1547845830-Classification_of_Quasithin_Groups_-_Volume_II__Aschbacher_

15.3. THE ELIMINATION OF Mf/CMf (V(Mf)) = Ss wr Z 2 1139

By 15.3.42, L+ is of Lie type and characteristic 2, and hence is described in

cases (a)-(c) of 1.1.5.3. By 15.3.41.2, Mt is a maximal S-invariant parabolic of

J+ and S+ acts on Yt with Yt s+ /02,w(Yt s+) s:! S3 x S3. Therefore either

L+ £:! L4(2) or L5(2), or L+ is of Lie rank 2 and defined over Fzn for n > 1 with

Y+ contained in the Borel subgroup B of Lover SL.

Assume the latter case holds. Then as Y+S/0 2 ,w(Y+S) £:! S 3 x S 3 , we conclude

from the structure of Aut(L+) that L+ ~ L 3 (2n) with n even. As 02 ,^3 (B) ::; CM(V)

by 15.3.2.1, and [V, Y+] ~ B, we conclude n = 2. Since O(L) = 1 by 15.3,33.3,

L ~ L 3 (4), so that m3(B) = 1, a contradiction.

Therefore L ~ L4(2) or L5(2). As Y+S/R ~ S3 x S3, we conclude that Sis

trivial on the Dynkin diagram of L. By 15.3.41.2, Mt is maximal in J+ subject to

F*(Mt) = 02(Mt), so L+ ~ L4(2) and Mt= Yts+ is the maximal parabolic

determined by the two end nodes. Therefore Yt s+ is irreducible on 02 (Yt s+) of

order 16, impossible as E4 ~ Vz = [Vz, Y+] ::;! Y+S.

This contradiction completes the proof of Theorem 15.3.35.

Finally we complete our analysis of Hypothesis 15.3.10 by eliminating the only
possiblity left in Theorem 15.3.35:

THEOREM 15.3.43. Y+S/R is not S3.

·PROOF. Assume otherwise; then case (ii) of 15.3.11.5 holds, so

(a) Y+ = 0

31

(M1) with IY+: 02(Y+)\ = 3.

In particular, case (2) of Hypothesis 15.3.10 holds, so that

(b) [V1, Y+] = 1 and Vz = [V2, Y+] ~ E4·

Further Na(Vi) :::; M by 15.3.11.3, so as L f:. M, [Vi, L] =/=-1. Therefore as V1 :::;

Cs(Y+) by (b):

(c) [V1, L] =/=- 1 and Cs(Y+) =/=-Cs(L).
Suppose Y+ ~ h. As IY+\3 = 3, case (1) of 15.3.7 holds and Y+ = 031 (M 2 ).
Thus Y = Y+ x Y-t. ~ A4 x A4 fort ET-S, contrary to 15.3.9. Therefore:

( d) Y+ is not isomorphic to A4.
Arguing as in the the proof of 15.3.37, one of the following holds:
(i) Y+ :::; e(J) = La.
(ii) L = Lo is of 3-rank 1 with L/02(L) ~ L 2 (2n), Lg(2m) with 2m = -8
mod 3, or L 2 (p) for some Fermat or Mersenne prime p.
(iii) L =Lo~ L3(2n), 2n = E mod 3, or L/0 2 (L) ~ L 3 (4). Further some y of
order 3 in Y+ induces a diagonal outer automorphism on L.

Suppose that Y+ does not induce inner automorphisms on L, Then as Y+S / R ~

S 3 , conclusion (iii) holds. As Y+S = SY+, Y+S acts on a Borel subgroup B of L
over Sn L, so by 15.3.33.1, B:::; ML. But then m3(M1) > 1, .contrary to (a).
Therefore Y+ induces inner automorphisms on L, and hence case (i) or (ii)

holds. As L f:. M, [L, Y+] =/=-1 by 15.3.11.4, so the projection YL of Y+ on L is

nontrivial. Now Ns(L) acts on YL, so Sn YL E Sylz(YL) and hence YL normalizes
02 (Y+0 2 (SnYL)) = Y+, so that YL:::; M 1 by15.3.11.4. Thus YL:::; 031 (M 1 ) = Y+,
so YL = Y+ since IY+\3 = 3.
As S normalizes Y+ and Y+::; L, S normalizes L, and hence L ::;! I. As Y+:::; L
and IE 1-l+,*' I= LS. Let J+ := I/C1(L). Now we obtain the following analogue