Finite REA-groups are solvable

Accepted Manuscript Finite REA-groups are solvable

Cai Heng Li, Lei Wang

PII: DOI: Reference:

S0021-8693(18)30691-4 https://doi.org/10.1016/j.jalgebra.2018.11.033 YJABR 16977

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Journal of Algebra

Received date:

12 February 2018

Please cite this article in press as: C.H. Li, L. Wang, Finite REA-groups are solvable, J. Algebra (2019), https://doi.org/10.1016/j.jalgebra.2018.11.033

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FINITE REA-GROUPS ARE SOLVABLE CAI HENG LI AND LEI WANG

Abstract. It is proved that finite REA-groups are solvable. A major step in the proof is based on an independent result about finite simple groups.

keywords. Solvable group, Simple group, REA-group 1. Introduction Two elements a, b of a group G are called fused or inverse-fused if a is conjugate under Aut(G) to b or b−1 , respectively. In the study of Cayley graph representations of complete multi-partite graphs, the concept of REA-group was introduced in [11]. Definition 1.1. A group G is called an REA-group (REA stands for relative elementary abelian), if there exists a subgroup N < G such that any two elements of G \ N are fused or inverse-fused. To emphasise this subgroup, G is sometimes said to be an REA-group relative to N . Examples of REA-groups include abelian p-groups, groups of order p3 , and lots of Frobenius groups. The main result of this paper is the following theorem. Theorem 1.2. Finite REA-groups are solvable. The proof of Theorem 1.2 is dependent on the following result about finite simple groups, which may be of independent interest. (Denote by ΔL(n, pe ) the extension of SL(n, pe ) by the automorphism φδ, where φ, δ are generators of field and diagonal automorphism of SL(n, pe ), respectively. Set PΔL(n, pe ) = ΔL(n, pe )/Z(SL(n, pe )). Theorem 1.3. Let G be an almost simple group, and let N be a proper normal subgroup in G. Then G \ N contains elements of different orders, and further either (i) G \ N contains elements of order divisible by two distinct primes, or  (ii) G = PΔL(2, 32 ) for some positive integer , G/N = C2 , and an element of G \ N has order 2+1 or 2+2 . One of our principle motivations for studying REA-groups comes from studying Cayley graph representations of certain graphs. Recall that a Cayley graph Γ = Cay(G, S) is called normal edge-transitive if the normaliser NAutΓ (G) is edgetransitive on Γ . In [11], it was shown that if a complete multi-partite graph is a normal edge-transitive Cayley graph of a group G, then G is an REA-group. We thus have the following consequence. Corollary 1.4. If a complete multi-partite graph is a normal edge-transitive Cayley graph of a group G, then G is solvable. Date: December 13, 2018. 2000 Mathematics Subject Classification. 20B05, 20D05. 1

2

CAI HENG LI AND LEI WANG

We remark that there is a family of groups, which have quite similar properties to / G REA-groups. A group G is called a Camina group if all elements of gG with g ∈ are conjugate to g, refer to [2, 10]. Recently, Guralnick and Navarro [7] initiated the study of groups G which have a normal subgroup N such that all elements of gN with g ∈ / N are conjugate to g, and proved that such normal subgroups N are solvable. Motivated by this result in [7] and Theorem 1.3, we would like to propose the following question. Question. Is there a non-solvable group N and an automorphism g of N such that all elements of gN have the same order? 2. A reduction In this section, we reduce Theorem 1.2 to Theorem 1.3. We first quote some properties of REA-groups, obtained in [11]. Lemma 2.1. Let G be an REA-group with respect to N . Then (i) N
G, and G/N ∼ = Cdp , where p is a prime and d is a positive integer; (ii) all elements of G \ N are of the same order pe for some integer e, and no element of G \ N centralises a non-identity element of N of order coprime to p. Lemma 2.2. Let G be an REA-group relative to N , and let R be a proper characteristic subgroup of G. Then R  N . Proof. Suppose for a contradiction that R  N . Let g1 ∈ R\N . Then g1 ∈ G\N . If R∪N = G, then by our assumption, N < R, and so R = G, which is a contradiction. Thus R ∪ N ⊂ G. Let g2 ∈ G \ (R ∪ N ). Since G is an REA-group relative to N , there exists θ ∈ Aut(G) such that g1 θ = g2 . Since R is a characteristic subgroup 2 of G, it follows that g2 ∈ R, which is a contradiction. Thus R  N . The socle of a group G is the product of all minimal normal subgroups of G. This is denoted by soc(G). The radical of G is the largest solvable normal subgroup of G, denoted by rad(G). Let G be a finite group with socle T  , where T is nonabelian simple, and   1. Then Aut(G) is isomorphic to a subgroup of Aut(T  ). Thus G and Aut(G) can be viewed as subgroups of Aut(T  ) = Aut(T ) S , and so each element of G and Aut(G) can be written as (t1 , . . . , t )π, where ti ∈ Aut(T ) and π ∈ S .

(2.1)

Take any g ∈ G. Then, by (2.1), g = (x1 , . . . , x )π, where xi ∈ Aut(T ) and π ∈ S . Assume that π = 1. Write π = π1 · · · πs , where the πi ∈ S are pairwise disjoint cycles. Let ki+1 = o(πi+1 ) + ki with k0 = 0, where 0  i  s − 1. Without loss of generality, we may assume that πi+1 = (ki + 1, ki + 2, . . . , ki+1 − 1, ki+1 ). Lemma 2.3. With the notation above, there exists h ∈ Aut(T  ) such that g h = g1 · · · gs gˆπ,

REA-GROUPS

3

where gi+1 = (1, . . . , 1, ti+1 , 1, . . . , 1) for some ti+1 ∈ Aut(T ) in the (ki + 1)th coordinate for 0  i  s − 1, and gˆ = (1, . . . , 1, xks +1 , . . . , x ). Proof. Set xˆi+1 = (1, . . . , 1, xki +1 , . . . , xki+1 , 1, . . . , 1)πi+1 , and gˆ = (1, . . . , 1, xks +1 , . . . , x ). π

j+1 Notice that for j = i, we have xˆi+1 = xˆi+1 . Also since gˆπi+1 = gˆ, it follows that

g = xˆ1 · · · xˆs gˆ. Let hki +1 = 1 and hki +j = xki +j xki +j+1 · · · xki+1 , where 2  j  ki+1 − ki . Set ˆ i+1 = (1, . . . , 1, hk +1 , . . . , hk , 1, . . . , 1). h i i+1 A straightforward calculation shows ˆ h

i+1 −1 xˆi+1 = (1, . . . , 1, h−1 ki +1 xki +1 hki +2 , . . . , hki+1 xki+1 hki +1 , 1, . . . , 1)πi+1 = (1, . . . , 1, xki +1 · · · xki+1 , 1, . . . , 1)πi+1 .

Denote by ti+1 the element xki +1 · · · xki+1 . Let gi+1 = (1, . . . , 1, ti+1 , 1, . . . , 1), ˆ h

i+1 where ti+1 lies in the (ki + 1)th coordinate. Then xˆi+1 = gi+1 πi+1 . Set ˆ s. ˆ1 · · · h h=h ˆ πi+1 = h ˆ j+1 with j = i, Then h ∈ Aut(T ) . We first note that gˆh = gˆ. Also since h j+1

ˆ h

j+1 we have xˆi+1 = xˆi+1 . Thus

ˆ

ˆ

xh1 1 · · · xˆshs )ˆ g = (g1 π1 ) · · · (gs πs )ˆ g = g1 · · · gs gˆπ. g h = (ˆ π

i+1 The last equality follows from gj+1 = gj+1 with j = i, and gˆπi+1 = gˆ. This completes the proof of Lemma 2.3.

