566 2. CLASSIFYING THE GROUPS WITH JM(T)J =^1
Set So = Cs(Sc). In any case, (t) = Z(Sc) and SK :::; So, so as IS : SKI :::; 2,
ISo : SKI :::; 4 and hence ZK E [SK, SK] :::; [So, So] :::; SK. Let Y be the cyclic
subgroup of index 2 in SK. Then fh(Y) = (zK) and [S, SJ :::; Y, so [So, So] :::; Y
and hence S2 1 ([S 0 , S 0 ]) = (zK)· ·However Sa :::; Cs(Sc) by 2.5.5.2, and hence
[S 0 , S 0 ] :::; [So, S 0 ], so tx = ZK and z = tzK # ZK· Therefore tzK t/--tG in view of
2.5.11.2.
We next show that K is a component of Cc(i) for each involution i E Sc. We
assume i is a counterexample and derive a contradiction: As z # ZK, 2.5.19.4 says
K ~ A 6 and K < Ki :sl Cc(i) with Ki ~ As and t induces a transposition on
Ki. But then CKJt) ~ S 6 , whereas S E Syl2(Gt) by 2.5.11.2, and no element of
S induces an outer automorphism in S 6 on K since H ~ PGL2(9) or M10. This
contradiction shows K is a component of Cc(i).
Next we claim that K :sl Cc(i) for each involution i of Cc,(K): For assume
u E Cc(i) with K # Ku. Then (K, Ku) = K x Ku as K is a component of
Cc(i), and i # t by 2.5.20. Now (i, t) is not Sylow in Ku(i, t) n Gt, so (i, t)
is not Sylow in Cc( (i, t)K). On the other hand as S is Sylow in Gt, we may
assume Cs 0 (i) E Sy[z(Cc,(K(t)), a contradiction as Sc is dihedral, semidihedral
or quaternion. This contradiction establishes the claim that K is normal in Cc(i)
for each involution i of Sc.
Now assume Sc is not Q 8 ; in this part of the proof we eliminate the shadows of
subgroups of PSL 2 (p) wr Z 2 • By our earlier remarks, either Sc is dihedral of order
at least 8, or H/Sc ~ M10 and Sc~ SD 16 • Recall from 2.5.5.1 that Sc n S 0 = 1,
so K # Kx and S 0 ~ S 0. Since K ~ L 2 (q) for q a Fermat or Mersenne prime or
9, we compute from the possibilities for H:::; Aut(K) that
K = (CK(j) : j an involution of S 0 ),
so that K ::::; Nc(Kx) by the claim in the previous paragraph. By symmetry Kx
acts on K, so [K, Kx] = 1, so K is not A6 since G is quasithin. Thus K ~ L 2 (p)
for p;::: 7 a Fermat or Mersenne prime. Let K+ := KKx, M+ := Nc(K+), and
S+ := S(x).
Next S+ ::::; MK+, and as we saw that tx = ZK EK, t E Kx. Then as K ~
L2(P) has one class of involutions, by a Frattini Argument, M+ = K+NM+ ( (t, tx) ).
Then as SE Syl2(Gt), S+ E Syl2(M+)· Also Cs(K+) =Sc n Sa= 1, and hence
Cs+(K+) = 1 so Cc(K+)) = O(Cc(K+)). As K = K 0 , Gt = KSCa,(K) and
O(Gt) = 1 by 2.5.18. Then as Kx ::::; Cc(K) ::::; Gt" since tx E' K, Kx is normal
in Cc(K). Thus Cc(K) = KxCc(K+) = KxO(Cc(K+)) = KxO(Nc(K)). Then
Cc,(K) = CKx(K)O(Nc(K)) = SK:O(Nc(K)), so
Gt= KSO(Nc(K)) = KSO(Gt) =KS= H::::; M+.
In particular K = 02 (Gt)·
We claim that tG n M+ is the set I of involutions in K U Kx. We saw earlier
that ZK = tx and K has one class of involutions, so I ~ tG n M+. Furthermore
we saw that z = tzK = ttx, so that the diagonal involutions in K+ are in zG, and
hence these involutions are not in tG by 2.5.11.3. Thus if the claim fails, there is
i := t^9 E S+ -I, such that either i induces an outer automorphism on K or Kx, or
Kx =Ki. In the latter case, CK+(i) =:Ki~ K, so Ki= KB since K = 02 (Gt)i