1 Introduction
The spectrumω(G) of a finite group G is the set of orders of its elements, and groups with equal spectra are said to be isospectral.
To solve the problem of recognition by spectrum for a given finite group G is to describe (up to isomorphism) the finite groups that are isospectral to G. If a finite group G has
a non-trivial normal solvable subgroup, then there are infinitely many finite groups isospectral to G, cf. [17, Lemma 1]. This implies that a finite group G with finite set of isospectral groups
must satisfy P≤G≤AutP, where P is a product of nonabelian simple groups, and the easiest situation is when G is just a nonabelian simple group. The problem of recognition
of nonabelian simple groups by spectrum has been studied extensively in recent years, and as a working hypothesis of this investigation, it was conjectured that a finite group G isospectral to a “sufficiently large” simple group S must be an almost simple group with socle S, i.e. S≤G≤AutS. In 2015, this conjecture was proved with the following precise meaning of the term “sufficiently large”:
S is a linear or unitary group of dimension larger than 44, or S is a symplectic or orthogonal group of dimension larger than 60, or
S is one of the sporadic, alternating and exceptional groups of Lie type other than J2, A6, A10, and D43(2) (see [14] and the references therein).
Thus for a vast class of simple groups, the initial problem of recognition by spectrum was reduced to a more specific problem of describing (up to isomorphism) almost simple groups with socle S that are isospectral to S,
and this is the problem we address in this paper.
For brevity, we refer to a finite almost simple group with socle S that is isospectral to S as admissible for S. Clearly we are interested in non-trivial admissible groups, i.e. other than S itself.
It is not hard to check that there are no non-trivial admissible groups for the alternating groups. The information collected in [7] allows to verify that the sporadic groups do not possess non-trivial admissible
groups either. One of the first examples of non-trivial admissible groups was discovered by Mazurov [16]: he showed that the finite groups isospectral to PSL3(5) are exactly PSL3(5) and its extension by the graph automorphism. Later Zavarnitsine [23] provided an example demonstrating that the number of admissible groups can be arbitrarily large: PSL3(73k) has exactly k+1 admissible groups, including itself,
and these are precisely extensions by field automorphisms.
Admissible groups are described for all exceptional groups of Lie type (see [27] for references), PSL2(q) (see [2]), PSL3(q) and PSU3(q) (see [23, 24]), classical groups over fields of characteristic 2 (see [11, 13, 26]), and symplectic and odd-dimensional orthogonal groups over fields of odd characteristic (see [12]). It is worth noting that for all of these groups, there is α∈AutS such that S0=〈S,α〉 is admissible and any other admissible group is conjugate in AutS to a subgroup of S0; in other words, any admissible group is a cyclic extension of S and
up to isomorphism there is a unique maximal admissible group. Below we will see that not all simple groups satisfy the latter property.
The main result of this paper is a description of admissible groups for linear and unitary groups over fields of odd characteristic (Theorems 1, 2 and 3).
Also we explicitly describe spectra of some almost simple groups with linear or unitary socle (Lemmas 3.3, 4.6, and 4.7).
In the rest of this section, we introduce the notation used in the theorems and then state our results.
Throughout this paper, p is a prime, F is the algebraic closure of the field of order p, H=GLn(F), with matrices acting on row vectors by right multiplication, and τ is the inverse-transpose map g↦g-⊤ of H. If q is a power of p, then Fq denotes the subfield of F of order q, λq denotes a fixed primitive element of Fq and φq denotes the standard Frobenius endomorphism of H of level q, i.e. the endomorphism induced by raising matrix entries to the qth power. We identify GLn(q) with CH(φq) and GUn(q) with CH(φqτ).
We write GLn+(q) instead of GLn(q) and GLn-(q) instead of GUn(q) and use a similar agreement for PGLn(q), SLn(q), and PSLn(q). For ε∈{+,-}, we shorten ε1 to ε in arithmetic expressions.
As usual, by (a1,…,as) we denote the greatest common divisor of positive integers a1,…,as, and by [a1,…,as] we denote their least common multiple. If a and b are positive integers, then
π(a) denotes the set of prime divisors of a, (a)b denotes the largest divisor c of a such that π(c)⊆π(b) and (a)b′ denotes the number
a/(a)b.
Let L=PSLnε(q) (q=pm), and define d=(n,q-ε). We write δ=δ(εq) to denote the diagonal automorphism of L induced by diag(λ,1,…,1), where λ is a primitive (q-ε)th root of unity in Fq2. We denote by φ the field automorphism induced by φp. The automorphism induced by τ is denoted by the same letter. The image of α∈AutL in OutL is denoted by α¯.
When n⩾3, the inverse-transpose automorphism is outer and OutL has the following presentation (we omit overbars for convenience):
〈δ,φ,τ∣δd=φm=τ2=[φ,τ]=1,δφ=δp,δτ=δ-1〉if ε=+,
〈δ,φ,τ∣δd=τ2=[φ,τ]=1,φm=τ,δφ=δp,δτ=δ-1〉if ε=-.
Theorem 1 is concerned with the extension by the inverse-transpose automorphism τ.
A criterion of admissibility of this extension is not very short, so it seems reasonable to write it up separately.
Theorem 1.
Let L=PSLnε(q), where n⩾3, ε∈{+,-} and q is odd, and let G=L⋊〈τ〉. Then either ω(G)=ω(L) or one of the following holds:
n=pt-1+2 with t⩾1,
q≡-ε(mod4) and 4pt∈ω(G)∖ω(L),
n=2t+1 with t⩾1,
(n,q-ε)>1 and 2(q(n-1)/2-ε(n-1)/2) is an element of ω(G)∖ω(L),
n=pt-1+1 with t⩾1 and 2pt∈ω(G)∖ω(L),
n is even, (n)2⩽(q-ε)2,
q≡ε(mod4) and qn/2+εn/2 is an element of ω(G)∖ω(L),
n is even, (n)2′>3,
(n,q-ε)2′>1 and 2[q(n)2-1,qn/2-(n)2+εn/2-(n)2] is an element of ω(G)∖ω(L).
Theorems 2 and 3 describe admissible groups in terms of admissibility of τ. As we have mentioned, the admissible groups for PSL3(q) and PSU3(q) were determined by Zavarnitsine [23, 24],
so we do not consider these groups. However, we include them in the statements of the theorems for completeness. Observe that two almost simple groups with socle S are isomorphic if and only if their images in OutS are conjugate.
Thus to describe admissible groups up to isomorphism, it is sufficient to describe their images in OutS up to conjugacy. We refer to α∈AutS as admissible if 〈S,α〉 is admissible.
Theorem 2.
Let L=PSLn(q), where n⩾3, q=pm and p is an odd prime.
Let d=(n,q-1), b=((q-1)/d,m)d, η=δ(d)2′, ϕ=φm/(b)2, and
ψ={φm/(b)2′if n≠4 or 12∤q+1,φ(m)3′if n=4 or 12∣q+1.
Suppose that L<G≤AutL. Then ω(G)=ω(L) if and only if G/L is conjugate in
OutL to a subgroup of 〈α¯〉, where α is one of the elements specified in Table 1.
Table 1
Conditions on the automorphism α of L=PSLn(q) in Theorem 2.
| Conditions on L | α=γβ=βγ |
| γ | β |
| n=pt+1, | (b)2>2 or (n)2<(p-1)2 | 1 | φm/2τη |
| n≠2u+2 | (m)2=2, (n)2<(p+1)2 | 1 | φm/2η |
| n≠pt+1, | τ not admissible, |ψ|>1 | ψ | 1 |
| b odd | τ admissible | ψ | τ |
| n≠pt+1, b even, p≡ϵ(mod 4) | n=ps+2u+1 | ψ | ϕ, | ϕiτ | | |
| n≠ps+2u+1,(n)2⩾(p-ϵ)2 | ψ | ϕ, | ϕiτ, | ϕ2jτη | |
| n≠ps+2u+1, (n)2<(p-ϵ)2, ϵ=+ | ψ | ϕ, | ϕiτ, | ϕ2jτη, | ϕτη |
| n≠ps+2u+1, (n)2<(p-ϵ)2, ϵ=- | ψ | ϕ, | ϕiτ, | ϕ2jτη, | ϕη |
| t,u>0, s⩾0 | | 2i∣(b)2, | 4j∣(b)2 | |
As an example of applying Theorem 2, let us consider admissible groups for PSL4(25). For this group, we have p=5, d=4, b=2, ϕ=φ, n≠pt+1 and n=p0+21+1. Thus
up to isomorphism there are two non-trivial admissible groups, namely, the extensions by φ and by φτ. Clearly they are both maximal.
Theorem 3.
Let L=PSUn(q), where n⩾3, q=pm and p be an odd prime. Let d=(n,q+1), b=((q+1)/d,m)d and
ψ={φ2m/(b)2′if n≠4 or 12∤q-1,φ2(m)3′if n=4 and 12∣q-1.
Suppose that L<G≤AutL. Then ω(G)=ω(L) if and only if n-1 is not a power of p and G/L
is conjugate in OutL to a subgroup of 〈α¯〉, where
α=ψ if
τ
is not admissible and |ψ|>1,
α=ψτ if
τ
is admissible, (n)2>2 and n⩾16,
α=ψφ(m)2′ if
τ
is admissible and either (n)2⩽2 or n⩽12.
