holds for (n,q){(3,3),(4,3)}.

If L=PSp4(5), then |g|{2,3,4,5,8,13} and dimϕ40. If |g|=8,13, then α(g)=2, whereas if |g|5, then α(g)5. In both cases, Lemma 2.4 yields a contradiction.

In view of the above, we are left to examine the groups L listed in Table 1 below. But again, in each case, in view of the data shown in table, we conclude by Lemma 2.4 that ϕ(g) cannot be almost cyclic. ∎

Table 1

Bounds for |g|, α(g) and dimϕ for some simple groups.

L|g|α(g)dimϕL|g|α(g)dimϕ
PSL3(5)31396PSL3(7)19496
PSL3(13)614672PSL4(4)176189
PSL4(5)316248PSL4(7)2561026
PSL4(9)4162912PSL4(11)6166650
PSL4(13)61613176PSU3(5)8328
PSU3(8)193105PSU3(11)373260
PSU3(17)323912PSU4(4)176220
PSU4(5)136272PSU4(7)4361074
PSU4(9)6462992PSU4(11)6466770
PSU5(3)615324PSU7(2)437320
PSp4(7)255126PSp4(9)415288
PSp4(11)615550PSp6(3)13678
PSp8(3)418780

5.2 Case (B)

We now consider case (B): gAd and all the constituents of the restriction of ϕ to L are Weil representations of L. We start by recording some arithmetic inequalities to be used when applying Lemma 2.4.

Lemma 5.2.

Let r be a prime, q=ra.

  1. Let n5. Then qn-1q-1-2>n(q-1). Furthermore, q3-1q-1-2>4(q-1) and q4-1q-1-2>7(q-1).

  2. Let n3 and q odd. Then qn-12>2n(q-1). Furthermore, q2-12>6(q-1) unless 3q11.

  3. Let n5 be odd. Then qn-qq+1>n(q-1). Furthermore, q3-qq+1>4(q-1) unless q=2,3,4.

  4. Let n6 be even. Then qn-1q+1>n(q-1). Furthermore, q4-1q+1>6(q-1) unless q=2.

Proof.

Elementary straightforward computations. ∎

Lemma 5.3.

Let L be one of the simple groups PSLn(q), n>2; PSUn(q), n>2; PSp2n(q), n>1, q odd. Let G=g,LAutL, where g is a p-element. Let ϕ be a cross-characteristic irreducible projective F-representation of G non-trivial on L, and suppose that p is coprime to |L|. Then ϕ(g) is not almost cyclic.

Proof.

Observe that if p is coprime to |L|, then p>3, and hence |g| divides |Φ|. This implies 3<|g|<q-1. Now, by Lemma 2.8, the minimum degree of a non-trivial irreducible projective representation of L is not less than the number at the left-hand side of items (1) to (4) of Lemma 5.2 unless (n,q){(3,2),(3,4),(4,2),(4,3)} in case (1), and (n,q){(4,2),(4,3)} in case (4). However, all these exceptions occur for q=r, that is, |Φ|=1, or q=4, and therefore are irrelevant here. Similarly, all the exceptions listed in cases (2), (3) and (4) of Lemma 5.2 are irrelevant here. Furthermore, by Lemmas 2.2 and 2.3, α(g)n if L is not symplectic, and α(g)2n in the symplectic case, with some exceptions described there. Again, as 2<|g|<q-1, all those exceptions are ruled out. By Lemma 5.2 and Lemma 2.4, it follows that ϕ(g) is not almost cyclic. ∎

We now suppose that p is a prime dividing |L| and distinguish two cases depending on the divisibility of q by p.

5.2.1 Case (p,q)=1

Lemma 5.4.

Let L=PSLn(q) with n>2, and let p>2 be such that p divides |L| and (p,q)=1. Let 1gAutL be a p-element, and let G=g,L. Suppose that gAd. Then n|g|<qn-1q-1-2, and g is not almost cyclic in any cross-characteristic projective irreducible representation of G that is non-trivial on L.

Proof.

Let H=GLn(q). As p>2, we have g(HΦ)/Z(H)AutL. By Lemmas 3.2 and 3.5,

|g|ηp(H)=pt|qe-1|pne(qe-1),

where pten<pt+1e. Observe that, since p is odd, the assumption gAd implies that |Φ|p=pm>1. This in turn implies q8, and hence qn-1q-1-2 is the minimum degree of a non-trivial cross-characteristic projective irreducible representation of L (Lemma 2.8). Notice that if e<n2, then n2e(qe-1)<qn-1q-1-2. Therefore, the statement is proven for n>2e.

Next, suppose that n2e, and observe that the assumption n>2 implies e>1. Since pten2e, p>2, it follows that pt=1. Let q1 be such that q1pm=qe. By Lemma 3.1, |qe-1|p=pm|q1-1|p, whence |g|ηp(H)pm(q1-1). We need to show that npm(q1-1)<qn-1q-1-2, and for this, it is enough to show that npm(q1-1)<qn-1+qn-2-2. Since en and e<p (note that p is a Zsigmondy prime for the pair (q,e)), it suffices to show that

2pm+1(q1-1)<qe-1+qe-2-2=q1pm-1+q1pm-2-2.

This is true except when m=1 and either p=3 or (p,q1){(5,2),(7,2)}. If p=5,7 and q1=2, then qe=25,27, whence q=2 and e=p, a contradiction as e<p. If p=3, then 2e<p yields e=2. Thus, we are left to consider the cases q2=q13 and n=3,4. Notice that 12(q1-1)<q13+5. It follows that, for n=3, the inequality npm(q1-1)<qn-1q-1-2 reduces to 9(q1-1)<q2+q-1 and, in fact, 9(q1-1)<12(q1-1)<q13+5q2+q-1 as q8. For n=4, the same inequality reduces to 12(q1-1)<q3+q2+q-1, which is true since 12(q1-1)<q13+5<q3+q2+q-1. This ends the proof. ∎

Lemma 5.5.

Let L=PSLn(q) with n>2, q odd. Let gAutL be a 2-element such that g21, and let G=g,L. Then g is not almost cyclic in any cross-characteristic projective irreducible representation of G that is non-trivial on L.

Proof.

Let H=GLn(q), and let Γ be the group generated by the inverse-transpose automorphism of H. Clearly, |Γ|=2. Then (HΦΓ)/Z(H)AutL, and therefore it suffices to prove the statement for HΦΓ in place of AutL. Obviously, η2(HΦΓ)2η2(HΦ), and by Lemma 3.6, η2(HΦ)=η2(H). Let m0 such that q=q02m, where q0 is not a square. By Corollary 3.3 and Lemma 3.7, we get η2(H)=2mη2(GLn(q0))2m+t(q0+1), where t is such that 2tn<2t+1 (notice that t1 as n>2). Hence, we have

η2(HΦΓ)2m+t+1(q0+1).

We now show that if n5 and (n,q)(5,3), then

n2m+t+1(q0+1)<qn-1q-1-2.

Under these restrictions, the statement of the lemma follows by Proposition 2.2 and Lemmas 2.4 and 2.8. Clearly, n2m+t+12m+2t+2 and

qn-1q-1-2qn-1+qn-2-1q2t-1+q2t-2-1,

so it suffices to prove that 2m+2t+2(q0+1)<q02m+t-2m+q02m+t-2m+1-1. Direct computation shows that this is true except when (m,t)=(0,2) and q07. So, suppose that m=0, t=2; we have to prove that 8n(q0+1)<q0n-1q0-1-2 for n=5,6,7. This is obviously true except when (n,q0)=(5,3).

Now, suppose that n=3. In this case, the statement will be proven if we show that 2m+4(q0+1)<q02m+1+q02m-1. Computation shows that this is true except when either m=0 and q013 or (m,q0)=(1,3).

Next, suppose that n=4. For q3, the statement will be proven once we show that

(5.2)62m+3(q0+1)<q032m+q02m+1+q02m-1.

Now, if m3, then 62m+3<332m-1 and 62m+3<32m+1. Direct computation for m2 shows that (5.2) holds except for m=0 and q0=3,5.

We finally deal with the exceptions arisen above. Let ϕ be a cross-characteristic projective irreducible representation of G non-trivial on L. In the cases listed in the following table, we easily get a contradiction by Lemma 2.4.

Lη2(L)dimϕLη2(L)dimϕ
PSL3(5)829PSL3(9)1689
PSL3(11)8131PSL3(13)8181
PSL4(5)8154PSL5(3)16119

Thus, we are left with the cases where L=PSL3(3), PSL3(7) and PSL4(3).

If L=PSL3(3), then η2(L)=8 and dimϕ11. Using GAP, we see that if |g|=4, then α(g)=2. By Lemma 2.4, we conclude that ϕ(g) is not almost cyclic. If |g|=8, we are left to consider the case 11dimϕ14. Again using GAP, we see that ϕ(g) is not almost cyclic.

