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Suppose there exists c E C such that Ki = c Ki. Show G = G . 1 Hint Let lG # n x E Irr(G). Show that w,(Ki) = 0 for some i. 14) (Brauer) In the notation of the previous problem, show that if G = G , then there exists c E Q such that 1K i = c n K i . Hint To show that two elements a, b E ZjC[G]) are equal, it suffices to prove that o,(a) = o,(b) for all x E Irr(G). 15) (Thompson) Let E be a Galois extension of Q with Galois group 9. A theorem of Siege1 (Ann. 46 (1945)p. 303,Theorem 111) asserts that if a # 1, then Use this to show that if x E Irr(G) then ~ ( xis) either zero or a root of unity for more than a third of the elements x E G.

U,} for V and {wl,. ,w,} for W . Let V 0 W be the @-spacespanned by the mn symbols ui @I w j . Wedefine u 0 w = aib,{ui 0 wj)E V 0 W. 1 1 c Note that not every element of V @I W has the form u 0 w for u E I/ and w E W (except in the special case that n or rn = 1). We define an action of G on V 0 W by setting (ui 0 W j ) g = V i g 0 W j g and extending this by linearity to all of V 0 W. The reader should check that if u E V ,w E W , and g E G, then ( u 0 w)g = ug @I wg. It follows that (xgl)gz = x(g,g,) for x E V 0 W and g i E G.

Proof Let V be a C[G]-module which affords x and let u l , u z , . ,u, be a basis for V. Let W = V @ V and define the linear map *: W -+ W by (ui @I uj)* = uj @I ui. Let and W, = {WEWlw* = w} WA= {WEWlw* = -w}. These subspaces are called the symmetric and antisymmetric parts of W , respectively. Since w=- w+w* 2 +-w-w* 2 ' it follows that W = Ws + WA. We claim that if w E W and g E G, then (we)* = w*g. It suffices to check this as w runs over the basis ui @I u j , and thus we need to show that .

### 381st Bomber Group

by Jeff

4.3