Subgroups of Type A 1 Containing a Fixed Unipotent Element in an Algebraic Group

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1 Journal of Algebra 23, (2000) doi:0.006/jabr , available online at on Subgroups of Type A Containing a Fixed Unipotent Element in an Algebraic Group Richard Proud and Jan Saxl Department of Pure Mathematics and Mathematical Statistics, University of Cambridge, 6 Mill Lane, Cambridge, CB2SB, United Kingdom and Donna Testerman Le Carroz, 867 Ollon, Switzerland Communicated by Gordon James Received December 6, 999 INTRODUCTION Let G be a simple algebraic group defined over an algebraically closed field of characteristic p>0. In [28, 4.], it was shown that any element u G of order p is contained in a closed subgroup of type A, provided that p is good. The corresponding result was also obtained for finite simple groups of Lie type over finite fields of good characteristic. These results have found many applications in investigations of subgroup structure of these simple groups. In this paper we extend the results to cover also the cases where the characteristic is not good. We also extend some of the proofs in [28]. The main result of this paper is the following. Main Theorem. Let G be a simple algebraic group defined over an algebraically closed field of characteristic p>0. Let σ be either a Frobenius morphism of G, orσ =. Let u G be a σ-invariant element of order p. Supported by Engineering and Physical Sciences Research Council Grant GR/L /00 $35.00 Copyright 2000 by Academic Press All rights of reproduction in any form reserved.

2 54 proud, saxl, and testerman Then u is contained in a closed σ-invariant subgroup of G of type A, except in the following cases: (i) G = G 2 p= 3, and u is an element of order 3 of type A 3 (ii) G = G 2, p = 3, and σ is a morphism involving the graph morphism of G, (iii) G = B 2 or G = F 4, p = 2, and σ is a morphism involving the graph morphism of G. We also obtain a somewhat more general result for semisimple algebraic groups defined over an algebraically closed field of characteristic p>0. Further, we prove a result for the corresponding finite groups. Since this paper has been under preparation, an important paper of Seitz [8] has come to our attention. In it, assuming that the characteristic is good, Seitz shows that each element of order p is contained in a canonical subgroup of type A, unique up to conjugacy in C G u.,. q-frobenius MORPHISMS In this section we introduce a special type of Frobenius morphism of a semisimple algebraic group and consider how it behaves with respect to restriction to certain semisimple subgroups. Let G be a semisimple algebraic group defined over an algebraically closed field K of characteristic p>0. Let σ G G be a surjective morphism of algebraic groups with finite fixed point subgroup G σ, that is, a Frobenius morphism of G. Note that σ will in fact be bijective; see [26, 7.]. Fix a σ-invariant Borel subgroup B of G and a σ-invariant maximal torus T of B. Then the opposite Borel subgroup B is σ-invariant, as are U = R u B and U = R u B.Let = G T denote the roots of G relative to T, and let denote the simple roots. For each α we may choose isomorphisms (of algebraic groups) x α a U α suchthat n α = x α x α x α N G T. The family x α α is called a realization of in G (see [23,.2] for more details). There then exists a permutation ρ of and powers q α of p α, suchthat σx α k =x ρα c α k q α, for some c α m and all k K (see [26,.2]). We have q α =q α c α c α =, and may take c α = for ±α, modifying the isomorphisms if necessary (see [8, 6.2C]). We will call σ a q-frobenius morphism if q α =q for all α. We can then write σ = gq, where qx α k =x α k q and gx α k =x ρα c α k, for all α and k K. Note that g is an algebraic automorphism of G, since some power of g is the identity map. When G is simple, we remark that any morphism involving a twisted graph morphism in bad characteristic is not a q-frobenius morphism (see Section 3).

