Irreducible Subgroups of Symplectic Groups in Characteristic 2

J. Austral. Math. Soc. 73 (2002), 85–95

IRREDUCIBLE SUBGROUPS OF SYMPLECTIC GROUPS IN CHARACTERISTIC 2

CHRISTOPHER PARKER and PETER ROWLEY

(Received 13 January 2000; revised 25 July 2001)

Communicated by R. B. Howlett

Abstract

Suppose that V is a finite dimensional vector space over a finite field of characteristic 2, G is the symplectic group on V and a is a non-zero vector of V . Here we classify irreducible subgroups of G containing a certain subgroup of O2.StabG hai/ all of whose non-trivial elements are 2-transvections.

2000 Mathematics subject classiﬁcation: primary 20CDE. Keywords and phrases: symplectic group, orthogonal group, 2-transvections, transvections.

1. Introduction

Let k denote the Galois ﬁeld GF.q/,whereq = 2m , and suppose V is a ﬁnite dimensional vector space over k. An involution g of GL.V / is a transvection (respectively a

2-transvection) of V if CV .g/ has codimension 1 (respectively 2) in V . A subgroup K of GL.V / is called a transvection subgroup if CV .K / has codimension 1, [V; K ] has dimension 1, and K is isomorphic to the additive group of k. Assume, additionally, that dim V = 2n where n ≥ 2and f is a non-degenerate alternating bilinear form on V . Letting G denote Sp.V /, the symplectic group on V deﬁned by f , we may now state our main result.

THEOREM 1.1. Let a be a non-zero vector of V and put H = StabG hai. Suppose that X is a subgroup of O2.H/ which satisﬁes .i/ |X|=q2n−2; and .ii/ no element of X acts as a transvection on V .

c 2002 Australian Mathematical Society 1446-8107/2000 $A2:00 + 0:00

85 86 Christopher Parker and Peter Rowley [2] If L is a subgroup of G which contains X and acts irreducibly on V , then L acts naturally on V as one of Sp.V /, .V /, O.V / or q = n = 2 and L =∼ Alt.6/ =∼ . /0 0 . ; / =∼ . / Sp4 2 or L 2 4 Sym 5 . We remark that the listed groups all satisfy the hypothesis of Theorem 1.1.The = = . / =∼ exceptional cases when q n 2 are caused by the non-simplicity of Sp4 2 Sym.6/ in the Alt.6/ case and by the fact that Sym.6/ has two classes of subgroups isomorphic to Sym.5/ in the second case. To clarify the second case further take V to be the doubly deleted GF.2/-permutation module for Sym.6/, the point stabilizer . / −. / . / . / Sym 5 acts on V as O4 2 , while the transitive Sym 5 when restricted to Alt 5 acts on V as SL2.4/.

Observe that, as the non-trivial elements of the elementary abelian 2-group O2.H/ are either transvections or 2-transvections, every non-trivial element of X must act as a 2-transvection on V . So Theorem 1.1 may be seen as a kindred spirit to the results of McLaughlin [3, 4], on irreducible linear groups which contain transvection subgroups and also to work of Dempwolff’s [1, 2]. In [1, 2] Dempwolff shows that an irreducible subgroup Y of SL.V / which is generatedby 2-transvections either contains transvections, has a normal abelian subgroup of odd order which has 1-dimensional homogeneous components on V which Y permutes transitively, contains a normal complex of Stellmacher elements, or a normal complex of roots involutions. Theorem 1.1 generalizes a result due to Timmesfeld for the case q = 2, which plays an important part in the proof of his Theorem 4.5 [9]. This more general theorem plays an equally vital role in the classification of symplectic amalgams [5]. The proof given here is modelled on Timmesfeld’s approach. We remark that Theorem 1.1 is also the principal content of [7, Satz 3.10]. However, the proof there calls upon the (lengthy) classification of groups generated by root involutions and then deals with the resulting configurations case by case. Our proof is direct and elementary—the only substantial results we use being the classification of groups generated by transvections due to McLaughlin [3, 4]. Our notation follows that of [8]. The following elementary result will be used in the proof of Theorem 1.1.