Returning now to the initial recognition problem, we state the following consequence of [21, Theorem 1] and the above results.
Corollary.
Let L=PSLnε(q), where n⩾45, ε∈{+,-} and q is odd. A finite group is isospectral to L if and only if
it is isomorphic to an almost simple group G with socle L and G/L=〈α¯〉, where α is an identity or is as specified in Theorems 2 and 3.
3 Extensions by field and graph-field automorphisms
In this section, we derive some formulas concerning orders of elements in extensions of PSLnε(q) by field or graph-field automorphisms. Following the lines of the proof of [24, Proposition 13], we will exploit a correspondence between σ-conjugacy classes of CK(σk) and conjugacy classes of CK(σ), where K is a connected linear algebraic group and σ is a Steinberg endomorphism of K, i.e. a surjective endomorphism with finitely many fixed points. Also we will use a slight modification of this correspondence inspired by [10, Theorem 2.1].
We begin with necessary notation and the Lang–Steinberg theorem. If K is a group and σ is an endomorphism of K, then we write Kσ to denote CK(σ).
Lemma 3.1 (Lang–Steinberg).
Let K be a connected linear algebraic group and let σ be a surjective endomorphism of K such that Kσ is finite. Then the map x↦x-σx from K to K is surjective.
Recall that F is the algebraic closure of the field of order p and φq is the endomorphism of GLn(F)
raising matrix entries to the qth power, where q is a power of p. An endomorphism σ of a linear algebraic group K is said to be a Frobenius endomorphism if there are an identification of K with a closed subgroup of GLn(F) and a positive integer k such that σk is induced by φq.
Clearly, if σ is a Frobenius endomorphism, then σ is a Steinberg endomorphism.
Lemma 3.2.
Let K be a connected linear algebraic group over F, let α be a Frobenius endomorphism of K and let
τ be an automorphism of K of order 2 that commutes with α. Let k be a positive integer, l∈{0,1} and let
β be the automorphism of Kαkτl induced by α.
Given g∈Kαkτl, choose z∈K such that g=z-αz and define ζ(g)=zτlz-αk. Then
ζ(g)∈Kα, (βg)k=z-1τlζ(g)z and the map
[βg]↦[τlζ(g)]
is a one-to-one correspondence between the Kαkτl-conjugacy classes in
the coset βKαkτl and the Kα-conjugacy classes in the coset τlKα.
Proof.
Observe that αk and αkτ are Frobenius endomorphisms of K.
Let g,z∈K, g=z-αz and h=zτlz-αk. Define f=τlh=zτiz-αk. Then
(3.1)gαkτl=g⇔τlz-αk+1zαkτl=z-αz⇔zατlz-αk+1=zτlz-αk⇔fα=f.
This if g∈Kαkτl, then f∈τlKα. Furthermore,
(βg)k=βkgαk-1…g=τlz-αkz=z-1fz.
Let g1=z1-αz1∈Kαkτl and g2=z2-αz2∈Kαkτl.
Suppose x-1αg1x=αg2 for some x∈Kαkτl. Then writing y=z1xz2-1, we have yα=y and
f1=z1(βg1)kz1-1=z1x(βg2)kx-1z1-1=yf2y-1.
Conversely, if f1=yf2y-1 with y∈Kα, then x=z1-1yz2 satisfies xαkτl=x and
x-1αg1x=z2-1y-1z1αz1-αz1z1-1yz2=z2-1y-1αyz2=z2-1αz2=αg2.
It follows that the map under consideration translates conjugacy classes to conjugacy classes and is injective. Furthermore, the conjugacy class [f]
does not depend on the choice of z in the equality g=z-αz.
Now let h∈Kα. By the Lang–Steinberg theorem, there is an element z∈K such that h-τl=zαkτlz-1, and therefore τlh=zτlz-αk. Then (3.1)
implies that g=z-αz lies in Kαkτl, with [βg] mapping to [τlh]. Thus the map is surjective, and the proof is complete.
∎
It should be noted that some special cases of Lemma 3.2 were proved in [8, Lemma 2.10] (l=0) and [10, Theorem 2.1] (l=1 and k=1).
We use Lemma 3.2 to establish the following result generalizing [24, Corollary 14].
Lemma 3.3.
Let L=PSLn(q) and U=PSUn(q), where n⩾3 and q=pm. Let k divide m, q=q0k and β=φm/k. Then for any i, we have the following:
ω(βδi(q)L)=k⋅ω(δi(q0)PSLn(q0)),
if k is even, then ω(βτδi(q)L)=k⋅ω(δi(-q0)PSUn(q0)),
if k is odd, then ω(βτδi(-q)U)=k⋅ω(δi(-q0)PSUn(q0)),
if k is odd, then ω(βτδi(q)L)=k⋅ω(τδi(q0)PSLn(q0)),
ω(βδi(-q)U)=k⋅ω(τδi(q0)PSLn(q0)).
In particular, ω(τδi(q)L)=ω(τδi(-q)U).
Proof.
Denote the center of H=GLn(F) by Z. In particular, we write |gZ| to denote the projective order of a matrix g∈H, that is, the order of g modulo scalars.
We apply Lemma 3.2 to H with α=φpm/k and l=0. Observe that Hαk=GLn(q) and Hα=GLn(q0). Let g∈GLn(q) and h=ζ(g) be the element of GLn(q0)
defined in Lemma 3.2.
Since (βg)k is conjugate to h in H, we see that
|βgZ|=k⋅|(βgZ)k|=k⋅|hZ|,
and also
deth=det(gβk-1…g)=(detg)q0k-1+…+1=(detg)(q-1)/(q0-1).
Let λ=λq and λ0=λq0. The condition gZ∈δiL is equivalent to
detg∈〈λi,λn〉=〈λ(i,n)〉.
Since λ(q-1)/(q0-1) is a primitive element of Fq0, it follows that
detg∈〈λ(i,n)〉⇔deth∈〈λ0(i,n)〉.
Thus (i) holds. Similarly, (ii) and (iii) follow from Lemma 3.2 with α=φpm/kτ and l=0.
Let k be odd and α=φpm/kτ. Then Hαkτ=GLn(q) and Hα=GUn(q0). Applying Lemma 3.2 with l=1, we construct from an element g∈GLn(q)
an element h∈GUn(q0) such that g=z-αz and h=zτz-αk for some z∈H.
As in the previous case, we deduce that |βτgZ|=k⋅|τhZ|. Furthermore, we have
detg=(detz)q0+1 and deth=(detz)q-1. Since (detg)q-1=1, it follows that (detz)(q0+1)(q-1)=1. Let λ=λq and λ0=λq0q0-1, and let
μ be a primitive (q0+1)(q-1)st root of unity in F. Then
detg∈〈λ(i,n)〉⇔detz∈〈μ(i,n)〉,
which is equivalent to deth∈〈λ0(i,n)〉.
Hence
(3.2)ω(βτδi(q)L)=k⋅ω(τδi(-q0)PSUn(q0)).
Similarly, taking α=φpm/k and l=1, we prove (v).
Applying (v) with k=1 and observing that in this case β acts on U in the same way as τ, we have
ω(τδi(q)L)=ω(τδi(-q)U).
Now this equality and (3.2) imply (iv).
∎
Note that for unitary groups, (iii) and (v) of Lemma 3.3 cover all possibilities for an element γ∈〈φ〉: if |γ|=2k, then 〈γ〉=〈φm/k〉;
while if |γ|=k with k odd, then 〈γ〉=〈φm/kτ〉 because (m+m/k,2m)=2m/k.
Thus Lemma 3.3 expresses the spectrum of an extension by a field or graph-field automorphism in terms of known spectra and the spectrum of the extension by graph automorphism, which we consider
in the next section.
4 Extension by graph automorphism
This section is largely concerned with matrices, so we need to define some of them. We denote by I the identity matrix whose size is clear from the context, and by Jk the k×k unipotent Jordan
block. As in the previous section, we will calculate the projective orders of matrices, that is, the orders modulo scalars.
Recall that 〈δ¯〉⋊〈τ¯〉 is a dihedral group of order 2(n,q-ε). It follows that for odd n, every τδi is conjugate to τ modulo L,
and hence
(4.1)ω(τL)=ω(∪i=1dτδiL)=ω(τPGLn(q)).
If n is even, then for any i, we have
(4.2)ω(τL)=ω(τδ2iL) and ω(τδL)=ω(τδ2i-1L).
As we saw in Lemma 3.3, the cosets τδi(q)PSLn(q) and τδi(-q)PSUn(q) have the same orders of elements, so it suffices to describe ω(τδi(q)PSLn(q)).
Since |gτ|=2|(gτ)2|=2|ggτ|, it follows that the elements of ω(τPGLn(q)) are exactly twice the projective orders of elements of
Γn(q)={ggτ∣g∈GLn(q)}.
