14.3. FIRST STEPS; REDUCING (VGi) NONABELIAN TO EXTRASPECIAL 1001LEMMA 14.3.25. Z(LT) n U = 1.
PROOF. Assume ZL := Z(LT) n U I 1. Set VH := (Zf); then VH:::; Zu, and
as usual VH E R2(H) by B.2.14. As ZL ::;! LT and M = !M(LT), Ca(VH) :S:
Ca(ZL) :::; M. As ZL I 1, Zu > Vi, so by 14.3.24, either d = 4 and one of
conclusions (i)-(iii) of 14.3.23.2 holds, or d = 6 and U is the natural module for
iI £::! G2(2). In any case, Zf; I 1 by 12.8.13.4.
Assume first that iI is not solvable. Then from the previous paragraph, F* (H)
is quasisimple, so there is K E C(H) with k = F*(H). As K is irreducible on U
and U > V in each case, K does not act on V. Then as Kn M :::; Mv by 14.3.3.6,
K "f; M. Thus as Ca(VH) :S: M, [VH,K] -=fa 1. Therefore KE Lt(G,T) by 1.2.10,
and then as k is not L 3 (2) from 14.3.24, K/0 2 (K) £::! L 2 (4) by 14.3.4.1. Thus
case (ii) of 14.3.23.2 holds, so that Z2 £::! Zf; :::; k; in particular K = [K, Zf;]. As
m(Zf;) = 1, Czi (U) is a hyperplane of Zf;, so Z 0 := (Zf; n Zu )V1 is a hyperplane
of Zu by 12.8.10.6. Thus Zf; induces transvections on VH with axis Z 0 n VH. This
is impossible, as Zf; induces inner automorphisms on k and we saw K = [K, Zf;].
Therefore iI is solvable. Hence by the first paragraph, case (i) of 14.3.23.2
holds, so d = 4, H • rv = 83 x 83, Ll • ::;! H, • and settmg • K .= • (Zu gH ), K · rv = 83,
iI = ktd', and Ki M. Then KnM:::; (KnT)CK(U), since the latter group ismaximal in KCH(U). Set H+ := H/CH(VH)· As in the previous paragraph, Zf;
induces transvections on V H with axis Zo n V H. By the first paragraph of the proof,CK(VH):::; M, so that CK(VH):::; (KnT)CK(U). Therefore K+ has the quotient
group k £::! 83 and CK(VH) :::; CK(U). Thus we conclude from the structure of
SQTK-groups generated by transvections (e.g., G.6.4) that K+ £::! 83, and hence
CK(VH) = CK(U) and [VH,K] is of rank 2. Indeed as Zo is a hyperplane of Zu
centralized by Zf;, Zu = [VH, K] xCzu(K) and Czu(K) ::;! H. Set H := H/Czu(K)and H^1 := H/CH(U); observe that CH([VH, K]) :::; CH(Zu ), and (J is a quotient of
U and so elementary abelian. As (J = (Vl) and V2 :::; fh(Z(T)), 02(H^1 ) = 1 by
B.2.13. As CK(U) = CK(VH):::; CK([VH,K]):::; CK(Zu), CK(U)^1 :::; 02(K^1 ) = 1.
Therefore CK(U) = CK(U), so K^1 £::! k £::! 83. Next [CH(U), K] :::; CK(U) =
CK(VH), so that [CH(U)+,K+] = 1. Then as EndK+([VH,K]) £::! F2, CH(U):::;
CH([VH, K]) :::; CH(Zu ), so CH(U)' :::; 02(H^1 ) = 1. Therefore CH(U) = CH(U),
and hence iI £::! H^1 • Next (J = (VH), while V = [V, Ll] and Li ::;! H^1 as H^1 £::! iI, so
we conclude (J = [U, L 1 ], contrary to 1 # [VH, K] :::; Cu(L1) since (J is eiementary
abelian. This contradiction completes the proof of 14.3.25. D
THEOREM 14.3.26. Assume Hypothesis 14.3.10. Then either Zu = V 1 so thatU is extraspecial, or G £::! H 8.
REMARK 14.3.27. If Hypothesis 14.3.1 did not exclude the possibility that
K/0 2 (K) £::! A 6 for some K E Lt(G,T), then 8p5(2) would also appear as a
conclusion in Theorem 14.3.26. Its shadow will be eliminated during the proof oflemma 14.3.31. Recall that the case leading to 8p 6 (2) was treated in Theorem
13.4.1.
Until the proof of Theorem 14.3.26 is complete, assume G is a counterexample.
Thus Vi< Zu. Then by 12.8.13.4, Zf; -=fa 1.
Recall V 2 = V 1 V{. As L/02(L) £::! L3(2) by Theorem 14.3.16, we may choose
l E CL(V[) with V = V 2 Vf. In particular Vi = 11;_^9 V{, so we may apply results