2

The following lemma reduces Theorem 1.2 to almost simple groups. Lemma 2.4. Assume that Theorem 1.3 holds. Then REA-groups are solvable. Proof. Suppose that G is a non-solvable REA-group relative to N . Since G/N is a p-group by Lemma 2.1, N is non-solvable. Note that rad(G) is a characteristic subgroup of G. By Lemma 2.2, we have rad(G)  N . By [11, Lemma 2.2], G/rad(G) is an REA-group relative to N/rad(G). Thus, to complete the proof of the lemma, we may assume that the radical rad(G) is trivial. Then soc(G) = M1 × · · · × Mr , where the Mi are non-isomorphic nonabelian characteristically simple groups. By Lemma 2.2, soc(G)  N . If r  2, then M2 × · · · × Mr is a characteristic subgroup of G. By [11, Lemma 2.2], G/(M2 × · · · × Mr ) is an REA-group relative to N/(M2 × · · · × Mr ). Thus, to complete the proof of this lemma, we may further assume that soc(G) = T  = T1 × · · · × T , where Ti ∼ = T is a nonabelian simple group, and  is a positive integer. Therefore, G  Aut(T ) S = Aut(T ) :S . We fix an element x ∈ G \ N.

4

CAI HENG LI AND LEI WANG

By Lemma 2.1, o(x) = pe with p prime and e  1. By (2.1), x can be written as x = (t1 , . . . , t )π, where ti ∈ Aut(Ti ) and π ∈ S . It is clear that o(π) = pf with f  e. Let y = (y1 , . . . , y ) ∈ T  . Then xy = (t1 , . . . , t )π(y1 , . . . , y ) = (t1 y1π , . . . , t yπ )π, belonging to G \ N because T   N . Suppose first that π = 1. Then each ti is a p-element, and o(ti )  pe . Since o(x) = pe , we may assume, without loss of generality, that o(t1 ) = pe . If t1 belongs to T1 , then we may choose y1 ∈ T1 such that t1 y1 is not of p-power order, and neither is xy, which is a contradiction. Thus t1 ∈ Aut(T1 ) \ T1 . Let Tˆ1 = T1 × 1 × · · · × 1 < T1 ×· · ·×T . Let H = Tˆ1 , x and L = Tˆ1 , xp . Then L
H. Since o(x) = o(t1 ) = pe , it implies that H is an almost simple group. Noting that |H:L| = p, we have (H \ L) ∩ N = ∅. By Lemma 2.1, all elements of H \ L have the same order pe , contradicting Theorem 1.3. Suppose now that π = 1. Write π = π1 · · · πs , where the πi ∈ S are pairwise disjoint cycles. Let ki = o(πi ) + ki−1 with k0 = 0, where 1  i  s. Without loss of generality, we may assume that πi = (ki−1 + 1, ki−1 + 2, . . . , ki − 1, ki ). x1 · · · xˆs+1 )π with By Lemma 2.3, there exists some h ∈ Aut(T  ) such that xh = (ˆ xˆi = (1, . . . , 1, xi , 1, . . . , 1), and xˆs+1 = (1, . . . , 1, tks +1 , . . . , t ), where xi ∈ Aut(T ) lies in the (ki−1 + 1)th coordinate. Noting that Gh ∼ = G and N h ∼ = N , it follows that Gh is an REA-group relative to N h . Thus, to prove the remainder of the lemma, we may replace G and N by Gh and N h , respectively. Therefore, xh ∈ G \ N , and so, for convenience, denote the element xh by x. That is to say, x has the form as x = (ˆ x1 · · · xˆs+1 )π. Since x and xy belong to G \ N , there exists α ∈ Aut(G) such that xα = xy or (xy)−1 . By (2.1), α can be written as α = (α1 , . . . , α )ρ−1 , where αi ∈ Aut(T ) and ρ−1 ∈ S . −1

−1

x1 · · · xˆs+1 )π)(α1 ,...,α )ρ = xπ ρ , where x ∈ Aut(T ) . Similarly, xy = Then xα = ((ˆ −1 x˜π and (xy)−1 = xˆπ −1 where x˜, xˆ ∈ Aut(T ) . Thus xπ ρ = xα equals xy = x˜π or −1 (xy)−1 = xˆπ −1 . It follows that π ρ = π  where = 1 or −1, and so (π ρ ) = (π  )ρ = −1 π, forcing π ρ = π  = π  . This implies that π1ρ = πi0 for some i0 . In any case, ρ induces a permutation of {{π1 , π1−1 }, . . . , {πs , πs−1 }}. Relabelling if necessary, we may assume that k = o(π1 ) = max{o(πi ) | 1  i  s}. We separate the argument into two cases. To complete the rest of the proof, we may take y1 = t and yi = 1 for 2  i  . Then y = (y1 , y2 , . . . , y ) = (t, 1, . . . , 1).

REA-GROUPS

5

Case 1: Assume that π1ρ = π1 . We compute x2 = (x1 , 1, . . . , 1, x1 , . . . )π 2 , and (xy)2 = (x1 , 1, . . . , 1, t, tx1 , . . . )π 2 . Furthermore, for the integer k defined above, a calculation shows that xk = (x1 , x1 , . . . , x1 , . . . ), (xy)k = (x1 t, tx1 , . . . , tx1 , . . . ) = ((tx1 )t , tx1 , . . . , tx1 , . . . ). Since o(x) = o(xy) = pe , from the above it follows that x1 and tx1 are of p-power order. If x1 belongs to T1 , then we may choose t ∈ T1 such that tx1 is not of p-power order, and so neither is (xy)k , which is a contradiction. Thus x1 ∈ Aut(T1 ) \ T1 . Now we readily check that −1

(xk )α = (x1 , . . . x1 , . . . )(α1 ,...,αk ,... )ρ = (xα1 1ρ , . . . xα1 kρ , . . . ).

Since xα = (xy) , we have (xk )α = (xy)k , and thus ((tx1 )t ) = (x1 t) = xα1 1ρ , that is, tx1 is conjugate to x1 or x−1 1 . It follows that both x1 and tx1 have the same order f p , where f is an integer. Since t is arbitrary, all elements of T1 x1 have the same order equal to pf . Case 2: Assume that π1ρ = πi0 for i0 = 1. Let m be the length of the orbit ρ  {π1 , π1−1 }ρ . Relabelling if necessary, we may assume πi+1 = πi+2 (reading the subscripts modulo m), where 0  i  m − 1. By our assumption, πi+1 = (ik + 1, ik + 2, . . . , (i + 1)k). Let x ˆm+1 · · · xˆs+1 , and π m+1 = x m+1 = πm+1 · · · πs . It is easily shown that π

j+1 = xˆi+1 , x xˆi+1 m+1

πj+1

π

m+1 m+1 = x ˆi+1 = xˆi+1 , y πj = y and y π = y, m+1 , x

where 0  j = i  m − 1 and 1 < j   m. It follows that x = (ˆ x1 π1 )(ˆ x2 π2 ) · · · (ˆ xm πm )(x  m+1 π m+1 ), and x2 π2 ) · · · (ˆ xm πm )(x π xy = (ˆ x1 π1 y)(ˆ m+1  m+1 ).