If n is even, then ω(τL) and ω(τδL) are analogously related to the projective orders of elements of
Γn□(q)={ggτ∣g∈GLn(q),detg∈(Fq×)2}
and
Γn⊠(q)={ggτ∣g∈GLn(q),detg∉(Fq×)2},
respectively.
A comprehensive treatment of the equation h=ggτ for a given matrix h is provided by Fulman and Guralnick in [9], and we use the terminology
and some results of this paper. First of all, it is helpful to note that h=ggτ yields
(4.3)xhx-1=(xgx⊤)(xgx⊤)τ
for any x∈GLn(q),
and hence h lies in Γn(q), Γn□(q) or Γn⊠(q) if and only if the whole conjugacy class [h] lies in the corresponding set. Thus we
may work not with individual matrices but with conjugacy classes. Recall that the conjugacy classes of GLn(q) are parametrized by collections of partitions
{λϕ∣ϕ is a monic irreducible polynomial over Fq}
such that |λz|=0 and ∑ϕdeg(ϕ)|λϕ|=n.
In this parametrization, the collection of partitions {λϕ} corresponds to the class [h] such that the multiplicity of ϕk as an elementary divisor of h
is equal to the multiplicity of parts of size k in λϕ. We denote the partition
corresponding under this parametrization to a class [h] and a polynomial ϕ by λϕ(h).
A criterion for a matrix h to lie in Γn(q) was obtained by Wall in [22]. In the same paper, Wall described the conjugacy classes of finite symplectic and orthogonal groups over fields of odd characteristic, and it turns out that the matrices of Γn(q) are very similar to symplectic and orthogonal ones.
Lemma 4.1.
If q is odd, then h∈Γn(q) if and only if h satisfies the following:
all even parts of λz-1(h) have even multiplicity,
all odd parts of λz+1(h) have even multiplicity.
Proof.
See [22, Theorem 2.3.1].
∎
Lemma 4.2.
Let h∈GLn(q) and let q be odd. Then h is conjugate to an element of Spn(q) if and only if h satisfies the following:
all odd parts of λz-1(h) and λz+1(h) have even multiplicity.
Proof.
See [22, p. 36, case (B) (i)].
∎
Lemma 4.3.
Let h∈GLn(q) and let q be odd. Then h is conjugate to an element of GOnε(q) for some ε if and only if h satisfies the following:
all even parts of λz-1(h) and λz+1(h) have even multiplicity.
Suppose that n is even and h is a unipotent element satisfying (ii). If λz-1(h) has odd parts, then
h is conjugate to an element of GOn+(q) and also to an element of GOn-(q). If there are no odd parts, then h is not conjugate to an
element of GOn-(q).
Suppose that n is even, |λz-1(h)|=|λz+1(h)|=0 and h satisfies (i). Then h is conjugate to an element
of only one of the groups GOn+(q) and GOn-(q).
Proof.
See [22, p. 38, Case (C) (i,i’)].
∎
Next we establish a necessary and sufficient condition for h to lie in Γn□(q) or Γn⊠(q). Let h∈GLn(q), let f be the characteristic polynomial of h, and suppose that
f=(z-1)n1(z+1)n2f0 with (f0,z2-1)=0. Then h is conjugate to a block diagonal matrix
(4.4)diag(h1,h-1,h0)
with blocks of dimension n1, n2 and n-n1-n2, respectively, corresponding to this factorization of f.
We refer to the matrix in (4.4) as the normal form of h.
Let h∈Γn(q). We may replace h by its normal form diag(h1,h-1,h0) with blocks of dimensions n1, n2 and n-n1-n2, respectively, and denote by V1 the subspace of Fqn spanned by the first n1 rows. Suppose that h=ggτ. Then hg=gτg=hτ. It follows that V1g is hτ-invariant and the characteristic polynomial of hτ on V1g is equal to (z-1)n1, and hence
V1g=V1. The same is true for other blocks, and so g is also block diagonal with blocks g1, g-1 and g0 of the same dimensions as h1, h-1 and h0, respectively (see also [9, Lemma 8.2]). Since hi=gigiτ for i=1,-1,0 and detg=detg1detg-1detg0, it suffices to consider the following special cases: f=(z-1)n, f=(z+1)n, and (f,z2-1)=1.
Recall that for odd q, the group Ω2nε(q) is the kernel of the spinor norm
θ:SO2nε(q)→Fq×/(Fq×)2.
The definition of the spinor norm in [20, pp. 163–165] implies the following way to calculate it (see also [3, Proposition 1.6.11]).
Lemma 4.4.
Let q be odd, n even, h∈SOnε(q) and let B be the matrix of the invariant symmetric bilinear form of SOnε(q).
Suppose that det(I-h)≠0. Then θ(h)≡det((I-h)B)(mod(Fq×)2).
Lemma 4.5.
Let q be odd, n even, h∈Γn(q) and let f be the characteristic polynomial of h.
If f=(z+1)n, then h∉Γn⊠(q).
Let f=(z-1)n. If λz-1(h) has no odd parts, then h∉Γn⊠(q); otherwise, h∈Γn□(q)∩Γn⊠(q).
Let (f,z2-1)=1. Then h is conjugate to an element of SOnε(q) for some unambiguously defined
ε
, and h∈Γn□(q) if and only if
-h is conjugate to an element of Ωnε(q).
Proof.
Let h=ggτ. Possibilities for g in the first two cases are found in [9, Section 8], and we use this result for calculations.
For brevity, we write x≡y to denote that x≡y(mod(Fq×)2).
(i) The group GLn(q)⋊〈τ〉 can be embedded into GL2n(q), and we consider the Jordan decomposition of gτ in the latter group.
So gτ=ug1τ=g1τu, where u is unipotent and (g1τ)2=-I; i.e. g1 is a skew-symmetric matrix.
Since the determinant of a skew-symmetric matrix is a square and detu=1, we see that detg is also a square.
(ii) Similarly, gτ=ug1τ=g1τu, where u is unipotent and (g1τ)2=I; i.e. g1 is a symmetric matrix.
Furthermore, h=u2 and the equality ug1τ=g1τu is equivalent to the condition that u preserves the bilinear form defined by g1. By Lemma 4.3,
if λz-1(u)=λz-1(h) has odd parts, then u is conjugate both to an element of SOn+(q) and to an element of SOn-(q), and so
we can choose g1 with any determinant, square or non-square. If λ(u) has no odd parts, then u is conjugate to an element of SOn+(q) only, and since
the multiplicities of even parts are even, n is divisible by 4. In this case detg1≡detI.
(iii) The fact that h is conjugate to an element of SOnε(q), where ε is unambiguously defined, follows from Lemmas 4.1 and 4.3.
By Lemma 4.4, we have θ(h)≡det(I-h)detB, where B is the matrix of the symmetric bilinear form preserved by h.
Observing that θ(-I)≡detB, we have
θ(-h)≡det(I-h)=det(I-ggτ)=det(g⊤-g)detgτ≡detg,
where the final equivalence holds because the determinant of the skew-symmetric matrix g⊤-g is a square. The proof is complete.
∎
We are ready to find ω(τPGLn(q)) and ω(τPSLn(q)).
Lemma 4.6.
Let q and n⩾3 be odd. Then
ω(τPSLn(q))=ω(τPGLn(q))=2⋅ω(Spn-1(q)).
Proof.
The first equality was established in (4.1). To prove the second one, we need to show that the set of the projective orders of
elements of Γn(q) is equal to ω(Spn-1(q)).
Let h∈Γn(q). Since n is odd, Lemma 4.1 implies that 1 is an eigenvalue of h. It follows that the projective order of h is equal to the ordinary order.
We show first that |h|∈ω(Spn-1(q)). Let diag(h1,h-1,h0) be the normal form of h as in (4.4) and denote the dimension of h1 by k. Then k is odd.
By Lemmas 4.1 and 4.2, the unipotent matrix h1 is conjugate to an element of SOk(q) and the matrix diag(h-1,h0) is conjugate to an element of Spn-k(q).
If k=1, then there is nothing to prove. If k>1, then |h1|∈ωp(SOk(q))=ωp(Spk-1(q)), and hence |h|∈ω(Spk(q)×Spn-k(q))⊆ω(Spn-1(q)).
It follows from Lemmas 4.1 and 4.2 that a semisimple matrix lies in Γn(q) if and only if it is conjugate to a matrix of the form diag(1,h′) with h′∈Spn-1(q), and thus ωp′(Γn(q))=ωp′(Spn-1(q)).
Let a∈ω(Spn-1(q)) and |a|p=pt>1. If a=2pt and n-1=pt-1+1 (i.e. the condition from (v) of Lemma 2.2 holds), we define h=diag(1,-Jn-1). If a=pt or n-1>pt-1+1, there is a semisimple matrix hs∈Spl(q), where n-1=pt-1+1+l, such that a=pt|hs|, and we define h=diag(Jpt-1+2,hs). It is easy to see that h∈Γn(q) and |h|=a. The proof is complete.
∎
Lemma 4.7.
Let q be odd and n⩾4 be even. Then
ω(τPGLn(q))=2⋅ω(PSpn(q))
and
ωp′(τPSLn(q))=2⋅ωp′(PΩn+(q))∪2⋅ωp′(PΩn-(q)).