3 subgroups of type A 55 Finally if G G n, n, are the simple components of G, we shall call Gσ-simple if, after possible renumbering, we have σg i = G i+, subscripts taken modulo n. For a more detailed account of the above, consult [26]. The following two lemmas are concerned with the restriction of a q- Frobenius morphism to semisimple subgroups of G, withsimple components of type A. For the first one, cf. [2, Proposition.3]. Lemma.. Let G be a semisimple algebraic group defined over an algebraically closed field K of characteristic p>0. Let σ be a q-frobenius morphism of G. Suppose X is a closed σ-invariant A -subgroup of G. Then σ X is a q-frobenius morphism of X. Proof. Let T B be as above. Considering a suitable power of σ = gq, we may assume that g =. Let B X be a σ-invariant Borel subgroup of X and T X a σ-invariant maximal torus of B X.AsC G T X is connected and σ-invariant, it contains a σ-invariant maximal torus x T of G, for some x G. Since T is σ-invariant, we have n = x σx N G T. Moreover, T X C G x T = x T.Fort T,wehaveσt = t q,sotx x is σ-invariant, and n N G TX x.if X T X = ±β, then β σ x t = qβ xn t for all x t T X. Since TX x = m, it follows from [26,.2(b)] that i n TX x =. Now apply [26,.2(c)]. Lemma.2. Let G be a semisimple algebraic group defined over an algebraically closed field K of characteristic p>0, with simple components G G n n. Let σ G G be a q-frobenius morphism of G for which G is σ-simple. Suppose X is a closed σ n -invariant A -subgroup of G. Then Y = X σx σ n X is a closed semisimple σ-invariant σ-simple subgroup of G, and σ Y is a q-frobenius morphism of Y. Proof. Let T B be as above. Clearly Y is σ-invariant, so it contains a σ-invariant Borel subgroup B Y and σ-invariant maximal torus T Y of B Y. Moreover, T Y x T, for some x G. Let x ±βi i = n,bea realization of Y T Y in Y, where ±β i corresponds to X i = σ i X.By [26,.2], we have σ X i X i X i+, where σx ±βi k =x ±βi+ k q i, for all k K, with q i a power of p (modifying isomorphisms if necessary). Define g i X i X i+ by g i x ±βi k =x ±βi+ k. Then g i is a bijective morphism of algebraic groups (abstractly, g i = σqi, and use the big cell [27, 5.7] for variety morphism). Moreover, gi = g i g g n g i+,so g i is an isomorphism of algebraic groups. Now T i = T Y X i is a maximal torus of X i.let x t T i. Then σ x t = σx gt q =g i x t q i, where σ = gq on G. Letφ = i x gi i σx g Ti x. Then φ is an algebraic automorphism of Ti x, satisfying φ tq =t q i, for all t T x i. Since Ti x = m, we must have q = q i, as required.

4 56 proud, saxl, and testerman 2. PRODUCTS OF A S In this section we handle the basic step of our induction. In the process, we provide a clarification of [28, 4.], together with some details of its proof, which were absent there. In fact, [28, 4.] is not quite correct as stated; however, it is valid for simple groups, as well as for semisimple groups G, under minor restrictions which are satisfied in applications. For example, they hold if G is a subsystem subgroup of a simple group in good characteristic, with σ the restriction to G of a Frobenius morphism of the simple group. Let G = X X 2 X n, n>, be a semisimple algebraic group defined over an algebraically closed field K of characteristic p > 0, withsimple components X i, i = n, of type A.Letσ G G be a Frobenius morphism of G. Choose a σ-invariant Borel subgroup B of G, and a σ- invariant maximal torus T of B. Let = ±α i be the corresponding set of roots, with ±α i corresponding to the component X i.let x ±αi be a realization of in G. By [26,.2] there is a permutation ρ of, and powers q i = p e i of p, suchthat σx ±αi k =x ±ραi c ±i k q i, for some c±i m with c i c i =, and all k K. We can modify the realization so that c i =. The following two lemmas are special cases of the above situation and are essentially the only ones that arise in our problem. Lemma 2.. Suppose ρα i = α i+, where α n+ = α. Let u G σ with o u =p. (i) If u X, a closed σ-invariant A -subgroup of G, then n n i= e i. (ii) If σ is a q-frobenius morphism (i.e., each q i = q), then u X, a closed σ-invariant A -subgroup of G, with σ X a q-frobenius morphism of X. Proof. (i) Suppose X is sucha subgroup. Choose a σ-invariant Borel subgroup B X of X, and a σ-invariant maximal torus T X of B X.Let x ±β be a realization of X = X T X in X. Now X is σ n -invariant, and σ n = q 0 = n i= q i on G relative to the realization x ±αi. By Lemma. we have σ n x ±β k =x ±β k q 0, for all k K, modifying the realization if necessary. But by [26,.2], σx ±β k =x ±β c ±β k q β, with qβ = p e, for some e, and c ±β m. Thus ne = n i= e i, as required. (ii) We have u = u σu σ n u, where u X satisfies o u =p and σ n u = u. There exists x X σ n with x u = x α λ, for suitable 0 λ F q n. Moreover x σx σ n x G σ, so it will be enoughto assume u = x α λ. Let X = x gx g n x x X, where σ = gq (see Section ). Consider the isogeny θ X X where θ x =x gx g n x, x X. Then X is a closed A -subgroup of G.

