By Konvisser M.W.

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**Additional info for 2-Groups which contain exactly three involutions**

**Example text**

B Hence 0 ,(A) cO ,(Op(CB(Q))) cO ,(B). P = p = P By symmetry, 0 ,(B) cO ,(A). °P,(B)== 1 p Hence 0 ,(A) = p p and 7T(F(A)) = 1T(F(B)) = {p L This completes the proof of Theorem B. 4. Let A be a maximal subgroup of a simple group G, S a subnormal subgroup of F*(A) such that CF*(A)(S);; S. Let 43 B ~ G be a subgroup of G containing 8. (a) 0 (B) n A = 1 for q q - E Then 1T(F(A))'. (b) [0 (B), OP(F*(A))] = 1, if P E 1T(F(A)). 1T(F*(A))! > 2 == 11T(F*(B)) I > 2, then each of the following ensures or = - that A = B.

2. 2 also. We may now apply Lemma 5. 6. if x E D#, CG(x) ;; CB(D). CB(D), CG(x) ;; X. 6, X is a uniqueness subgroup for D, and also A, and this is a contradiction. 6. Let A = n (A) and B = n (Z (F(X) )) as usual. 12 q -01- Then (i) (ii) CB(A)= 1 and lA I =p3 o 0 A contains a subgroup D of order p2 such that if o E = CB(D), then lE I = q and ~G(E);; X (iii) There exists d Proof. E D such that CG(d);; X. 5, if D is any subgroup of A iCB(D)I~q. o or order p2, If B is non-abelian of order q3, then A and so A/CA (B);; GL(2, q).

1 it follows that any Sylow 2-sub- group of U lies in F*(X) and so U;; F*(X). 3(a), U = [t, U] <]<] E(X). Thus U ~ E(X). 3(b) is not applicable here since if a component K of U is not a component of E (X), then [K, t] = 1. But U = [t, U] is just the product of components not centralized by t. Now assume that U = [t, U] is a p-group where p is an odd prime and that X is a minimal counter example. First if 0p(X) "* 1, let X = X/Op(X). By induction [t, U]<]

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