If n>4, then
ωp~(τPSLn(q))=2⋅ωp~(Ωn+1(q)).
The set ωp~(τPSL4(q)) consists of all multiples of p dividing p(q±1) if p>3, and it consists of these multiples together with 18 if p=3.
Proof.
It suffices to find the sets of the projective orders of elements of Γn(q) and Γn□(q). For brevity, we denote these sets by
ω(PΓn(q)) and ω(PΓn□(q)), respectively. Also we denote by Z the center of GLn(q).
Let h∈Γn(q) and diag(h1,h-1,h0) be the normal form of h. Since n is even, it follows from Lemma 4.1 that the dimension of h1 is even too. We denote this dimension by 2k and define h2=diag(h-1,h0). Then h1 is conjugate to an element of SO2k+(q) and h2 is conjugate to an element of Spn-2k(q).
Let k>0. Then |h1|∈ωp(SO2k+(q))=ωp(Sp2k-2(q)), and hence
|h|∈ω(Sp2k-2(q)×Spn-2k(q))⊆ω(Spn-2(q)).
Using Lemmas 2.2 and 2.3, it is easy to check that
ωp′(Spn-2(q))⊆ωp′(PΩn+(q))∪ωp′(PΩn-(q))⊆ω(PSpn(q))
and
ωp~(Spn-2(q))⊆ωp~(Ωn+1(q))⊆ωp~(PSpn(q)).
Thus the projective order of h lies in the required sets.
Let k=0. Then h=h2 and |hZ|∈ω(PSpn(q)). It follows that
(PΓn(q))⊆ω(PSpn(q)).
Suppose, in addition, that h∈Γn□(q). Denote the dimension of h-1 by 2l. By Lemma 4.5, we deduce that h-1∉Γ2l⊠(q), and so
-h0 is conjugate to an element of Ωn-2lε(q) for some ε. Observe that |hZ|=|-hZ|.
Assume that |-h-1|>1. Then
|-hZ|=|-h|=|-h-1|⋅|-h0|.
Since |-h-1|∈ωp(Sp2l(q))=ωp(Ω2l+1(q)), we have
|hZ|∈ωp~(Ω2l+1(q)×Ωn-2lε(q))⊆ωp~(Ωn+1(q)).
Furthermore, if n=4, then either |hZ|=9 and p=3, or |hZ| lies in p⋅Ω2ε(q) and, therefore, divides p(q-ε)/2.
If |-h-1|=1, then either l=0 and h=h0, or l>0 and |-hZ|=|-h0|. In either case, |-hZ|∈ω(PΩnε(q))⊆ω(Ωn+1(q)). In particular, if n=4 and the order |-hZ| is a multiple of
p, then it divides p(q±1)/2.
Now we prove the reverse containments. If h is semisimple, then h∈Γn(q) if and only if h is conjugate to an element of Spn(q),
and hence
ωp′(PSpn(q))⊆ω(PΓn(q)).
Let a∈ω(PSpn(q)), |a|p=pt>1. Lemma 2.2 implies that there is a semisimple matrix hs∈Spl(q), where n=pt-1+1+l, such that a=pt|hs|, and we define h=diag(-Jpt-1+1,-hs). It is easy to see that h∈Γn(q) and
|hZ|=|-hZ|=|-h|=a.
Thus
ωp~(PSpn(q)⊆ω(PΓn(q)),
and the assertion concerning τPGLn(q) follows.
Let h∈Ωnε(q) be semisimple and let diag(h1,h-1,h0) be the normal form of h, with h1, h-1 of dimension k and l, respectively. If k>0, then h∈Γn□(q) by Lemma 4.5 (ii).
If k=0, then applying (i) or (iii) of Lemma 4.5 according as l>0 or l=0, we deduce that -h∈Γn□(q). In any case, |hZ|∈ω(PΓn□(q)).
Suppose n>4 and a∈ωp~(Ωn+1(q)), or n=4 and a=pc with c dividing (q±1)/2, or n=4, p=3 and a=9. Let |a|p=pt. By Lemmas 2.2 and 2.3, there are ε∈{+,-} and a semisimple matrix hs∈Ωlε(q), where n=pt-1+1+l, such that a=pt|hs|.
Defining h=diag(-Jpt-1+1,-hs), we see that h∈Γn□(q) and |hZ|=|-hZ|=|-h|=a. The proof is complete.
∎
In contrast to the sets of the projective orders of elements of Γn(q) and Γn□(q), the corresponding set for Γn⊠(q), where n is even, is not in general
closed under taking divisors. So we do not give an explicit description of the set ω(τδPSLn(q)) for even n. However, we derive some properties of this set.
Lemma 4.8.
Let q be odd and n⩾4 even. Then ω(τδPSLn(q))⊈ω(PSLn(q)).
If k=2s, where s⩾0, and q=q0k, then k⋅ω(τδ(q0)PSLn(q0))⊈ω(PSUn(q)). In particular,
ω(τδ(-q)PSUn(q))⊈ω(PSUn(q)).
Proof.
Recall that ω(τδPSLn(q)) consists of the projective orders of matrices of Γn⊠(q) multiplied by 2.
There is an element h∈SOnϵ(q)∖Ωnϵ(q) whose projective order is equal to (qn/2-ϵ)/2. By Lemma 4.5, it follows that -h∈Γn⊠(q), and so we obtain that
qn/2±1∈ω(τδPSLn(q)).
Assume that n>4 and l=(n-2)/2 is odd. Then Lemma 2.2 together with the existence of primitive divisors rl(±q) implies that
p(ql±1)∈ω(PSpn(q))∖ω(Ωn+1(q)).
Applying Lemma 4.7, we see that
(4.5)2p(ql±1)∈ω(τPGLn(q))∖ω(τPSLn(q))⊆ω(τδPSLn(q)).
Assume now that n=4 and q≡ϵ(mod4). The numbers p(q+1) and p(q-ϵ) lie in ω(PSp4(q)) and do not divide p(q±1)/2 (if q-1 divides (q+1)/2, then q=3),
and hence
(4.6)2p(q+1),2p(q-ϵ)∈ω(τδPSL4(q)).
We prove the first assertion of the lemma and the second assertion for k=1 together. To this end, suppose that ω(τδPSLn(q))⊆ω(PSLnε(q)) and as usual define d=(n,q-ε).
Then qn/2+εn/2∈ω(PSLnε(q)). By Lemma 2.7, this is equivalent to (n)2>(q-ε)2.
It follows that l=(n-2)/2 is odd and
(d)2=(q-ε)2,
and so (ql-ε)/d is odd. Then 2(ql+ε) does not divide (qn-2-1)/d and, therefore,
2p(ql+ε)∉ω(PSLnε(q)). If n>4, this contradicts (4.5). If n=4, then q≡-ε(mod4) and (4.6) implies that
2p(q+ε)∈ω(τδPSL4(q)) yielding a contradiction.
Suppose that k>1 and k⋅ω(τδ(q0)PSLn(q0))⊆ω(PSUn(q)). Observe that (q+1)2=2 and choose ϵ so that q0≡ϵ(mod4).
Let n/2 be odd. By assumption, we have a=k(q0n/2-ϵ)∈ω(PSUn(q)). Since a is a multiple of rn/2(ϵq0) and rn/2(ϵq0)∈Rn/2(q)=Rn(-q),
it follows that a divides
c=qn-1(q+1)(n,q+1).
This is a contradiction because
(a)2=(qn/2-1)2>(c)2=(qn-1)24.
Thus l=(n-2)/2 is odd and by (4.5) and (4.6), we have
2p(q0l-ϵ)∈ω(τδPSLn(q0)).
Then a=2kp(q0l-ϵ)∈ω(PSUn(q)), and hence a divides p(qn-2-1)/(n,q+1). This contradicts (a)2=(qn-2-1)2.
The final assertion follows from the second one and Lemma 3.3.
∎
We close this section with a proof of Theorem 1 and one of its consequences.
Proof of Theorem 1.
Recall that L=PSLnε(q), d=(n,q-ε) and consider the difference ω(τL)∖ω(L).
We analyze separately two cases according as n is odd or even.
Let n be odd. By Lemma 4.6, the difference under consideration is equal to 2⋅ω(Spn-1(q))∖ω(L).
We consider the numbers from Lemma 2.2 defining ω(Spn-1(q)) in turn.
Let a=2pt, where pt-1+1=n-1. Any element of ω(L) that is a multiple of pt
divides pt(q-ε)/d. Since d is odd, 2a divides pt(q-ε)/d if and only if 4 divides q-ε.
Thus 4pt∈ω(τL)∖ω(L) if and only if n=pt-1+2 for some t⩾1 and q≡-ε(mod4).
Let
a=pt[qn1±1,qn2±1,…,qns±1],
where pt-1+1+2n1+2n2+…+2ns=n-1,
or
a=[qn1±1,qn2±1,…,qns±1],
where s⩾2 and 2(n1+n2+…+ns)=n-1.
Then 2a divides pt[q2n1-1,q2n2-1,…,q2ns-1,q-ε] or [q2n1-1,q2n2-1,…,q2ns-1,q-ε], respectively.