5 subgroups of type A 57 Write T = T X and T X = θt. Then x ±β, where x ±β = θx ±α,is a realization of X T X in X. Moreover σx ±β k =x ±β k q, for all k K. Now there exists t T with t x α k =x α λk, k K. Since λ qn = λ, wehaveσ n t = tz, z Z X. Set a = t σt σ n t G and y = θx α X. Then a y = u and σa = az. It follows that a X is the required subgroup. Note that Lemma 2. (ii) is a special case of the converse of (i). In fact, we remark that the converse is true in general, but the proof is more involved, and we do not give it here. In the next lemma, we consider the other extreme, where ρ fixes each α i. Lemma 2.2. Suppose ρ =. Let u = n i= u i, where u i X i σ and o u i =p. Then u X, a closed diagonal σ-invariant A -subgroup of G, if and only if q = q 2 = = q n = q. Moreover if such an X exists, then σ X is a q-frobenius morphism of X. Proof. Suppose X is sucha subgroup. Let n = 2, the general case being essentially the same. We may assume G is of adjoint type (cf. paragraph 4 of Proof of Theorem 5.), so G = X X 2 is a direct product of algebraic groups. Choose a σ-invariant Borel subgroup B X of X, and a σ-invariant maximal torus T X of B X.Let x ±β be a realization of X = X T X in X. By [26,.2] we have σx ±β k =x ±β c ±β k q, for all k K, with q some power of p, and we may take c β =. Now T X x T, for some x G, with n = x σx N G T. Let x t T X, t T. Then σ x t = x t q.we have T = T T 2, where T i = T X i is a maximal torus of X i, so can write (uniquely) t = t t 2, t i T i. Since σt i = t qi i, we obtain t q tq 2 = n t q tq 2 2, and so n t q i i =t q i. The projection pr i Xx X i is an isogeny, so pr i TX x is a maximal torus of X i contained in T i, hence equal to T i. Since T i = m,we obtain q i = q (cf. Proof of Lemma.2). Conversely, assume n = 2 by induction. There exists x i X i σ with u i = x i xαi λ i, for suitable λ i F q, i = 2. Set a = x x 2 G σ. Then u = a x α λ x α2 λ 2, so it is enoughto consider x α λ x α2 λ 2. Without loss of generality, there is a central isogeny ψ X X 2 (for definition, see [2, 22]). Modifying ψ by an inner automorphism, if necessary, we may assume ψb = B 2 and ψt = T 2, where B i = B X i, T i = T X i. (Then ψb = B 2 by the uniqueness of opposite Borel subgroups.) Then ψx ±α is another realization of X 2 T 2 in X 2, so there exists µ m with ψx ±α k =x ±α2 µ ± k for all k K (see [23,.2.]). Replacing ψ by i t ψ for suitable t T 2, we may further assume that µ = λ λ 2. Define X = x ψx x X, a closed A -subgroup of G. Consider the isogeny θ X X where θ x =x ψx), x X. Define T X = θt θb. Then x ±β, where x ±β = θx ±α, is a realization of X T X in X. Since λ q i = λ i,