(2.2)

Arguing as in Case 1, we calculate that (ˆ x1 π1 y)k = ((tx1 )t , tx1 , . . . , tx1 , 1, . . . , 1), and (ˆ xi+1 πi+1 )k = (1, . . . , 1, xi+1 , xi+1 , . . . , xi+1 , 1, . . . , 1). By (2.2), we conclude that k x1 π1 )k (ˆ x2 π2 )k · · · (ˆ xm πm )k (x  xk = (ˆ m+1 π m+1 ) , and k x1 π1 y)k (ˆ x2 π2 )k · · · (ˆ xm πm )k (x  (xy)k = (ˆ m+1 π m+1 ) .

It follows that xk = (x1 , . . . , x1 , . . . , xi , . . . , xi , . . . , xm , . . . , xm , . . . ), and (xy)k = (((tx1 )t ) , (tx1 ) , . . . , (tx1 ) , . . . , xi , . . . , xi , . . . , xm , . . . , xm , . . . ). By Case 1, x1 ∈ Aut(T1 ) \ T1 . Let wik+j = xi+1 , where 1  j  k. Then xk = (w1 , . . . , wk , . . . , w(i−1)k+1 , . . . , wik , . . . , w(m−1)k+1 , . . . , wmk , . . . ). Noting that α = (α1 , . . . , α )ρ−1 , we have α

−1

(i+1)k+1 αmk (xk )α = (w1α1 , . . . , w(i+1)k+1 , . . . , wmk , . . . )ρ α((i+1)k)ρ α(mk)ρ = (w1αρ1ρ , . . . , w((i+1)k) ρ , . . . , w(mk)ρ , . . . ).

6

CAI HENG LI AND LEI WANG

Since xα = (xy) , we obtain (xk )α = (xy)k . By comparing the (ik + 1)th coordinates of (xk )α and (xy)k for 0  i  m − 1, we have α

α

ρ

ρ

(ik+1) ((m−1)k+1)  ((tx1 )t ) = w1αρ1ρ , . . . , xi+1 = w(ik+1) ρ , . . . , xm = w((m−1)k+1)ρ .

(2.3)

ρ  Recall that πi+1 = πi+2 (reading the subscripts modulo m), and πi+1 = (ik + 1, ik + 2, . . . , (i+1)k). Then (ik+1)ρ = (i+1)k+i where i ∈ {1, . . . , k}. Let θi+1 = α(ik+1)ρ where 0  i  m − 1. Recall that wik+j = xi+1 where 1  j  k. It is easily shown that α(ik+1)ρ α((m−1)k+1)ρ θi+1 θm w1αρ1ρ = xθ21 , . . . , w(ik+1) ρ = xi+2 , . . . , w((m−1)k+1)ρ = x1 .

By (2.3), we have θ

i+1 ((tx1 )t ) = xθ21 , . . . , xi+1 = xi+2 , . . . , xm = xθ1m .

Set θ = θm θm−1 · · · θ1 . Then θ θm−1 ···θ1

xθ1 = x1m

= (xm )θm−1 ···θ1 = · · · = (xθ21 )

m−1

m

= ((tx1 )t ) .

That is to say, tx1 and x1 are conjugate or inverse-conjugate. So tx1 and x1 have the same order equal to pf , where f is an integer. Since t is arbitrary, all elements of T1 x1 have the same order pf . Let H = T1 , x1  and L = T1 , xp1 . Then T1  L  H. Since x1 ∈ Aut(T1 ) \ T1 , we have T1 < H. Noting that T1 = Inn(T1 ) by the above identification, H is almost simple. By Cases 1-2, all elements of T1 x1 have the same order pf . Let Ω = {i | gcd(i, p) = 1 and 1  i < pf }. Clearly, H \ L = ∪i∈Ω T1 xi1 . Since T1 xi1 = (T1 x1 )i , it follows that all elements of T1 xi1 have the same order pf . The same is true for H \ L. This is a contradiction. Therefore, G is solvable. This completes the proof of Lemma 2.4. 2 3. Examples in Theorem 1.3 In this section, we treat the examples appeared in part (ii) of Theorem 1.3. For a positive integer n and a prime divisor p, denote by np and np the p-part and p -part of n, respectively, namely, n = np np with gcd(p, np ) = 1. Let φ be the field automorphism of SL(2, 3e ) induced by the Frobenius automor× phism of F3e , that is, ω → ω 3 , where ω  = F3e . Let δ be the diagonal automorphism ω 0 of SL(2, 3e ) induced by diag(ω, 1) = . 0 1 Lemma 3.1. Let G = ΔL(2, 3e ) with e = 2 for some positive integer , and N < G such that G/N = C2 . Then all elements of G \ N are 2-elements. 

Proof. Let T = SL(2, 32 ) and σ = φδ. Then G = T :σ and N = T :σ 2 . We find that o(σ) = 2+1 . Suppose that there exists g ∈ G \ N such that o(g) is not a 2-power. Put m = o(g). Let p be an odd prime divisor of m. Set μ = g m2 and x = g m/p . Then μ ∈ G \ N and x ∈ T . Write μ = σ n t , where n is odd, and t ∈ T . Then there exist integers a, b such that na + 2+1 b = 1, and so na ≡ 1(mod 2+1 ). A calculation shows μa = σ na t = σt, where t ∈ T . Set   α11 α12 a τ = μ = σt = φδt, where t = with α11 α22 − α12 α21 = 1. α21 α22

REA-GROUPS

Then

7

a

(3.4) xτ = xμ = x. Noting that τ = σt and σ ∈ G \ N , we have τ ∈ G \ N . Since p divides |T |, it follows   that p | 32 + 1, p | 32 − 1, or p = 3. We divide our analysis into three cases. 



i

Case 1: Suppose that p | 32 +1. Then p  | 32 −1, and so p  | 32 −1 for i  . Thus +1  p is a primitive divisor of 32 − 1. View ΓL(2, 32 ) as a subgroup of GL(2+1 , 3). Then, by [8, p.187, Theorem 7.3], we have CG (x)  CΓL(2,32 ) (x)  CGL(2+1 ,3) (x) = GL(1, 32 +1

+1

).





It follows that x ∈ GL(1, 32 )  GL(2, 32 ). However, since τ ∈ / GL(2, 32 ), we conclude that τ does not centralise x, which contradicts (3.4).   1 ζ Case 2: Suppose that p = 3. By Sylow’s Theorem, we may assume x = , 0 1 . A calculation shows where ζ ∈ F× 32   1 ω −1 ζ 3 τ φδt −1 −1 φ −1 t. x = x = x = t (δ x δ)t = t 0 1   1 ω −1 ζ 3 Then tx = t, and so 0 1       1 ω −1 ζ 3 α11 α12 α11 α12 1 ζ = . 0 1 α21 α22 α21 α22 0 1 A computation shows α21 ω −1 ζ 3 = 0, α22 ω −1 ζ 3 = α11 ζ and α21 ζ = 0. −1 2 Since ζ = 0, and α11 α22 − α12 α21 = 1, it follows that α21 = 0 and ω = (α11 ζ) , and hence ω is a square in F32 , which is a contradiction. 



Case 3: Suppose that p | 32 − 1. Let f be the integer such that (32 − 1)p = pf .   −1  32 −1   0 ω f 0 Let ω0 = ω p . Let h = . Set H = h. Since gcd(32 − 1, 32 + 1) = 2, 0 ω0 we conclude that H is a Sylow p-subgroup of T . By Sylow’s Theorem, we may f −1 matrices, assume that x = hip , where gcd(i, p) = 1. Since xφ and δ are diagonal   −3 f −1 ω1 0 ip −1 φ φt φδt τ φ we have t x t = x = x = x = x. Let ω1 = ω0 . . Then x = 0 ω13 So     −1    −3 α11 α12 α11 α12 ω1 0 0 ω1 φ = x t = tx = . α21 α22 α21 α22 0 ω13 0 ω1 A calculation shows α11 ω1−1 = α11 ω1−3 , α12 ω1 = α12 ω1−3 , α21 ω1−1 = α21 ω13 and α22 ω1 = α22 ω13 . Since α11 α22 − α12 α21 = 1, there exists some αij = 0 where 1  i, j  2. It follows 2 that ω14 = 1, which is impossible. This completes the proof. 



2

−1 = 2k + 1 where k  0. It is easy to see (32 − 1)2 = 2+2 . Then (32 − 1)2 = 32+2 Recall from Lemma 3.1 that σ = φδ. Let σ  = σdiag(ω k , ω k ) where ω = F× . 32

8

CAI HENG LI AND LEI WANG

Then o(σ  ) = 2+1 . Set  = T :σ  , where T = SL(2, 32 ) with   1. G   ∼ Notice that for t ∈ T , we conclude that tσ = tσ , and so G = G. Thus, for  convenience, we may assume that G = G. Set τ = σ  diag(ω k , ω −k ). By the previous paragraph, we conclude that  2k+1  0 ω 2 G = T :τ  and N = T :τ , where τ = φ . 0 1

Let ω0 = ω 2k+1 . Set

 x=

ω0−1 0 0 ω0





 0 1 . −1 0

and y =

Lemma 3.2. Let H = x, y, τ . Then H is a Sylow 2-subgroup of G, and H ∩ N = x, y, τ 2 . Proof. We readily check that xy = x−1 , xτ = x3 , y τ = xy, and |H| = 22+4 . It follows that H is a Sylow 2-subgroup of G. Since N = T :τ 2 , we have H ∩N = x, y, τ 2 . 2 Set τ1 = τ m xn and τ2 = τ m xn y, where gcd(m, 2) = 1 and n  0. Lemma 3.3. Each element of G \ N has order equal to o(τ1 ) = 2+1 or o(τ2 ) = 2+3 . Proof. Let μ = diag(ω0 , 1). Then τ = σ  diag(ω k , ω −k ) = φμ. We easily check 3m −1 that τ m = φm μ 2 . By Lemma 3.2, each element of H \ N has the form as n(32m −1) 32m −1 − 3m −1 2

of τ1 or τ2 . A calculation shows that τ12 = diag(ω0 (3m −2n)(32m −1) 2(3m +1)

diag(ω0

(2n+1)(32m −1) 2(3m +1)

, ω0

 τ12

τ22



n(32m −1) 3m −1

, ω0

), τ22 =

), and further,   n(32 m −1) 32 m −1 − 3m −1 2

= diag(ω0

 (3m −2n)(32 m −1) 2(3m +1)

= diag(ω0

 n(32 m −1) 3m −1

, ω0

), and

 (2n+1)(32 m −1) 2(3m +1)

, ω0

).



Since m is odd, we compute (32 m − 1)2 = 2+2 , (3m − 1)2 = 2 and (3m + 1)2 = 4.  Noting that o(ω) = 32 − 1, we have o(ω0 ) = 2+2 . Set   n(32 m −1) 32 m −1 − 3m −1 2

x 1 = ω0

 n(32 m −1) 3m −1

and x2 = ω0

.

Then x1 = 1 and x2 = −1, or x1 = −1 and x2 = 1, according to whether n is odd or  o(τ12 )

+1

 32 m −1 2(3m +1)

= 2, and so o(τ1 ) = 2 . Also since o(ω0 ) = 8, we not. It follows that 2 +3 have o(τ2 ) = 8, and hence o(τ2 ) = 2 . This implies that each element of H \ N has order equal to o(τ1 ) = 2+1 or o(τ2 ) = 2+3 . By Lemma 3.1, G \ N consists of 2-elements. By Sylow’s Theorem, each element of G \ N is conjugate to an element of H \ N . Thus the lemma holds. 2 Lemma 3.4. Let G = G/Z(G) and N = N/Z(G). Then G/N = C2 , and each element of G \ N has order equal to o(τ1 ) = 2+1 or 12 o(τ2 ) = 2+2 .

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Proof. Let τ 1 , τ 2 be the images of τ1 and τ2 in G, respectively. Since o(τ1 ) = 2+1 ,  / Z(G), we conclude that o(τ1 ) = o(τ 1 ) = 2+1 . A calculation shows and τ12 ∈ +2  τ22 = (τ22 )4 = diag(−1, −1) ∈ Z(G). Thus o(τ 2 ) = 12 o(τ2 ) = 2+2 . By Lemma 3.3, 2 each element of G \ N has order 2+1 or 2+2 . This completes the proof. 4. Proof of Theorem 1.3: alternating or sporadic socles In this section, one more reduction is given for the proof of Theorem 1.3. We first treat alternating groups and sporadic groups. Lemma 4.1. Let G be an almost simple group which has a socle an alternating group or a sporadic group. Let N be a proper normal subgroup of G. Then either (i) G \ N contains elements of order divisible by two distinct primes, or (ii) G = M10 ∼ = PΔL(2, 9), and N = A6 . Proof. Let T be the socle soc(G) of G. Then N  T . First, let T = An with n  5. Assume G = Sn , acting naturally on {1, 2, . . . , n}. Let σ = (12), and x = (345). Then σx ∈ G \ N has order 6. If G = Sn , then T = A6 , and G = PGL(2, 9), PΓL(2, 9) or PΔL(2, 9). Suppose that G = PGL(2, 9). Then N = PSL(2, 9) ∼ = A6 . For this case, G \ N contains elements of order 10. Suppose that G = PΓL(2, 9). Then N = PSL(2, 9), PΣL(2, 9), PΔL(2, 9) or PGL(2, 9). If N is one of the first three groups, then G \ N contains elements of order 10, and if N = PGL(2, 9), then G \ N contains elements of order 6. Suppose that G = PΔL(2, 9). By the Atlas [6], N = T = A6 , and each element of G \ N has order 4 or 8. Thus the lemma holds for T = An . Next, let T be a sporadic simple group. Then Out(G) = 1 or 2, see [6]. By our assumption, Out(T ) = C2 , and so N = T is one of the following groups: M12 , M22 , Mc L, HS, Suz, J2 , J3 , Fi22 , Fi24 , HN, He, O’N. Let M be a subgroup of G. Let G = M12 .2. Then N = M12 . By the Atlas [6], we may choose M ∼ = A4 × S3 . Thus M \ (M ∩ N ) contains = S4 × S3 such that M ∩ N ∼ an element g of order 6, and so g ∈ G \ N . Thus part (i) holds for G = M12 .2. Similarly, for G = HS.2, J3 .2, O’N.2, Mc L.2, or Suz.2, we have the following table: G HS.2 J3 .2 O’N.2 Mc L.2, Suz.2 M S8 × C2 PSL(2, 17) × C2 J1 × C2 M11 × C2 M ∩N S8 PSL(2, 17) J1 M11 It follows that M \ (M ∩ N ) contains elements of order divisible by two distinct primes. The same is true for G \ N . Thus part (i) holds for the groups G in the above table. Let G = M22 .2. Then N = M22 . By the Atlas [6], we may choose M ∼ = S6 ∼ such that M ∩ N = A6 . Then G \ N contains an element of order 6. Finally, the above argument works equally well for G = Fi22 .2, Fi24 .2, J2 .2, HN.2, or He.2, which provides the following table: G Fi22 .2, Fi24 .2 J2 .2 HN.2, He.2 M S6 S5 S12 M ∩N A6 A5 A12 Therefore, G \ N contains elements of order divisible by two distinct primes.