The remaining possibility is a=ql±1, where l=(n-1)/2. If a=ql+εl, then 2a divides
qn-1-1d=(ql+εl)ql-εld
since d is odd. Let a=ql-εl. Assume that l is not a 2-power. Then (l)2⩽l/3, and hence l+2(l)2<n.
It follows that L has an element of order
c=[ql-εl,q2(l)2-1,q-ε].
Since 2(a)2=2(ql-εl)2⩽(q2l-1)2=(q2(l)2-1)2,
we have that 2a divides c. Finally, assume that l is a 2-power. Then the 2-exponent of L is equal to (q2l-1)2 and any element of ω(L) that a multiple of (q2l-1)2
divides (q2l-1)/d. Clearly, 2a divides (q2l-1)/d if and only if d divides ql+εl, which is equivalent to d=1.
Thus 2(ql-εl)∈ω(τL)∖ω(L) if and only if n-1=2t and d≠1.
Let n be even. By Lemma 4.7, the set ωp~(τL) is equal to 2⋅ωp~(Ωn+1(q)) if n>4, it consists of divisors of p(q±1) if n=4 and p>3, and
it consists of divisors of p(q±1) together with 18 if n=4 and p=3. We can consider the numbers from Lemma 2.2 defining ωp~(Ωn+1(q) for n⩾6 and
the numbers defining ωp~(τL) for n=4 together.
If a=pt and n=pt-1+1, then 2a∉ω(L).
Let
a=pt[qn1±1,qn2±1,…,qns±1],
where s⩾2 and pt-1+1+2n1+2n2+…+2ns=n. Then we have that 2a divides pt[q2n1-1,q2n2-1,…,q2ns-1]∈ω(L).
Let a=pt(qn1-ϵ)/2, where pt-1+1+2n1=n. If ϵ=εn1, then 2a divides pt(qn1-εn1)∈ω(L), while if ϵ=-εn1, then
2a divides
ptq2n1-1d=pt(qn1+εn1)qn1-εn1d.
Thus
ωp~(τL)⊆ω(L),
whenever n≠pt-1+1.
To handle ωp′(τL)=2⋅ω(PΩ2n+(q))∪2⋅ω(PΩ2n-(q)), we consider the numbers from Lemma 2.3 that define ωp′(PΩ2n±(q)).
Let
a=[qn1±1,qn2±1,…,qns±1],
where s⩾3 and 2(n1+n2+…+ns)=n. In this case, we have that 2a divides [q2n1-1,q2n2-1,…,q2ns-1]∈ω(L).
Let a=(qn/2-ϵ)/(4,qn/2-ϵ). If ϵ=εn/2, then 2a divides
qn/2-εn/2∈ω(L).
If ϵ=-εn/2 and (4,qn/2+εn/2)=4, then 2a divides
qn-1(q-ε)d=(qn/2+εn/2)(qn/2-εn/2)(q-ε)d
because d/2 divides (n/2,q-ε)=((qn/2-εn/2)/(q-ε),q-ε). If ϵ=-εn/2 and (4,qn/2+εn/2)=2, then Lemma 2.7
implies that 2a=qn/2+εn/2 does not lie in ω(L) if and only if (n)2⩽(q-ε)2.
Assuming the last inequality, the condition (4,qn/2+εn/2)=2 is equivalent to q≡ε(mod4).
Finally, let a=[qn1-ϵ1,qn2-ϵ2]/e, where 2(n1+n2)=n and with e=2 if (qn1-ϵ1)2=(qn2-ϵ2)2 and e=1 if (qn1-ϵ1)2≠(qn2-ϵ2)2.
We define n0=n/(2n1,2n2). We may assume that (qn1-ϵ1)2⩾(qn2-ϵ2)2.
If ϵ2=εn2, then 2a divides [q2n1-1,qn2-εn2]∈ω(L) since
2(a)2⩽2(qn1-ϵ1)2⩽(q2n1-1)2.
Similarly, we obtain that if ϵ1=εn1 and (qn1-ϵ1)2=(qn2-ϵ2)2, then 2a divides [q2n2-1,qn1-εn1].
Let ϵ2=-εn2 and ϵ1=-εn1. If e=2, then 2a divides
[q2n1-1,q2n2-1]q-ε.
If e=1, then n1 and n2 have opposite parity,
therefore, n/2 is odd. It follows that n0 is odd too, and 2a divides [q2n1-1,q2n2-1]/(n0,q-ε).
Let ϵ2=-εn2 and ϵ1=εn1. We may assume that (qn1-εn1)2>(qn2+εn2)2. If n1≠(n1)2, then
n1+2(n1)2+2n2<n
and 2a divides [qn1-εn1,q2n2-1,q2(n1)2-1].
If 2(qn1-εn1)2⩽(q2n2-1)2, then 2a divides [q2n2-1,qn1-εn1]. If qn2+εn2 divides q2n1-1, then 2a divides q2n1-1 too. Finally, if (n0,q-ε)=1, then 2a divides [q2n1-1,q2n2-1].
Thus it remains to consider the case when n1=(n1)2, 2(qn1-εn1)2>(q2n2-1)2, qn2+εn2 does not divide q2n1-1 and
(n0,q-ε)≠1. The two first conditions yield n1>(n2)2. In particular, (n/2)2=(n2)2 and n0=n/(2n2)2=(n)2′.
Then the third condition is equivalent to n2 not being a 2-power. Observe that
(n)2′=n(2n2)2=n1+n2(n2)2=n1(n2)2+(n2)2′.
It follows that (n)2′ is a sum of a non-identity 2-power and an odd number greater than 1 and also ((n)2′,q-ε)≠1.
In particular, (n)2′>3 and (n,q-ε)2′≠1.
Conversely, suppose that n=2t⋅l, where t⩾1, l⩾5 is odd and (n,q-ε) is divisible by an odd prime r. Writing n1=(n)2=2t, n2=n/2-n1=2t-1(l-2) and a=[qn1-1,qn2+εn2], we see that
(qn1-1)2=2t-1(q2-1)2>(n2)2(q+ε)2⩾(qn2+εn2)2
and hence 2a∈ω(τL).
Assume that 2a∈ω(L) and let
c=[ql1-εl1,…,qls-εls]f,
where l1+…+ls=n, be a number from Lemma 2.1 that 2a divides. Since
2(a)2=2(qn1-1)2=(q2n1-1)2, it follows that at least one of the numbers l1,…,ls is a multiple of 2n1. Also r2n2(εq) divides a, and hence at least one of them is a multiple of 2n2. Observing that
[2n1,2n2]=2t+1(l-2)>2tl=n, we deduce that those are different numbers, which yields s=2, l1=2n1 and l2=2n2. Then f=(n/(2n1,2n2),q-ε), and so (f)r>1. Since (r,q+ε)=(r,n1)=(r,n2)=1, it follows that
(c)r=(q-ε)r(f)r<(q-ε)r=(a)r.
This is a contradiction, therefore, 2a∉ω(L), and the proof is complete.
∎
An interesting consequence of Theorem 1 is that whenever τ is admissible as an automorphism of PSLnε(q), it is also admissible as an automorphism of PSLnε(q1/k)
for every odd k.
Lemma 4.9.
Let k be odd and q=q0k. If ω(τL)⊆ω(L), then
ω(τPSLnε(q0))⊆ω(PSLnε(q0)).
In particular, if ε=+, τ and β=φm/k are admissible for L, then βτ is also admissible.
Proof.
Assume that ω(τPSLnε(q0))⊈ω(PSLnε(q0)). This implies that the numbers n and q0 satisfy the conditions of one of items (i)–(v) of Theorem 1.
Since k is odd, it follows that (n,q0-ε) divides (n,q-ε) and (q-1)2=(q0-1)2, and thus the numbers n and q satisfy the same conditions.
This contradicts the hypothesis that ω(τL)⊆ω(L).
To prove the second assertion, it suffices to show that ω(βτL)⊆ω(L). By Lemma 3.3, the proved containment and admissibility of β, we see that
ω(βτL)=k⋅ω(τPSLn(q0))⊆k⋅ω(PSLn(q0))=ω(βL)⊆ω(L),
and the proof is complete.
∎
5 Admissible groups
In this section, we prove Theorems 2 and 3. As we have mentioned, the theorems with n=3 were proved in [23, 24], and so we assume that n⩾4.
Throughout this section, q=pm is odd, L=PSLnε(q) and G is a group such that L<G≤AutL. Also, we fix the numbers d=(n,q-ε) and b=((q-ε)/d,m)d.
We begin with lemmas that hold for both linear and unitary groups. We say that a subgroup of OutL is admissible if it is the image of an admissible group.
Lemma 5.1.
If G∩InndiagL>L, then ω(G)≠ω(L). In particular, admissible groups of OutL are abelian
and any non-trivial admissible subgroup of the group 〈δ¯〉⋊〈τ¯〉 is conjugate in this group to 〈τ¯〉.
Proof.
By Lemma 2.1, if |G∩InndiagL|/|L|=i>1, then G has an element of order (qn-εn)i/(q-ε)d, which does not lie in ω(L).
