15.3. THE ELIMINATION OF Mr/CMr(V(Mr)) = Ss wr Z2 1135Again we will divide the proof into two cases: m 3 (Y+) = 2 and m 3 (Y+) = 1.
We eliminate the first case in the next theorem:THEOREM 15.3.35. Case (2) of Hypothesis 15.3.10 holds, Y+S/R ~ S 3 , Y+ =
0 3' (Mr), Y+ < Y, and R = Cs(Vi)...
Until the proof of Theorem 15.3.35 is complete, assume I is a counterexample.
Therefore case (i) of 15.3.11.5 holds, so:LEMMA 15.3.36. (1) Y+S/0 2 ,w(Y+S) ~ S 3 x·S3.
(2) R = Cr(V).
The next lemma eliminates the shadow of G ~ A 10 , where L ~ A 6 • It also
eliminates the shadows of G ~ L wr Z 2 , for various groups L of Lie rank 2 over F 2.LEMMA 15.3.37. One of the following holds:
(1) Y+ :S Lo.
(2) L = Lo ~ L2(2n) or U3(2n) with n even, or L3(2). Further Y+ = YLYc
where YL := 02 (Y+ n L), Ye:= 02 (CY+(L)), IYLl3 = 3 = IYcl, and Y+/02(Y+) ~
Eg.
(3) L = Lo, with L ~ L3(2n), n even, or L/0 2 (L) ~ L 3 (4). Further YL :=
02 (Y+ n L) =f=. 1 and Y+ = YL(y) with y of order 3 inducing a diagonal outer
automorphism on L.PROOF. By 15.3.34, m 3 (L) ~ 1. Thus if L < L 0 , then Lo = 0
31
(I) by 1.2.2,
so (1) holds. Therefore we may assume L = L 0. By 15.3.12.1, YL =f=. 1.Suppose first that m3(L) = 1. Then IYLl 3 = 3. Further by 15.3.33.3 and
1.1.5.3, one of the following holds: Lis L 2 (2n), Lis L~(2m) with 2m = -8 mod 3,
or Lis L 2 (p) for some Fermat or Mersenne prime p. Then as YL =f=. 1 and SL acts
on YL, n is even in the first case; in the second case, either L ~ L 3 (2), or mis even
and L ~ U 3 (2m); and in the third case, p = 5 or 7, so that Lis L2(4) or L3(2) and
so appears in previous cases. Now if Lis L2(2n) or U3(2m), then Mr acts on the
Borel subgroup B of L containing SL, so B =ML by 15.3.33.1 and maximality of B
in L. Thus B acts on Y+ by 15.3.11.4. Hence Y+ induces inner automorphisms on
L. This also holds if Lis L 3 (2) since there Out(L) is a 2-group. Then by 15.3.12.2,
Y+ = YLYc with IYLl3 = 3 = IYcb and Y+/02(Y+) ~ Eg. Then (2) holds.
Finally suppose m 3 (L) = 2. Again by 15.3.33.3, Lis described in 1.1.5.3 with
O(L) = 1, and then by A.3.18, either(i) L = e(I), or
(ii) L ~ L3(2n) with 2n = f. 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.In case (i), Y+.:::; L, so that (1) holds. In case (ii), Y+ = YL(y)0 2 (Y) is S-
invariant of 3-rank 2, so f. = +1 and hence (3) holds, completing the proof of the
lemma.. D
The next lemma rules out conclusion (3) of 15.3.37, and eliminates the firstappearance of a shadow of Aut(He), where L/Z(L) ~ L 3 (4).
LEMMA 15.3.38. Ifm3(L) = 2, then Y+ :SL.
PROOF. Assume otherwise, and let J+ := I/Cr(L). Then case (3) of 15.3.37
holds, and in particular some element y of order 3 in Y+ induces a diagonal outer
automorphism on L. If L/0 2 (L) ~ L 3 (4), let n := 2; in the remaining cases L ~