LEMMA 1.2. Let V be a vector space of dimension 2n, n ≥ 1 which is equipped with a non-degenerate alternating bilinear form f .PutG = Sp.V /.Ifa and b are non-zero vectors of V with f .a; b/ 6= 0, then

G =hO2.StabGhai/; O2.StabGhbi/i:

PROOF. See [6]. [3] Irreducible subgroups of symplectic groups 87 2. Orbits on points and vectors and a quadratic form

From now on we assume the situation depicted in Theorem 1.1.IfL contains a transvection subgroup, then we may avail ourselves of McLaughlin’s results ([3, 4]) to conclude, since L ≤ Sp.V /, that L must be isomorphic to one of Sp.V /; O.V /; Sym.2n + 1/,orSym.2n + 2/. Of these groups only Sp.V / and O.V / can contain an elementary abelian 2-group as large as X and so Theorem 1.1 holds in this case. Henceforth, therefore, we shall suppose that L contains no transvection subgroups.

⊥ LEMMA 2.1. .i/ [a ; X]=hai; .ii/ [V; X]=a⊥;

.iii/ CV .X/ =hai; and .iv/ Let Z be a subgroup of index q in X and let U =ha; ci be a 2-dimensional ⊥ subspace of a .IfZ centralizes U, then CV .Z/ = U.

⊥ ⊥ PROOF. Since O2.H/ ≥ X, hai=[a ; O2.H/]≥[a ; X] and part (i) follows as X contains no transvections. Now suppose that [V; X] < a⊥. Then |[V; X]| ≤ q2n−2. v ∈ \ ⊥ ={[v; ]| ∈ } 2n−2 So, letting V a , Î x x X contains at most q vectors. If

| 2n−2 |= Î =[ ; ] [v; ]= ∈ Î q , then V X and hence, by part (i), x a,forsomex X. But then [V; x]≤h[a⊥; x]; [v; x]i ≤ hai; which implies that x is a transvection on V , | | < 2n−2 | |= 2n−2 a contradiction. Thus Î q .SinceX q , there must exist distinct x1; x2 ∈ X with [v; x1]=[v; x2], whence [v; x1x2]=0. This then implies that x1x2 is a transvection on V . Hence we infer that [V; X]=a⊥, and we have (ii). Using ⊥ ⊥ ⊥ part (ii) gives CV .X/ =[V; X] = .a / =hai; so proving (iii). For part (iv), since Z centralizes a and c,wehave[V; Z]≤a ⊥ ∩ c⊥ = U ⊥ with |U ⊥|=q2n−2.Now,as|Z|=q 2n−3, we may argue as in part (ii) to obtain ⊥ ⊥ [V; Z]=U . Hence CV .Z/ =[V; Z] = U. By hypothesis L acts irreducibly upon V and so haL i=V . Thus we can ﬁnd an image hbi of hai (under L) with b ∈ V \ a⊥. Select the vector representative b of hbi . ; / = ½ ∈ = {h x + ½ i| ∈ } ; so as f a b 1. Then, for k we put Ç½ b a x X and note that \ ⊥ each Ç½ is an X-orbit which consists of points (1-spaces of V )inV a .

⊥ LEMMA 2.2. X has q regular orbits on the points in V \ a .TheseX-orbits are ½ ∈ Ç½, k.

⊥ PROOF. Suppose hvi is a point of V \ a whichisﬁxedbyx ∈ X #. Then, by Lemma 2.1 (ii), [hvi; x]≤[V; X]∩hvi=a⊥ ∩hvi=0: Hence

[V; x]=[a⊥ +hvi; x]=[a⊥; x]≤hai; 88 Christopher Parker and Peter Rowley [4] a contradiction as X contains no transvections. Therefore the X-orbits of points of V not contained in a⊥ are regular. So, by counting, we see that X has exactly q orbits on the points in V \ a⊥. To complete the proof of the lemma we must show that for

½;¼ ∈ = Ç ½ = ¼ ½;¼ ∈ Ç = Ç k, Ç½ ¼ implies that .Let k be such that ½ ¼. Then b + ½a = bx + ¼a for some x ∈ X.So[b; x]=b + bx = .¼ + ½/a ∈hai and consequently [V; x]≤hai. Since no element of X # is a transvection on V , we must have x = 1andthen½a = ¼a. Hence ½ = ¼.