Thus admissible groups of OutL can be embedded into the image of the group generated by φ and γ, which is abelian. The group 〈δ¯〉⋊〈τ¯〉
is dihedral, and so all of its non-trivial subgroups that intersect trivially with 〈δ¯〉 are conjugate to 〈τ¯〉 or 〈δ¯τ¯〉, and in the latter case
we may assume that n is even. But Lemma 4.8 says that 〈δ¯τ¯〉 is not admissible.
∎
Lemma 5.2.
Suppose that n⩾5 and |G/L| is odd. Then ω(G)=ω(L) if and only if n-1 is not a p-power, G/L is conjugate in OutL to a subgroup of 〈φ¯〉 and |G/L| divides ((q-ε)/d,m)d.
Proof.
If ε=+, the assertion is proved in [11, Propositions 6,7], and if ε=-, it is proved in [13, Proposition 6].
∎
Lemma 5.3.
Let n=4 and |G/L| is odd. Then ω(G)=ω(L) if and only if 12 divides q+ε and G/L is a 3-group.
Proof.
Let ω(G)=ω(L) and r∈π(G/L). Then G contains a field automorphism of order r, and hence
by Lemma 3.3, ω(G) includes r⋅ω(PSL4ε(q0)), where q=q0r. It follows that rr4(q0),rr3(εq0)∈ω(L).
Since r4(q0)∈R4(q), we see that r divides a=(q2+1)(q+ε)/d and, in particular, it does
not divide q-ε. If r≠3, then r3(εq0)∈R3(εq), and so r divides c=(q3-ε)/d, which is a contradiction because (a,c)=1.
Let r=3. Then 3 divides q+ε. Since 3p(q0-ε) is an element of ω(L), we have that 3(q0-ε) divides (q2-1)/d, and thus 4 divides q+ε.
Conversely, let 12 divide q+ε and G/L be a 3-group. We may assume that G is the extension of L by a field automorphism of order dividing (m)3.
To prove that ω(G)⊆ω(L), it suffices to check k⋅ω(PSL4ε(q0))⊆ω(PSL4ε(q)) with q=q0k for
every divisor k of (m)3. Note that 12 divides q0+ε, (q+ε)3=k(q0+ε)3 and (4,q0-ε)=2=d. Since p≠3,
the set ω(PSL4ε(q0)) consists of divisors of (q02+1)(q0+ε)/2, (q03-ε)/2, q02-1 and p(q02-1)/2.
Observing that
k(q02+1)(q0+ε)∣(q2+1)(q+ε)
and
k(q03-ε),k(q02-1)∣q2-1,
we obtain the desired containment.
∎
By Lemmas 5.2 and 5.3, it follows that up to conjugacy OutL has only one maximal admissible subgroup of odd order, and we can take this subgroup to be 〈ψ¯〉, where
ψ∈〈φ〉 has order (b)2′ if n>4 or q≢1(mod12), and order (m)3 if n=4 and q≡-ε(mod12).
Let η=δ(d)2′ and S2 be the Sylow 2-subgroup of OutL generated by η¯, φ¯(m)2′ and τ¯.
Lemma 5.4.
If ω(G)=ω(L), then G/L is conjugate in OutL to a subgroup of 〈ψ¯〉×S2.
Proof.
By Lemma 5.1, the group G/L is abelian, and hence is the direct product of its Hall 2′-subgroup A1 and Sylow 2-subgroup A2. Since A1 is admissible, it is conjugate to a subgroup of 〈ψ¯〉. Replacing G/L by a conjugate if necessary, we may assume that A1⩽〈ψ¯〉.
Note that η¯ centralizes ψ¯. Indeed, if ε=+, then η¯ψ¯=η¯q0, where q=q0|ψ|.
Since |ψ| is odd, we have
(q-1)2=(q0-1)2,
and hence (d)2 divides q0-1. If ε=-, then
η¯ψ¯=η¯q0, where q2=q0|ψ|. Now (q+1)2<(q2-1)2=(q0-1)2, and again (d)2 divides q0-1. It follows that the whole group S2 centralizes ψ¯.
Thus A2 is conjugate in COutL(A1) to a subgroup of S2, and the whole group G/L is conjugate to a subgroup of 〈ψ¯〉×S2.
∎
The structure of the group S2 varies according to whether linear or unitary groups are under consideration, so in the rest of this section we consider the cases ε=+ and ε=- separately.
We begin with the case of unitary groups, in which S2=〈η¯〉⋊〈φ¯(m)2′〉.
Lemma 5.5.
Let ε=- and 1<G/L≤S2. If ω(G)=ω(L), then G/L is conjugate to a subgroup of 〈φ¯(m)2′〉.
Proof.
We may assume that n is even. If m is odd, then S2=〈η¯〉⋊〈τ¯〉 and by Lemma 5.1, the group G/L is conjugate to 〈τ¯〉.
Let m be even. Then |η¯|=2 and S2=〈η¯〉×〈φ¯(m)2′〉. Suppose G≰〈φ¯(m)2′〉. Then
G contains φm/kη for some k>1 dividing (m)2. Let q=q0k. By Lemmas 3.3 and 4.8 together with Lemma 4.2, we have
ω(φm/kηL)=k⋅ω(τδd/2(q0)PSLn(q0))=k⋅ω(τδ(q0)PSLn(q0))⊈ω(L),
which is a contradiction.
∎
Thus if ε=- and τ is not admissible, then S2 has no non-trivial admissible subgroups.
Lemma 5.6.
Suppose ε=-, τ is admissible, k>1 divides (m)2 and β=φm/k. Then
β is admissible if and only if (n)2⩽2 or n=4,8,12. If β is admissible and γ∈〈φ〉 is admissible and has odd order, then γβ is also admissible.
Proof.
We show first that ω(βL)⊆ω(L) if and only if (n)2⩽2 or n=4,8,12.
By Lemma 3.3, the set ω(βL) is equal to k⋅ω(τPSLn(q0)), where q=q0k.
Let n be odd. Since τ is admissible and q≡1(mod4), Theorem 1 implies that n-2 is not
a power of p. Then using Lemmas 2.2 and 2.6, it is not hard to check that k⋅ω(Spn-1(q0))⊆ω(Spn-1(q)).
Now applying Lemma 4.6 yields
k⋅ω(τPSLn(q0))=2k⋅ω(Spn-1(q0))
⊆2⋅ω(Spn-1(q))=ω(τL)⊆ω(L).
Let n be even. Observe that (d)2=(q+1)2=2. By the admissibility of τ and Theorem 1, it follows that n-1 is not a p-power.
Also it is not hard to verify that k⋅ω(Ωn+1(q0))⊆ω(Ωn+1(q)) (cf. [12, Theorem 1]).
Applying Lemma 4.7, we see that k⋅ωp~(τPSLn(q0))⊆ωp~(τL) for n>4. Since k(q0±1) divides q-1, the same is true for n=4.
Thus it remains to examine when 2k⋅ωp′(PΩn±(q0)) is a subset of ω(L), and we consider the numbers from Lemma 2.3 in turn.
Let
a=[q0n1±1,q0n2±1,…,q0ns±1],
where s⩾3 and 2(n1+n2+…+ns)=n. Then 2ka divides
[q2n1-1,q2n2-1,…,q2ns-1]∈ω(L).
Let
a=(q0n/2-ϵ)(4,q0n/2-ϵ).
If n/2 is even, then 2ka divides qn/2-1∈ω(L).
Suppose that n/2 is odd. Then rn/2(ϵq) divides a and lies in Rn/2(q)=Rn(-q), and so 2ka∈ω(L) if and only if
2ka divides
c=qn-1(q+1)d=(qn/2-1)(qn/2+1)(q+1)d.
Clearly, (a)2′ divides (c)2′. Since (d)2=2, we see that (c)2=(q-1)2/2.
If (4,q0-ϵ)=4, then 2k(a)2=k(q0-ϵ)2/2=(q-1)2/2. If (4,q0-ϵ)=2, then 2k(a)2=2k⩽(q-1)2/2.
Finally, let a=[q0n1-ϵ1,q0n2-ϵ2]/e, where 2(n1+n2)=n, and with e=2 if (qn1-ϵ1)2=(qn2-ϵ2)2 and e=1 if (qn1-ϵ1)2≠(qn2-ϵ2)2.
Suppose that n/2 is odd. We may assume that (qn1-ϵ1)2⩾(qn2-ϵ2)2.
If n2 is even, then 2ka divides [q2n1-1,qn2-1]. Let n2 be odd. Then n1 is even. If n1≠(n1)2, then
2ka divides [qn1-1,q(n1)2-1,q2n2-1]. If n1=(n1)2 and n2=1, then 2ka divides q2n1-1. If (n1)=(n1)2 and n2>1, then
by the admissibility of τ, we have (n,q+1)2′=1 and, therefore, (n/(2n1,2n2),q+1)=1. So [q2n1-1,q2n2-1] lies in ω(L), and it is divisible by 2ka.
Suppose that n/2 is even and n/2⩾8. We can take n1 and n2 to be odd coprime numbers larger than 1.
Also we take ϵ1=+ and ϵ2=-. Then
a=[q0n1-1,q0n2+1]
and 2k(a)2=k(q02-1)2=(q2-1)2. Since a is a multiple of
both rn1(q0) and rn2(-q0) and rni(±q0)∈R2ni(-q) for i=1,2, it follows that 2ka∈ω(L) if and only if 2ka divides c=[q2n1-1,q2n2-1]/(n/(2n1,2n2),q+1).
But we have (c)2=(q2-1)2/2<(a)2 because n/2 is even. Thus ω(βL)⊈ω(L).
We are left with the cases n=4,8,12. If n=4, then 2a divides q02-1, and hence 2ka divides q2-1.
If n=8, then a divides [q03±1,q0±1] or (q04-1)/2, and so 2ka divides q6-1 or q4-1.
If n=12, then a divides [q05±1,q0±1], [q04±1,q02±1], or (q06-1)/2, and 2ka divides q10-1, q8-1, or q6-1.
We established that the condition for ω(βL) to be a subset of ω(L) depends only on n and not on q0, and thus β is admissible if and only if ω(βL)⊆ω(L).
Next, suppose that β is admissible, γ∈〈φ〉 is admissible and has odd order l and let q0=q1l. To prove that γβ is admissible, it suffices to show the inclusion ω(γβL)⊆ω(L). By Lemma 4.9, if we regard τ as an automorphism of PSUn(q1k), it is still be admissible, and so by the above result, it follows that
k⋅ω(τPSLn(q1))⊆ω(PSUn(q1k)).
By Lemma 3.3, we have
ω(γβL)=lk⋅ω(τPSLn(q1))⊆l⋅ω(PSUn(q1k))=ω(γL)⊆ω(L),
as desired.
∎
We are now in a position to prove Theorem 3.
Proof of Theorem 3.
Suppose that ω(G)=ω(L). By Lemmas 5.4 and 5.5, it follows that G/L is conjugate to a subgroup in 〈ψ¯〉×〈φ¯(m)2′〉.
If n-1 is a p-power, Lemmas 4.7, 5.2 and 5.3 imply that any non-trivial element of G/L is not admissible. If n-1 is not a p-power, then
ψ is admissible, and applying Lemma 5.6 completes the proof.
∎
Now we assume that ε=+, and so we have S2=〈η¯〉⋊(〈φ¯m2′〉×〈τ¯〉).
Given an element 1≠β∈〈φ(m)2′〉, we define βϵ with ϵ∈{+,-} by setting β+=β and β-=βτ.
Lemma 5.7.
Let ε=+, k>1 divide (m)2 and q=q0k. Then for any ϵ∈{+,-}, the following hold:
If n is odd or k does not divide (q-1)/d, then
k⋅ω(PSLnϵ(q0))⊈ω(L).
If n is even, k divides (q-1)/d and n=1+pt-1, then
k⋅ω(PSLnϵ(q0))∖{kpt}⊆ω(L).
If n is even, k divides (q-1)/d and n-1 is not a p-power, then
k⋅ω(PSLnϵ(q0))⊆ω(L).
In particular, if β=φm/k, then βϵ is admissible if and only if n is even, k divides (q-1)/d and n-1 is not a p-power.
If βϵ is admissible and γ∈〈φ〉 is admissible and has odd order, then γβϵ is also admissible.
Proof.
If n is odd, then 2rn(ϵq0) is not an element of ω(L) since rn(ϵq0)∈Rn(q)
and (qn-1)/(q-1) is odd. If n is even and k does not divide (q-1)/d, then krn-1(ϵq0)∉ω(L). Indeed, otherwise the fact that rn-1(ϵq0)∈Rn-1(q) implies that
k divides (qn-1-1)/d. But (qn-1-1)2/(d)2=(q-1)2/(d)2, and thus (i) follows.
Let n be even and suppose that k divides (q-1)/d. We claim that ka∈ω(L) for all a∈ω(PSLnϵq0)) except a=pt for n=1+pt-1.
The number a divides one of the numbers in items (i)–(v) of Lemma 2.1, and we consider these possibilities in turn.
If we have a=(q0n-1)/(q0-ϵ)(n,q0-ϵ), then ka divides qn/2-1. Similarly, if a=[q0n1-ϵn1,q0n2-ϵn2]/(n/(n1,n2),q0-ϵ],
where n1+n2=n and both n1, n2 are even, then ka divides [qn1/2-1,qn2/2-1]. If a=[qn1-ϵn1,qn2-ϵn2]/(n/(n1,n2),q0-ϵ], where n1 and n2 are odd,
then ka divides the element c=[qn1-1,qn2-1]/(n/(n1,n2),q-1]. Indeed, if q0≡-ϵ(mod4), then
k(a)2=k⩽(c)2,
while if q0≡ϵ(mod4), then
k(q0-ϵ)2=(q-1)2⩾(k)2(n)2,
which implies that (q0-ϵ)2⩾(n)2 and k(a)2=(q-1)2/(n)2=(c)2. Also, a divides c by Lemma 2.5.
Finally, if a=[q0n1-ϵn1,…,q0ns-ϵns], where s⩾3 and n1+…,+ns=n,
or a=pt[q0n1-ϵn1,…,q0ns-ϵns], where s⩾2 and 1+pt-1+n1+…+ns=n, we have that ka divides [qn1-1,…,qns-1] or pt[qn1-1,…,qns-1],
respectively.
Let β=φm/k. By Lemma 3.3, we have ω(βϵL)=k⋅ω(PSLnϵ(q0)). Thus if
βϵ is admissible, then n is even, k divides (q-1)/d and n-1 is not a p-power. Conversely, if all these three conditions are satisfied, then by the above
ω(βk1L)=(k/k1)⋅ω(PSLn(q0k1))⊆ω(L)
for every divisor k1 of k, and hence βϵ is admissible.
To prove the final assertion, it suffices to check ω(γβϵL)⊆ω(L) for admissible γ and βϵ.
Let |γ|=l and q0=q1l. Observing that (q-1)2=(q1k-1)2, we see that k divides (q-1)/d if and only if
it divides (q1k-1)/(n,q1k-1). Thus the admissibility of βϵ yields
k⋅ω(PSLnϵ(q1))⊆ω(PSLn(q1k)).
Applying Lemma 3.3,
we have
ω(γβϵL)=lk⋅ω(PSLnϵ(q1))⊆l⋅ω(PSLn(q1k))=ω(γL)⊆ω(L),
and the proof is complete.
∎
Lemma 5.8.
Let ε=+, let n be even, k>1 divide (m)2, β=φm/k and q=q0k.
If α=βϵηj is admissible, then α¯ is conjugate in the group S2 to either β¯ϵ or β¯ϵη¯,
with q0≡-ϵ(mod4) and 2k⩽(q-1)2/(d)2 in the latter case. Furthermore, if q0≡-ϵ(mod4) and 2k⩽(q-1)2/(d)2, then the following statements hold:
βϵη is admissible if and only if n cannot be represented as 1+pt-1+2u with t,u⩾1 and, in addition, k=2 whenever
n-1 is a p-power,
If βϵη is admissible and γ∈〈φ〉 is admissible and has odd order, then γβϵη is admissible.
Proof.
Note that η¯β¯ϵ=η¯ϵq0, and so
(5.1)β¯ϵη¯=η¯-1β¯ϵη¯=β¯ϵη¯1-ϵq0.
Suppose that α=βϵηj is admissible and α¯ is not conjugate in S2 to β¯ϵ. Then (5.1) implies that (q0-ϵ,(d)2) does not divide j, or in other words
(5.2)(j)2<(q0-ϵ,d)2.
Since η=δ(d)2′, applying Lemma 3.3 yields
(5.3)ω(αL)=k⋅ω(δj(d)2′(ϵq0)PSLnϵ(q0)).
The index of δj(d)2′(ϵq0)PSLnϵ(q0) in PGLnϵ(q0) is equal to (j(d)2′,q0-ϵ). There is an element of order q0n-1-ϵ in the group PGLn(q0), and hence writing a=(q0n-1-ϵ)/(j(d)2′,q-ϵ), we see that ka∈ω(αL). Since a is a multiple of rn-1(ϵq0)∈Rn-1(q) and lies in ω(L), it follows that
a divides c=(qn-1-1)/d. If q0≡ϵ(mod4), then by (5.2), we deduce that
(a)2=k(q0-ϵ)2(j)2=(q-1)2(j)2>(q-1)2(d)2=(c)2.
Thus q0≡-ϵ(mod4). Then (q0-ϵ,d)2=2, and so j is odd. In particular, (5.1) shows that α is conjugate to βϵη.
Also, (a)2=2k⩽(c)2=(q-1)2/(d)2, and the first assertion is proved.
Next, let α=βϵη, q0≡-ϵ(mod4) and 2k⩽(q-1)2/(d)2. Observing that
(q-1)2=k(q0+ϵ)2, we deduce from the last inequality that 2(d)2⩽(q0+ϵ)2, and hence
α¯2=β¯2η¯1+ϵq0=β¯2.
Since 2k⩽(q-1)2/(d)2, it follows from Lemma 5.7 that β2 is admissible if and only if k=2 or n-1 is not a p-power.
