113 HW 7_Math 113

# 113 HW 7_Math 113 - M H Curtis Tekell Dr Poirier Exercise 1...

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Curtis Tekell Dr. Poirier October 24, 2010 Exercise 1. If G is a group, and | G : Z ( G ) | = n , then every conjugacy class of G has at most n elements. Proof. Let K be the conjugacy class of some x G . Then | K | = | G : C ( x ) | , where C ( x ) is the centralixer of x . As as Z ( G ) C ( x ) , we have | G : Z ( G ) | = | G : C ( x ) || C ( x ) : Z ( G ) | , and thus | K | = | G : C ( x ) | = | G : Z ( G ) | | C ( x ) : Z ( G ) | = n | C ( x ) : Z ( G ) | n . ± Exercise 2. For any groups { G i } n 1 , Z ( n Y 1 G i ) = n Y 1 Z ( G i ) . Proof. As multiplication is deﬁned componentwise, it is clear that an element of Q n 1 G i is central if and only if each component is central, i.e. Z ( Q n 1 G i ) = Q n 1 Z ( G i ) . ± Exercise 3. Let { G i } n 1 be groups and let G = Q n 1 G i . Let N = { i } n 1 and let , I ( N and J = N \ I . Deﬁne G I = ± { g i } n 1 : g i = 1 ⇐⇒ i J ² . (a) G I ± Q i I G i (b) G I E G and G / G I ± G J . (c) G ± G i × G J . Proof. (a) The map f : G I Q i I G i deﬁned by { g i } n 1 7→ { g i } i I is clearly an isomorphism. 1

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113 HW 7_Math 113 - M H Curtis Tekell Dr Poirier Exercise 1...

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