0.4. AN OUTLINE OF THE PROOF OF THE EVEN TYPE THEOREM 495Gz. At this point our recognition theorems show that G is G 2 (3), LH3), or U4(2).
The shadow of the Harada-Norton group F5 also arises to cause complications.
We have reduced to the case where U is abelian. In this difficult case, we show
that only G ~ Ru arises. Our approach is to use a modified version of the amalgam
method on a pair of groups (LT, H), where H E H(LzT) with H i. M. Using
the fact that U is abelian, we can show that (VGz) is abelian, and hence conclude
that [V, VB] = 1 if V n VB i=- 1. In the context of the amalgam method, this
shows that the graph parameter b is odd and greater than 1. Then we show that
q(H/CH(U), U) :::; 2, which eventually leads to the elimination of all choices for
L/02(L), V, H/CH(U), and U other .than the 4-tuple leading to the Ru example.
We have completed the outline of our treatment of quasithin groups in the
main case, when there is LE .Cj(G, T) with L/0 2 (L) quasisimple. The case where
L/02(L) is not quasisimple is handled fairly easily in section 13.1. That leaves:
0.3.5. The case .C1(G, T) empty. In Part 6 we handle the case .C1(G, T) = 0.
Part of the analysis here has some similarities to the F 2 -case just discussed, and
leads to the groups Jz, Ja,^3 D4(2), the Tits group^2 F 4 (2)', U 3 (3), M 12 , L 3 (2), and
A5.
To replace the uniqueness subgroup (L, T), we introduce the partial order :S
on M(T) defined in section A.5, choose M E M(T) maximal with respect to :S,
and set Z := fh(Z(T)) and V := (ZM). Then by A.5.7, for each overgroup X ofT
in M with M =: CM(V)X, we obtain M = !M(X). The case where Ca(Z) is not
a uniqueness subgroup is relatively easy, and handled in the last section of Part 6;
in this case G ~ L 3 (2) or A5. The.case where Ca(Z) is a uniqueness subgroup is
harder; the subcase where m(V) = 2 and AutM(V) ~ L2(2) presents the greatest
difficulties, and is handled in Part 5-in tandem with the cases where V is the
natural module for L/0 2 (L) ~ Ln(2) for n = 4 and 5. The elimination of these
cases completes the proof of our Main Theorem.
0.4. An Outline of the Proof of the Even Type Theorem
Assume in this section that G is a simple QTK-group of even type, but G is
not of even characteristic. We outline our approach for showing G is isomorphic to
the smallest Janko group J 1.
As G is of even type, there is an involution z E Z(T) and a component L of
Ca(z). As G is quasithin of even type, the possibilities for L are few. Our object
is to show that L is a standard subgroup of G: That is, we must show that L
commutes with none of its conjugates, Na(L) = Na(Ca(L)), and Ca(L) is tightly
embedded in G. This last means that Ca(L) is of even order, but if g E G-Na(L)
then Ca(L) nCa(LB) is of odd order. Once this is achieved, the facts that z E Z(T)
and that L is highly restricted will eventually eliminate all configurations except
L ~ L 2 (4) and Ca(z) = (z) x L, where G ~ J 1 via a suitable recognition theorem.
Here are some details of the proof. We first observe that if i is an involution
in CT(L) and IS: CT(i)I :::; 2 for some SE Sylz(Ca(i)), then Lis a component of
Ca(i): For Lis a component at least of Cca(i)(z), and hence contained in KKz
for some component K of Ca(i) by "L-balance" (see I.3.1). Now the hypothesis
that IS: CT(i)I :::; 2, together with the restricted choices for K, leads to L = K as
desired.