550 2. CLASSIFYING THE GROUPS WITH IM(T)I = 1Let 02 := G 2 /Ca 2 (K). By 2.4.35, S 1 K 9:! Aut(A5) and hence G2 9:! Aut(A5).
In particular IB1I = 2^5 , so as CR(K) 9:! Z4 and IB1I = ITl/2 = 2^7 , it follows that
CR(K) E Sylz(Ca 2 (K)). Then by Cyclic Sylow 2-Subgroups A.1.38, Ca 2 (K) =
O(G 2 )CR(K). Recall that z = z1z2 with z1 E K, so that Ca 2 (K) ::::; Ca(z).
However by 2.3.9.5, z inverts O(G 2 ), so O(G2) = 1, completing the proof of (2). D
LEMMA 2.4.37. zf n R = (zf n RK) u (zf n R'K) with lzf n RKI = 5.PROOF. Recall A 1 E A(T) is defined in 2.4.27.2. Further by 2.4.27.2, T ii:iduces
the 4-group((Q, Qx), (A1, Al))
of permutations on A(T). Thus y := x or xr acts on A 1 , so SA := R(y) is of index
2 in T and BA normalizes Ai. As z E Ai, H 1 := Na(A1) E 'He by 1.1.4.3. Now
NH(A 1 ) contains L 2 1:_ M. Also Q::::; R = J(S), so by 2.3.8.5c, Co2(M)(R) ::::; R.
Then R E f3 by 2.3.8.5b, so as usual H 1 E r. Then as IBAI = ISi, Hi E r* by
2.3. 7.1, so we may apply the results of this section to Hi in the role of "H". In
particular we conclude from 2.4.29^2 that H 1 induces ot(2) on Ai. Therefore for
each A E A(T), A= A^1 x A^2 with Ai 9:! E 4 and A^1 # u A^2 # = zf n A. By 2.4.36.1,
R = RK x R'K, so A = (An RK) x (An R'K) with An RK 9:! An R'K 9:! E 4. Thus
as all involutions in RK are in zf, A^1 =An RK and A^2 =An R'K. Therefore aseach involution in R is in a member of A(T), the lemma holds. D
We are now in a position to obtain a contradiction, and hence complete the
proofofTheorem 2.4.1. By 2.4.36, B := Ca 2 (K) = CR(K) 9:! Z4 and R = RKxR'K.
Let G2 := G2/B; then R = RK(u), where u EU - (zz). By 2.4.35, u induces a
transposition on K, so R = (u) x RK 9:! Z2 x D 8 is Sylow in RK 9:! 85.
Next each involution in R - K is either a transposition or of cycle type 23 ,
and there are a total of 6 involutions in R - K. Further u E zf and u is a
transposition, so as x induces an outer automorphism on RK, ux is of type 23.
Thus~:= zf n (R-K) is oforder 6m, where m := lzf nuBI. However by 2.4.37,
~ is s-conjugate to zf n RK of order 5.
This contradiction finally completes the proof of Theorem 2.4.1.2.5. Eliminating the shadows with I'Q empty
The groups occurring in the conclusion of Theorem 2.1.1 have already appeared
in Theorems 2.2.5 and 2.4.1, so from now on we are working toward a contradiction.
We have also dealt with the most troublesome shadows, although a number of other
shadows are still to appear.
By Theorem 2.4.1, we may assume r5 is empty: that is no member of r 0 iscontained in 'He. In 2.5.3, we will produce a component K of H, consider the
various possibilities for K listed in 2.3.9.7, and analyze the structure of Cs((K^8 )),
where S E Sylz(H). Eventually we eliminate all configurations, completing the
proof of Theorem 2.1.1.We continue to assume that G is a counterexample to Theorem 2.1.l. Therefore
as the groups in Theorem 2.4.1 are conclusions of Theorem 2.1.1, in the remainderof the section we assume that
rg = 0.
(^2) As mentioned earlier, our use of 2.4.29 here to exclude As-blocks is essentially eliminating
the shadow configuration.