# Characters of finite groups (Mathematics lecture notes) by Walter Feit By Walter Feit

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Is 0 or 1 . 1 l- Proof. The multiplicity of X in Xi X j and the multiplicityof Xi in XXj are both equal to 1/INI 1:;(\\ Xi(G)Xj(G)x(G). If X is linear then XXj is irreducible. Thus Xi is a con- stituent of XX j if and only if Xi:::; XX j in which case the multiplicity of Xi in XX j is 1. 7) Let ~ be a representation of ® which affords the character e . Let S; be the kernel of eo Then (i) le(G)1 :5 e(1) for G E: ®. \) is in the center of ®/S). n in the center of ®/~). Thus in particular {Gle(G) :::: e(l)} <1 ® and {Glle(G)1 :::; e(1)} <1 ®.

Proof. It may be assumed that ~ is unitary and ~ :;: if ~ is of the first kind. X(G2) is the trace of ~(G) ~(G) ~(G)~(G-l)'. Thus if ~(G) :;: a .. (G) then ii 1) v(X) ::: 1(\1 6 6 a .. (G) a .. 9). , 6.. 6(;" 1,J I ~I \21 61' a .. (G) a .. (G-l) 1,J IJ 0(iii a 11.. (G) a 11.. (G-l) = 1 22 CHARACTERS OF FINITE GROUPS Suppose that tl is of the second kind. 3) there f exists U unitary such that U :::: -U and U- 1 ~{G)U = ~(G) f 1 for G E: (~. Since UU = -U this implies that if U = (u ) then ij a ..

Element G. it follows easily that s 1 1 1 =1 if and only if condition (i) is satisfied. For any n-tuple (G , ... ,G ) define HI' G ... G. • , n - 1. Thus C H A RAe T ERS 0 F FIN I T E G R0 UPS 50 1 x 6 H , ... ,H 1 n-l X. (H)··· X. 0 n - 1 times this yields that 1 By the Frobenius reciprocity theorem s = 1 if and only if for j ;;t! n1;:l c h ::; 1. Thus 1 there exists i such that c .. = O. 1J In other words s = 1 if and only if condition (ii) is satisfied. Thus the statement that s = 1 is equivalent to both statements (i) and (ii).