1547845830-Classification_of_Quasithin_Groups_-_Volume_II__Aschbacher_

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1184 16. QUASITHIN GROUPS OF EVEN TYPE BUT NOT EVEN CHARACTERISTIC


LEMMA 16.4.3. {1) KE b..(K').


{2) L' = [L',z].

PROOF. Part (I) is a consequence of 16.4.2.1. By 16.4.2.1, z E Z(T) n Tc :::;
Nr 0 (K'). Thus (2) follows from 16.4.2.1 and the fact that b.. is symmetric. D


LEMMA 16.4.4. {1) Assume R is of order 2. Then Cr 0 (R) = (z) is also of order

2, m 2 (RTc) = 2, and RTc is dihedral or semidihedral. If furthermore Z(L) f-I,


then z E Z(L).

(2) If Tc is cyclic, then K = TcO(K) and CK(z) =Tc.


PROOF. Assume R is of order 2. Then by 16.4.2.1, Cr 0 (R) is also of order 2,

so that Cr 0 (R) = (z). Hence by Suzuki's lemma (cf. Exercise 8.6 in [Asc86a]),


RTc is dihedral or semidihedral, so that (I) holds.

Next assume Tc is cyclic. Then by Cyclic Sylow 2-Subgroups A.1.38, K =

TcO(K). Further as Gz :::; H = Na(K), Co(K)(z) :::; O(Gz) = 1 by (El), so that

(2) holds. D

Now we begin the process of obtaining restrictions on H, and in particular on

the Sylow 2-subgroup R of NK1(K).


LEMMA 16.4.5. Assume pis an odd prime with mp(L) > 1, and i is an involu-


tion in K. Then either

(1) L =OP' (Ca(i)), or

(2) p = 3, 031 (Ca(i)*) ~ PGL3(2n) or L~'^0 (2n), with 2n = E mod 3, and


L = 031 (LCK(i)). In particular, 031 (H*) ~ PGL3(2n) or L~'^0 (2n).

PROOF. Recall L ::::! Ca(i) by Remark 16.3.3, and O(L) = 1 by (EI). Thus as
Lis in the list of (E2), Ca(i) satisfies conclusion (I) or (2) of A.3.18, so the lemma
holds. D


The next lemma eliminates the shadow of L2(p^2 ) (p a Fermat or Mersenne

prime) extended by a field automorphism, and the shadow of 87. These groups are

quasithin, and have a 2-central involution with centralizer Z2 x PG L2 (p), but the

groups are neither simple nor of even type.


LEMMA 16.4.6. If L ~ L 2 (q), q odd, then no involution in R induces an outer

automorphism in PGL 2 (q) on L.

PROOF. Let r denote an involution in R with L(r) ~ PGL 2 (q). Recall q is a

Fermat or Mersenne prime or 9 by (E2). Further if q f-9, Aut(L) ~ PGL 2 (q), so


either H ~ PGL2(q), or q = 9 and H ~ Aut(PGL 2 (9)) ~ Aut(A 6 ).

If q f-9, let Ro := RnLK; while if q = 9, let Ro be the subgroup of R inducing
automorphisms in 85. Then R = R 0 (r). If Ro f- 1 there is an involution ro in Ro,


and L = (CL(r), CL(ro)) :::; H' by 16.4.2.5, contrary to 16.4.2.2. Thus Ro = 1, so

R = (r) is of order 2. Hence by 16.4.4.1, (z) = Cr 0 (R) is also of order 2. Choose
Tso that Nr(R) E 8yh(NH(R)).


Let E := (r, z) and TE := Cr(E). As RTL is dihedral, CrL (r) =: (v) is of

order 2. Therefore as Cr 0 (R) = (z), TE nLKR =: V = (v,z,r) ~ E 8 , and either


H ~ PGL2(q) and TE = V, or H ~ Aut(A 6 ) and TE ~ E4 x Z4. Further

TE E 8yl2(Ca(E)) by 16.4.2.6. As RTL is dihedral, rv E rNTL(TE). From the

structure of Aut(L), Z(T) = (v), so Z(T) :::; (v)Tc n TE = (v, z), and hence

Z(T) = (z, v)..
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