872 i3. MID-SIZE GROUPS OVER F2
B.5.1. Now by Theorem B.5.1, either conclusion (3) holds or U E Irr +(L, Ve), in
which case B.4.2 says U satisfies one of conclusions (1), (2), or (4), using I.1.6.4 in
the latter case. Furthermore in any case Z ::; R2(Mc) = Ve, and (ZL) = UGz(L)
by B.2.14. Thus in conclusion (4), since Zn U = Gu(L) by B.4.8.2, we conclude
that [Z, L] = 1. D
LEMMA 13.1.11. For X E X, if X ::; Xi ::; Mc and Xi acts irreducibly on
X/0 2 ,<J?(X), then Xi is contained in a unique conjugate of Mc.
PROOF. Suppose Xi ::; Mg and let p be the odd prime in n(X), PE Sylp(X),
and Mg := Mg/0 2 (Mg). We apply A.1.32 with Mg, f(g, P, p, pin the roles of
"G, R ,P, r, p". The second case of A.1.32.2 does not arise as X^9 is not of order
p. Thus f(g = X, so X = 02 (X) ::; 02 (X9) = X9, and hence g E Na(X) =Mc,
so Mc= Mg. D
LEMMA 13.1.12. Let u EU#, Gu:= Ga(u), Mu:= CMJu), and pick u so that
Tu:= Gr(u) E Syl2(Mu)· Then
(1) Mu is irreducible on X/02,<J?(X) for each XE X.
(2) IT: Tul ::; 2.
{3) Gu ::; Mc.PROOF. Let
u := {u Eu#: ITul < ITI}.If u E Z then Tu = Tis irreducible on X/02,<J?(X) for each X E X, while Gu ::;
Mc= !M(Ca(Z)) by(+), so that (1)-(3) hold. Thus we may assume that U is
nonempty, and it suffices to establish (1)-(3) for u E U. In particular Mc is not
transitive on U#, so Gu(L) =/= 1 in case (1) of 13.1.10, and U is the sum of two
irreducibles for L* in case (3).
Let u E U, and recall from 13.1.4.5 that X centralizes R2(Mc) = Ve andU = [v;,, L], so that X ::; Mu. As X E B+(G, T), X ~ EP2 or p1+^2 for some
prime p > 3. Set M;}- := Mu/GMu(X/02,<J?(X)); thus M;}-::; GL2(p). To prove
(1), it will suffice to show that M;t is nonabelian; as Ga(X)::; 02,F(L)::; Ga(U),
it also suffices to show GL(u)* is nonabelian. Indeed L/02,F(L) ~ SL2(q) for
q = 5 or 7, so if GL(u)* contains a 4-group, then the preimage of this 4-group in
CL(u)/02,F(X) is the nonabelian group Qs, which again suffices. To prove (2), it
will suffice to show for each orbit 0 of LT on U that IOI = 2 mod 4.Assume case (1) of 13.1.10 holds. By I.2.3.1, U is a quotient of the rank-6
extension Uo of U/Gu(L), so m(Cu(L)) = 1 or 2. However if m(Cu(L)) = 1 or
T ::; L, then all members of U# are 2-central in Mc. Therefore m(Cu(L)) = 2,
so U = Uo and hence U admits an F4-structure by I.2.3.l. Further T* 1:. L*, so
L*T* ~ 85 • Then LT has two orbits on U: one of length 2 in Gu(L), and one of
length 30 on U -Gu(L). In either case, luLT\ = 2 mod 4, so that (2) holds by theprevious paragraph. Also (Tun L) E Syl 2 (L), so (1) also holds by the previous
paragraph.Assume case (3) of 13.1.10 holds. We saw U is the sum of two natural modules,
so L is transitive on
u I
lEirr+(L,U)