Download PDF by I. Martin Isaacs: Character Theory of Finite Groups

By I. Martin Isaacs

Very good textual content ways characters through jewelry (or algebras). as well as ideas for making use of characters to "pure" crew concept, a lot of the booklet makes a speciality of homes of the characters themselves and the way those homes replicate and are mirrored within the constitution of the crowd. difficulties keep on with each one bankruptcy. Prerequisite a first-year graduate algebra path. "A excitement to read."—American Mathematical Society. 1976 version.

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G k be representatives of the conjugacy classes of G. The first orthogonality relation asserts that k Chapter 2 22 This system of k2 equations may be replaced by the single matrix equation lGlI = XDXT, where I is the identity matrix and the superscript denotes transpose. Since a right inverse for a square matrix is necessarily also a left inverse, this yields lGlI = D X T X . We now write this as a system of equations and obtain Since 1 G I/ I X i 1 = I C(gi)1, this yields which is the desired result.

2) (a) Let x be a character of G. Show that x is afforded by a representation X such that all entries of X(g) for all g E G lie in some field F E C with ( F : Q ( < a. (b) Let E = elni/",where n = I G 1 and let x be a character of G. ) Let CJ be an automorphism of the field Q [ E ] and define xu: G 4 C by f ( g ) = x(g)d. Show that xu is a character and that xu E Irr(G) iff x E Irr(G). 3) Let x be a character of G. Define det affording x and set (det x ) ( g ) = det x: G 4 C as follows. Choose 3E w.

Then ICl(g)l = IG: C(g)I divides I G : P 1, which is a prime power. 9 that the simple group G is abelian and the proof is complete. I We now obtain some strong results about the degrees of the irreducible characters of a group G . The fact is that x( 1)I I G : Z(x)I for x E Irr(G). We shall first prove the weaker statement that the irreducible character degrees divide the group order. This proof is much less complicated and serves to motivate the stronger proof. 11) THEOREM Let x E Irr(G). Then x( 1)I IG I Proqf From the first orthogonality relation we have IGI = Cx(s)x(g-').

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