# ⊥ For each x ∈ X we deﬁne a 2-dimensional subspace Tx of a by

Tx =[V; x]=ha; [b; x]i = hai+h[b; x]i:

# ⊥ x LEMMA 2.3. For x ∈ X we have Tx ∩ b =h[b; x]+ f .b; b /ai.

⊥ ⊥ ⊥ PROOF. Since dim b = 2n − 1andhai6≤b ,dim.Tx ∩ b / = 1. Now f .b; [b; x]+ f .b; bx /a/ = f .b; b + bx / + f .b; bx / f .b; a/ = f .b; b/ + f .b; bx / + f .b; bx / = 0;

x ⊥ x and so [b; x]+ f .b; b /a ∈ Tx ∩ b . Thus, as [b; x]+ f .b; b /a 6= 0, Lemma 2.3 holds.

⊥ LEMMA 2.4. Suppose T is a 2-dimensional subspace of a which contains hai. Then # .i/ there exists x ∈ X such that Tx = T ; and # .ii/ XT := {x ∈ X | Tx = T }∪{1} is a subgroup of X of order q.

⊥ PROOF. Let T be a 2-dimensional subspace of a containing hai, and suppose that # ⊥ {x ∈ X | Tx = T } has at least q distinct elements. Then, since dim.T ∩ b / = 1, # Lemma 2.3 implies there exists x1; x2 ∈ X with x1 6= x2 such that

x1 x2 [b; x1]+ f .b; b /a =[b; x2]+ f .b; b /a:

So b + bx1 + b + bx2 = . f .b; bx1 / + f .b; bx2 //a, whence

− − x x 1 x x x 1 x x b + b 1 2 = . f .b; b 1 / + f .b; b 2 //a 2 = . f .b; b 1 / + f .b; b 2 //a ∈hai:

−1 This forces x1x2 to be a transvection of V and so we conclude that # |{x ∈ X | Tx = T }| ≤ q − 1: So there must be at least .|X|−1/=.q − 1/ = .q2n−2 − 1/=.q − 1/ 2-dimensional # subspaces of V of the form Tx ,forsomex ∈ X . However, this is also the number of 2-dimensional subspaces of a⊥ which contain hai and so (i) holds. Moreover, # x2 |{x ∈ X | Tx = T }| = q − 1. Since, for x1; x2 ∈ XT , [b; x1x2]=[b; x1] +[b; x2]∈

T we infer that XT is a subgroup of X of order q. [5] Irreducible subgroups of symplectic groups 89

⊥ LEMMA 2.5. If T is a 2-dimensional subspace of a containing hai, then

.i/ |X=CX .T /|=q; and .ii/ X has two orbits on the points of T , with one of length q and one consisting of hai.

= ∈ # . / =[ ; ]⊥ = ⊥ = ⊥ PROOF. By deﬁnition T Tx for all x XT . Thus CV x V x Tx T . ⊥ ⊥ Hence CV .XT / = T . Select a 2-dimensional subspace U of a containing hai so as ⊥ ⊥ ⊥ ⊥ ⊥ ⊥ U + T = a .SinceT 6= a , U ∩ T =hai. Because CV .XU / = U we then deduce that CT .XU / =hai and therefore XU has orbits of size 1 and q on the points of T .SinceX=CX .T / embeds into GL2.q/, this gives the lemma.

⊥ LEMMA 2.6. Suppose that hci≤a is in the same L-orbit as hai and hci6=hai. Then .i/ the non-zero vectors of ha; ci are all in the same L-orbit; and

.ii/ NL .hai/=CL .hai/ has order q − 1.

PROOF. Put T =ha; ci. By assumption hci=haig for some g ∈ L. Applying g g Lemma 2.5 (i) to the pair T , hci gives |X =CX g .T /|=q.SinceX and X act g ∼ differently on T , we deduce that hX; X i=ChX;X g i.T / = SL2.q/, and this gives the result. An immediate consequence of Lemma 2.6 is

LEMMA 2.7. If L is transitive on the points of V , then L is transitive on V #.

LEMMA 2.8. Suppose that L is not transitive on the points of V . Then there exists a 2-dimensional subspace T of a⊥ containing hai such that haiL ∩T ={hai}. Moreover, ⊥ for U =hb; T ∩ b i; NX .U/ = 1.

⊥ PROOF. Suppose no such 2-dimensional subspace of a exists. Then all points in a⊥ are in haiL by Lemma 2.6.Lethdi be a point of V not in a⊥. Since dim V ≥ 4, there exists 0 6= c ∈ a⊥ ∩ d⊥ whence hci∈haiL .Andthenhdi∈haiL .SoL is transitive on the points of V , a contradiction. We now let T be such a 2-dimensional subspace and prove that, for U =hb; T ∩b⊥i, ⊥ ⊥ NX .U/ = 1. Let 1 6= x ∈ X. We claim that [b; x]6∈T ∩ b .Forif[b; x]∈T ∩ b , then hbi≤h[b; x]; bi=hbx ; bi≤b⊥: Since hbx ; bi is a 2-dimensional subspace by Lemma 2.1 (ii), we may apply Lemma 2.6 (with b in place of a) to conclude that h[b; x]i ∈ hbiL . Hence, as hai and hbi are in the same L-orbit, h[b; x]i ∈ haiL ∩ T = {hai}, a contradiction which establishes the claim. Since b 6∈ T , U is a 2-dimensional ⊥ ⊥ X subspace and U ∩ a = T ∩ b . If there exists 1 6= x ∈ NX .U/, then, since hbi is a regular X-orbit, hb; bx i=U and consequently [b; x]∈U ∩ a⊥ = T ∩ b⊥, contrary ⊥ to [b; x]6∈T ∩ b . Therefore, NX .U/ = 1. 90 Christopher Parker and Peter Rowley [6]

LEMMA 2.9. Either L is transitive on the points of V or L has two orbits on the points of V . In the latter case we have [

h iL ⊥ X L ⊥ h + i \ = Ç : a \ a =hbi = Ç0 and a b a ½ ½∈k#

PROOF. Suppose that L is not transitive on the points of V . Then, by Lemma 2.8, there exists a 2-dimensional subspace T with hai≤T ≤ a⊥ and haiL ∩ T ={hai}. ⊥ Setting U =hb; T ∩ b i we also have NX .U/ = 1byLemma2.8. Since the elements of X # all have order 2, it follows that no two points of U \ a⊥ are in the same X-orbit. Hence the q points of U \ a⊥ may be chosen as representative of the X-orbits on \ ⊥ ∈ h ig =h i = g V a .Letg L be such that a b , and set Y X . Then, by LemmaS 2.5 (ii),

h iY ={h i} Ç Y has orbits of length 1 and q on the points of U and so, as b b , ½∈k# ½ is contained in an L-orbit. h i≤ ⊥ h i6∈h iL h ; i For c a ,if c a , then we may use theS above argument with a c in h i place of T to show that c is in the same L-orbit as ½∈k# ÇS½. Thus L has two orbits

h iL ⊥ L ⊥ \ = Ç h + i \ = Ç on the points of V with a a 0 and a b a ½∈k# ½, and the lemma is proven.

So far we have been dealing with points of V . The next lemma tells us about the orbits of L on the vectors of V . As will shortly be seen this plays an important role in the remainder of the proof of Theorem 1.1.

LEMMA 2.10. Suppose that L is not transitive on the points of V . Then .i/ the vectors of ha + bi are in distinct L-orbits; and .ii/ the vectors of hai# are in a single L-orbit.

PROOF. Clearly, if q = 2 there is nothing to prove, so we may assume q > 2. Again we choose a 2-dimensional subspace T with hai≤T ≤ a⊥ and haiL ∩ T ={hai}.Let # x ⊥ x ∈ X be such that T = Tx . Then, by Lemma 2.3, c =[b; x]+ f .b; b /a ∈ T ∩ b . Put U =hb; ci. Then T and U are both subspaces of c⊥ which contain hci.Let g g g ∈ L be such that hai =hbi, and set Y = X .PutK =hCX .c/; CY .c/i and ⊥ 2n−3 W = c =hci. By Lemma 2.5 (i), |CX .c/|=|CY .c/|=q and clearly K acts . / = h i≤ ≤ ⊥ . / = upon W . From Lemma 2.8 NX U 1andso,as c U c , CCX .c/ W 1. . / = = = . / . / . / Likewise CCY .c/ W 1. Put K K CK W .ByordersCX c and CY c are, respectively, the largest normal 2-subgroups of the stabilizer, in K,ofa +hci and b +hci.Sincef .a +hci; b +hci/ = 1, using Lemma 1.2 we deduce that K =∼ Sp.W /.

Because L is assumed to contain no transvection subgroups, the structure of CG.c/ ∼ and q > 2 imply that O2.CL .c// = 1. So CK .W / = 1 and hence K = Sp.W /. Since Sp.W / acts transitively upon the non-zero vectors of W and haiL ∩ T ={hai}, it follows that the vectors of hai# are in the same L-orbit, so (ii) holds. If (i) were [7] Irreducible subgroups of symplectic groups 91 false, then there would exist an element in L acting non-trivially upon hci. Hence, as ∼ K = Sp.W / and O2.CL .c// = 1, StabL .hci/ = Z × K with Z 6= 1andZ centralizing W . Noting that |Z| divides .q − 1/ and Z must act non-trivially on V=c⊥, we see that

V =[V; Z]×CV .Z/ with dim[V; Z]=2andK inducing the full symplectic group on CV .Z/.SinceK is a simple group (recall q > 2) and K centralizes c ∈[V; Z],it follows that K centralizes [V; Z].ButthenK , and hence L, contains a transvection subgroup. Thus we conclude that (i) also holds.

LEMMA 2.11. Suppose that L has two orbits on the points of V . Then for ½; ¼ ∈ k;¼6= 0, the vectors ½a + ¼b and ½¼a + b are in the same L-orbit.

PROOF. Since all the non-zero vectors of hai are in the same L-orbit by Lem- ma 2.10 (ii), ag = ¼a for some g ∈ L.Sog stabilizes hai and hence leaves a⊥ invariant. Consequently g leaves invariant haiL\a⊥ =hbiX , using Lemma 2.9. So for a g1 gx x suitable x ∈ X, gx stabilizes hbi.Setg1 = gx, and notice that a = a = ¼a = ¼a. Now 1 = f .a; b/ = f .ag1 ; bg1 / = f .¼a; bg1 / = f .a;¼bg1 /.Since¼bg1 ∈hbi,we − must have ¼bg1 = b by the original choice of b.Sobg1 = ¼ 1b and then

− .½a + ¼b/g1 = ½ag1 + ¼bg1 = ½¼a + ¼¼ 1b = ½¼a + b; which gives the result.

For the moment we suppose that L has two orbits on the points of V .Forv ∈ V deﬁne Q : V → k by ( 0ifv ∈ aL ∪{0} Q.v/ = ½2 if v ∈ .½a + ½b/L :

In view of Lemma 2.10, Q is well deﬁned. Observe that for v = ¼a +b, ¼ 6= 0, v is in √ √ L ¼a + ¼b by Lemma 2.11 and therefore Q.v/ = ¼. Also for all ¼ ∈ k, v ∈ V , we have Q.¼u/ = ¼2 Q.v/. It is immediate from its deﬁnition that Q is L-invariant.

LEMMA 2.12. Assume that v ∈ aL . Then for w ∈ V and Þ; þ ∈ k

Q.Þv + þw/ = Þ2 Q.v/ + þ2 Q.w/ + Þþ f .v; w/:

PROOF. Clearly, if þ = 0 then we are done—so we assume þ 6= 0. Without loss of generality we may suppose that v = a. Suppose first that w ∈ a⊥. In this case we need to show that Q.Þa + þw/ = þ2 Q.w/ = Q.þw/. This means we need to show that Þa + þw is in the same L-orbit as þw.Ifw ∈hai, then Q.Þa + þw/ = 0 = Q.þw/. Therefore, we may also assume w 6∈ hai and so hw; ai is a 2-dimensional subspace ⊥ of a . By Lemma 2.5 (i) X=CX .hw; ai/ has order q and hence there exists x ∈ X such 92 Christopher Parker and Peter Rowley [8] that [þw;x]=Þa. Thus þw + þwx = Þa, which is to say that þw = .Þa + þw/x and we are done. Next we suppose that w 6∈ a⊥. Thenwehavetwocases,thefirstbeingwhenw ∈ aL . Here we have w = −bx for some − ∈ k .− 6= 0/, x ∈ X by Lemma 2.9. Hence, as X fixes a, Þa + þw = Þa + þ−bx = .Þa + þ−b/x : Therefore, by Lemma 2.11, Þa + þw is in the same L-orbit as Þþ− a + b, whence Q.Þa + þw/ = Þþ− .Now f .a;w/ = f .a;−bx / = f .a;−b/ = − f .a; b/ = −. Therefore, Þ2 Q.a/ + þ2 Q.w/ + Þþ f .a;w/ = Þ20 + þ20 + Þþ− = Þþ− = Q.Þa + þw/; as required. Turning to the second case, w 6∈ a⊥, again using Lemma 2.9 we have w = −bx +−½a,forsome½, − ∈ k (− 6= 0 6= ½), x ∈ X. So, by Lemma 2.11, Þa +þw and þ−.Þ+þ−½/a+b are in the same L-orbit and hence Q.Þa+þw/ = þ−.Þ+þ−½/. Because w and −b + −½a are in the same L-orbit, we obtain Q.w/ = −2½.Since Q.a/ = 0and f .a;w/ = −, Þ2 Q.a/ + þ2 Q.w/ + Þþ f .a;w/ = þ2− 2½ + Þþ− = Q.Þa + þw/: This verifies the equation in the final case and so the lemma holds.

LEMMA 2.13. Q is a quadratic form on V .

PROOF. Let v;w ∈ V and Þ; þ ∈ k. So we must show that Q.Þv + þw/ = Þ2 Q.v/ + þ2 Q.w/ + Þþ f .v; w/. In view of Lemma 2.12 we only need examine the case when v and w are not in the same L-orbit as a. Also, we may assume that w 6∈ ⊥ ½ ∈ # w = x + ½ a . Then, by Lemma 2.2, w ∈ Ç½ for some k . Hence b a for some x ∈ X. Therefore, w + ½a = bx ∈ a L and so, using Lemma 2.12, Q.Þv + þw/ = Q.Þv + þ½a + þw + þ½a/ = Q.Þv + þ½a/ + Q.þw + þ½a/ + f .Þv + þ½a;þw+ þ½a/ = Þ2 Q.v/ + þ2½2 Q.a/ + Þþ½ f .v; a/ + þ2 Q.w/ + þ2½2 Q.a/ + þ2½f .w; a/ + Þþ f .v; w/ + Þþ½ f .v; a/ + þ2½f .a;w/+ þ2½2 f .a; a/ = Þ2 Q.v/ + þ2 Q.w/ + Þþ f .v; w/: This proves the lemma.

LEMMA 2.14. If L has two orbits on the points of V , then L acts naturally on V as one of .V / and O.V /.

PROOF. From Lemma 2.13 L preserves the quadratic form Q and so L is a subgroup ∼ of O.V; Q/.= O .V //.NowX = O2.StabO.V;Q/hai/ and the orbits of L and O.V; Q/ upon the points of V are the same. Hence L ≥hXO.V;Q/i≥.V /. Thus Lemma 2.14 holds. [9] Irreducible subgroups of symplectic groups 93 3. Proof of Theorem 1.1

