972 i3. MID-SIZE GROUPS OVER F2PROOF. We claim first that Q ::::] Gz. Let Qz := 02(Gz)· By G.2.2 with (z),
(t, z), 1 in the roles of "1/i, V, L", tJ := (iP•) ~ Di(Z(Oz)) and tJ E R2(Gz)· Now
Cc,(U)/Cc.,t(U) is of order at most 2, so by 13.9.7, 00 .(U) is a 2-group, and
hence Oz= Caz (U). Let a; := Gz/Qz, so that 02(G;) = 1 and a; ~ GL(U).
If Q ~ Qz, then O = J(Oz) by 13.9.6, so that Q::::] Gz, as claimed. Thus we mayassume Q 1:. Qz, so in particular m(U) < m(J(T)) = 4 by 13.9.6. Further using the
identification a, no element ofT induces a transvection on 0; so if IQ: QnQzl = 2,
then tJ ~ Cz;(OnOz) ~ Cr(O), and then 0 ~ Cr(U) =Oz by the first paragraph,
contrary to assumption. Thus [Q*[ > 2, so m(U) > 2 and hence m(U) = 3. Then
a; ~ GL(U) = L 3 (2), with Sylow group T* oforder at least 4 and 02(G;) = 1, so
we conclude a; = GL(U). Then Gt,z has order divisible by 3, contrary to 13.9.7,
completing the proof of the claim.By the claim, Q ::::] Gz. In particular as t E Q, we have U ~ Q ~ Qz. Now
00 .(0) ~ 00 .(U) =Oz, so as 0 is self-centralizing in T, we conclude 00 .(0) = O.
Hence G~ := Gz/Q lies in the orthogonal group 0(0) ~ ot(2) with Sylow group
T' ~ E 4. As T < Gz by 13.9.5, we conclude that either (1) holds, or a; ~ 83 x 83.In the latter case Gz is transitive on the involutions in Q - (z). This is impossible
as An Q ~ Es contains an element of .6.i, and hence t is fused into zG in Gz,
contrary to 13.9.2.5. This completes the proof of (1).
Let b E B-Q. As B ::::] T and m([O, b]) = 2, [b, Q] = BnQ ~Es. Similarly for
a EA - Q, [Q, a] =An Q ~Es. Thus a and b interchange the two Qs-subgroups. Qi and Q2 of Q. Now T/Q ~ E4 has three subgroups Ei/Q, 1 ~ i ~ 3, of order
2, with Ei := NT(Qi). Then Qz -I Ei, since (1) shows that Qz/Q centralizes an
a-invariant subgroup of Gz/Q of order 3, whereas Ei/Q does not. Furthermore
Ei is not AQ or BQ as we saw these subgroups interchange Qi and Q 2. Let
AQ =: E2; then CAQ([Q, a]) ~ Em and [Q, a] =· [Q, i] for each i E AQ - Q.
Therefore A= CAQ([Q,a]) and BQ -I AQ. Thus BQ = E 3. As M = Nc(A) and
T = CM(z) < Gz by 13.9.5, A is not normal in Gz; therefore A 1:. Qz as A is weaklyclosed in T by 13.9.2.3. Thus BQ = E 3 = Qz, and then B ::::] Gz, as B is weakly
closed in T by 13.9.2.3. This completes the proof of 13.9.8. D
LEMMA 13.9.9. (1) B is the natural module for K/ B ~ 0.4(2).
(2) zK and tK are of order 5 and 10, respectively, and afford the set of singular
and nonsingular points in the orthogonal space B.
(3) zG n M = .6.i u .6.3 u .6.5.
(4) tG n M = .6.2 U .6.4 U .6.5.(5) G has two classes of involutions with representatives z and t.
PROOF. First Q = (s)[J(T),s] with [J(T),s] = CJ(T)(s)(t). This allows us to
calculate that T has four orbits ri, 1 ~ i ~ 4, on the set I' of 18 involutions inQ - (z): I'i := {t, tz} ~ .6.2, I'2 ~ .6.i of order 4, I' 3 := sT of order 8 containing
s E .6.5, and I'4 := B n .6.4 of order 4. On the other hand, from 13.9.8.1, Gz has
two orbits on r: ri of length 6, and r^2 of length 12, with t E ri as (£'; = Z(T). As
t tf_ zG ~ .6.i ~ I'2, we conclude
ri = ri u I'4 =tan Q and r^2 = r2 u I'3 = z^0 n Q - {z}. · (*)
In particular .6.4 ~ tG. Next M/0^2 (M) ~ Ds; let Mo be the subgroup of M ofindex 2 with Mo/0^2 (M) ~ Z4. Then .6.i U .6. 2 U .6. 4 is the set of involutions in Mo,
so each is in zG U tG. Hence (5) follows from Thompson Transfer.