If L=PSL3(7), then η2(L)=16 and dimϕ55. Using GAP, we see that, for any 2-element g such that g21, we have α(g)=2, and hence again, by Lemma 2.4, ϕ(g) is not almost cyclic.

Finally, if L=PSL4(3), then we have η2(L)=8 and dimϕ26. If |g|=8, then α(g)=2; if |g|=4, then α(g)3. In both cases, by Lemma 2.4, ϕ(g) is not almost cyclic. This ends the proof. ∎

Let L=PSUn(q), n>2, and let p be a prime such that (p,q)=1 and p divides |L|. Let Φ=Gal(𝔽q/𝔽r), and set pm=|Φ|p. Thus, pm<rpmq. Let gAutL be a p-element, and set G=L,g. Recall that, by Propositions 2.2 and 2.3, L,g is generated by n conjugates of g. Finally, set Φ2=Gal(𝔽q2/𝔽r) so that AutL=AdΦ2.

Lemma 5.6.

Let p>2, m>0 and e2(mod4). Let 1gAutL be a p-element. Set G=L,g. Then n|g|<qn-qq+1 if n is odd, n|g|<qn-1q+1 if n is even, and ϕ(g) is not almost cyclic in any cross-characteristic irreducible projective representation ϕ of G non-trivial on L.

Proof.

Let H=GUn(q). As e=ep(q)2, p is coprime to q+1=|Z(H)|. It follows that |Ad/L| is coprime to p. So, (HΦ2)/Z(H)AutL contains a Sylow p-subgroup of AutL.

It is known (see e.g. [16, Lemma 3.5]) that H contains a subgroup isomorphic to H1:=GLn/2(q2), where n2 is the integral part of n2, and |H:GLn/2(q)| is coprime to p. Moreover, H1 can be chosen to be invariant under Φ2, and therefore H1Φ2 contains a Sylow p-subgroup of HΦ2. So, by Lemma 3.5, |g|ηp(H1). Set l=n2 (that is, l=n2 if n is even, l=n-12 if n is odd). As p>2, we have |Φ|p=|Φ2|p>1; in particular, qrp8.

Note that |q2e-1|p=|qe-1|p if e is odd, and ep(q2)=ep(q)2 if e is even. So, in both cases, |(q2)ep(q2)-1|p=|qe-1|p. Therefore, by Lemma 3.2, we get |g|n2|qe-1|p, whence |g|n2|qe-1|pl(ql-1) (since el).

Suppose first that n is even. In order to prove the statement, by Propositions 2.2 and 2.3, Lemma 2.4 and Lemma 2.8, it will be enough to show that n|g|<qn-1q+1. For this purpose, it suffices to show that (2l)l(q+1)<ql+1. This is true if l>2 unless l=3, q4; l=4, q3; 5l8, q=2. However, all these exceptions are ruled out since qrp. Next, suppose that l=2 so that n=4 and e=1, that is, p(q-1). Then 8(q+1)<q2+1 unless q8. As qrp, this forces q=8. However, as p(q-1), this in turn implies p=7, a contradiction.

Suppose that n is odd. In this case, it will be enough to show that n|g|<qn-qq+1. For this purpose, it suffices to show that (2l+1)l(q+1)<ql+1 (since qn-1<qn-q). This is true if l>2 unless l=3, q5; l=4, q3; 5l9, q=2. However, all these exceptions are ruled out since qrp. If l=2, that is, n=5, we have 10(q2-1)<q5-qq+1 unless q3. If l=1, that is, n=3, we have |g|(q-1), and the inequality 3(q-1)<q(q-1) holds unless q3. As q8, we are done. ∎

Lemma 5.7.

Let L=PSUn(q) with n>2. Let 1gAutL be a p-element for some prime p>2 such that ep(q)2(mod4), and let G=g,L. Then g is not almost cyclic in any cross-characteristic projective irreducible representation of G that is non-trivial on L unless n>3 and |Φ|p=1 (and hence gAd).

Proof.

First of all, we observe that, by Lemmas 3.2 and 3.5,

|g|ηp(GLn(q2)Φ2)=ηp(GLn(q2))=pt|(q2)e-1|p,

where e=ep(q2) and pten<pt+1e. By the definition of e, e=2e and, since e is even, we obtain that |g|pt|qe-1|p=pt|qe/2+1|p.

Assume first that n=3. Then ep(q)=2, that is, p divides q+1. We have dimϕq2-q by Lemma 2.8 (3). Thus, the statement will be proven once we show that 3|g|q2-q (Proposition 2.3 (3)). Since p(q+1), we have e=2, whence |g|pt|q2-1|p=pt|q+1|ppt(q+1). If p>3, then q9; moreover, pt3<pt+1 implies pt=1. It follows that 3|g|3(q+1)q2-q, and we are done. If p=3, then pt3<pt+1 yields pt=3 and |g|3(q+1). Now, 3|g|9(q+1)q2-q if q11; thus, we are only left to examine the cases q=5,8. In both cases, we get a contradiction with Lemma 2.4.

Next, assume that n>3. In this case, we will show that n|g|qn-(-1)nq+1-1. Let us write q=q0pm, where pm=|Φ|p. Then, by Lemma 3.1 (since e<p),

|g|pt+m|q0e/2+1|ppt+m(q0e/2+1)npm(q0e/2+1)npm(q0p-12+1).

If m1 and (n,q)(4,8),(5,8), then n2pm(q0p-12+1)+1q0n/2pm. Hence,

(n|g|+1)(q+1)q0n/2pm(q0pm+1)q0npm-2,

whence n|g|qn-(-1)nq+1-1. If L{PSU4(8),PSU5(8)}, then p=3, e=2 and t=1, whence |g|3|8+1|3=27. Since

427454=84-18+1-1and5273640=85+18+1-1,

we are done. ∎

Lemma 5.8.

Let L=PSUn(q) with n>2 and q odd. Let gAutL, gAd, be a 2-element such that g21, and set G=g,L. Suppose (n,q){(3,3),(4,3)}. Then g is not almost cyclic in any cross-characteristic projective irreducible representation of G that is non-trivial on L.

Proof.

Set H=GUn(q), H1=GLn(q2) and q=q02m, where 2m=|Φ|2. We claim that α(g)η2(G)<dimϕ for every cross-characteristic projective irreducible representation ϕ of G non-trivial on L, whence the statement by Lemma 2.4. Clearly, we may replace G with HΦ2. By Lemmas 3.6 and 3.7,

η2(H1Φ2)=η2(H1)=2t|q2-1|2,where 2tn<2t+1.

Since HH1, it follows that η2(H)2t|q2-1|22n(q+1).

Suppose that n5. Then, by Proposition 2.2, α(g)n, and by Lemma 2.8, dimϕqn-qq+1. Now, if either n7 or n=5,6 and q5, we have

2n2(q+1)<qn-qq+1,

and we are done. If (n,q)=(6,3), then η2(H1)=16 and dimϕ182, and we are done. If (n,q)=(5,3), then, again using the GAP package, η2(G)=16, and moreover, α(g)5 if |g|=4, α(g)3 if |g|=8 and α(g)=2 if |g|=16. Since dimϕ60, we are done.

Next, suppose that n=4. Then dimϕq3-q2+q-1 and, by Proposition 2.3, α(g)6. Since 68(q+1)<q3-q2+q-1 if q9, we are left to consider the cases q=5,7. For q=5, we find that η2(G)=8, whence the claim since dimϕ104. For q=7, we find that η2(G)32 and dimϕ300, whence the claim.

Finally, suppose that n=3. Then

dimϕq2-q,α(g)3andη2(H1)4(q+1).

Since 34(q+1)<q2-q for q17, we are left to consider the cases q=5,7,9,11,13. For q=5, we find that η2(G)=8, dimϕ20. Moreover, if |g|=4, then α(g)3; if |g|=8, then α(g)=2. Thus, in both cases, ϕ(g) is not almost cyclic.

For q=7, we find that η2(G)=16 and dimϕ42. If |g|=4 or 8, then α(g)3; if |g|=16, then α(g)=2. Thus, we are done with the case q=7. Next, for q=9, we have η2(G)=16 and dimϕ72; for q=11, η2(H1)=16 and dimϕ110; for q=13, η2(H1)=28=16 and dimϕ156. It follows that, also in these cases, we are done by Lemma 2.4. This ends the proof. ∎

Let L=PSp2n(q) and H=Sp2n(q), q odd and n>1. Let pm=|𝔽q:𝔽r|p. Recall that CSp2n(q) denotes the conformal symplectic group. It is well known that Ad=CSp2n(q)/Z, where Z is the group of scalar matrices in GL2n(q). Furthermore, |Ad:L|=2 as q is odd.

Lemma 5.9.

Let L be as above, with p odd. Let 1gAutL be a p-element, and set G=g,L. Let ϕ be a cross-characteristic irreducible projective representation of G such that dimϕ>1. Suppose that gL. Then ϕ(g) is not almost cyclic.

