1547845830-Classification_of_Quasithin_Groups_-_Volume_II__Aschbacher_

2.5. ELIMINATING THE SHADOWS WITH r~ EMPTY 565

We now eliminate all possibilities for K remaining in 2.5.16 except for the one

corresponding to the most stubborn remaining shadow discussed earlier:

LEMMA 2.5.21. fl 9:! Aut(A5).

PROOF. First Ko = K by 2.5.20, so H = KS. Assume fl is not Aut(A 6 ).

Then by 2.5.16, either K 9:! L2(P) for p ~ 7 a Fermat or Mersenne prime, or

fl 9:! PGL2(9) or Mio· Therefore either Sis dihedral, or fl 9:! M 10 and S 9:! SD 16.

Hence by 2.5.5.1, Sc is cyclic or dihedral, unless possibly Sc 9:! Qs or SD 16 when

fl 9:! M10. In each case [fl : .K[ ~ 2.

Assume that Sc is of order 2, so that Sc= (t). As [H: .K[ ~ 2, [S: SK[~ 4, so

S /SK is abelian and hence [S, SJ ~ SK. Also S is dihedral or semidihedral of order

at least 8, so fh([S, SJ) = (zK)· Therefore Di([S, SJ) = (zK)· Then x centralizes

ZK, so by 2.5.13.2, z = ZK and tx = tz. Thus all involutions in Kare in z^0 , and all

involutions in tK are in t^0. Choose (U,Hu) E U*(H). Then by 2.5.12, there is a

4-subgroup E of SK such that U = Ns(E) and Hu= NH(E). Since Sc= (t) is of

order 2, NH(E) = NH(F) 9:! Z2 x S4, where F := ESc = 02(NH(E)) = 02(Hu) 9:!

Es. In particular Ns(F) = U E Syl 2 (Na(F)) by 2.5.10.2. If px E F^8 , then by a

Frattini Argument, we may take x E NT(F), contradi<;ting Ns(F) E Syh(Na(F)).

Thus px rJ. F^8.

Assume first that S 1:. KSc. H =KS is transitive on Es-subgroups of KSc,

so px 1:. KSa. But all involutions in M10 are in E(Mio), so if K 9:! A 6 then

fl 9:! PGL2(9). Thus fl 9:! PGL2(q) for q a Fermat or Mersenne prime or 9. But

x acts on Z(S) = (z, t) ~ KSc, so as px 1:. KSc by the previous paragraph,

ex rJ. KSc fore EE - (z). As ex rJ. KSc and fl 9:! PGL2(q), O(CK(ex)) =j:. 1, so

since K is a component of Gt by 2.5.18, 1 =/:-O(CK(ex)) ~ O(Ca((ex,t))). Hence

Ca(ex) rJ. He by 1.1.3.2, contradicting ex E z^0.

Therefore S ~ KSc, so H = K x Sc, and hence S = SK x Sa. This rules

out cases (2) and (3) of 2.5.16 in which S is nontrivial on the Dynkin diagram of

K, so K 9:! L 2 (p) for p > 7 a Fermat or Mersenne prime. We saw earlier that

tE ~ t^0 , so there is g E G with t9 E F-(t, z). As Sc= (t) is oforder 2, Cat(K) =

0( Ca. (K) )Sc by Cyclic Sylow 2-Subgroups A.1.38. By 2.5.18, Gt = K SC at (K)

and O(Ca.(K)) = O(Gt) = 1, so Gt= KSc = H. Thus F ~ Gf = H^9 = K^9 Sf:y,

so AutKg(F) 9:! S 3 ; and hence (AutK(F),AutKg(F)) is the parabolic in GL(F)

stabilizing (zG n F) =Kn F = E. As this group is transitive on F - E of order 4

. and S ~ Gt, we conclude [Na(F) : Ns(F)[2 ~ 4, contradicting our earlier remark

that Ns(F) E Syh(Na(F)). Therefore [Sc[ > 2.

Suppose next that Sc is abelian. From remarks at the start of the proof, either

Sc is cyclic, or possibly Sc 9:! E4 when fl 9:! Mio· By 2.5.11.1, Z(S) n Sc = (t),

so Sc 1:. Z(S), and hence S 1:. KSa. Indeed as [S: SK[ ~ 2, [S: SKSc[ = 2 and

Cs(Sc) = SKSc. Thus conjugating by x, also [S: Cs(Sc)I = 2, so [S: Cs(Sc)I ~

2. Hence as S is dihedral or semidihedral of order at least 16, while Sc 9:! Sc is
   abelian of order at least 4 by 2.5.5.1, we conclude Sc is cyclic and Sc :SK. Since
   Sc Sc= Sc x Sc by 2.5.5.1.2 we conclude Sc x Sc~ Sc x Y, where Y is the cyclic
   subgroup of index 2 in SK, and Cs(Sc) =Sc x Y. This is impossible, as
   Cs(Sc) = Cs(Sc)x 9:! Cs(Sc) =Sc x SK,
   and SK is nonabelian.

This contradiction shows that So is nonabelian. So again by our initial remarks,

either So is dihedral of order at least 8, or H/So 9:! M10 and Sc 9:! Qs or SDrn.