In the light of Lemma 2.14 from now on we may assume that L is transitive on the points of V , whence, by Lemma 2.6, L is transitive on V #. We recall there is a one-to- one correspondence between vectors c of V and symplectic transvections −c of Sp.V / where −c is defined by −c.v/ = v + f .v; c/c (v ∈ V ). Hence L acts transitively (by conjugation) upon the symplectic transvections. Thereforeif L contains a transvection it must contain them all and, in particular, will then contain a transvection subgroup. So we conclude that L contains no transvections. For c a non-zero vector of V we g g g1 define Xc to be X where g ∈ L is such that a = c.Ifa = c for g1 ∈ L, then g g1 g g1 g g1 X X ≤ O2.StabG hci/ and if X 6= X , then X X intersects the transvection subgroup of O2.StabG hci/ non-trivially. So Xc is well defined; note that Xa = X.We now complete the proof of Theorem 1.1, arguing by induction on n and starting with n = 2. When q = 2wehaveG =∼ Sym.6/ and it is fairly straightforward to calculate in Sym.6/ to deduce that L = Alt.6/ or 0 L.2; 4/. Thus we may assume q > 2. We 2 ⊥ note that |Xa |=|X|=q .LetT be a 2-dimensional space of a containing hai, ∈ \h i = . / and let t T a . From Lemmas 2.4 (ii) and 2.5 (i) we have that XT CXa T 1+2 has order q.PutYat =hXa ; Xt i. Then Yat ≤ StabG T ∼ q .SL2.q/ × .q − 1//.

Recall that O2.StabG .T // is a indecomposable GF.q/ StabG .T /-module. Hence, 20 ∼ since Yat ≤ O .StabG .T // and L contains no transvections, Yat = q × SL2.q/ with ∩ . . // = ∈ Yat O2 StabG T XT having order q. Evidently Xa Syl2 Yat. Furthermore . / 2. − / h i ½ 0 | ½ ∈ . / NYat Xa has order q q 1 , stabilizes a and induces 01 GF q on Xa . ⊥ For 2-dimensional subspaces T1, T2 of a containing hai with T1 6= T2 we must have 6= = + = ⊥ XT1 XT2 for otherwise XT1 XT2 centralizes T1 T2 a . Thus =h . / | =h; i ⊥i.≤ h i/ N NYat Xa T t a a subspace of a StabG a induces a 2-transitive action upon the points of Xa , and consequently N=CN .Xa / ∼ 2 = SL2.q/ × .q − 1/.So,asL has no transvections, N ∼ q .SL2.q/ × .q − 1//.But this contradicts the structure of StabG hai and therefore we have veriﬁed the theorem for n = 2. From now on we assume that n ≥ 3. Again we consider a 2-dimensional subspace ⊥ T of a which contains hai, and let t ∈ T \hai.PutXT;t =hx ∈ Xt |[V; t]=T i,and . / | |= 2n−4 let KT;t be a complement to XT;t in CXt T . By Lemmas 2.4 and 2.5 (i) K T;t q . We next show that

2n−4 3.1. .i/ |KT;t O2.H/=O2.H/|=q ; ⊥ .ii/ KT;t O2.H/ centralizes T =T ;and ⊥ .iii/ no element of KT;t induces a transvection on a =hai. ⊥ ⊥ Suppose that 1 6= x ∈ KT;t ∩ O2.H/. Then [a ; x]≤hai and [t ; x]≤hti.Since x is not a transvection on V this gives [V; x]=T which implies x ∈ K T;t ∩ XT;t = 1, a 94 Christopher Parker and Peter Rowley [10]

⊥ ⊥ contradiction. Therefore part (i) holds. Part (ii) follows from [T ; K T;t ]≤[t ; Xt ]≤ ⊥ hti≤T . Finally suppose that 1 6= x ∈ KT;t operates as a transvection on a =hai,and let W be the preimage of Ca⊥=hai.x/. Then [W; x]≤hai. Assume that [W; x]=hai holds. Since x is a 2-transvection on V and x ∈ Xt ,thisgives[V; x]=T and so ⊥ x ∈ KT;t ∩ XT;t = 1, a contradiction. Therefore [W; x]=0andsoW = CV .x/ ≤ t . ⊥ ⊥ ⊥ ⊥⊥ Hence CV .x/ = W = t ∩ a = T .Butthen[V; x]=T = T which again gives ⊥ the impossible x ∈ KT;t ∩ XT;t = 1. Thus (iii) holds. Put a = a =hai and =h . / | ∈ \h i ⊥ h ii: L0 CXt T t T a and T a 2-dimensional subspace of a containing a

