Question

Problem 3 (10 points). Continuation of Problem 1. [1] (Work 5 points, answer 5 points.) Let X1, X2, X3, X4 be the irreducible characters of D5, where X1 is the trivial character, and X2, X3 are the charac- ters of representations that you found in Problem 2. Compete the character table. Work: Character Table: Char C(1) C(x) C(y) C( ) 1 1 X1 1 X2

Answer 1

[2]. Lec dj = din . Let d; = dini & diLd. L ds Lda There are total of (5+3) = 4 irreducible representatio One dimensional Representations. • Trivial representation - sending all elements to the 1x1 matrix (1). of the of The representation sending all elements in <a> (derived subgroup) to 9. & all elements outside <a> oot1.). Two dimensional Representations : Group | Matrix as I complex unitary element unitary I reznik (:)

Solution I G= Ds. <x, y 1 xS= 1, y = 1, y = xty [] listing down all the conjugacy classes of Dr. The conjugacy class of e is just ney. .. To find conjugacy class of x" by x. When we conjugate x by yxm we get a = yy x x x Note that since each resuction ym has orden 2. (yrm) = yxm. Thus the conjugacy class of each an contains n e xh. Thus we obtain ha,x), 4 x x3y To find conjugacy class of y, we compute that when we conjugate y by zm we get. & when we conjugate y by yem we get 1 y yy x = y + m since. n can ve 1,2,3 & 4 , 2m is either... 2,4, I or 3 since we have to take am mod 5 This all reflections form a single conjugacy class - hy, yx, x², yx, yx" } . These are the four conjugacy classes... -listed above cu = 415. c) WE- ** ) = {x,xy CExty = {X², X ccyg = 2 y, yx, y x,y x², yz 1.