16.2. NORMALITY AND OTHER PROPERTIES OF COMPONENTS 1175PROOF. Let g E Na(Lu), and let t be an involution in Lu. By 16.2.6.1, Lis a
component of Ca(t), so as Ca(Lu) :::; Ca(t), Lis a component of Ca(Lu). Since
g E Na(Lu), LB is also a component of Ca(Lu). If L =/= £9, then Ca(Lu) contains
three isomorphic components Lu, L, and LB, contrary to A.1.34.2. Thus L = LB,
so g normalizes LLu = L 0. Therefore Na(Lu) :::; Na(Lo) = H, so the lemma holds
as u EH. D
In the next few lemmas, we will show that L is tightly embedded in G. Recall
that a subgroup Kofa finite group G is tightly embedded in G if K has even order,
but Kn KB is of odd order whenever g E G - Na(K).
LEMMA 16.2.8. (1) If g E G such that (z)Lu n ( (z)Lu)B has even order, then
gEH.
(2) Either L is tightly embedded in G, or L/0 2 (L) ~ Sz(8) and 02 (L) =/= 1.
(3) If Xis a nontrivial 2-subgroup of (z)Lu, then Na(X):::; H.
PROOF. Observe that (3) is a special case of (1). Assume the hypotheses of
(1). Then there is an involution t E (z)Lu n ((z)Lu)B, so by 16.2.6, Land LB are
both components of Ca(t). Then LB normalizes L so that LB :::; H by 16.2.7, and
hence LB is a component of CH(t). Applying I.3.2, LB lies in a 2-component of H,
which is a member of C(H), so that by A.3.7, either LB E {L, Lu} or [LLu, LB]= 1.
The latter case is impossible, for since L/0 2 (L) is not U 3 (8), case (1.a) of A.1.34
holds, so that or' (H) = LLU for a suitable odd primer; while in the former, either
g or gu-^1 lies in Na(L), so g EH by 16.2.7. Thus (1) holds.
Now if Lu n LuB has even order, then g E H by (1). Hence if g tf. Na(L),
then LuB = L, so that 1 =/= Lu n L :::; 02 (L). Then we conclude from 16.2.5 that
L/0 2 (L) ~ Sz(8), so that (2) holds. D
LEMMA 16.2.9. (1) Let p be a prime divisor of 2n - 1 if L/0 2 (L) ~ Sz(2n) or
L 2 (2n), and let p := 3 if L ~ L2(r) for odd r. Then Lo= QP' (H).
(2) Loi. HB for g E G - H.
PROOF. Observe if L ~ L 2 (r) for r odd that 3 divides the order of some 2-local
subgroup of L. Then part (1) follows as case (a) of A.1.34.1 holds." If Lo :::; HB then
Lo= QP' (Lo) :::; QP' (HB) =Lg by (1), so that g E Na(Lo) = H, and (2) holds. D
When analyzing a tightly embedded subgroup Kofa group G, one focuses on
the conjugates KB such that NKY(K) is of even order. (See e.g. the definition of
Ll(K) in Section 4.) In our present setup, ·we need a slightly stronger condition,
which we establish in the next lemma:
LEMMA 16.2.10. (1) The strong closure of TL in NT(L) with respect to G
properly contains TL U Tl';.
(2) There is g E G - H such that jLB n NH(L)l2 > 1.
PROOF. Set A 1 := TL, A2 := T}';, and assume A1 U A2 'is strongly closed in
NT(L) with respect to G; we check that the hypotheses of Lemma 3.4 of [Asc75]
are satisfied. First if Af n Aj =/= 1 for some i, j, then A;{B n A~ =/= 1 for some choice of
v, w E {l, u-^1 }; therefore wgv-^1 E Na(A 2 ) :::; H by 16.2.8.1, and hence also g E H
as u EH. Thus the subgroup H 0 of H generated by all such elements g plays the
role of the group "H" in 3.4 of [Asc75]. Next as H permutes {L, Lu}, A 1 U A 2 is
strongly closed in T with respect to H. Of course NT(Ai) = NT(L), so hypothesis
(*) of 3.4 of [Asc75] is satisfied.