1186 16. QUASITHIN GROUPS OF EVEN TYPE BUT NOT EVEN CHARACTERISTIC
is transitive on V - (vz), which is impossible since v tfi z^0 by the claim. Therefore
To = (z), so ToTLR =To x D as vz = ur.
Suppose that H* ~ PG L 2 ( q). Then [T : D [ = 2, so since we saw that z^0 n D =
0, z tfi 02 (G) by Thompson Transfer, contrary to the simplicity of G.
Therefore H* ~ Aut(A6), so TE ~ E4 x Z4. Now (s, t) acts on TE andCv((t,s)) = (vz), so we conclude that (vz) = <l?(TE)· Next D = TL(u) and
T = TETL with TE abelian, so that D :::;I T. As vz tfi D and vz E <l?(TE),
T/D ~ Z 4. Then since we saw that zG n D = 0, z ~ 02 (G) by Generalized
Thompson Transfer A.1.37.2, contrary to the simplicity of G, completing the proof
of 16.4.6. D
In the next lemma, we deal with the only case (other than those eliminated
in 16.4.6 and the case L ~ L 2 (2n)) where for some involution r in R, Lr is not
generated by its p-elements asp varies over those odd primes such that mp(L) > 1.
LEMMA 16.4.7. If L* ~ M 22 , then no involution in R induces an outer auto-
morphism on L with CL*(r) ~ Sz(2)/Em.
PROOF. Assumer is a counterexample, and set LR:= 02 (NL(02(Lr))). Recall
Lr= 02 (CL(r)). Then R = R 0 (r), where Ro := RnLK. From the structure of the
extension of L* by a 2-group in I.2.2.6a, Lr is isomorphic to Z5/Em if [Z(L)[ < 4,
but isomorphic to Z 5 /Q 8 Ds if Z(L) ~ Z4.
We first show that Ro = 1, so we assume that Ro # 1 and derive a contradiction.
Then as Lr:::; H', [Lri CR 0 (r)]:::; Lr nK' =: Y :::;I Lr, and Y* # 1asCH·(L;)=1.Thus Y* contains the unique minimal normal subgroup 02(L;) of L;, so 02 (Lr):::;
Y :::; K'. It follows from 16.4.2.5 that LR :::; H'. Further as O(L') = 1 by (El),
L' = 031 (H') by A.3.18, and hence LR:::; L'. But then 02 (Lr):::; K' n L':::; Z(L'),impossible as m2(0 2 (Lr)) 2 3 by the first paragraph, while Z(L') is cyclic by
16.1.2.2.
Thus Ro = 1, so R = (r) is of order 2. So by 16.4.4.1, CTa (R) = (z) is of
order 2, and ToR dihedral or semidihedral. Thus if Z(L) # 1, (z) = fh(Z(L)).
Conversely if z EL, then Z(L) # 1 and (z) = 01 (Z(L)). By 16.4.3.2, L' = [L',z].Again we establish symmetry between L, z, r and L', r, z, by showing that
the action of z on L' is the same as that of r on L: ·First RTo is dihedral or
semidihedral and isomorphic to a Sylow group of H / L, so as Lr :::; H' and Lr
is irreducible on 02(Lr)/<l?(02(Lr)) ~ Em, we conclude that 02(Lr) :::; L' and
02(Lr) n K' :::; <l?(02(Lr)). As z centralizes Lr, we conclude from 16.1.5.5 that z
induces an outer automorphism of L' with CL'/o 2 (L')(z) ~ Sz(2)/Em, establishing
the symmetry.In particular as [Out(L')[ = 2,· z ~ [H',H']. But if Z(L) ~ Z4, then we saw
that z E [02(Lr)), 02(Lr]; so [Z(L)[:::; 2, and hence 02(Lr) ~Em from our earlier
discusion. Further we saw 02 (Lr):::; L'.
Set E := (r, z), V := 02(Lr)E, and choose g E G with Kg= K' and NT(K'):::;
Tg. Set M := (NH(V),NH'(V)) and M+ := M/CM(V); then NL(V)+ ~ 85.
Assume for the moment that Z(L) # 1, so that Z(L) = (z) is of order 2. Then
Vis a 6-dimensional indecomposable for NL(V)+, so by symmetry between rand
z, also NL'(V)+ is indecomposable on V, and hence Mis irreducible on V when
Z(L) # 1. Now assume that Z(L) = 1. Then V = (z) E9 02(Lr)(r) as ai:i NL(V)+-
module, and 02 (Lr) (r) is a 5-dimensional indecomposable with trivial quotient.