Suppose n=1+pt-1+2u with t,u⩾1. Then PGLnϵ(q0) has an element of order pt(q02u-1), and so ka∈ω(αL), where a=pt(q02u-1)/(d,q0-ϵ)2′.
Since ka is a multiple of kpt(q02u-1)2=pt(q2u-1)2, it lies in ω(L) only if it divides pt(q2u-1)/d, which is not the case.
To prove (i), it remains to verify that ω(αL)⊆ω(L) whenever n cannot be represented as 1+pt-1+2u with t,u⩾1. Since
((d)2′,q0-ϵ)=(n,q0-ϵ)2′=(n,q0-ϵ)/2, it follows from (5.3) that
ω(αL)=k⋅ω(2PSLnϵ(q0)∖PSLnϵ(q0)),
where 2PSLnϵ(q0) is the unique subgroup of PGLnϵ(q0) that contains PSLn(q0) as a subgroup of index 2. Denote ω(2PSLnϵ(q0)∖PSLn(q0))
by ω.
If n=pt-1+1, then any element of GLn(q) whose order is divisible by pt is conjugate to a scalar multiple of the n×n unipotent Jordan block, therefore,
pt∉ω. By Lemma 5.7, we have
k⋅(ω∩ω(PSLnϵ(q0)))⊆ω(L),
and hence we are left with elements of ω∖ω(PSLnϵ(q0)).
By Lemma 2.1, it follows that this difference consists of some divisors of the following numbers:
2(q0n-1)/((q0-ϵ)(n,q0-ϵ)),
[q0n1-ϵn1,qn2-ϵn2]/(n/(n1,n2),q0-ϵ), where n1,n2 are odd, n1+n2=n,
2pt(q0n1-ϵn1)/(n,q0-ϵ), where n1>0 and pt-1+1+n1=n.
Let a=2(q0n-1)/((q0-ϵ)(n,q0-ϵ)). Then
k(a)2=2k(q0n-1)2(q0-ϵ)2(q0-ϵ,n)2=(qn-1)22=(qn/2-1)2,
and so ka divides qn/2-1∈ω(L).
Let a=2[q0n1-ϵn1,qn2-ϵn2]/(n/(n1,n2),q0-ϵ), where n1, n2 are odd and n1+n2=n. Consider the number c=[qn1-1,qn2-1]/(n/(n1,n2),q-1) lying in ω(L).
By Lemma 2.5, both numbers (qni-1)/(n/(n1,n2),q0-ϵ), i=1,2, divide c. Also, k(a)2=2k⩽(q-1)2/d2=(c)2, and hence ka divides c.
Let a=2pt(q0n1-ϵn1)/(n,q0-ϵ), where n1>0 and pt-1+1+n1=n. Note that n1 is even and by hypothesis, (n1)2≠n1. Since
k(a)2=2k(q0n1-1)2(q0-ϵ)2=(qn1-1)2=(q(n1)2-1)2,
we see that ka divides pt[qn1/2-1,q(n1)2-1]∈ω(L).
To prove part (ii), take admissible γ and α. We may assume that n-1 is not a p-power because otherwise γ=1 by Lemmas 5.2 and 5.3.
Then by the equality
α¯2=β¯2
and Lemma 5.8, to prove that γα is admissible, it suffices to check the inclusion ω(γαL)⊆ω(L).
Denote |γ| by l and let q0=q1l. Observe that
q1≡q0≡-ϵ(mod4)
and 2k divides (q1k-1)2/(n,q1k-1)2.
By Lemma 3.3, we have
ω(γαL)=lk⋅ω(δ(d)2′(ϵq1)PSLnϵ(q1)).
Using the fact that (d,q1-ϵ)2′=(q1-ϵ)/2 and the above results, we deduce that
k⋅ω(δ(d)2′(ϵq1)PSLnϵ(q1))=k⋅ω(2PSLnϵ(q1)∖PSLnϵ(q1))⊆ω(PSLn(q1k)).
Thus
ω(γαL)⊆l⋅ω(PSLn(q1k)))=ω(γL)⊆ω(L),
and the proof is complete.
∎
We are now ready to describe admissible 2-subgroups of OutL in the case of linear groups, and then prove Theorem 2. Recall that b=((q-ε)/d,m)d and define ϕ=φm/(b)2.
Lemma 5.9.
Let ε=+ and let 1<G/L≤S2=〈η¯〉⋊(〈φ¯(m)2′〉×〈τ¯〉). Then
ω(G)=ω(L) if and only if G/L is cyclic and up to conjugacy in S2 is generated by the image of one of the following elements:
τ
if b is odd and
τ
is admissible,
β±, where 1≠β∈〈ϕ〉, if b is even and n-1 is not a p-power,
βτη, where 1≠β∈〈ϕ2〉, if (b)2>2 and n cannot be represented as 1+pt-1 or 1+pt-1+2u with t,u⩾1,
ϕ-κη, where p≡κ(mod4), if (p-κ)2>(n)2 and n cannot be represented as 1+pt-1 or 1+pt-1+2u with t,u⩾1,
φm/2τη, if (b)2>2,
n=1+pt-1 for some t⩾1 and n cannot be represented as 2+2u with u⩾1,
(φm/2)-κη, where p≡κ(mod4), if (m)2=2,
(p-κ)2>(n)2,
n=1+pt-1 for some t⩾1 and n cannot be represented as 2+2u with u⩾1.
Proof.
Observe that b is even if and only if all the numbers n, m and (q-1)/d are even.
Suppose that n is odd. Then S2=〈φ¯(m)2′〉×〈τ¯〉. By Lemma 5.7, any non-trivial element of 〈φ(m)2′〉 is not admissible
and thus the only potentially admissible subgroup of S2 is 〈τ¯〉.
Suppose that n is even but at least one of m and (q-1)/d is odd. It follows from Lemmas 5.7 and 5.8 that every admissible subgroup of S2 is contained in 〈η¯〉⋊〈τ¯〉. Then by Lemma 5.1, it is conjugate to a subgroup of 〈τ¯〉, and we are done in the case of odd b.
Suppose now that all the numbers n, m and (q-1)/d are even. In particular, (q-1)2>(d)2=(n)2 and so τ is not admissible by Theorem 4.7. Then by Lemma 5.1, we obtain that all non-trivial elements of 〈η¯〉⋊〈τ¯〉 are not admissible. It follows that every admissible subgroup of S2 is isomorphic to a subgroup of 〈φ¯(m)2′〉,
and hence is cyclic.
Let α¯∈S2∖〈η¯〉⋊〈τ¯〉 be admissible. We may assume that α=βϵηj, where
β=φm/k for some 1<k=2l⩽(m)2. By Lemmas 5.7 and 5.8, admissibility of α yields k⩽(q-1)2/(d)2 or, equivalently, β∈〈ϕ〉.
Moreover, all conjugates of β¯ϵ are admissible whenever n-1 is not a p-power, and this gives the automorphisms in (ii).
It remains to consider the case when α¯ is not conjugate to β¯ϵ. In this case by Lemma 5.8, we may assume that α=βϵη,
(5.4)2k⩽(q-1)2(d)2
and q0≡-ϵ(mod4), where q=q0k. Moreover, we have k=2 whenever n-1 is a p-power. Note that (q-1)2=(m)2(p-κ)2, where p≡κ(mod4).
Suppose that (q-1)2/(d)2⩽(m)2 or, in other words, (p-κ)2⩽(n)2. Then k⩽(m)2/2, so q0≡1(mod4) and condition (5.4) is equivalent to β∈〈ϕ2〉. Thus (b)2>2 and α=βτη with β∈〈ϕ2〉. In particular, if n-1 is a p-power, then α=φm/2τη.
Suppose that (q-1)2/(d)2>(m)2 or, equivalently, (p-κ)2>(n)2. Then condition (5.4) holds for all β∈〈ϕ〉. Also, we have q0≡1(mod4) if k<(m)2 and q0≡κ(mod4) if k=(m)2. Thus α=βτη with β∈〈ϕ2〉 or α=ϕ-κη. In particular, if n-1 is a p-power, then α=φm/2τη when (b)2=(m)2>2 and α=ϕ-κη when (b)2=(m2)=2.
Conversely, let α be one of the resulting automorphisms and let the associated conditions be satisfied. Then by Lemma 5.8, α is admissible unless n cannot be represented as 1+pt-1+2u.
If n=1+pt-1 and n=1+ps-1+2u, then we have pt-1=ps-1+2u or, equivalently, ps-1(pt-s-1)=2u, and so s=1 and n=2+2u. Thus we obtain the automorphisms in (iii)–(vi), and the proof is complete.
∎
Proof of Theorem 2.
Let ω(G)=ω(L). By Lemma 5.4, it follows that G/L is conjugate to a subgroup of 〈ψ¯〉×S2. Admissible subgroups of 〈ψ¯〉 and S2 are described in Lemmas 5.2, 5.3 and Lemma 5.9, respectively. These descriptions together with Lemmas 5.7 and 5.8 imply that the product of an admissible subgroup of 〈ψ¯〉 and an admissible subgroup of S2 is also admissible, and this completes the proof.
∎
