568 2. CLASSIFYING THE GROUPS WITH [M(T)[ =^1
PROOF. This follows from 2.5.12.3. DLEMMA 2.5.24. <I>(Sc) =/:-1.PROOF. Assume (Sc) = 1, define E 1 and E2 as in 2.5.23, and set Qi :=
02(NH(Ei)). Now S/SK So! E4 since fI So! Aut(A5) by 2.5.21, so that (S) :::;
SK Sc. Let Y denote the cyclic subgroup of SK of index 2. Then Y:::; [SK, SJ :::;
[S, SJ :::; (S). Since Y = (S) ~ (S), (S) :::; Y x Sa. Then using the Dedekind
Modular Law, (S) = Y x o, where o := (S)nSc. In particular as (Sc) = 1,
((S)) = (Y) = (zK), so by 2.5.13.2, z = ZK EK and tx = tzK = tz.
Next Cs(Y) = S 1 , where S 1 is the modular subgroup M15 (see p. 107 in
[Asc86a]) of S. Thus
S+ := r21(Cs((S))) = D1(Cs 1 (c)) is either So(z) or So(z)
where So is the preimage in S of the subgroup generated by the transposition in
fI So! Aut(A 6 ) centralizing :Y. Thus as Sc n S 0 = 1 by 2.5.5.1 while x acts on
S+, we conclude as usual that m(Sc) :::; 2, with S+ = So(z) =Sc x S 0 in case of
equality.
Suppose the latter case holds. Then m( Sc) = 2, and S 0 contains an element
inducing a transposition on K. Thus A(S) = {Q1, Qz}, and Qi = ScS 0 Ei So!
E 32. Further S is transitive on A(S), so by a Frattini Argument, we may take
x E NT(S) n NT( Qi) for each i, and hence Ns(Ei) < Ns(Ei)(x), so Ns(Ei) ¢,
Sylz(Nc(Qi)). But by 2.5.23, (Ns(Ei), NH(Ei)) E U*(H), whereas 2.5.10.2 says
Ns(Ei) is Sylow in Nc(Qi)· This contradiction eliminates the case m(Sc) = 2.
Therefore m(Sc) = 1, so as we are assuming Sc is elementary abelian, in
fact Sc = (t) is of order 2. Suppose first that Ef :::; KSc. Then x acts on
S := E1Sc(E1Sc)x. If x does not normalize E1Sc then A(S) = {E1Sc, E2Sc},
so Sis transitive on A(S-), and again by a Frattini Argument we may replace x by
x' E NT(S) n NT(SaE 1 ), and assume x acts on E 1 Sc. Thus x acts on E 1 Sc and
hence on Cs(E 1 Sc) = Q 1 , allowing us to obtain a contradiction as in the previous
paragraph.
Thus Ef i SaSK. We showed z = ZK, so Ef ~ z^0. Therefore by 2.5.22.1,
ex induces a transposition on K So! A 6 for some e E E 1 - (z). Now some conjugate
v of ex in SKex centralizes SK, so Qi =Sc x Ei(v) So! E15, and Sis transitive on
A(S) = {Q1, Qz}, so by a Frattini Argument we may choose x E N.T(S) n NT(Qi),
leading to the same contradiction as in the two previous paragraphs. D
LEMMA 2.5.25. tx = ZK and z = tzK.
PROOF. Assume otherwise. Then by 2.5.11.1, z = ZK, tx = tzK, and (t) =
Z(S) n Sc, so W) = Z(S) n S 0. But SK n S 0 is normal in S, so if 1 =/:-SK n S 0
then 1 =/:-Z(S) n SK n S 0 , contradicting tx = tzK. Hence SK n S 0 = 1. Thus
[SK,ScJ :::; SK n S 0 = 1, so S 0 ::; Cs(SK) =:So, and hence S 0 is isomorphic
by 2.5.5.1 to a subgroup of So So! E4, whereas Sc is not elementary abelian by
2.5.24. D
LEMMA 2.5.26. m2(So) = 1.
PROOF. Assume m 2 (Sc) > 1. In the first few paragraphs of the proof, we will
establish the claim that K is a component of Cc(i) for each involution i E Sc.
Assume otherwise; by 2.5.18, i =/:- t, and by 2.5.19.2, Cs 0 (i) = (i, t). Further