Proof.

By Lemma 3.5, ηp(AutL)ηp(H). As gL, the assumption p>2 implies q8, and hence, as q is odd, q27. By Lemma 2.4, it suffices to show that 2n|g|<qn-12dimϕ.

Suppose first that e is odd. Then H contains a subgroup H1GLn(q), and |H|p=|H1|p (again, see [35]). Thus, by the above, |g|ηp(H1). The latter, by Lemma 3.2, equals pt|qe-1|p, where pten<pt+1e. If n>2, it follows from Lemma 5.4 that n|g|<qn-1q-1-2. Since 2(qn-1q-1-2)<qn-12 for q7, we are done. So, let n=2. Then e=1 and |g|ηp(H1)=|q-1|p. Thus, we have 4|g|4|q-1|p, and one easily checks that 4|g|<q2-12 for q>7.

Next, let e be even. Then it is known that Sp2n(q) contains a Sylow p-subgroup of GL2n(q) (again, see [35]). Write 2n=ae+b, where 0b<e, and as usual, let q=q0pm, pm=|Φ|p. It is easy to observe that |GL2n(q)|p=|GLae(q)|p. Therefore, we have |GLae(q)|p=|Spae(q)|p, whence |Sp2n(q)|p=|Spae(q)|p. Thus, we get |g|ηp(GLae(q)).

Suppose first that 2n=ae. Observe that e=ep(q)=ep(q0pm)=ep(q0) by Lemma 3.1 (1), and hence p divides q0e-1. It then follows by Lemmas 3.1 (2) and 3.2 that |g|ηp(GLae(q))=pt+m|q0e-1|p, where pta<pt+1. We wish to show that ae|g|2npt+m|q0e-1|p4n2epm(q0e-1)<qae/2-12. To this purpose, it suffices to show that

4n2pm(q02n-1)<q0npm-12.

Now, 8n2pmq02n<q0npm holds for any p3, any n2 and any q03 unless p=3, m=1 and one of the following occurs:

  1. q0=3 and n6,

  2. q0=5 and n3,

  3. q0=7 and n=2,

  4. q0=9 and n=2.

As pq, cases (i) and (iv) are ruled out. On the other hand, in all the other cases, we still have 2n|g|qn-12. Namely, in case (ii), we have

2n|g|=4|g|4η3(GL4(53))=36<1252-12forn=2,
6|g|6η3(GL6(53))=162<1253-12forn=3.

In case (iii), we have 2n|g|=4|g|4η3(GL4(73))=108<3432-12.

Finally, suppose 2n>ae. Then

2n|g|=2naeae|g|2nae(qae/2-1)2<qn-12for everyn2.

Lemma 5.10.

Let G=Sp2n(q) and H=CSp2n(q), where q is odd.

  1. G has two inequivalent irreducible 𝔽-representations of degree qn-12; they do not extend to H.

  2. Let char𝔽2. Then G has two inequivalent irreducible 𝔽-representations of degree qn+12. They do not extend to H.

  3. If ϕ is an irreducible 𝔽-representation of H with dimϕ>1, then we have dimϕqn-1.

Proof.

(1) and (2) It is well known that, over the complex numbers, G has two inequivalent irreducible representations of degree qn-12 and two inequivalent irreducible representations of degree qn+12 (see [8]). Moreover, it follows from [8, Proposition 1.4] that the representations of equal degrees can be obtained from each other by twisting by an involution hH, and hence none of them extends to H. The reductions modulo an odd prime distinct from of these representations remain irreducible and inequivalent (see [34]). As for the representations of degree qn-12, the reduction modulo 2 remains irreducible, whereas the reduction modulo 2 of the representations of degree qn+12 has a composition factor of degree qn-12. This implies that the irreducible 𝔽-representations of G do not extend to H. Moreover, it is clear that H has an irreducible 𝔽-representation of degree qn-1, whose restriction to G splits into two inequivalent irreducible representations of G of degree qn-12.

(3) By [12, Theorem 2.1], it is known that if τ is an irreducible 𝔽-representation of G of degree greater that qn+12, then dimτ(qn-1)(qn-q)2(q+1) . It follows from Clifford’s theorem that the same is true for ϕ, provided dimϕ>qn+1.

Now, suppose dimϕqn+1. Again by [12, Theorem 2.1], it follows from Clifford’s theorem that the composition factors of ϕ|G are of degree qn±12. On the other hand, by the above, dimϕ=qn±1. Taking into account that if char𝔽=2, there are no irreducible representation of G of degree qn+12, item (3) follows. ∎

Lemma 5.11.

Let L=PSp2n(q), where n2 and q is odd. Let 1gAutL be a 2-element such that g21, and let G=g,L. Suppose that (2n,q)(4,3). Then g is not almost cyclic in any cross-characteristic projective irreducible representation ϕ of G that is non-trivial on L.

Proof.

Let G1=CSp2n(q)Φ. Then we have G1GL2n(q)Φ. By Lemmas 3.6 and 3.7,

η2(AutL)η2(G1)η2(GL2n(q)Φ)=η2(GL2n(q))2t+1(q+1),

where 2tn<2t+1. For n>2, by Lemma 2.4, we get that ϕ(g) is not almost cyclic once we show that 2n|g|<qn-12. Now,

2n|g|2n2t+1(q+1)4n2(q+1)and4n2(q+1)<qn-12

for all n3 and q odd unless (n,q){(3,3),(3,5),(3,7),(4,3),(4,5),(5,3),(6,3)}. For L as listed in Table 1, by Lemma 2.4, we conclude that ϕ(g) is not almost cyclic.

If L=PSp6(3), then |g|8 and dimϕ13. If |g|=4, then α(g)3, and hence, by Lemma 2.4, ϕ(g) is not almost cyclic. If |g|=8, then α(g)=2; hence we only need to examine the representations ϕ such that 13dimϕ14. Using the GAP package, we see that ϕ(g) is not almost cyclic.

If L=PSp12(3), we have |g|η2(GL12(3))=32, dimϕ364, α(g)12. Thus, if |g|=4,8,16, we are done by Lemma 2.4. If |g|=32, then gL (since η2(PSp12(3))=16). So, we can refine the bound using Lemma 5.10, and again, we are done by Lemma 2.4.

Suppose now that n=2 and q5. Then it is enough to show that 5|g|q2-12 (Proposition 2.3 (6)). By the above, 5|g|522(q+1) and 20(q+1)<q2-12 for all q>41. Assume q41. Then we may refine the previous bound, obtaining that 5|g|5η2(GL4(q))q2-12 unless q=5,7,9,11,17,31. In these exceptional cases, we get the following.

L|g|α(g)dimϕL|g|α(g)dimϕ
PSp4(5)=4312PSp4(9)=16=240
PSp4(5)=8=212PSp4(11)8360
PSp4(7)8324PSp4(17)323144
PSp4(7)=16=224PSp4(31)645480
PSp4(9)8340

Hence, by Lemma 2.4, we are only left to consider the following instances:

  1. L=PSp4(5) and |g|=8. Then gL, whence dimϕ24. Thus, ϕ(g) is not almost cyclic by Lemma 2.4.

  2. L=PSp4(7) and |g|=16. In this case, gL, whence dimϕ48 (again by Lemma 5.10), which implies, by Lemma 2.4, that ϕ(g) is not almost cyclic.∎

5.2.2 Case pq

In this subsection, we examine the case where p divides q. Although, for p odd, we might refer to the earlier paper [5], the approach chosen in the present paper (for all p) appears to be more efficient than in [5] (and may lead to a shorter proof of the results in [5]).

First, let H=GLn(q), and let Φ be the group of field automorphisms of H. Suppose that t,m are such that pt<npt+1 and |Φ|p=pm. Write q=q0pm. Observe that ηp(H)=pt+1. Indeed, pt+1 is exactly the order of a unipotent element of GLn(q) consisting of a single Jordan block of size n. Furthermore, note that ηp(H)=pt+1 also for H=SLn(q),GUn(q) and SUn(q).

Next, we need the following arithmetic lemma.

Lemma 5.12.

Let p be a prime, and let q be a p-power. Let n be an integer greater than 4, and suppose that t, m and q0 are defined as above. If p=2, then n2t+m+2qn-qq+1 unless q=2 and 5n9. If p is odd, then npt+m+1qn-qq+1. Furthermore, n2t+m+2qn-1q-1-2 unless q=2 and n=5,6.

Proof.

Since pt<npt+1, we have npt+m+2p2t+m+3 and

qn-qq+1qn-2=q0pm(n-2)ppm(pt-1).

Hence, it suffices to prove that p2t+m+3ppm(pt-1), which is equivalent to prove that 2t+m+3pm(pt-1). Notice that 2t+m+32m(2t-1) holds unless one of the following cases occurs: (i) t=0; (ii) t=1 and m=0,1,2; (iii) t=2 and m=0,1; (iv) t=3 and m=0. Direct computations in each case yield the statement. ∎

Lemma 5.13.

