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**Extra resources for 2-Groups which contain exactly three involutions**

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Then 7 X'2(h) = {o),(b)2 if b ~ 2(B). if bE 2(8). X2 (1f As 2 t IBI, there exists a field automorphism ~ of the field of 12(B)I-th roots of unity such that ;,(b)iJ = ),(b)2. (b)X2(1) {o;,(b)2 if b ¢ 2(B), if bE 2(B). (1)2 = x(l)blj;(l) and a = where bx(1). d. xixg EO Irr G. 9 Theorem (Burnside, R. Brauer). Let X EO Irr G be a faithful character of G. If X takes only k d~fJerel1t values 011 G, then all irreducible characters of G appear as irreducible components ill some i, where ~ i ~ k - 1. ° X(g)2 = alj;(g) = bX(l)lj;(g) = X(1)(O"(g) - a(g)) = X(1)x(g2).

H;j, k = 1, ... , l1i' 9 E G) is an algebraic number field with (Ko : 0) < 00, in which all the matrix representations Di can be realized. 6 0< Ch-I,h,g b) K. Then Ko = O(aJ~(g)li X(g-l) i= O. X(l) Proof. a) 9 conjugate to for some y E IGI L. Ch-I,h,gIK 12 hEG h = = K 2 I hl IGI hence can be written over the field L XElrrG Ix(hW x (g-1). X(l) "X(g-l) " Ix. (h)12 L. L. X(1) hEG XElrrG = IGI " L. 1). 3. We come back to this question in § 38. of KG into two-sided, indecomposable ideals Bi (called p-blocks) plays a central role.

339). 6 this would imply ;( 1) :::;: IG : A I = 2 for all nf. S. Obviously this forces IGIG'I ): 4. 2 would imply by Burnside's basis theorem for p-groups that G is cyclic; Huppert I, p. ) As I1J : : ; 28, certainly only nj = 2, 4 are possible. Suppose that (IGIG'I = X E Irr G. Hence the type (1, 1, 1, 1,2,2,2,4) cannot appear. Do the 6 types listed above really belong to groups? In most cases we can see this easily by reference to examples in § 7 and § 8. (1, ... , 1) is the degree pattern of all 7 abelian 32 groups of order 32; (1, ..