844 12. LARGER GROUPS OVER F2 IN .C'f (G, T)LEMMA 12.7.16. If U = R1, then G ~ M24.
PROOF. Assume that U = R 1. By 12.7.2.2, [U, V] = V1, so V induces a
group of transvections on U with center .V 1. Also m(V*) = 2 or 4 by 12.7.15.4.
-a - - -
Thus if ~ 1 , ... , ~s are the orbits of Kon V 1 z, then by G.3.1, U = U1 EB··· EB Us
and K* = Ki x · · · x K;, where Ui := (~i) is of dimension n ;:::: 3, Ki is generated
by the transvections in K* with centers in ~i, [Ki, UjJ'= 0 for i =f. j, and (as
02 (Ki) = 1by12.7.15.1) Ki acts faithfully as GL(Ui) on Ui. Observe in particularthat Li :::; K*. Now each preimage Ki contains a member of C(H) by 1.2.1.1, so
by 1.2.1.3, s ::=; 2; and in case of equality, H* ~ L 3 (2) wr Z2. Therefore s = 1,
as T* acts on Li and Li/02(Li) ~ Z3 by 12.7.15.4. Thus by Theorem C (A.2.3),
K* = GL(U) ~ Ln(2), n = 3, 4, or 5. Then K is transitive on tJ#, so (U) = l.
By 12.7.15.4, LiT is a rank one parabolic of K.
If n = 5, then CK(Vi)* ~ L4(2)/El6, so as Xis faithful on V1, m3(Na(V1)) > 2,
contrary to Na(V1) an SQTK-group.
Thus n = 3 or 4. As we saw V*:::; CK·(U/V1) ~ E 2 n-1 and m(V*) = 2 or 4,
we conclude m(V*) = 2. Next
m( Q n U) = m(U) - m(U) = n + 1 - 2 = n - 1.
As V 1 :::; VnU:::; Cv(U) = Cv(R 1 ) = V1, we conclude Vi= Cv(U) = VnU. Thus
m((Q n U)V/V) = n -1-m(Vi) = n - 3 ::=:: 1.
Now [Q, U] :::; Q n U, so that m([W, U]) :::; 1 for W any noncentral chief factor
for Lon Q/V. However U = R 1 does not induce transvections on any nontrivial
irreducible for A 6 ; hence [U, Q] ::=:: V, and Lis an A5-block. In particular, L1 has
exactly three noncentral 2-chief factors.Suppose n = 4. Then as Li = 02 (P) for some rank one parabolic P of K*,
L 1 has one noncentral chief factor W on the natural module U, and two such factors
on 02 (Li). We conclude [Qz, L 1 ] :::; U, so that [Qz, K] :::; U. Thus by the Thompson
Ax B-Lemma, 02 (K)/U is faithful on Cu(0 2 (KT)/U), so as K is irreducible on
U, 02 (KT) centralizes U. Then as H^1 (K*, U) = 0 by I.1.6, U = [U, K] EB (z). This
is impossible as z E [V, R1] by 12.7.2.2, while UCT(V) = R1 by hypothesis, so that
ZE [V,U]
Therefore n = 3, so U ~ E 16. Assume Mv = L. Then Vi ::=:: Z(T), so by B.2.14,
U E R2(Gz), and hence Ca,(U) = Ca,(U). However we saw that 4 = IV*I =IV:
Cv(U)I and Cv(U) =Vi is of index 16 in V. Thus Mv ~ 86 and U tJ. R2(Gz), sothat Ca(U)/Ca(U) =f. 1. Therefore from the action of H* = GL(U), Ca(U)/Ca(U)
is the full group of transvections on U with center z, and affords the K*-module
dual to U. Recall that L is a A 6 -block, while CT(L) = 1 by 12.7.12. Then by
12.7.6.2, V = 02 (M) and M = LT, so that M/V ~ S 6 • Therefore ITI = 210 ,
so as IT*I = 8 = ICT(U)/CT(U)I and IUI = 16, CT(U) = U. As T normalizes. U, Na(U) E 'He by 1.1.4.6, so U = Ca(U). Hence as AutK(U) = CAut(U)(z),
H =Na, (U) = K with Qz ·= 02 (K) <;>f order 27. In particular, K has 2-chief series
1 < (z) < U < Qz.
As Qz/U is dual to U, K is transitive on ( Qz/U)# and zJ#. As V n Qz i U, there
are involutions in Qz - U. It follows that <I>(Qz) = 1, so Qz ~ 21 +6 ~ D~. Now Gz