Let L{PSLn(q),PSUn(q)}, n5. Let gAutL be a p-element such that g21, and let G=g,L. Then ϕ(g) is almost cyclic for some cross-characteristic projective irreducible representation ϕ of G that is non-trivial on L if and only if G=PSU5(2).2, dimϕ=10 and |g|=16.

Proof.

Recall (see Lemma 2.8) that dimϕqn-1q-1-2 if L=PSLn(q), whereas dimϕqn-qq+1 if L=PSUn(q) and n is odd, and dimϕqn-1q+1 if L=PSUn(q) and n is even. Also, notice that

qn-qq+1qn-1q+1qn-1q-1-2.

If p3, then ηp(G)pt+m+1, where t and m are defined as above. By Lemma 5.12, we get n|g|npt+m+1dimϕ. Thus, ϕ(g) is not almost cyclic by Lemma 2.4. If p=2, then η2(G)2t+12t+m+2. Likewise, we may apply Lemma 5.12 obtaining that ϕ(g) is not almost cyclic unless L is one of the following groups: PSL5(2), PSU5(2), PSL6(2), PSU6(2), PSU7(2), PSU8(2) and PSU9(2). Thus, we need to examine these groups in detail. Moreover, in order to bound efficiently α(g), we make use of the package GAP whenever necessary.

(1) L=PSL5(2). Here |g|16 and dimϕ29. If |g|=4, then α(g)3; if |g|=8, then α(g)=2. Thus, if |g|8, by Lemma 2.4, ϕ(g) is not almost cyclic. If |g|=16, then α(g)=2. This means that we have to examine the representations ϕ such that 29dimϕ30. Using the GAP package, again we see that ϕ(g) is not almost cyclic.

(2) L=PSU5(2). Here |g|16 and dimϕ10. If |g|=4, then α(g)3. Thus, ϕ(g) is not almost cyclic by Lemma 2.4. If |g|=8, then α(g)=2, and hence we need to examine the representations ϕ such that 10dimϕ14. If |g|=16, then α(g)=2, and we need to examine the representations ϕ such that 10dimϕ30. Using the GAP package, we see that ϕ(g) is almost cyclic if and only if G=PSU5(2).2, dimϕ=10 and |g|=16, as claimed.

(3) L=PSL6(2). In this case, |g|16, α(g)3 and dimϕ61. Thus, by Lemma 2.4, we conclude that ϕ(g) is not almost cyclic.

(4) L=PSU6(2). In this case, |g|16 and dimϕ21. If |g|=8 or 16, then α(g)=2, whereas if |g|=4, then α(g)3. Thus, if |g|=4 or 8, by Lemma 2.4, ϕ(g) is not almost cyclic. If |g|=16 (gL), we need to examine the representations ϕ such that 21dimϕ30. Using the GAP package, we find that ϕ(g) is never almost cyclic.

(5) L=PSU7(2). Here |g|16 and dimϕ42. If |g|=4, then α(g)4, whereas if |g|=8 or 16, then α(g)=2. In both cases, by Lemma 2.4, ϕ(g) is not almost cyclic.

(6) L=PSU8(2). Here |g|16, α(g)4 and dimϕ85. By Lemma 2.4, ϕ(g) is not almost cyclic.

(7) L=PSU9(2). In this case, |g|32 and dimϕ170. If |g|16, then ϕ(g) is not almost cyclic by Lemma 2.4. If |g|=32, then we may consider g16. The latter belongs to L, and L is generated by 9 conjugates of g16. By Lemma 2.4, dimϕ9d, where d equals the dimension of an eigenspace of ϕ(g16). Suppose that ϕ(g) is almost cyclic. Then it readily follows that either 1 or -1 appears as an eigenvalue of ϕ(g16) with multiplicity 16. Thus, we are led to conclude that dimϕ9d916<170, a contradiction. ∎

Lemma 5.14.

Let L{PSL3(q),PSL4(q)} with LPSL3(2),PSL3(4),PSL4(2). Let gAutL be a p-element such that g21, and let G=g,L. Then g is not almost cyclic in any cross-characteristic projective irreducible representation ϕ of G that is non-trivial on L.

Proof.

Let L=PSL3(q). If q=3, then |g|=3, α(g)3, dimϕ11, whence 113(3-1)=6. Thus, ϕ(g) cannot be almost cyclic. Now, assume that q5.

Let t and m be defined as above and suppose that p>2. Then n=3 implies t=0, whence ηp(G)pm+1. Therefore, 3ηp(G)3pm+1pm+2, whilst

dimϕq2+q-1q2p2pm.

Thus, it is enough to show that m+223m2pm. This clearly holds for all m0. If p=2, then η2(G)2m+3. Hence (see Proposition 2.3), we need to show that 4η2(G)2m+522m+1q2+q-1. Now, m+52m+1 provided m>1. For m=0,1, we get the inequality 42m+364q2+q-1, which holds for any even q>4.

Next, let L=PSL4(q). If q=3, then |g|9 and dimϕ26. If |g|=3, then α(g)4, whereas if |g|=9, then α(g)=2. In both cases, ϕ(g) cannot be almost cyclic by Lemma 2.4. Now, assume that q5.

If p>3, then ηp(G)pm+1. Hence,

4ηp(G)4pm+1pm+2p3pmq3+q2+q-1dimϕ,

and we are done. If p=3, then η3(G)3m+2, and hence

4η3(G)43m+23m+433m+1dimϕfor allm1.

If m=0 (and q>3), we still have 4η3(G)36q3dimϕ. Finally, let p=2. Then η2(G)2m+3. By Proposition 2.3, α(g)6, whence

6η2(G)2m+6232mdimϕfor allm2.

Furthermore, if m=0,1 (and q>4), then 6η2(G)9683dimϕ.

Finally, let L=PSL4(4). In this case, |g|=4,8, α(g)3 and dimϕ83, and hence ϕ(g) cannot be almost cyclic by Lemma 2.4. ∎

Lemma 5.15.

Let L{PSU3(q),q>4;PSU4(q),q>3}. Let gAutL be a p-element such that g21, and let G=g,L. Then g is not almost cyclic in any cross-characteristic projective irreducible representation ϕ of G that is non-trivial on L.

Proof.

Let L=PSU3(q). If p>2, then ηp(G)pm+1. Since 3ηp(G)3pm+1 and dimϕq2-q3q3ppm, it suffices to show that m+1pm, which clearly holds for all m0. If p=2, then η2(G)2m+3. Hence, it will be enough to show that 4η2(G)2m+521+2m2qq2-q. Now, m+42m provided m3. If 0m2, then 4η2(G)128q2-q for all q16. If q=8, then 4η2(G)=32dimϕ. Therefore, we are done.

Now, let L=PSU4(q). If p>3, then ηp(G)pm+1. Whence

4ηp(G)4pm+14p2pm4q2q3-q2+q-1dimϕ

(note that m+12pm for all m0). If p=3, then η3(G)3m+2. Hence,

4η3(G)43m+24323mdimϕfor allm0.

If p=2, then η2(G)2m+3. It follows that

6η2(G)62m+3622m+16q2dimϕfor allm1.

If m=0 (and q>4), then 6η2(G)48q3-q2+q-1.

Finally, let L=PSU4(4). In this case, η2(G)=16, α(g)=2 and dimϕ51; hence ϕ(g) cannot be almost cyclic by Lemma 2.4. ∎

Lemma 5.16.

Let L=PSp2n(q), where n2 and q is odd. Let 1gAutL be a p-element, and let G=g,L. Let ϕ be a cross-characteristic projective irreducible representation of G that is non-trivial on L. Then ϕ(g) is almost cyclic if and only if L=PSp4(3) and the following occurs:

  1. 2,3: g=3a,3b and dimϕ=4; g=3c and dimϕ=6; g=3d and dimϕ=4,5; |g|=9 and dimϕ=4,5,6;

  2. =2: g=3a,3b and dimϕ=4; g=3c and dimϕ=6; g=3d and dimϕ=4; |g|=9 and dim=4,6.

Proof.

First, note that ηp(G)pt+m+1, where pt<2npt+1. Thus, we aim to show that

2nηp(G)2npt+m+1qn-12,that is,4npt+m+1qn-1.

Now, 4npt+m+12p2t+m+2p2t+m+3-1 and qn-1pnpm-1. Therefore, it will suffice to consider the inequality 2t+m+3npm.

If gL, we may refer to [5]. In particular, by [5, Theorem 1.1 and Lemma 4.2], ϕ(g) is almost cyclic if and only if L=PSp4(3) and (1) or (2) holds. So, suppose that gL. This implies in particular that we may assume m>0.

If t1, then 2t+m+3pt+m-1npm unless t=m=1 and p=3,5. For p=3, we get 3<2n9 and 2nη3(G)54n33n-12. For p=5, we get 5<2n25 and 2nη5(G)250n55n-12. Thus, we are done.

Finally, suppose that t=0, i.e., 2np. In this case, we have to consider the inequality

2nηp(G)pm+2pnpm-12.

Since 2pm+2pm+3-1pnpm-1 is equivalent to m+3npm, we only need to observe that m+323m holds for all m1, thus proving the statement. ∎

6 Some low-dimensional classical groups

In this section, we deal with the low-dimensional simple groups L which were left out in either or both of the two previous sections. These groups, which need a separate treatment, are the following: PSL3(2), PSL3(4), PSL4(2), PSU3(3), PSU4(2)PSp4(3), PSU4(3). Observe that, with the exception of PSU3(3), they all have an exceptional Schur multiplier (by this, we mean that the order of the multiplier is divisible by the defining characteristic of the group in question). As always, G=g,L, where gAutL. When needed, we use the notation of the GAP package to denote G and the conjugacy class of g. Also, in view of Lemma 2.10, we may assume g21.

6.1 L=PSL3(2)PSL2(7)

Here |g|{3,4,7,8} and α(g)=2. Using the GAP package, we find that ϕ(g) is almost cyclic if and only if one of the following occurs:

  1. |g|=3 and either 7 and dimϕ=3,4, or =7 and dimϕ=2,3,4,

  2. |g|=4 and either 7 and dimϕ=3,4, or =7 and dimϕ=2,3,4,5,

  3. |g|=7 and either 3,7 and dimϕ=3,4,6,7,8, or =3 and dimϕ=3,4,6,7, or =7 and dimϕ=2,3,,7,

  4. |g|=8 and either 7 and dimϕ=6,7,8, or =7 and dimϕ=2,3,,7.

6.2 L=PSL3(4)

In this case, |g|{3,4,5,7,8}. If gL and has order 3, then α(g)3; otherwise α(g)=2.

Suppose first that g is a 2-element. Then, using the GAP and MAGMA packages, we find that ϕ(g) is almost cyclic if and only if one of the following occurs:

  1. 3,7: |g|=8 and either G=L.21, dimϕ=6, or G=L.23, dimϕ=8,

  2. =3: |g|=4, G=L, dimϕ=4; |g|=8 and either G=L.21, dimϕ=6, or G=L.22, dimϕ=4,6, or G=L.23, dimϕ=6,8,

  3. =7: |g|=8 and either G=L.21, dimϕ=6, or G=L.23, dimϕ=8, or g8a, dimϕ=10.

Next, suppose that |g|=3. Then, again using the GAP and MAGMA packages, we find that ϕ(g) is almost cyclic if and only if

  1. =3, gL and dimϕ=4.

Finally, suppose that |g|=5,7. Here, in principle, we have to examine representations of dimension 4, 6, 8 and 10, where 4-dimensional representations only occur for =3. Furthermore, the Sylow 5-subgroups as well as the Sylow 7-subgroups are cyclic; hence, when =5 and =7, respectively, we can refer to [6, Lemma 2.13 and Corollary 2.14]. As a result, we get that ϕ(g) is almost cyclic if and only if one of the following occurs.

  1. |g|=5 and either =3 and dimϕ=4, or dimϕ=6,

  2. |g|=7 and either =3 and dimϕ=4, or dimϕ=6,8.

6.3 L=PSL4(2)

In this case, |g|{3,4,5,7,8} and either dimϕ{7,8,13,14} or dimϕ16. If |g|=3, then gL and α(g)4, and hence we only have to examine the representations ϕ of dimension 7 and 8. If |g|=4 and gL, then α(g)3, and hence we only have to examine the representations ϕ such that 7dimϕ9. In all the other cases, α(g)=2. Using the GAP package, we find that ϕ(g) is almost cyclic if and only if one of the following occurs:

  1. dimϕ=7 and either g3a,5, or |g|=4 and gL,

  2. dimϕ=7,8 and |g|=7,8.

6.4 L=PSU3(3)

Using the GAP package, we find that α(g)=2 for |g|=7,8, and α(g)3 for |g|=3,4. Suppose first that p3. The following holds.

If G=L, then ϕ(g) is almost cyclic if and only if

  1. |g|=7,8, dimϕ=6,7 and 2,3,

  2. |g|=7,8, dimϕ=6 and =2.

If G=L.2, then p=2 and ϕ(g) is almost cyclic if and only if

  1. |g|=8, dimϕ=6,7 and 2,3,

  2. |g|=8, dimϕ=6 and =2.

Next, assume that p=3. In this case, |g|=3, α(g)3 and dimϕ6. So, we only need to examine the representations ϕ of degree 6. Note that gL. Again, using the GAP package, we obtain that ϕ(g) is not almost cyclic.

6.5 L=PSU3(4)

This group was left out by assumption in Lemma 5.15. In this case, η2(G)=16, α(g)=2 and dimϕ12. If |g|=4, then, by Lemma 2.4, ϕ(g) cannot be almost cyclic. If |g|=8, we only need to examine the representations ϕ such that 12dimϕ14. If |g|=16, we only need to examine the representations ϕ such that 12dimϕ30. Using the GAP package, we find that ϕ(g) is almost cyclic if and only if |g|=16 and dimϕ=12.

6.6 L=PSU4(2)PSp4(3)

In this case, |g|{3,4,5,8,9}. If |g|=3, then gL and α(g)4. If |g|=4 and gL, then α(g)=2 (in which case, we only have to examine the representations ϕ such that 4dimϕ6). If |g|=4 and gL, then α(g)3 (in which case, we only have to examine the representations ϕ such that 4dimϕ9). If |g|=8, then gL and α(g)=2 (hence, we only have to examine the representations ϕ such that 4dimϕ14). Finally, if |g|=5,9, then α(g)=2. Using the GAP package, we find that ϕ(g) is almost cyclic if and only if one of the following cases occurs:

  1. G=L and

    1. 3, g3a,3b,3d or |g|=4,5,9 and dimϕ=4,

    2. =3, g3a,3b,4a or |g|=5,9 and dimϕ=4,

    3. 2,3, g3d,4b or |g|=5,9 and dimϕ=5,

    4. =3, g3c,3d,4b or |g|=5,9 and dimϕ=5,

    5. 3, g3c or |g|=5,9 and dimϕ=6.

  2. G=L.2 and

    1. =3, g4d or |g|=8 and dimϕ=4,

    2. =3, |g|=4,8 and dimϕ=5,

    3. 3, |g|=4,8 and dimϕ=6.

Clearly, if we view L as PSU4(2), we must add the extra condition 2 to items (a), (e) and (h), whereas if we view L as PSp4(3), we must ignore items (b), (d), (f) and (g). Also observe that, in item (c), the assumption 2 is due to the fact that PSp4(3) does not have 2-modular irreducible representations of degree 5.

6.7 L=PSU4(3)

In this case, dimϕ6. Observing that if gL, then g is a 2-element, and using the GAP package, we get the following evidences for the values of α(g) and the almost cyclicity condition on ϕ(g):

|g|α(g)dimϕ|g|α(g)dimϕ
3gL44(3-1)=87gL=22(7-1)=12
4gL=22(4-1)=68gL=22(8-1)=14
4gL44(4-1)=128gL=22(8-1)=14
5gL=22(5-1)=89gL=22(9-1)=16

So, we only have to examine the representations ϕ such that 6dimϕ16. It turns out that if either =0 or (p,)=1, then ϕ(g) is almost cyclic if and only if dimϕ=6 and

  1. G=L, g3b or |g|=5,7,8,9,

  2. G=L.22, g4d or |g|=8 and 2,3.

Next, suppose that =p. Then, for =2,5,7, we are left to examine the following possibilities:

  1. =2, G{L,L.22}, |g|=4,8 and dimϕ=6,

  2. =2, G=L.21, |g|=4,8 and dimϕ=12,

  3. =5, G=L, |g|=5 and dimϕ=6,

  4. =7, G=L, |g|=7 and dimϕ=6.

Considering (1), we see that if |g|=8, then, since dimϕ=6, ϕ(g) must have a single non-trivial block of size 5. Thus, ϕ(g) is almost cyclic. So, assume that |g|=4. Assume first that gL. There are two classes of elements of order 4 in L, namely, 4a,4b. If g4a, w.l.o.g., we may assume that g lies in a maximal subgroup M of L isomorphic to PSU3(3). We find that ϕ restricted to M remains irreducible and ϕ(g) is the sum of two Jordan blocks of size 3. So, ϕ(g) is not almost cyclic. If g4b, w.l.o.g., we may assume that g lies in a maximal subgroup N of L isomorphic to A7. We find that ϕ restricted to N remains irreducible and ϕ(g) is the sum of two Jordan blocks of size 4 and 2. So, again, ϕ(g) is not almost cyclic.

Next, assume that gL and L,g=L.22. There are two classes of such elements, namely, 4c,4d. If g4c, w.l.o.g., we may assume that g lies in a maximal subgroup P of L isomorphic to PSU3(3).2. We find that ϕ restricted to P remains irreducible and ϕ(g) is the sum of two Jordan blocks of size 3. So, ϕ(g) is not almost cyclic. If g4d, w.l.o.g., we may assume that g lies in a maximal subgroup Q of L isomorphic to S7. We find that ϕ restricted to Q remains irreducible and ϕ(g) is the sum of two Jordan blocks of size 1 and one of size 4. So, ϕ(g) is almost cyclic.

The instances in (2) can be ruled out, as a 2-modular 12-dimensional representation of G when restricted to L splits into two 6-dimensional components intertwined by g. Now, ϕ(g) is almost cyclic if and only if g2 is cyclic on each component, but this cannot be by [4, Proposition 2.14].

Finally, observe that the Sylow 5-subgroups as well as the Sylow 7-subgroups are cyclic. It follows that, in instances (3) and (4), by [6, Lemma 2.13 and Corollary 2.14], ϕ(g) is almost cyclic (in fact, cyclic in instance (4)).

We conclude that, for any value of , ϕ(g) is almost cyclic if and only if dimϕ=6 and one of the following occurs:

  1. G=L and either g3b or |g|=5,7,8,9,

  2. G=L.22 and either g4d or |g|=8.

Remark.

Note that, in the following cases, ϕ|L is an irreducible Weil representation of L, and g is a semisimple element. Thus, they have been already described in [6].

  1. G=PSL3(2), dimϕ=5, =7 and |g|=7,

  2. G=PSL3(2), dimϕ=7 and |g|=7,

  3. G=PSU3(3), dimϕ=6 and |g|=7,8,

  4. G=PSU3(3), dimϕ=7, 2 and |g|=7,8,

  5. G=PSU4(2), dimϕ=5 and g3d or |g|=5,9,

  6. G=PSU4(2), dimϕ=5, =3 and g3c,

  7. G=PSU4(2), dimϕ=6, 3 and g3c or |g|=5,9,

  8. G=PSp4(3), dimϕ=4 and |g|=4,5,

  9. G=PSp4(3), dimϕ=5, 2 and g4b or |g|=5.

7 The groups Sp2n(q), n>1, q even; Ω2n+1(q), n>2, q odd; PΩ2n±(q), n>3

As shown in [11, Theorem 2.16], if L is one of the simple groups Sp2n(q) (n2 and q even), Ω2n+1(q) (n3 and q odd) or PΩ2n±(q) (n4), then |g|qn+1q-1 for all gAutL.

Lemma 7.1.

Let L=Sp2n(q), where n2, q is even and (n,q)(4,2). Let gAutL, where g21, and let G=L,gAutL. Let ϕ be a cross-characteristic projective irreducible representation ϕ of G that is non-trivial on L. Then ϕ(g) is almost cyclic if and only if one of the following occurs.

  1. G=Sp4(4), |g|=17 and dimϕ=18.

  2. G=Sp6(2) and

    1. dimϕ=7 and either g3a,4c, or |g|=5,7,8,9,

    2. dimϕ=8 and |g|=7,8,9.

Proof.

First, recall that

dimϕ(qn-1)(qn-q)2(q+1)andα(g)2n+1for anygAutL.

Suppose that ϕ(g) is almost cyclic. By Lemma 2.4, this implies that

(qn-1)(qn-q)2(q+1)(2n+1)(|g|-1)(2n+1)qn+1-q+1q-1.

The above inequality holds if and only if (2n,q){(4,4),(4,8),(6,2),(6,4),(8,2),(10,2),(12,2),(14,2)}. Thus, almost cyclicity can only occur for these values of the pair (2n,q). However, it is easy to check that ϕ(g) cannot be almost cyclic unless (2n,q)=(4,4),(6,2) by Lemma 2.4.

If L=Sp4(4), then dimϕ18 and |g|{3,4,5,8,17}. If |g|=4,8,17, then α(g)=2; if |g|=3,5, then α(g)3. If |g|<17, then 18>α(g)|g|, and hence ϕ(g) is not almost cyclic. If |g|=17, then gL, and using the GAP package, we see that ϕ(g) is almost cyclic if and only if dimϕ=18, whence item (1) of the statement.

If L=Sp6(2), then |g|{3,4,5,7,8,9} and dimϕ7. If |g|=3, then α(g)4; if |g|=4, then α(g)3; finally, if |g|5, then α(g)=2. Thus, ϕ(g) is not almost cyclic unless, possibly, when dimϕ16. Using the GAP package, we see that ϕ(g) is almost cyclic if and only if one of the instances listed in item (2) of the statement occurs. ∎

Remark 7.2.

The group Sp4(2) is not simple. However, the commutator subgroup L=Sp2(2) is simple and isomorphic to PSL2(9). Using the package GAP and assuming that g is not involution, we find that ϕ(g) is almost cyclic if and only if G=L and the following holds:

  1. dimϕ=3,4,5, 2,3 and |g|=3,4,5,

  2. dimϕ=6, 2 and |g|=5,

  3. dimϕ=3,4, =2 and |g|=3,4,5,

  4. dimϕ=2,3,4, =3 and |g|=3,4,5.

Lemma 7.3.

Let L=Ω2n+1(q), where n3 and q is odd, or L=PΩ2n-(q), where n4. Let gAutL, where g21, and let G=L,gAutL. Then ϕ(g) is not almost cyclic in any cross-characteristic projective irreducible representation ϕ of G that is non-trivial on L.

Proof.

Suppose that L=Ω2n+1(q), and assume that ϕ(g) is almost cyclic. By Lemma 2.8, if q=3 and n4, then dimϕ(3n-1)(3n-3)8, whence the bound

(3n-1)(3n-3)8(2n+1)3n+1-22,

which holds if and only if n=4. If q>3, then, again by Lemma 2.8, we have dimϕq2n-1q2-1-2, whence the bound

q2n-2q2+1q2-1(2n+1)qn+1-q+1q-1,

which holds if and only if n=3 and q=5,7. By the above, we are left to examine the cases where (n,q){(3,3),(3,5),(3,7),(4,3)}. In each of these cases, it is routine to check that ϕ(g) is not almost cyclic by Lemma 2.4.

Now, suppose that L=PΩ2n-(q), and assume that ϕ(g) is almost cyclic. By Lemma 2.8 (8),

dimϕ(qn+1)(qn-1-q)q2-1-1,

provided (n,q)(4,2),(4,4),(5,2),(5,3). In the first case, the following bound must be met:

(qn+1)(qn-1-q)-(q2-1)q2-12nqn+1-q+1q-1.

However, this only holds if either n=4 and q8, or n=6,7 and q=2. Thus, we are left to examine the possible exceptional cases (n,q)=(4,2),(4,3),(4,4),(4,5),(4,7),(5,2),(5,3),(6,2),(7,2). In each individual case, from the available data, we get dimϕ>α(g)(|g|-1), a contradiction. ∎

Lemma 7.4.

Let L=PΩ2n+(q), where n4. Let gAutL, where g21, and set G=L,gAutL. Let ϕ be a cross-characteristic projective irreducible representation of G non-trivial on L. Then ϕ(g) is almost cyclic if and only if L=Ω8+(2), dimϕ=8 and one of the following occurs:

  1. G=L and either g3a,5a,8b or |g|=7,9,

  2. G=SO8+(2) and either g4f or |g|=8.

Proof.

Suppose that ϕ(g) is almost cyclic. If q3 and (n,q)(4,2), then

dimϕ(qn-1)(qn-1-1)q2-1.

It follows that the bound

(qn-1)(qn-1-1)q2-12nqn+1-q+1q-1

must be met. This happens if and only if either n5 or q=2 and n=6,7. If q>3, then dimϕ(qn-1)(qn-1+q)q2-1-2, and so the bound to be met is

(qn-1)(qn-1+q)-2(q2-1)q2-12nqn+1-q+1q-1.

This holds if and only if n=4 and q=4,5,7,8, and in each case, knowledge of α(g) and |g| yields a contradiction by Lemma 2.4.

Finally, let us deal with the case when L=Ω8+(2). Here |g|{3,4,5,7,8,9} and dimϕ8 (see Lemma 2.8). (Indeed, L does have a (unique) projective representation of degree 8 for any characteristic , and this is not an ordinary representation of L).

Suppose that gSO8+(2). If |g|=3, then α(g)4; if |g|=4, then α(g)3; if |g|5, then α(g)=2. Since either dimϕ=8 or dimϕ28, applying the usual bound, we are only left to consider the case when dimϕ=8.

Let G=L. If p, then, using the GAP package, we find that ϕ(g) is almost cyclic if and only if g is one of elements described in the statement.

Now, suppose that =p=7. Since the Sylow 7-subgroups are cyclic, for =7, we can refer to [6, Lemma 2.13 and Corollary 2.14], and we find that ϕ(g) is almost cyclic.