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This completes the proof of the lemma.

2

In the remainder of this paper, we complete the proof of Theorem 1.3 for almost simple groups of Lie type. This will consist of a series of lemmas. For convenience, we make the following hypothesis: Hypothesis 4.2. Let G be an almost simple group of Lie type over Fpe . Let N be a proper normal subgroup of G such that each element of G \ N has order being a power of r, where r is a prime. Remark 4.3. Under Hypothesis 4.2, no element of G \ N centralises a non-identity element of N of order coprime to r. The following lemma shows that G \ N does not contain a field automorphism. Lemma 4.4. Assume Hypothesis 4.2. Then G \ N does not contain any field automorphism. Proof. Let soc(G) = T . Then T  N . Suppose, by way of contradiction, that G \ N contains a field automorphism σ of r-power order. Assume first that T is a Chevalley group. Let H be a subgroup of T , which is defined over Fp . By definition, σ centralises H. Since |H| is not an r-power, this is a contradiction by Remark 4.3. Assume now that T = 2 B2 (22e+1 ), 2 F4 (22e+1 ) or 2 G2 (32e+1 ), where e  1. By [3, p.234], σ centralises 2 B2 (2), 2 F4 (2) or 2 G2 (3), which is impossible by Remark 4.3. Assume finally that T = 2 A2m (p2e ), 2 A2m+1 (p2e ), 2 Dn (p2e ), 2 E6 (p2e ) or 3 D4 (p3e ), where m, e  1, and n  4. By [12, p.12], Bm (pe )  2 A2m (p2e ), Cm+1 (pe )  2 A2m+1 (p2e ), Bn−1 (p )  2 Dn (p2e ), F4 (pe )  2 E6 (p2e ) and G2 (pe )  3 D4 (p3e ). e

For T = 2 A2m (p2e ) or 2 A2m+1 (p2e ), σ centralises either Bm (p) or Cm+1 (p). Similarly, for T = 2 Dn (p2e ), 2 E6 (p2e ) or 3 D4 (p3e ), σ centralises Bn−1 (p), F4 (p) or G2 (p). It follows that σ centralises at least one non-identity element of T of order coprime to r, which contradicts Remark 4.3. This completes the proof. 2 5. Classical groups We treat classical groups in this section. We first fix some notation. As usual, En is the n×n identity matrix, and AT is the transpose of A. We write diag(a1 , a2 , . . . , an ) for the matrix A = (aij )n×n with aii = ai for all i and aij = 0 for i = j. We write antidiag(a1 , a2 , . . . , an ) for the matrix A = (aij )n×n with ai,n−i+1 = ai for all i and aij = 0 otherwise. Let G be an almost simple classical group, defined over a field Fpe . Let soc(G) = T . Following [1], we shall use δ and δ  for generators of diagonal automorphism, γ and τ for generators of graph automorphism, and φ or ϕ for a generator of field automorphism of T . We abuse notation and use the same symbols to denote both their images in Out(T ) and, when they exist, specific matrices in GL(n, pe ) that induce them by conjugation. 5.1. Linear groups, unitary groups, and symplectic groups. Let Ω be a quasi-simple group, and PΩ the quotient by scalar matrices.

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(i) For an element g of Ω, denote by g the image of g in PΩ. (ii) For a subgroup H of Ω, denote by H the image of H in PΩ. Lemma 5.1. Let T  N
G, where T = PSL(n, pe ). Then either (i) G \ N contains elements of order divisible by two distinct primes, or  (ii) G = PΔL(2, 32 ), G/N = C2 , and G \ N consists of elements of order 2+1 and 2+2 , where   1. Proof. Assume first that n  3. Then Out(T ) = δ:(γ×φ) ∼ = C(n,pe −1) :(C2 ×Ce ). By [1, p.34], we may choose δ to be induced by diag(ω, 1, . . . , 1), and γ the duality automorphism g → (g −1 )T of T , where ω = F× pe . Then G \ N contains an element i j i j σ of the form φ δ or φ δ γ, where i, j are integers. Set    En−2 0 H= | M ∈ SL(2, p) . 0 M Then H  SL(n, pe ), and hence H  T . Clearly, H ∼ = SL(2, p). Let =H∼     En−2 0 0 1 α= with A = . (5.5) 0 A −1 0 Then α ∈ H. If σ = φi δ j , then σ centralises H. If σ = φi δ j γ, then σα centralises H. Since |H| is not a power of a prime, there exist h1 , h2 ∈ H such that σh1 , σαh2 ∈ G \ N have order divisible by two distinct primes. Assume now that n = 2. Suppose further that all elements of G \ N have r-power order, where r is a prime. We pick an element σ ∈ G \ N . By Lemma 4.4, we may assume σ = φa δ, where δ is induced by diag(ω, 1), and a is a non-negative integer. For this case, we have that p is an odd prime. Let o(φa ) = f . Since gcd(a, e) = e/f , we may assume a = e/f . Let   ω 0 a . τ1 = φ 0 1 p2a −1

pe −1

Then we calculate that τ12 = φ2a diag(ω pa −1 , 1), and further, τ1f = diag(ω pa −1 , 1). Since o(ω) = pe − 1, we have o(τ1f ) = pa − 1 and so o(τ1 ) = f (pa − 1). Thus as σ is the image of τ1 in PΓL(2, pe ), we conclude that o(σ) = f (pa − 1), which is divisible by 2. It follows that r = 2, and hence o(σ) is a 2-power. Write f = 2 where  is a non-negative integer. If  = 0, then σ = δ, and so o(σ) = pe −1. Thus PGL(2, pe )  G, but PGL(2, pe )  N for σ ∈ G \ N . Take σ  ∈ PGL(2, pe ) with o(σ  ) = pe + 1. Then σ  ∈ G \ N . Since gcd(pe − 1, pe + 1) = 2, we have pe = 3, and so e = 1, which is impossible. Thus  is a positive integer. Set     0 ω 0 1 a = τ1 . τ2 = φ −1 0 −1 0 a

A straightforward calculation shows τ22 = φ2a diag(−ω p , −ω), and furthermore, a

pe −1

pe −1

τ2f = diag((−ω p ) p2a −1 , (−ω) p2a −1 ). Since o(ω) = pe − 1, we have o(τ2f ) = p2a − 1. Let τ be the image of τ2 in PΓL(2, pe ). a

pe −1

pe −1

It is easily shown that τ ∈ G \ N . Since (τ2f )p +1 = diag((−ω) pa −1 , (−ω) pa −1 ) ∈ Z(GL(2, pe )), it implies that o(τ f ) = pa + 1, and hence o(τ ) = f (pa + 1).