3.2. L0 acts irreducibly upon a.

⊥ ⊥ Suppose that W is an L0-invariant subspace with hai < W < a .Letc ∈ a be a vector not in W , and put U =ha; ci. Then, by Lemma 2.1 (i), [ ⊥ ∩ ; . /]≤ ∩h i= : c W CXc U W c 0 h i≤ ⊥ ∩ ≤ . . // 6∈ ⊥ ∩ =hi Therefore a c W CV CXc U . Hence as c W , c W a by Lemma 2.1 (iv)andsodimW = 1. Thus every proper non-zero L0-invariant subspace ⊥ ⊥ of a has dimension one. Since W is L0-invariant, this forces W = W , whence dim V = 4. However we have dim V > 4,andso(3.2) holds.

Together (3.1) and (3.2) imply that L0 actingonthe2.n−1/-dimensional symplectic = . / space a satisﬁes the hypotheses of the theorem. Therefore, by induction, L0 CL0 a is isomorphic to one of O.a/; .a/,Sp.a/,Alt.6/,or0 L.2; 4/ with q = n − 1 = 2 in the latter two cases. The orthogonal cases are impossible as for every 1-dimensional subspace T of a there is a corresponding subgroup of order q2n−4 centralizing T ⊥=T which is not the case in the orthogonal groups when we select a non- . / = . / =∼ . / isotropic 1-dimensional subspace (stabilizer Sp2n−4 q ). Thus L0 CL0 a Sp a or = . / =∼ . / = − = L0 CL0 a Alt 6 and q n 1 2. Now L containing no transvections forces = . / =∼ . / L0 to have index q (or 4 in the L0 CL0 a Alt 6 case) in H both of which are = . / =∼ 0 . ; / ∼ 40 . ; / impossible. Therefore, we have that L0 CL0 a L 2 4 and L0 2 L 2 4 with O2.L0/ a ‘natural’ GF.2/0 L.2; 4/ module. However, this means that L0 is a subgroup of index 12 in H ∼ 21+4 Sp.4; 2/, and, since it has order coprime to 11, has an orbit of length at most 10 on its twelve right cosets in H. Byconsideringanorbiton which the O2.L0/ acts non-trivially we ﬁnd that L0 must have a subgroup of index 10, which intersects O2.L0/ in a subgroup of order 8 and projects to Sym.4/ in L0=O2.L0/. . / = . / 6=∼ 0 . ; / This contradicts the module structure of O2 L0 and thus L0 CL0 a L 2 4 . This completes the proof of Theorem 1.1.

References

[1] U. Dempwolff, ‘Some subgroups of SL.n; 2m /, I’, Results Math. 4 (1981), 1–21. [11] Irreducible subgroups of symplectic groups 95

[2] , ‘Some subgroups of SL.n; 2m /, II’, preprint 41, (Universitat¨ Kaiserslautern, 1982). [3] J. McLaughlin, ‘Some groups generated by transvections’, Arch. Math. 18 (1967), 364–368.

[4] , ‘Some subgroups of SLn.F2/’, Illinois J. Math. 13 (1969), 108–115. [5] C. Parker and P. Rowley, Symplectic amalgams, Springer Monographs in Math., (Springer, New York, 2002). [6] S. Smith, ‘Spin modules in characteristic 2’, J. Algebra 77 (1982), 392–401. [7] G. Stroth, ‘Endliche Gruppen, die eine Maximale 2-lokale Untergruppe besitzen, so daß Z.F ∗.M// eine TI-Menge in G ist’, J. Algebra 64 (1980), 460–528. [8] M. Suzuki, Group theory I, Grundlehren Math. Wiss. 247 (Springer, New York, 1982). [9] F. G. Timmesfeld, ‘Finite simple groups in which the generalized ﬁtting subgroup of the centralizer of some involution is extraspecial’, Ann. of Math. (2) 107 (1978), 297–369.

School of Mathematics and Statistics Department of Mathematics University of Birmingham UMIST Edgbaston P.O. Box 88 Birmingham B15 2TT Manchester M60 1QD United Kingdom United Kingdom e-mail: [email protected] e-mail: [email protected]