i4.6. ELIMINATING L2(2) WHEN (VG1) IS ABELIAN 1035Kz = KiK2, we obtain J = [J, Ki]. As m5(G2):::; 2 and m 5 (K 2 ) = 1, m 5 (J):::; 1,
with J = 0
51
(Oo 2 (K2)) :=:;I G2 in case of equality.Now either Qi does not normalize J, so that C.2.4.1 holds, or Qi normalizes J,
so that C.2. 7.3 holds. Set J+ := (JQ^1 ), and observe that B := J+ n Gi, 2 normalizes
K2 and hence also Ki.
Suppose first that Ki i. J. Then an element k of Ki of order 5 induces
an outer automorphism of order 5 on J, since we saw that J = 05 ' (0 02 (K 2 ))
when 5 E 7r(J). Inspecting C.2.4.1 and C.2.7.3 for cases where Jj0 2 (J) admits an
outer automorphism of order 5, we conclude J/0 2 (J) is L 2 (2n) or SL 3 (2n) with 5
dividing n, k induces a field automorphism on J / 02 ( J), and B is a Borel subgroupor a minimal parabolic of J+, respectively. But then B does not normalize Ki,
contrary to the previous paragraph.Thus Ki :::; J, so that m5(J) = 1 and J :=:;I G 2. Now we examine the list of
C.2.7.3 for those J of 5-rank 1 with a subgroup Ki normalized by B, such that
Ki/02(Ki) ~ Z5 or L2(16). We conclude that Ki/0 2 (Ki) ~ Z 5 , J is an L 2 (2m)-block, and B is a Borel subgroup of J. But then as Z :::; B :::; Gi, 2 :::; Oo(Z),
Z :::; Z(B) = Z(J) using the structure of an L2(2m)-block; so J :::; Oo(Z) = Gi,
contrary to Ji. Gi,2· This contradiction completes the proof of 14.6.15. D
We will see shortly in 14.6.17 that the group T 0 in the following result can play
the role of "To" in (Ul) in the first subsection.
LEMMA 14.6.16. Let To:= NT(Hi). Then [T: Toi= 2 and No 1 (To)= T.
PROOF. From 14.6.12, IT: To[= 2 and T = NH(T 0 ). Further p = 3 by 14.6.15,
and in particular case (4) of 14.6.14 does not hold.
Suppose case (2) or (3) of 14.6.14 holds. Then Gi = KzT and B := N kz ( CJH)is a parabolic subgroup of k z with uni potent radical Q H and fI = BT = N 01 ( Q H).
Thus QH is weakly closed in Twithrespect to Gi byl.2.5, so No 1 (To):::; No 1 (QH) =
H, and hence No 1 (To)= NH(To) = T.
Finally assume case (1) of 14.6.14 holds. Then k = Ki x K2 where Ki :=
02 (Hi) and Ki and K 2 are the two T 0 -invariant subgroups of k of order 3. Thus
X := 02 (No 1 (To)) acts on ki and hence X centralizes k. Then as Ok(T 0 ) = 1and m3(Gi) = 2, X is a 31 -group. However as case (1) of 14.6.14 holds, Gi is a
{2, 3}-group, so again we conclude that No 1 (To) = T. D
We can now determine H*, and show that the set U(H) of involutions discussed
in the first subsection is nonempty.
LEMMA 14.6.17. (1) UK is a 4-dimensional orthogonal space over F 2 for H* =O(UK) - rv = 0 +
4 (2).
(2) z, i UHl·
( 3) Let u E UK with u nonsingular in the orthogonal space UK and centralizedby NT(Hi). Then u E U(H).
(4) Oo(u) E 'I, so I* is nonempty.
(5) m( (V02(H2))) = 4.
PROOF. Set To := NT(Hi) and let ui E UH,i - Z with ui E Z(To). We first
show that ui E U(H) in the sense of Subsection 1. By choice of ui, (UO) and
(Ul) are satisfied, and (U3) holds by 14.6.16. Next fort ET - To, ui E UH,2, so