aa_hw_7_Math 113 - Homework 7 October 25, 2010 Introduction...

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Introduction to Abstract Algebra Homework 7 October 25, 2010 Problem 1 Given a group G with | G : Z ( G ) | = n , we wish to show that all conjugacy classes have at most n elements. We know that for some x G , Z ( G ) C G ( x ) G, and in a previous homework we have shown that if K H G , then | G : K | = | K : H || G : H | . Thus, we find that | G : Z ( G ) | = | Z ( G ) : C G ( x ) || G : C G ( x ) | . By proposition 6 (p. 123), the number of elements in the conjugacy class of x is the index of C G ( x ) in G , which we have shown above to divide the index of the center of G in G . The center has index n , so the number of elements in every conjugacy class divides n , and is thus at most n . Problem 2 Let G 1 ,G 2 ,...,G n be groups. We show that Z ( G 1 × ··· × G n ) = Z ( G 1 ) × ··· × Z ( G n ) . Let ( g 1 ,...,g n ) Z ( G 1 ×···× G n ) and let ( x 1 ,...,x n ) be an arbitrary element in G 1 ×···× G n . Then by definition ( x 1 ,...,x n )( g 1 ,...,g n ) = ( g 1 ,...,g n )( x 1 ,...,x n ) ( x 1 g 1 ,...,x n g n ) = ( g 1 x 1 ,...,g n x n ) . Note that the above requirement on the element in position i is met by all elements g i Z ( G i ), for all x i G i . Thus, the right hand side above is just the cross product Z ( G 1 ) × ··· × Z ( G n ). Next we show, that G 1 × ··· × G n is abelian if and only if G i is abelian for i = 1 ,...,n . We first show the if direction. Assume that G 1 × ··· × G n is abelian. Let ( g 1 ,...,g n ) and ( x 1 ,...,x n ) be elements of G 1 × ··· × G n . Then, by assumption, ( x 1 ,...,x n )( g 1 ,...,g n ) = ( g 1 ,...,g n )( x 1 ,...,x n ) ( x 1 g 1 ,...,x n g n ) = ( g 1 x 1 ,...,g n x n ) . From this, we see that for any i in the range, with x,g G i , that xg = gx , so all G i are abelian, for i = 1 ,...,n . Conversely, assume that G i is abelian for i = 1 ,...,n . Then for g,x G i : xg = gx . In the following, let x i ,g i G i , we then find ( x 1 g 1 ,...,x n g n ) = ( g 1 x 1 ,...,g n x n ) ( x 1 ,...,x n )( g 1 ,...,g n ) = ( g 1 ,...,g n )( x 1 ,...,x n ) . From this we see, that if we consider arbitrary elements x,g G 1 × ··· × G n , then xg = gx , so G 1 × ··· × G n is abelian. 1/ ??
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Introduction to Abstract Algebra Homework 7 October 25, 2010 Problem 3 Let G 1 ,G 2 ,...,G n be groups, and let G = G 1 × ··· × G n . Let I and J be proper, nonempty subsets of { 1 , 2 ,...,n } such that I J = . Define G I to be the set of elements of G that have the identity of G j in positions j for all j J . (a) We prove that G I is isomorphic to the direct products of the groups G i for i I . For convenience we denote this group × i I G i . Define φ : G I × i I G i , s.t. φ maps an element in G I to the element in × i I G i where all elements on positions in J are removed. Consider two arbitrary
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aa_hw_7_Math 113 - Homework 7 October 25, 2010 Introduction...

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