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By the above discussion, G \ N contains two elements σ and τ , which have orders f (pa − 1) and f (pa + 1), respectively. Since gcd(pa − 1, pa + 1) = 2, we conclude that pa = 3, and so p = 3 and a = 1. Noting that a = e/f , we have e = f = 2 . Thus   T = PSL(2, 32 ), N = T, σ 2  = T.C2−1 and G = PΔL(2, 32 ) = T, σ = T.C2 . By Lemma 3.4, the lemma follows. 2 Lemma 5.2. Assume Hypothesis 4.2. Then T is not a unitary group. Proof. Suppose T = PSU(n, pe ) with n  3. Then Out(T ) = δ:φ = C(n,pe +1) :C2e , where φe = γ, and γ is the duality automorphism g → (g −1 )T of T , see [1, p.34]. Then G \ N contains an element φi δ j , where i, j are integers. Set    En−2 0 | M ∈ SU(2, p) . L= 0 M Then L  SU(n, pe ), and so L  T . Clearly, L ∼ =L∼ = SU(2, p). By [1, p.34], we e p −1 may choose δ to be induced by diag(ω , 1, . . . , 1), where ω = F× p2e . It follows i i that δ centralises L. Thus φ = 1. If i is even, then φ centralises L. So does φi δ j , which is a contradiction by Remark 4.3. Thus i is odd. Let α be as in (5.5). Then α ∈ SU(n, pe ), and so α ∈ T . Thus G \ N contains an element φi δ j α. Notice that for any g ∈ L, we have g φ = g γ = g α . It follows that φα centralises L. So does 2 φi δ j α, again a contradiction by Remark 4.3. Thus T is not a unitary group. Lemma 5.3. Assume Hypothesis 4.2. Then T is not a symplectic group. Proof. Suppose T = PSp(2n, pe ) with n  2. Assume that p is an odd prime. By [1, p.34], Out(T ) = δ × φ = C2 × Ce , and δ can be chosen to be induced by diag(ωEn , En ), where ω ∈ F× pe . Set    (A−1 )T 0 H= | A ∈ GL(n, p) . 0 A Then H  Sp(2n, pe ), and so H  T . By definition, H ∼ = GL(n, p), and hence |H| is not a power of a prime. It is easily shown that G \ N has an element φi δ j which centralises H. This is a contradiction by Remark 4.3. Thus p = 2. By [1, p.34], Aut(T ) = T :γ ∼ = T.C2e and γ 2 = φ. Let σ = γ i ∈ G\N , where i  1. By [3, p.234], γ centralises a subgroup Sz(2) of T . So does σ, again a contradiction. Thus T is not a symplectic group. 2 5.2. Orthogonal groups. Let V be an orthogonal space over a field Fpe , where p is a prime. Let Q be a quadratic form on V , and let β be its symmetric bilinear form with the form matrix B. (a) A vector v is singular if Q(v) = 0, and non-singular otherwise. (b) A pair of vectors (u, v) is called a hyperbolic pair if Q(u) = Q(v) = 0, and β(u, v) = 1. The plane u, v is called a hyperbolic plane. Lemma 5.4. Let U be a hyperbolic plane in V . Then (i) for p = 2, there exists some u ∈ U with Q(u) = 0; (ii) for p  3, there exist two v, w ∈ U such that β(v, v) and β(w, w) are square and non-square numbers, respectively.

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Proof. Let (u1 , v1 ) be a hyperbolic pair in U . Then {u1 , v1 } is a basis of U . Suppose first that p = 2. Let u = u1 + v1 . Note that Q(u) = Q(u1 ) + Q(v1 ) + β(u1 , v1 ). Since u1 and v1 are singular vectors, we have Q(u) = 1, as in part (i). Suppose now that p is odd. Let F× pe = ω. Let v = u1 + v1 and w = ωu1 + v1 if 2 is a square in Fpe , and v = ωu1 + v1 and w = u1 + v1 otherwise. A straightforward calculation shows that β(v, v) and β(w, w) are square and non-square, respectively, as in part (ii). 2 Definition 5.5. Let v ∈ V be non-singular. We define the reflection rv : V → V by rv (x) = x −

β(v, x) v. Q(v)

Let U1 , U2 , U be subspaces of V . Write U1 ⊥ U2 if β(u1 , u2 ) = 0 for all u1 ∈ U1 and u2 ∈ U2 , and call U non-singular if Q(u) = 0 for u ∈ U \ {0}. By [9, Proposition 2.5.3], V = u1 , v1  ⊥ · · · ⊥ un , vn  ⊥ W, where (ui , vi ) is a hyperbolic pair, and W = 0, x or x, y is non-singular. Then dim V = 2n + , where = 0, 1 or 2. In what follows, we may identify GOε (2n + , pe ) with GOε (V ), and write each element of GOε (2n + , pe ) as a matrix with respect to the ordered basis {u1 , v1 , . . . , un , vn , xˆ, yˆ}

(5.6)

where ε ∈ {o, +, −}, xˆ = x if dim W = 1 or 2, and xˆ = 0 otherwise, and yˆ = y if dim W = 2, and yˆ = 0 otherwise, unless specified otherwise. Lemma 5.6. Assume Hypothesis 4.2. Then T is not an orthogonal group of odd dimension. Proof. Suppose T = Ω(2n + 1, pe ) with n  3. By Lemma 5.3, we may assume that p is an odd prime. By [1, p.34], Out(T ) = δ × φ = C2 × Ce . Then G \ N contains an element δ i φj , where i, j are integers. By Lemma 5.4, there exist two elements u, v ∈ u1 , v1  such that β(u, u) and β(v, v) are square and non-square in Fpe , respectively. By [9, Proposition 2.6.3], we may choose δ = ru rv , namely, δ = diag(A, E2n−1 ), where A ∈ GL(2, pe ). Set    E2 0 | M ∈ Ω(2n − 1, p) . H= 0 M Then H ∼ = Ω(2n − 1, p) is a subgroup of T . It is easily shown that δ i φj centralises H. This is a contradiction by Remark 4.3. Thus the statement follows. 2 Lemma 5.7. Assume Hypothesis 4.2. Then T = PΩ+ (2n, pe ) with n  5. Proof. Assume T = PΩ+ (2n, pe ) with n  5. Suppose that p = 2. By [1, p.37], Out(T ) = φ × γ = Ce × C2 . By Lemma 5.4, there exists some x in u1 , v1  with Q(x) = 0. By [9, Proposition 2.7.3], we may choose γ = rx , namely, γ = diag(A, E2n−2 ) where A ∈ GL(2, 2e ). Let    E2 0 + | C ∈ Ω (2n − 2, 2) . H= 0 C

