l108 is. THE CASE ..Cf(G, T) = 0We next construct a subgroup Yi of H with Yi of order 3, Yi :::l Mi, and
Mi = (Min Mc) Yi.
Suppose first that V = V(M). Then by A.5.3.3, CM(V) ::; CM 1 (Vi) n M::; H,
so as H = Y 0 f', Yo::; Hand H = Y 0 T. Therefore Yo :::l YQT = fI =Mi. Further
Y 0 =/:-1 as 02 (H) =/:-1, so as Yo has order 3 and CM(V) ::; CM 1 (Vi), we conclude
that Yo has order 3. In this case let Yi be the preimage in Mi ofYQ, so that Yi= Yo
has order 3 and Mi= YiT, so that Mi= Yi(Mi n Mc) and Yi :::l Mi.
Suppose instead that V < V(M). Then by our earlier discussion, M = H(M n
Mc) with IM : Mn Mel = 3. Thus M = Yo(M n Mc), and Yo is the unique f'-
invariant subgroup of M of order 3 not contained in Mn Mc. Let Re := 02(M n
Mc), and define Y as in 15.2.3. As IM : M n Mel = 3, Y = [0^2 (Mo), Re] <
02 (M 0 ) ~ E 9 , so case (iii) of 15.2.3.1 holds, and Y ~ Y = [0^2 (M 0 ), Re]. By the
uniqueness of Yo mentioned above, Y = Y 0. Therefore YCM(V) = (YonH)CM(V).
Now Re acts on Yon H, Y R~CM(V) = Y R~ x CM(V) by 15.2.3.5, and Y =
[Y, R~] since Y ~ Y = [Y, Re]· Thus Y ::; H, so as JM : Mn Mel = 3 is prime,
H = Y(H n Mc)· Then as we saw Mi = H(Mi n Mc), Mi = Y(Mi n Mc)· As
Y :::l M and Mi = fI, Y :::J Mi. As Yi Mc:::::: CM 1 (Vi) :::::: CM(Vi), Y =/:-1, so as
Y* has order 3 and Cy(V) ::; CM 1 (Vi), we conclude Y has order 3. In this case, let
Yi be the preimage in Mi ofY, so that Yi= Y has order 3, and Mi= Yi(MinMc)·
This completes the definition of Yi in our second case. ·
Now in either case we have the hypotheses of case (b) of 14.1.17, with Mi, Mc,
Ri, Yi in the roles of "M, Mi, R, Yo". We claim Yi = [Yi, Ri]: For otherwise
Ri is normal in YiM01Afc = Mi, whereas 02(Mi) = 1 by B.2.14. Set Y2 :=
02 ((Rf^1 )) = 02 ((Rf^1 T)), so that Y:i plays the role of "Y" in 14.1.17. Since Riis
normal in Min Mc and Mi= Yi(Mi n Mc), we have Y 2 = 02 ((Rf'1^1 )) normal in
Mi, so that Mi= Nc(Y2) as Mi EM. To complete the proof, we will show that
Y2 :::l M, so that M = Na(Y2) =Mi, contrary to our choice of Mi=/:-M.
Since Yi = [Yi,Ri] is of order 3, Y2 =Yi~ Z3. But by 14.1.17.3, Cy 2 (Vi)+
centralizes Rt, soy;+~ Y2 ~ Z3. Moreover °Y:iCM 1 (Vi) = YiCM 1 (Vi), so arguing
as above when we showed Y::; H, we conclude Y2::; H. Further Y 2 + = [Yi+,RtJ.
Suppose first that V < V(M). Then by construction Y ::; Yi and Y = Yi =
[Yi,Ri], so that 02(Y) = Cy(Vi) as Y* ~ Z3. Then as Ri acts on Y, Y = [Y,Ri]. _
Thus Z3 ~ Y 2 + = [Yi+, Ri] :::::: [Y+, Ri] = y+ ~ Z3, so y;+ = y+. Then °Y:i =
02 (Y202(Mi)) = 02 (Y0 2 (Mi)) = Y :::l M by 15.2.3, completing the proof in this
case.
Thus we may assume that V = V(M). Here we saw that CM(V) ::; H, so
CM(V) ::; Mi = Na(Y2), and [R, CM(V)] ::; 02(H) n CM(V) ::; 02(CM(V)) ::;
02(M), so that R centralizes CM(V). Further R centralizes O(H), so case (ii)
of 15.2.3.1 does not hold, since there Y ~ 3i+^2 , in which case involutions not
inverting 02 (M) do not centralize Cy(V). Therefore a Sylow 3-subgroup P of M
is abelian by 15.2.3.1. Choose P with X := P n Y 2 E Syh(Y 2 ). Then P centralizes
X and normalizes CM(V). Now we saw CM(V) ::; CM 1 (Vi) and Y 2 :::l Mi, so
(XCM(V)) = Y2. Hence P acts on Y 2 so HP = M acts on Y 2 , completing the
~~ D
LEMMA 15.2.6. Define Re and Yasin 15.2.3. Then there exists a T-invariant
subgroup Yi:= 02 (Yi) of Y such that