Next, suppose that =p=5. We observe that L=Ω8+(2) contains three classes of elements of order 5 and three classes of maximal subgroups isomorphic to Sp6(2). Each of these symplectic groups contains elements from a single conjugacy class of elements gL of order 5, and each element of order 5 of L is contained in a symplectic group. (Note that the same holds for elements of order 9, considering the three classes 9a, 9b, 9c of elements of order 9 and the three classes of maximal subgroups isomorphic to A9.)

Now, considering the group 2.L, we see that the elements of the class 5a belong to a maximal subgroup M of type 2×Sp6(2) (a single class). The restriction of ϕ to M decomposes as the sum of a representation of degree 1 and a representation ψ of degree 7. Since ψ(g) is almost cyclic (see Lemma 7.1), it follows that ϕ(g) is almost cyclic.

On the contrary, the elements of the classes 5b and 5c belong to a maximal subgroup N of type 2.Sp6(2) (non-split central extensions: there are two classes of such subgroups). The representation ϕ of degree 8 of 2.L restricts irreducibly to N, and working in N, we see that ϕ(g) is not almost cyclic.

Finally, suppose that =p=3. The group L has five classes 3a, 3b, 3c, 3d, 3e of elements of order 3, and three classes 9a, 9b, 9c of elements of order 9. Each of the classes 3a, 3b, 3c and similarly each of the classes 9a, 9b, 9c is contained in a unique maximal subgroup isomorphic to Sp6(2), and conversely each of these symplectic groups contains elements from a single conjugacy class of elements of order 3 and 9 of L, respectively.

On the other hand, if g belongs to one of the classes 3d, 3e, then α(g)=3. This yields a contradiction, by Lemma 2.4.

Next, considering the group 2.L, we see that the elements of classes 3b, 3c, 9b, 9c belong to a maximal subgroup M of type 2.Sp6(2) (there are two classes of such subgroups). The representation ϕ of degree 8 of 2.L restricts irreducibly to M, and working in M, we obtain that ϕ(g) is almost cyclic if and only if g has order 9.

On the other hand, the elements g of the classes 3a,9a belong to a maximal subgroup N of type 2×Sp6(2) (a single class). The restriction of ϕ to such subgroups decomposes as the sum of a representation of degree 1 and a representation ψ of degree 7. Since ψ(g) is almost cyclic, ϕ(g) is also almost cyclic.

Next, let G=L,g=SO8+(2). Then p=2, and using the GAP package, one can check that ϕ(g) is almost cyclic if and only if either |g|4f or |g|=8, as claimed.

Finally, suppose that gSO8+(2), and hence |g|=3,9. Then α(g)8; moreover, either dimϕ50 or dimϕ105 (see [1, 19]). If 3, then use of the GAP package shows that ϕ(g) is not almost cyclic. If =3, we need to consider representations of degree 28 or 48. Applying the usual bound, we can easily see that ϕ(g) is not almost cyclic if |g|=3. Now, let gL be an element of order 9 belonging to a class C. Then the power maps in [1] show that g3 belongs to the class 3d of L, and therefore every element x3d is the cube of some yC. We find that we can choose three elements in 3d, say x1,x2,x3, which generate L. Let g1,g2,g3 be three elements of C such that xi=gi3 for i=1,2,3. Thus, the group H=g1,g2,g3 contains g13,g23,g33=L, whence H=L.3.

The above implies that α(g)3 for all gC. Thus, by Lemma 2.4, ϕ(g) cannot be almost cyclic. ∎

8 The exceptional groups of Lie type

In this section, we deal with the groups AutL, where L is a finite simple exceptional group of Lie type. The notation for the groups L is standard. Note, however, that our notation for q in the case of twisted groups is such that E62(q)E6(q2), D43(q)D4(q3), B22(q)B2(q), F42(q)F4(q) and G22(q)G2(q). We also recall that the structure of AutL can be found in [22, Table 5.1.B]. For a thorough and detailed reference, however, see [9, Section 2.5 (Theorem 2.5.12)].

The maximal order of an element in AutL can be easily deduced from [11, 20]; these data are summarised in Table 2. Lower bounds for the dimension of a non-trivial cross-characteristic projective representation of L are found in [18, Table 1, p. 14]. Finally, we recall that, for any non-trivial element gAutL, the minimal number α(g) of L-conjugates of g sufficient to generate g,L is provided by Proposition 2.2 (2).

Table 2

Maximal order of an element in AutL.

LConditions|g|
B22(22e+1)e1(2e+1)(22e+1+2e+1+1)
G2(q)q=re,r3e(q2+q+1)
G2(q)q=3e2e(q2+q+1)
G22(32e+1)e1(2e+1)(32e+1+3e+1+1)
D43(q)q=re3e(q3-1)(q+1)
F4(q)q(q3-1)(q+1)
F42(22e+1)e1(2e+1)(24e+2+23e+2+22e+1+2e+1+1)
E6(q)q2(q3+1)(q2+q+1)
E62(q)q(q+1)(q2+1)(q3-1)
E7(q)q(q+1)(q2+1)(q4+1)
E8(q)q(q+1)(q2+q+1)(q5-1)

Lemma 8.1.

Let L be a simple exceptional group of Lie type. Let 1gAutL be a p-element, and let G=g,L. Then g is almost cyclic in some cross-characteristic projective irreducible representation ϕ of G that is non-trivial on L if and only if one if one of the cases listed in Table 3 below occurs.

Proof.

Suppose that ϕ(g) is almost cyclic.

If L=B22(22e+1), where e2, then α(g)5. By Lemma 2.4, Table 2 and [18, Table 1, p. 14], the bound

2e(22e+1-1)5((2e+1)(22e+1+2e+1+1)-1)

must be met. However, the bound only holds for e6. So, we may assume 2e6. If gL, then |g|22e+1+2e+1+1, and hence we can refine the bound to 2e(22e+1-1)5(22e+1+2e+1). However, this only holds for e=2. This means that, for e=3,4,5,6, it suffices to examine gL. But, for these values of e, |g|2e+1, which yields 2e(22e+1-1)>10e, a contradiction.

If L=B22(23), then |g|{2,3,4,5,7,13} and dimϕ8. If |g|=2, then α(g)=3, whereas if |g|>2, then α(g)=2. We readily get a contradiction if |g|=2,3,4. So, we may assume |g|5, which implies gL. Inspection of the Brauer characters shows that we are left to consider the following cases: (i) |g|=5,7,13, =5 and dimϕ=8; (ii) |g|=13, 2 and dimϕ=14; (iii) |g|=13, =13 and dimϕ=16,24. Using the GAP package, we obtain item (1) of the statement. If L=B22(25), then |g|{2,4,5,25,31,41} and dimϕ124. Since α(g)3, we get a contradiction.

Next, suppose that L=G2(q). If q5, then a detailed check shows that we get a contradiction by Lemma 2.4.

Next, let L=G2(3). Then |g|{2,3,4,7,8,9,13} and either dimϕ=14 or dimϕ27. If |g|3, then α(g)4; otherwise α(g)=2. Thus, we readily get a contradiction unless |g|=8,9,13 and dimϕ=14. Using the GAP package, we find that ϕ(g) is almost cyclic only for |g|=13.

Finally, let L=G2(4). Then |g|{2,3,4,5,7,8,13,16} and either dimϕ=12 or dimϕ64. If |g|4, then α(g)=2, whereas α(g)4 if |g|3. We readily get a contradiction unless dimϕ=12 and |g|7. Using the GAP package, we find that ϕ(g) is almost cyclic if and only if |g|=13,16.

Now, suppose that L=D43(q), where q=re. Ruling out this group requires a rather delicate analysis, which we need to develop in detail. Here α(g)7, and we have to consider the bound r3e(r2e-1)7(3e(r4e+r3e-re-1)-1), which only holds when q has one of the following values: q=2e, where e7; q=3e, where e4; q=5e, where e2; q=7,11,13,17,19.

Suppose first that g is semisimple. From the description of the maximal tori of L given in [21], we can extract the following table, which gives, for each value of q, an upper bound s(q) for the maximal order of a semisimple p-element of AutL:

qs(q)qs(q)qs(q)
2133734241
560171818243
964811111171328393
16673178323319769
2539000127530713321321
643873781648112814449

It is easy to verify that, for q7, we get a contradiction by Lemma 2.4.

Next, suppose that g is unipotent. Then ηp(L)=p if p3, while η2(L)=8 and η3(L)=9 (see [27]). Thus, we readily get a contradiction for q3.

Let L=D43(2). Then |g|{2,3,4,7,8,9,13} and dimϕ25. If |g|4, then α(g)=2, whereas if |g|=2,3, then α(g)4. In both cases, we get a contradiction.

At this stage, we are left to examine the cases q=3,4,5 and g semisimple.