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Then H ∼ = Ω+ (2n − 2, 2) is a subgroup of T . It is easily shown that G \ N contains an element γ i φj which centralises H. This is not possible by Remark 4.3. Suppose that p is an odd prime. By [1, p.34],  (δ × δ  ):(γ × φ) ∼ = D8 × Ce if pen ≡ 1 (mod 4), Out(T ) = otherwise. δ × γ × φ ∼ = C2 × C2e For this case, we reorder the basis of (5.6) into {u1 , . . . , un , vn , . . . , v1 }. Then B = antidiag(1, . . . , 1). In the following, we write each element of GO+ (2n, pe ) as a matrix with respect to this ordered basis. By Lemma 5.4, there exist two elements u, v in u1 , v1  such that β(u, u) and β(v, v) are respectively square and non-square in Fpe . By [1, p.35], we may choose δ to be induced by diag(ωEn , En ), γ = ru , and δ  = ru rv , where ω ∈ F× pe . Notice that for i = j, ui , vi  is orthogonal to uj , vj . Then, by Definition 5.5, we readily check that γ and δ  have the form ⎡ ⎤ a11 0 0 a12 ⎢ 0 En−1 0 0 ⎥ ⎢ ⎥ ⎣ 0 0 En−1 0 ⎦ 0 a22 a21 0 where aij ∈ Fpe . Let B  = antidiag(1, . . . , 1) ∈ GL(n − 1, p). Let M  GL(n − 1, p), of index 2. Set L = diag(1, B  (D−1 )T B  , D, 1) | D ∈ M . It is easily shown that LT BL = B. By [9, Proposition 2.5.7], L  Ω+ (2n, pe ), and so L  T . By definition, φ centralises L. A calculation shows that γ, δ and δ  centralise  L. It follows that G \ N contains an element δ i (δ  )i γ j φk which centralises L. By 2 Remark 4.3, this is a contradiction since L ∼ = M . Thus T = PΩ+ (2n, q). Lemma 5.8. Assume Hypothesis 4.2. Then T is not an orthogonal group of minus type with even characteristic. Proof. Suppose T = Ω− (2n + 2, 2e ) where n  3. By [1, p.37], Out(T ) = ϕ ∼ = C2e . By [9, Proposition 2.8.2], we may choose ϕ to be φ followed by conjugation by   E2n 0 0 J   1 ζ where φ is a field automorphism of order e, and J = with ζ ∈ F× 2e . Set 0 1    A 0 + | A ∈ Ω (2n, 2) . H= 0 E2 Then H ∼ = Ω+ (2n, 2) is a subgroup of T . It is easily shown that G \ N contains an element ϕk , which centralises H. This is impossible by Remark 4.3. Thus the lemma follows. 2 In the remainder of this section, we will deal with the case that either T = PΩ+ (8, pe ) with p prime or T = PΩ− (2n, pe ) with p an odd prime. Since their outer automorphism groups are more complicated, we use another method to treat such groups.

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Let g be a simple Lie algebra over the complex field. We denote by Φ and Π respectively the sets of roots and fundamental roots relative to a fixed ordering of the additive group generated by the roots. Then, as in [5], one can replace the complex field by an arbitrary finite field Fpe after choosing a Chevalley basis {ha , a ∈ Π; ea , a ∈ Φ}. Define xa (t) = exp(t ad ea ), where a ∈ Φ and t ∈ Fpe . Set T = xa (t) | a ∈ Φ, t ∈ Fpe . Chevalley shows that, with a few exceptions, T is simple, see [4] or [5]. Later, Steinberg constructed families of groups by taking fixed points of T of a product of a diagram and a field automorphism, which are also simple unless a few exceptions and also denoted generically by T . For convenience in what follows, we use P and Q to denote the additive groups generated by all the fundamental roots and all the fundamental weights of g, respectively. Lemma 5.9. Assume Hypothesis 4.2. Then T = PΩ+ (8, pe ). Proof. Suppose T = PΩ+ (8, pe ). By [1, p.37],  γ, τ  × φ = S3 × Ce Out(T ) = ((δ × δ  ):γ, τ ) × φ = S4 × Ce

if p = 2, otherwise.

Then G \ N contains an element η of the form δ0 γ0 φ0 , where δ0 ∈ δ, δ  , γ0 ∈ γ, τ  and φ0 ∈ φ. Since T ∼ = D4 (pe ), we may assume that T = D4 (pe ). By definition, γ and τ are induced by the symmetries ργ and ρτ of the Dynkin diagram D4 , respectively. Let a be a simple root such that aργ = a and aρτ = a. Set H = xa (t), x−a (t) | t ∈ Fp . By [4, Proposition 12.2.3], we have x±a (t)γ = x±aργ (t) = x±a (t) and x±a (t)τ = x±aρτ (t) = x±a (t). It follows that both τ and γ centralise H. By definition, φ centralises H. If δ0 = 1, then η centralises H. Thus δ0 = 1. By [4, p.98], δ0 is induced by some Fpe -character χ of P , that is, xs (t)δ0 = δ0−1 xs (t)δ0 = xs (χ(s)−1 t), where s is any root of D4 , and t ∈ Fpe . By [4, Lemma 11.1.3], there exists some Fpe -character ψ of Q with ψ(a) = χ−1 (a), and by [4, Theorem 7.1.1], ψ gives rise to an element h(ψ), belonging to T . Let σ = h(ψ)η. Then σ ∈ G \ N is of r-power order. For any t ∈ Fp , we easily check that xa (t)σ = xa (((χ(a)ψ(a))−1 t)φ0 ) = xa (t), x−a (t)σ = x−a (((χ(−a)−1 ψ(−a)−1 )t)φ0 ) = x−a (((χ(a)ψ(a))t)φ0 ) = x−a (t). It follows that σ centralises H. Since H ∼ = SL(2, p) or PSL(2, p), this is a contradic+ tion by Remark 4.3. Thus T = PΩ (8, q). 2 Lemma 5.10. Assume Hypothesis 4.2. Then T is not an orthogonal group of minus type with odd characteristic.

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Proof. Suppose T = PΩ− (2n, pe ) where n  4. By [1, p.37],  if pen ≡ −1 (mod 4), δ:ϕ ∼ = D8 × Ce Out(T ) = δ × ϕ ∼ = C2 × C2e otherwise. Then G \ N contains an element δ i ϕj , where i, j are integers. By Lemma 4.4, we conclude that δ i = 1. Since T ∼ = 2 Dn (p2e ), we may assume that T = 2 Dn (p2e ). Let g be of type Dn with n  4. Let a be a simple root of g such that a is adjacent to three other simple roots of g. Set H = xa (t), x−a (t) | t ∈ Fp . By [4, Proposition 13.6.3], we have H  T . By definition, ϕ centralises H. It is known that δ i is induced by a self-conjugate Fp2e -character χ of P . Notice that for the root a, we have χ(a) ∈ Fpe . By [4, p.313, Hartley’s Lemma], there exists a self-conjugate Fp2e -character ψ of Q such that ψ(a) = χ(a)−1 , and ψ gives rise to an element h(ψ), belonging to T . Let σ = h(ψ)δ i ϕj . Then σ ∈ G \ N . For any t ∈ Fp , we may readily check that j

xa (t)σ = xa (((ψ(a)−1 χ(a)−1 )t)ϕ ) = xa (t), j x−a (t)σ = x−a (((ψ(−a)−1 χ(−a)−1 )t)ϕ ) j = x−a (((ψ(a)χ(a))t)ϕ ) = x−a (t). It follows that σ centralises H, a contradiction by Remark 4.3. Thus T = PΩ− (2n, pe ). 2 6. Exceptional groups Let T  G  Aut(T ), where T is a nonabelian simple group. In this section, we consider the case where T is an exceptional group of Lie type. Following [1], we continue to use δ for a generator of diagonal automorphism, γ for a generator of graph automorphism, and φ or ϕ for a generator of field automorphism of T . By abuse of notation, we use the same symbols to denote their images in Out(T ). Note first that if T is one of the following groups F4 (pe )(p = 2), G2 (pe )(p = 3), 3 D4 (p3e ), E8 (pe ), 2 B2 (22e+1 ), 2 G2 (32e+1 ) and 2 F4 (22e+1 ), then the only outer automorphisms of T are field automorphisms, refer to [13]. By Lemma 4.4, each field automorphism of T of prime-power order centralizes at least one non-identity element of T of coprime order. Thus we only need to treat the case where T is another exceptional group of Lie type. Lemma 6.1. Assume Hypothesis 4.2. Then T = 2 F4 (2) , F4 (2e ) or G2 (3e ). Proof. First, suppose T = 2 F4 (2) . For this case, G = T.C2 , N = T = 2 F4 (2) , and r = 2. By the Atlas [6], there exists a subgroup M ∼ = C13 :C12 such that M is contained in G but not in N . It follows that there exists σ ∈ G \ N of order 4 centralising an element of N of order 3. This is a contradiction. Thus T = 2 F4 (2) . Next, suppose T = F4 (2e ) or G2 (3e ) with e  1. By [4, p.225], Aut(T ) = T :γ ∼ = T.C2e and γ 2 = φ, where φ is a field automorphism of order e. In view of [3, p.234], γ fixes 2 F4 (2) or 2 G2 (3) pointwise, according to whether T = F4 (2e ) or G2 (3e ). It follows that G \ N contains an element γ i , which centralises 2 F4 (2) or 2 G2 (3). This is a contradiction by Remark 4.3. Thus the statement follows. 2