6 58 proud, saxl, and testerman we have σx ±β k =x ±β k q for all k K. Moreover u = x β λ X, and so we are done. 3. TWISTED CASES AND A 3 ELEMENTS IN G 2 In this section we consider the exceptional cases stated in the Introduction. Let G be a simple algebraic group of type B 2, G 2,orF 4 defined over an algebraically closed field K of characteristic p = 2 3, or 2, respectively. Fix a Borel subgroup B of G, a maximal torus T of B, and a realization x α α of = G T in G. Letg G G be the twisted graph morphism corresponding to the symmetry of the Dynkin diagram of G. The symmetry extends to a permutation ρ of which interchanges long and short roots. Then gx α k =x ρα c α k p α, for all k K, where p α =ifα is long, p α =p if α is short. Moreover c α =±, with c α = for ±α simple. Let σ = gq G G, where qx α k =x α k q, q = p e, e 0. Note that σ is not a q-frobenius morphism of G. See [27, 0] for more details. Lemma 3.. There are no closed σ-invariant A -subgroups of G. Proof. We have σ 2 = pq 2 = p 2e+ on G relative to the given realization. Suppose X is a closed σ-invariant A -subgroup of G. Choose a σ-invariant Borel subgroup B X of X and a σ-invariant maximal torus T X of B X.Let x ±β be a realization of X = X T X in X. Since X is σ 2 -invariant, Lemma. gives σ 2 x ±β k =x ±β k p2e+, for all k K (modifying if necessary). But [26,.2] gives σx ±β k =x ±β c ±β k q β, for some qβ = p f, and so 2f = 2e +, a contradiction. However it is still possible that G σ contains (finite) A -subgroups, over fields of characteristic p, overlying unipotent elements. We now consider the various possibilities. If G is of type B 2 and σ G G is defined as above, then G σ = 2 B 2 2 2e+ of order r 2 r 2 + r, r = 2 2e+. Since 3 does not divide this order, it follows that G σ does not contain any subgroups isomorphic to SL 2 2 = Sym 3. Now let G be of type G 2,sothatG σ = 2 G 2 3 2e+. By [7, 8.5] the Sylow 2-subgroups of G σ are elementary abelian of order 8. It follows that G σ does not contain any subgroups isomorphic to SL 2 3 i or PGL 2 3 i, i. Indeed SL 2 3 has one Sylow 2-subgroup which is isomorphic to Q 8, and PGL 2 3 = Sym 4 has three Sylow 2-subgroups which are isomorphic to D 8. However, we have the following result, where G 2 a denotes the class of subregular unipotent elements of G. Lemma 3.2. Let G be a simple algebraic group of type G 2 defined over an algebraically closed field of characteristic p = 3, and let σ G G be

7 subgroups of type A 59 the morphism defined above. Let u G 2 a G σ. Then u lies in a subgroup of G σ isomorphic to PSL 2 3 2e+. Moreover, such a subgroup contains representatives of the two G σ -classes in G 2 a G σ. Proof. Let u G 2 a G σ. Then C G u =V t, where V = C G u is unipotent and t is an involution (see [0, 3.], for example). Let y = vt C G u, v V, be an involution. Then y t is dihedral of order 2o yt, and so y is conjugate to t by an element of v. Thus by Lang s theorem there exists an involution s C G u σ. Now by [29], s is a representative of the unique class of involutions in G σ, withcentralizer isomorphic to s PSL 2 3 2e+. Thus u C G s σ = C Gσ s = s PSL 2 3 2e+, and so u PSL 2 3 2e+ by the Jordan decomposition. For G of type G 2, we are left withthe class A 3. A representative of this class is u = x 2a+b x 3a+2b, with a b simple and a short. We have C G u =U and u G σ (see [7, Table 2;, Table B], for example). In particular, u is semiregular and A 3 G σ gives rise to a single class in G σ withrepresentative u. Lemma 3.3. Let G be a simple algebraic group of type G 2 defined over an algebraically closed field K of characteristic 3. Let u = x 2a+b x 3a+2b A 3. Then u does not lie in any subgroup of G isomorphic to PSL 2 3. Proof. Suppose u H G, where H = PSL 2 3 = Alt 4. Then H = V u, where V H is the Klein 4-group. An element of H either centralizes V, and hence lies in V, or via conjugation permutes the involutions of V transitively. Let t V. Since u/ V,wehave u t = t 2, t, and V = t t 2. Now t and t 2 are commuting semisimple elements of G, which is simply connected, and so V T, a maximal torus of G (see [9, 2.; 24, II, 5.]). We claim that N G V =N G T () Since dim T =2, we have V = O 2 T, the subgroup of T generated by its involutions, and so N G T N G V. In the notation of [3, 7.], we have V = h χ h χ 2, where χ a =, χ b =, χ 2 a =, and χ 2 b = (the remaining involution is h χ χ 2 ). Suppose x r k C G V, r +, k K. Then χ r =χ 2 r =, since h χ i x r k h χ i = x r χ i r k by [3, 7.]. Writing r = λ a + λ 2 b, gives = λ = λ 2, and so λ i 0 2 4, a contradiction as there are no roots of this form. It follows that C G V = T (see [24, II, 4.(b)]). Since V is a diagonalizable group, we have N G V = C G V = T, by rigidity [8, 6.3]. Thus if x N G V, then xtx N G V = T, and so () holds. In particular, u N G T. We next show N G T has a unique class of elements of order 3 (2)