Let L=D43(3). Then |g|{2,4,7,8,13,73}. Since dimϕ216, we get a contradiction unless |g|=73. In this case, using the GAP package, we find that α(g)=2, which yields a contradiction.

Let L=D43(4). Then |g|{3,5,7,9,13,27,241}. Since dimϕ960, we get a contradiction unless |g|=241. Finally, let L=D43(5). Then |g|{2,3,4,7,8,9,31,601}. Since dimϕ3000, we get a contradiction unless |g|=601.

Notice that, in the instances |g|=241 for L=D43(4), and |g|=601 for L=D43(5), the exceptional element g belongs to L and is a generator of the cyclic torus T5 (in the notation of [21]). Since g is regular semisimple, we have α(g)3 by Lemma 2.7, whence a contradiction.

Finally, we list in Table 3 the exact occurrences of almost cyclic elements for the simple group F42(2), the so-called Tits group, as well as for the simple groups G=G2(2) and G=G22(3). These results are based, as usual, on Lemma 2.4 and on direct computations with GAP. All the remaining exceptional groups are ruled out using Lemma 2.4. The relevant computations are left to the reader. ∎

Table 3

Occurrences of almost cyclic elements in exceptional groups of Lie type.

Gdimϕ|g|Gdimϕ|g|
B22(23)all1413B22(23)587,13
G2(2)all6,77,8G2(2)33all
G2(2)all6,78G2(3)all1413
G2(4)all1213G2(4).2all1216
G22(3)27,87,9G22(3)2,799
G22(3)22allG22(3)243,7,9
G22(3)287,9G22(3)27,89
G22(3)26,89

9 Proof of the main theorem

The following theorem answers with full details the question raised in the present paper, thus also proving the main theorem quoted in the introduction.

Theorem 9.1.

Let L be a finite simple group of Lie type, let gAutL be a p-element for some prime p, and let G=L,g. Let ϕ be an irreducible projective representation of G over an algebraically closed field F of characteristic , different from the defining characteristic of L. Suppose that ϕ is non-trivial on L; furthermore, assume that g21, LPSL2(q) and ϕ(g) is almost cyclic. Then either ϕ|L is an irreducible Weil representation of L, where L{PSLn(q)(n3),PSUn(q)(n3),PSp2n(q)(n2,q𝑜𝑑𝑑)} and g is semisimple, or L is of exceptional type and (dimϕ,,|g|) are given in Table 3, or one of the following occurs:

  1. L=PSL3(2), dimϕ=2, =7 and either |g|=3,4,7 or gL, |g|=8,

  2. L=PSL3(2), dimϕ=3,4 and either |g|=3,4,7 or gL, |g|=8,

  3. L=PSL3(2), dimϕ=5, =7 and either |g|=4 or gL and |g|=8,

  4. G=PSL3(2), dimϕ=6, =7 and |g|=7,

  5. G=PSL3(2).2, dimϕ=6,7 and |g|=8 with gL,

  6. L=PSL3(2), dimϕ=8 and either 3,7 and |g|=7, or 7, gL and |g|=8,

  7. G=PSL3(4), dimϕ=4, =3 and |g|=3,4,5,7,

  8. G=PSL3(4), dimϕ=6 and |g|=5,7,

  9. G=PSL3(4), dimϕ=8 and |g|=7,

  10. G=PSL3(4).21, dimϕ=6 and |g|=8,

  11. G=PSL3(4).22, dimϕ=4,6, =3 and |g|=8,

  12. G=PSL3(4).23, dimϕ=6, =3 and |g|=8,

  13. G=PSL3(4).23, dimϕ=8 and |g|=8,

  14. G=PSL3(4).23, dimϕ=10, =3 and g8a,

  15. G=PSL4(2), dimϕ=7 and either g3a or |g|=5,7,

  16. G=PSL4(2), dimϕ=8 and |g|=7,

  17. G=PSL4(2).2, dimϕ=7 and |g|=4,8,

  18. G=PSL4(2).2, dimϕ=8 and |g|=8,

  19. G=PSU3(3).2, dimϕ=6 and |g|=8,

  20. G=PSU3(3).2, dimϕ=7, 2 and |g|=8,

  21. G=PSU3(4), dimϕ=12 and |g|=16,

  22. G=PSU4(2), dimϕ=4, 3 and g3a,3b,3d or |g|=4,5,9,

  23. G=PSU4(2), dimϕ=4, =3 and g3a,3b,4a or |g|=5,9,

  24. G=PSU4(2), dimϕ=5 and g4b,

  25. G=PSU4(2).2, dimϕ=4, =3 and g4d or |g|=8,

  26. G=PSU4(2).2, dimϕ=5, =3 and |g|=4,8,

  27. G=PSU4(2).2, dimϕ=6, 3 and |g|=4,8,

  28. G=PSp4(3), dimϕ=4 and g3a,3b,3d or |g|=9,

  29. G=PSp4(3), dimϕ=5, 2 and g3d or |g|=9,

  30. G=PSp4(3), dimϕ=6 and g3c or |g|=5,9,

  31. G=PSp4(3).2, dimϕ=6 and |g|=4,8,

  32. G=PSU4(3), dimϕ=6 and either g3b or |g|=5,7,8,9,

  33. G=PSU4(3).22, dimϕ=6 and either g4d or |g|=8,

  34. G=PSU5(2).2, dimϕ=10 and |g|=16,

  35. G=Sp4(2), dimϕ=3,4,5 and 2,3,

  36. G=Sp4(2), dimϕ=6, 2 and |g|=5,

  37. G=Sp4(2), dimϕ=3,4 and =2,

  38. G=Sp4(2), dimϕ=2,3,4 and =3,

  39. G=PSp4(4), dimϕ=18 and |g|=17,

  40. G=PSp6(2), dimϕ=7 and either g3a,4c or |g|=5,7,8,9,

  41. G=PSp6(2), dimϕ=8 and |g|=7,8,9,

  42. G=PΩ8+(2), dimϕ=8 and either g3a,5a,8b or |g|=7,9,

  43. G=PSO8+(2), dimϕ=8 and either g4f or |g|=8.

Moreover, in each of the cases listed above, ϕ(g) is indeed almost cyclic.

Proof.

Suppose that L is a classical group, and assume furthermore that L is neither orthogonal nor symplectic of even characteristic. By Lemma 5.3, we may assume that p divides |L|.

Assume first that g is a semisimple element. Observe that if gAd, then ϕ(g) is not almost cyclic unless L{PSU3(3),PSU4(3),PSp4(3)} (see Lemmas 5.45.9 and 5.11). Now, suppose that gAd. Then (see the argument following Lemma 5.3) either ϕ|L has an irreducible constituent which is not a Weil representation, or ϕ|L is an irreducible Weil representation of L. It follows that ϕ(g) is not almost cyclic unless either p>2 and L{PSL3(2),PSL3(4),PSL4(2)} (see Lemmas 5.1 and 5.5), or L{PSU4(2),PSU4(3)} or L=PSp4(3) (see Lemmas 5.1 and 5.11).

Next, assume that g is unipotent, without any restriction to ϕ. Then ϕ(g) is not almost cyclic unless L{PSL3(2),PSL3(4),PSL4(2),PSp4(3),PSU5(2)} (see Lemmas 5.135.16). The above listed exceptions are analysed in Section 6. The case L=PSL3(2) is dealt with in Section 6.1, yielding items (1) to (6) of the statement. The case L=PSL3(4) in Section 6.2, yielding items (7) to (14) of the statement. The case L=PSL4(2) is dealt with in Section 6.3, yielding items (15) to (18) of the statement. The case L=PSU3(3) is dealt with in Section 6.4, yielding items (19) and (20). The case L=PSU3(4) is dealt with in Section 6.5, where it is found that ϕ(g) is never almost cyclic unless ϕ is a Weil representation of dimension 12 and |g|=16, thus yielding item (21). The case L=PSU4(2)PSp4(3) needs some extra care. If we view L as PSU4(2), the Weil representations of L have dimension 5 and 6. Thus (see Section 6.6), we get items (22) to (27). If we view L as PSp4(3), the Weil representations have dimension 4 and 5. This group is dealt with in Lemma 5.16 if g is unipotent (that is, a 3-element), and in Section 6.6 if g is semisimple, yielding items (28) to (31). The case L=PSU4(3) is dealt with in Section 6.7, yielding items (32) and (33). The case L=PSU5(2) is dealt with in Lemma 5.13, yielding item (34) of the statement.

Finally, assume that L is a classical group, either orthogonal or symplectic of even characteristic. These groups are dealt with in Section 7. If L is symplectic, then we get items (35) to (41) of the statement by Lemma 7.1 and Remark 7.2. Lemmas 7.3 and 7.4 deal with the orthogonal groups, yielding items (42) and (43) of the statement. ∎


Communicated by Christopher W. Parker


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Received: 2018-08-14
Revised: 2019-09-19
Published Online: 2019-10-16
Published in Print: 2020-03-01

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