6.2. IDENTIFYING M22 VIA L2(4) ON THE NATURAL MODULE 681field automorphism on Lg /0 2 (Lg) and Vg; and of type 3 if F is of neither of the
first two types. By 6.2.2, there exist 4-subgroups of V of field type, and of courseZs is of central type.
LEMMA 6.2.5. (1) Let~ denote the number of orbits of Mv on the 4-subgroups
of V of type 3. Then~= 0 or 1 for IMv : Li = 6 or 2, respectively. The orbits are
of length 0 or 20, respectively.
(2) Mv is transitive on 4-subgroups of V of each type.
(3) For each 4-subgroup u of v, u^0 n v = uLT.
(4) V is the unique member of v^0 containing Zs.
(5) If g E G with V n Vg noncyclic, then [V, Vg] = 1.
(6) V is the unique member of v^0 containing any hyperplane of V.
PROOF. As V is the natural module for L, L preserves an F 4 -space structure
VF 4 on V, in which the central 4-subgroups are the five I-dimensional subspaces
of VF 4 , and Mv ~ AutaL(V)(L) = I'L(VF 4 ). In particular, Lis transitive on 4-
subgroups of central type, and there are 30 4-subgroups not of central type, which
form an orbit under AutaL(V)(L). This orbit splits into three orbits of length 10under L, and AutaL(V) (L) induces 83 on this set of orbits. By 6.2.2, Mv f':! 85 or
I'L2(4), so it follows that (1) and (2) hold.
By 6.2.4, we can choose a representative U for each orbit so that NT(U) E
8yb(Na(U)). Now T = NT(U) iff U is of central type, so groups of central type
are not fused to groups of field type or type 3. Similarly if NT(U) < T, then
IT : NT(U)I = 2 or 4 for U of field type, or type 3, respectively, so distinct M-
orbits are not fused in G. Thus (3) holds.By (3) and A.1.7.1, Na(Zs) is transitive on G-conjug?-tes of V containing Zs;
then as Na(Zs) ~ Na(V) by 6.2.1, (4) holds. As Vis a self-dual F2L-module and
Lis transitive on V#, Lis transitive on hyperplanes of V, so (4) implies (6).
Assume the hypotheses of (5), and let Ube a 4-subgroup of V n Vg; then by (3)
and A.1.7.1, Na(U) is transitive on G-conjugates of V containing U. Furthermorefor U of each type, Auta(U) S:! 83 S:! AutMv(U), so that Na(U) = Ca(U)NMv(U);
we conclude that Ca(U) is transitive on the G-conjugates of V containing U. Thus
if Ca(U) ~ Na(V), then V = Vg and (5) is trivial. If Ca(U) i Na(V), then U
is of field type by 6.2.4.1, so (V, Vg) is abelian by 6.2.4.2, completing the proof of
(5). D
LEMMA 6.2.6. Assume A:= Vg n Na(V) and U := V n Na(Vg) are of index
2 in Vg and V, respectively. Then either
(1) A and U/Cu(Vg) are of order 2, CA(V) and Cu(Vg) are of field type, and
(V, Vg) is a 2-group, or
(2) AS:! E 4 , Ai L, Y := (V, Vg) S:! 83/Q§, V n Vg = [A, U] is of order 2, and
02(Y) ~ 02 (Y).
PROOF. Without loss, we may assume A~ T. First B := [A, U] ~An U,
so B -=/= Zs by 6.2.5.4, and hence A ¢'. 8yb(L). Also A =/=-1, as otherwise V ~
Ca(A) ~ Na(Vg) by 6.2.5.6, contrary to hypothesis.
Suppose first that A is of order 2. Then Ao := CA (V) is of codimension 2
in Vg, so as V ~ Ca(Ao) but V i Na(Vg), we conclude from 6.2.4.l that Ao is