8 60 proud, saxl, and testerman First W = N G T /T = Dih 2.Soifx N G T has order 3, then xt is ut or u T. Suppose x = u s, s T (the former case is simpler and is omitted) and consider the map φ T T, where φt = utu t, t T. Clearly φ is a morphism of varieties and is injective since u is semiregular. By dimensions we have φ onto as well. Thus s = utu t, for some t T, and so x = tu t. But u is conjugate to u by h χ χ 2 T, and so (2) holds. Finally, using [6, 2], we have elements n 2 n 3 N G T satisfying n 2 x r k n 2 = x r k and n 3 x r k n 3 = x w3 r k for all r k K (3) where w 3 W has order 3. Since the x r k generate G, and Z G =, Eqs. (3) imply o n 2 =2, o n 3 =3, and n 2 n 3 =. But u is conjugate to n 3 in G by 2, contradicting the fact that u is semiregular, thus proving the lemma. We now consider the implications of this result. Let τ = q or τ = gq = σ, as defined above. Then G τ = G 2 q or G τ = 2 G 2 3q 2, respectively. Corollary 3.4. (i) If u A 3 then u does not lie in any closed (algebraic) A -subgroup of G. (ii) Let τ = q or τ = gq and let u A 3 G τ. Then u does not lie in any ( finite) A -subgroup of G τ. Proof. (i) Suppose u X, a closed A -subgroup of G. Since u is semiregular, X = PSL 2 K, so X contains a subgroup Y = PSL 2 3. Choose v Y witho v =3. Then u is conjugate to v in X, contradicting Lemma 3.3. (ii) Suppose u lies in a (finite) A -subgroup of G τ. Since u is semiregular, it must lie in a subgroup isomorphic to PSL 2 q, q = 3 i, i. This group has two conjugacy classes of elements of order 3, which are fused by an outer automorphism. Therefore, every element of order 3 in PSL 2 q lies in a PSL 2 3 -subgroup. This contradicts Lemma 3.3. Finally let G be of type F 4 and let σ G G be as above. Then G σ = 2 F 4 2 2e+. By [9, 2, Corollary 2], G σ has two classes of involutions withrepresentatives u = α 0 and u 2 = α 2 in the notation of [20, Table II]. Moreover u is non-central and u 2 is central; that is, u 2 lies in the centre of a Sylow 2-subgroup of G σ, while u does not (also see [2, VII, 7.]). Lemma 3.5. In the notation above, u lies in a subgroup of G σ isomorphic to SL 2 2 2e+. However, u 2 does not lie in any subgroup of G σ isomorphic to SL 2 2.

9 subgroups of type A 6 Proof. For u see [4, 2.2; 2, II, (2.2)]. In particular, u inverts an element t G σ of order 3. So N = N Gσ t = C Gσ t u = SU 3 2 2e+ 2 by [4,.2(); 20, 3.2]. All involutions in N inverting t are conjugate in N to u by [, 9.8(i)]. Since G σ has a unique class of elements of order 3 (see [20, Table IV]), the statement about u 2 follows. 4. DISTINGUISHED CLASSES IN BAD CHARACTERISTIC In this section, for a simple algebraic group G, we determine the distinguished classes of elements of order p, where p is a bad prime for G. Lemma 4.. Let G be a simple classical algebraic group of type B l, C l,or D l defined over an algebraically closed field of characteristic 2. Suppose G contains a distinguished involution. Then G is of type C 2 and has a unique class of distinguished involutions. Moreover, if σ is a q-frobenius morphism of G, orσ =, and u G σ, then u X, a closed σ-invariant A - subgroup of G. Proof. For eachtype, it is enoughto consider one group in the isogeny class (see Proof of Theorem 5., for example). Moreover, by [2, 23.6], there exists a (purely inseparable) isogeny ψ O 2l+ K Sp 2l K, so we need not consider type B l at all. Let G = Sp V be of type C l, withdim V = 2l, l 2. By [, Sect. 7] we have l = 2, and G has a unique class of distinguished involutions with representative v = relative to A = where A is the matrix of the symplectic form on V. Moreover, C G v is a maximal connected unipotent subgroup of G, so no splitting occurs in G σ. Thus by Lang s theorem, we may take σ to be the qth-power map on matrix elements. Then σv = v and v lies in a closed σ-invariant subgroup of G of type A A, so we are done by Lemma 2.2. Finally, if G is of type D l, there are no such classes by [, Sect. 8]. Lemma 4.2. Let G be a simple exceptional algebraic group defined over an algebraically closed field of bad characteristic p. Ifu G is distinguished of order p, then G is of type G 2, p = 3, and u G 2 a or u A 3. Moreover, if σ is a q-frobenius morphism of G, orσ =, and u G 2 a G σ, then u X, a closed σ-invariant A -subgroup of G.

10 62 proud, saxl, and testerman Proof. For type G 2, use [7, ] to obtain the given classes. For the subregular class G 2 a, we may take G σ representatives x a+b x 3a+b and x a+b x 3a+b µ, where µ F q is a non-square (with a b simple and a short). Both lie in the closed σ-invariant semisimple subgroup H = U ± a+b U ± 3a+b of G of type A A, so we may apply Lemma 2.2. For type F 4, use [, 9, 22], and for types E n, n = 6 7 8, use [, 5, 6], to see that there are no distinguished classes of order p 5. MAIN THEOREM We are now in a position to prove the main result of this paper, as stated in the Introduction. In the process we obtain a more general result for semisimple algebraic groups. Theorem 5.. Let G be a semisimple algebraic group defined over an algebraically closed field K of characteristic p>0. Let σ G G be a q- Frobenius morphism of G, orσ =. Let u G σ with o u =p. Ifp = 3 and G has a simple component of type G 2, assume that the projection of u into this component does not lie in the class A 3. Then u X, a closed σ-invariant A -subgroup of G, with σ X a q-frobenius morphism of X. Proof. We use induction on dim G 3, the case where dim G =3, where G is of type A, being immediate. If σ =, the result follows from [28, 4.], Lemmas 4. and 4.2, so assume σ is a q-frobenius morphism of G. LetG G 2 G n be the simple components of G. First suppose G is σ-simple, with n 2. Then u = u σu σ n u, where u G satisfies o u =p and σ n u = u. Moreover, σ n G is a q n -Frobenius morphism of G. By induction, u X, a closed σ n - invariant A -subgroup of G. Thus u Y = X σx σ n X, and so we are done in this case by Lemmas.2 and 2.(ii). Next consider the general case G = H H 2 H t, where H i is the product of components in an orbit of σ on the simple components of G, and so is a closed σ-simple semisimple (and possibly simple) subgroup of G. Then u = u u 2 u t G, where for i t, u i H i σ and o u i =p (removing any superfluous H i if necessary). By above, assuming the result holds for simple groups, we have u i X i, a closed σ-invariant A -subgroup of H i, and so we are done by Lemmas. and 2.2. We are thus reduced to the case where G is simple. Relative to some maximal torus T of G, and realization x α α of = G T, wehave σ = gq G G. If u is not distinguished, there exists a semisimple element s C G u. Since C G u is connected and σ-invariant, it contains a σ-invariant torus S. Now u C G S, which is connected and σ-invariant, so contains a σ-invariant maximal torus x T of G, for some

