By 11.3.7.4, V =Ai. 02 (H). Therefore we have Hypothesis E.2.8 with H, T,
HnM in the roles of "H, T, M". Leth EH-Mand set I:= (V, Vh). By 11.3.7,
K ~ L 2 (2n) with V E Syl 2 (K), so I = K*. Thus I is in the set I(H, T, V)
defined in Definition E.2.4. Therefore by E.2.9.1, 02(H) acts on I, so K = 02 (!),
and I :::! Has Tacts on V and H =KT.
Next by E.2.11, the hypotheses of E.2.10 are satisfied with I, Mn I, T n I, V
in the roles of "H, M, T, V". Notice also that B := Cv(VH) = V n 02(H) plays
the role of "B" in E.2.10, and by that result P := BBh is a normal 2-subgroup of
I. Since Vn1/h :_::; Z(I) and K = 02 (!), VnVh = 1 since CvH(K) = 1by11.3.7.3.
Therefore P =Bx Bh and B = Cp(V*) by E.2.10.2. By E.2.10.7, P = 02(I), and
by G.1.7, Pis a sum of j natural modules for K/02(K) ~ L2(q), so P = [P, K]
and hence P = 02 (K). Therefore Bh acts faithfully on V, with B = Cv(Bh) of
corank n in V and m(Bh) = m(B) = jn. Thus Bh is a group of transvections with
axis B, so L ~ SL 3 (q), j = 2, and Bis T-invariant of rank 2n; hence B =Vi and
P = f3h = R 2 • Thus Pis of rank 4n with P n V = Vi- As Vin Z :.::; Vi, and
V 1 = [V 1 , Do] by 11.3. 7.5, V K := ( (Zn P)H) = (V 1 H). As P is a sum of two natural
modules for K* ~ L 2 (2n), VK is a natural submodule of P of rank 2n and
Therefore L = [L, VK] centralizes 02 (LT)/V, so Lis an SL3(q)-block. Thus L/V is
a covering group of SL 3 (q), so from the list of Schur multipliers in I.1.3, either V =
02 (£) = CL(V), or q = 4 and 02 (£/V) i= 1. However in the latter case (R2nL)/V
does not split over 02 (£)/V by I.2.2.3b, whereas P = R2 and PV/V ~ P/Vi ~ Eq2
is T-invariarrt. Thus V = CL(V) = 02(L), and then as P = J(PV) ~ Eq4,
P = 02(L2).
Recall X is a Cartan subgroup of L acting on TL and we may take Do :.::; X
by 11.3.7.5. Notice that in the shadow L 4 (q), the Cartan group D of H n Mis not
contained in the derived subgroup L of M, and DX is a group ofrank 3 for primes
dividing q - 1. Indeed, Do i. L, so it remains to show that the unnatural inclusion
Do :.::; L leads to a contradiction. This is accomplished by studying the action of X
on P and V.
Assume n is even. Then the subgroup D3 of Do of order 3 is contained in X.
However Cp(D3) = 1 = Cv(D3) as Pis the sum of natural modules for K and
V = [V*,D3], whereas Cx(L/V) E Syl3(02,z(L)) is the only subgroup of X of
order 3 having no fixed points on V, and it centralizes PV/V.
Thus n is odd, so D = Do :.::; X and T = TL. Now CT(L) = 1 by 11.3.7,
and H^1 (L/V, V) = 0 by l.1.6, so that V = 02 (LT) by C.1.13.b. Thus T =TL E
Syl2(L). Set Gp := Na(P), Gp := Gp/P, and Y := (£ 2 ,I). We've seen that
P = 02(L2) = 02(K) with L2 ~ k ~ L2(q) and T ~ Eq2· As K E C(Gp, T),
K:.::; K+ E C(Gp) by 1.2.4, and if K < K+, then the embedding of Kin K+ is
described in A.3.12. As k ~ L 2 (2n) with n odd and T is abelian, we conclude
K = K+ E C(Gp). Similarly L2 E C(Gp). Next L 2 -/= k since K i. M, so
y = k x L2 ~ nt(q). As ITI = q^2 = IYl2, y = 0
21
(Gp). Asp is the sum of two
copies of the natural module for k, and V and P/V are natural modules for L 2 ,
Pis the orthogonal module for Y. As X :.::; Nc(P) and mp(Na(P)) :.::; 2 for each
prime divisor p of q - 1, Xis a Cartan subgroup of Y.
We next show: