16.3. SHOWING L IS STANDARD IN G 1177In case (ii), CL 0 (i) ~ Dp+E x Dp+E, where p = E mod 4 and E = ±1. But 3 divides
p + E as p is a Fermat or Mersenne prime, so C La ( i) contains E ~ E 9. This isimpossible, since i E T£, so using 16.2.9.1,
E::::; 0
31
(CH9(i))::::; CLg(i)::::; CL9(i) X Lug~ Dp-E X L2(p).Similarly in case (i), CL 0 (i) ~ L2(2n) x L2(2n), and by 16.2.8.3 and 16.2.9.1,
1-:/= 02 (CL 0 (i))::::; Oq' (Hg)= Lg,
for any prime divisor q of 2n - 1. Since L2(4) ~ L 2 (5), we may assume n > 1, so
such primes q exist and mq(CL 0 (i)) = 2 while mq(CLg(i)) = 1. This contradiction
completes the proof of Theorem 16.2.4.16.3. Showing L is standard in GIn Theorem 16.3.7 of this section, we will show that the component L is in
standard form in G, in the sense of [Asc75]: that is Ca(L) is tightly embedded in
G, Na(L) = Na(Ca(L)), and L commutes with none of its conjugates.
To show that L is standard, we show that L is "terminal" in the sense of
[GLS99], as defined earlier. The next two lemmas show that if Lis terminal then
Lis standard. The proof of the first lemma makes use of the normality of Lin Gz
which we established in Theorem 16.2.4.LEMMA 16.3.1. Ca(L) contains at most one component isomorphic to L, and
no component G-conjugate to L.PROOF. The first assertion follows from A.1.34.2 with Na(L) in the role of
"H". Assume that Lg is a component of Ca(L). Then by the first assertion,
8(L) := {L"': x E G and L"' is a component of Na(L)} = {L, Lg}.
Since L is not SU 3 (8), case (1.a) of A.1.34 holds, so that LLg = or' (Na(L))
for a suitable odd prime r, and hence U = or' (Ca(L)). It follows that L =
or' (Ca(L9)), so that 8(L) = 8(U) = 8(L)g. Thus g E Na(8(L)) =: N, andhence a Sylow 2-subgroup of N is transitive on 8(L) of order 2. This is impossible,
as by Theorem 16.2.4, T::::; Na(L) ::::; Na(8(L)) = N so that T fixes 8(L) pointwise
but is also Sylow in N. D
LEMMA 16.3.2. Assume that L is a component of Ca(t) for each involution
t E Ca(L). Then Lis standard in G.
PROOF. Assume the hypothesis of the lemma. We first observe that Ca(L)
contains no conjugate of L, verifying the third condition in the definition of "stan-
dard form". For if Lg ::::; Ca(L), then L9 is a component of Ca(i) for each involutioni E L by hypothesis, so as Ca(L) ::::; Ca(i), L9 is a component of Ca(i), contrary
to 16.3.1.
Set H := Na(L), X := Ca(L), and assume that X n Xg is of even orderfor some g E G. We will show that g E H, which will suffice: For then since
Ca(L) is of even order and H::::; Na(Ca(L)), Ca(L) is tightly embedded in G and
Na(Ca(L)) = H, verifying the remaining conditions for L to be in standard form.
Finally assume that g €}!. H. Thus L9 -:/= L, while as X n X9 is of even order,
there is an involution t E X n Xg. By hypothesis, L and L^9 are components of
Ca(t), so as Lg i= L, L9::::; Ca(L), contrary to our first observation. D