11 subgroups of type A 63 x G (and S x T, since C G x T = x T ). Since T is also σ-invariant, we have n = x σx N G T. Now C G S = x T U α α, for some closed subsystem of. Define G = T U α α and H = G = U α α.letρbe the symmetry of the Dynkin diagram of G corresponding to the graph automorphism g (possibly trivial); see [26,.6]. Now nt = w W = N G T /T, and wρ =.Let be a base of. Then wρ is another one, so there is w W = N G T /T = N G T G /T with wρ = w. Thus w wρ =,sow wρ corresponds to a symmetry ρ of the Dynkin diagram of H. Letg H H be the graph automorphism of H corresponding to ρ. Choose n N G T suchthat n T = w. Then for α,wehave i n i ng U α =U ρ α = g U α. Thus i n i ngx α =x ρ α c α, for some c α m, whereas g x α =x ρ α. Since is a linearly independent subset of X T, there exists t T with i t i n i ngx α =g x α, for all α. Set ψ = g i ti n i ng Aut alg H. Then ψx α =x α, for all ±α (see [25, 5.2]). Then ψ is a field automorphism of H by [25, 5.7]; also see [4, Proposition 6]. Thus ψ is the identity map, and so i n g = i y g, where y = n t G. By Lang s theorem, there is a G with x y = σ x a x a. For α, define iso y α = i xa x α a xa U α x H. Relative to the realization y α α,we have σy α k =i σ xa n i n g x α k q =i xa g x α k q =y ρ α d α k q, for some d α m, and all k K. Thus σ x H is a q-frobenius morphism, and we are done by induction in this case. Now let u be distinguished. If p is bad, we are done by Lemmas 4. and 4.2, so assume p is good. We may assume G is of adjoint type. Indeed, let ψ = ad G G = G ad be the adjoint representation of G. Set T = ψt.forα, define x α = ψx α a U α = ψu α. Then x α α is a realization of G T in G. Define σ G G by σ ψx =ψ σx, x G. Now u = ψu G σ and o u =p, so by assumption u X, a closed σ-invariant A -subgroup of G. Define X = ψ X. Then X is simple of dimension dim X, so is of type A. Moreover, u X (since ψ X =XZ G ) and X is σ-invariant, as required. We may assume C G u is connected. Given that u is distinguished, we have C G u disconnected if and only if u is not semiregular. (Here u is semiregular if C G u is unipotent.) For this we need the structure of the component group A u =C G u /C G u. See [5, 3.], for example; the possibilities are Z2 i, Sym 3, Sym 4,orSym 5. When G is exceptional, we have i =. For the cases Z 2 and Sym 3, argue as in claim 7 of [28, 4.]. The cases Sym i, i = 4 5, only occur for a single class in types F 4 and E 8, respectively. Moreover, in bothcases, there exists v G σ where u = gvg, g G, with σ acting trivially on the component group A v =C G v /C G v ; see [6, Lemma 70; 22, (2.3)]. (These cases are also dealt with in [28, 4.], but we provide here an alternative proof modelled on [22, (2.3)].) Thus we

12 64 proud, saxl, and testerman can write C G v = a i C G v, for some a i C G v σ by Lang s theorem. Now C G u = ga i g C G u, and h = g σ g C G v. The coset ga i g C G u is σ-invariant if and only if h a i =, bars denoting images in A v. Note that a non-σ-invariant coset cannot contain any σ-invariant elements, whereas each σ-invariant coset has a σ-invariant representative b i by Lang s theorem. Moreover, distinct elements a i a j of C A v h give rise to distinct cosets. Therefore C Gσ u = ai C A v h b i C G u σ where b i ga i g C G u. Thus if r denotes the order of C A v h, then r divides C Gσ u. But r and r divides A v = 2 3 3or , respectively, and so u commutes witha non-identity σ-invariant semisimple element of G, since p is good. We are then done by induction (cf. the nondistinguished case above). Finally we consider the case Z2 i, i>, which only occurs for G classical of type B C, ord. We claim there exists v G σ where u = gvg, g G, with σ acting trivially on A v, so we can then argue as in the cases above. Let V be a finite dimensional vector space over K, endowed witheither a non-degenerate symplectic or quadratic form. It will be convenient to work first in G = I V, the full classical group of V,sothat G = Sp V or O V, respectively. The claim will then follow on applying the adjoint representation to G = G, on noting that every element of G which is distinguished but not semiregular is obtained from a like element in G.Let σ be a q-frobenius morphism of G and extend naturally to G; so σ acts trivially on Z O V in case G = O V. Note that the case of triality in D 4 does not arise here, as the component groups in that case are either trivial or Z 2 (see [5, Chap. 3], for example). Therefore σ is induced by an element of ƔL V, the group of semilinear maps of V, as described in [3, Sect. 2], for example. Write V = V V r,an orthogonal direct sum of non-degenerate subspaces of different dimensions, all even dimensional for G = Sp V, and all odd dimensional for G = O V. From the definition of σ in [3], it is clear that we may take each V i to be σ-invariant. Let H = g G gv i = V i = I V I V r, where each I V i =Sp V i or O V i, depending on whether G = Sp V or O V, respectively. Note that each I V i = g H g V j = j i (and H) is σ-invariant. Let v i I V i be a σ-invariant regular unipotent element. Then v = v v r is a σ-invariant regular unipotent of H. By [28, 3., 3.2], every distinguished unipotent element of G is conjugate in G to an element of this form. Write L = i Z I V i C G v. Then it follows from [24, IV, 2.26(ii)] that C G v =C G v L. Note that σ acts trivially on L. Since A v =C G v /C G v is a subgroup of C G v /C G v, it follows that σ acts trivially on A v, as required.

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