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Recall that g is a simple Lie algebra over the complex field. In this section, we assume g to be of type E6 or E7 . For convenience in what follows, we continue to use P and Q to denote the additive groups generated by all the fundamental roots and all the fundamental weights of g, respectively. Lemma 6.2. Assume Hypothesis 4.2. Then T = E6 (pe ). Proof. Suppose T = E6 (pe ). By [13, p.172],  δ:(γ × φ) = C3 :(C2 × Ce ) if 3 | pe − 1, Out(T ) = otherwise. γ × φ = C2 × Ce Then G \ N contains an element η of the form δ i γ j φk , where i, j, k are integers. Let g be of type E6 , and let ργ be a symmetry of the Dynkin diagram of g. Let a be a simple root of g such that aργ = a. Set H = xa (t), x−a (t) | t ∈ Fp . By [4, Proposition 12.2.3], we may choose γ to be induced by ργ . Then x±a (t)γ = x±aργ (t) = x±a (t) where t ∈ Fpe . It follows that γ centralises H. By definition, φ also centralises H. Therefore, δ i = 1. By [4, p.98], we know that δ i is induced by an Fpe -character χ of P . By [4, Lemma 11.1.3], there exists some Fpe -character ψ of Q such that ψ(a) = χ(a)−1 , and by [4, Theorem 7.1.1], ψ gives rise to an element h(ψ), belonging to T . Let σ = h(ψ)η. Then σ ∈ G \ N . For any t ∈ Fp , we easily check that k

xa (t)σ = xa (((ψ(a)χ(a))−1 t)φ ) = xa (t), k x−a (t)σ = x−a (((ψ(−a)χ(−a))−1 t)φ ) k = x−a (((ψ(a)χ(a))t)φ ) = x−a (t). It follows that σ centralises H, a contradiction by Remark 4.3. Thus T = E6 (pe ). 2 Lemma 6.3. Assume Hypothesis 4.2. Then T = 2 E6 (p2e ). Proof. Assume T = 2 E6 (p2e ). By [13, p.173], Out(T ) = δ:φ = C(3,pe +1) :C2e . Then G \ N contains an element δ i φj , where i, j are integers. By Lemma 4.4, we conclude that δ i = 1. Let g be of type E6 . Let ργ be a symmetry of the Dynkin diagram of g. Let a and a be the non-consecutive simple roots of g such that aργ = a. It is known that δ i is induced by a self-conjugate Fp2e -character χ of P . By [4, p.313, Hartley’s Lemma], there exists a self-conjugate Fp2e -character ψ of Q such that ψ(a) = χ(a)−1 , and ψ gives rise to h(ψ), belonging to T . Let σ = h(ψ)δ i φj . Then σ ∈ G \ N . Set H = xS (t), x−S (t) | t ∈ Fp , where e e xS (t) = xa (t)xa (tp ) and x−S (t) = x−a (t)x−a (tp ). By [4, Proposition 13.6.3], H is a subgroup of T . Since aργ = a, we have χ(a) = e e χ(a)p and ψ(a) = ψ(a)p . For any t ∈ Fp , we readily check that j

e

j

xS (t)σ = xa (((ψ(a)χ(a))−1 t)φ )xa (((ψ(a)χ(a))−p t)φ ) = xS (t), e j j x−S (t)σ = x−a (((ψ(a)χ(a))t)φ )x−a (((ψ(a)χ(a))p t)φ ) = x−S (t). It follows that σ centralises H. Since H ∼ = SL(2, p) or PSL(2, p), this is a contradic2 2e tion by Remark 4.3. Thus T = E6 (p ). 2

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CAI HENG LI AND LEI WANG

Lemma 6.4. Assume Hypothesis 4.2. Then T = E7 (pe ). Proof. Suppose T = E7 (pe ). By [13, p.177], Out(T ) = δ × φ = C(2,pe −1) × Ce . Then G \ N contains an element δ i φj , where i, j are integers. By Lemma 4.4, we conclude that δ i = 1. Let g be of type E7 . Let a be any fundamental root of g. Set H = xa (t), x−a (t) | t ∈ Fp . By definition, φ centralises H. Arguing as in Lemma 6.2, there exists some h(χ) ∈ T such that h(χ)δ i φj centralises H. Noting that h(χ)δ i φj belongs to G \ N , we obtain a contradiction by Remark 4.3. Thus 2 T = E7 (pe ). 7. Proofs of Theorems 1.2 and 1.3 

Proof of Theorem 1.3: By Lemma 3.3, for G = PΔL(2, 32 ) and N is a subgroup of G of index 2, each element of G \ N has order 2+1 or 2+2 . On the other hand, if G is an almost simple group and N
G such that either  G = PΔL(2, 32 ) or N is not of index 2, then there exist elements of G \ N of order divisible by two distinct primes, refer to Lemma 4.1 and Lemmas 5.1-6.4. 2 The assertion of Theorem 1.2 follows from Theorem 1.3 and Lemma 2.4. References [1] J. N. Bray, D. F. Holt, C. M. Roney-Dougal, The Maximal Subgroups of the Low-Dimensional Finite Classical Groups, Cambridge University Press, Cambridge, 2013. [2] A. R. Camina, Some conditions which almost characterize Frobenius groups, Israel J. Math. 31 (1978) 153-160. [3] R. W. Carter, Simple groups and simple Lie algebras, J. London Math. Soc. 40 (1965) 193-240. [4] R. W. Carter, Simple Groups of Lie Type, Wiley, London, 1989. [5] C. Chevalley, Sur certains groupes simples, Tˆ ohoku Math. J. (2) 7 (1955) 14-66. [6] J. H. Conway, R. T. Curtis, S. P. Norton, R. A. Parker, R. A. Wilson, Atlas of Finite Groups, Oxford University Press, Oxford, 1985. [7] R. M. Guralnick, G. Navarro, Squaring a conjugacy class and cosets of normal subgroups, Proc. Amer. Math. Soc. 144 (2016) 1939-1945. [8] B. Huppert, Endliche Gruppen I, Springer, Berlin, 1967. [9] P. B. Kleidman, M. W. Liebeck, The Subgroup Structure of the Finite Classical Groups, Cambridge University Press, Cambridge, 1990. [10] M. L. Lewis, On p-group Camina pairs, J. Group Theory. 15 (2012) 469-483. [11] C. H. Li, L. Wang, Relative elementary abelian groups, and a class of edge-transitive Cayley graphs, J. Aust. Math. Soc. 100 (2016) 241-251. [12] G. M. Seitz, Topics in the theory of algebraic groups, Group representation theory, EPFL Press, Lausanne, (2007) 355-404. [13] R. A. Wilson, The Finite Simple Groups, Springer, 2009. Department of Mathematics, South University of Science and Technology, Shenzhen, Guangdong 518055, P. R. China E-mail address: [email protected] Department of Mathematics, Yunnan University, Kunming 650091, People’s Republic of China E-mail address: [email protected]