Problem Set 7 Solutions

# Problem Set 7 Solutions - Problem Set 7 Solutions Igor...

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Igor Rapinchuk 1. Suppose | G | = p n . Let S 1 ,...,S k be the orbits of G on S. Then | S | = | S 1 | + ··· + | S k | Suppose that G does not have a ﬁxed point. Then the size of every orbit is > 1 . On the other hand, | S i | = [ G : G x i ] where x i S i . It follows that | S i | = p n i for some n i > 0 , in particular, | S i | is divisible by p. But then | S | is also divisible by p. A contradiction. 2. Let G be the dihedral group of order 2 n, G = { 1 ,x,. ..,x n - 1 ,y,xy,. ..,x n - 1 y } where x n = 1 , y 2 = 1 , and yxy - 1 = x - 1 . We have x i x j ( x i ) - 1 = x j and ( x i y ) x j ( x i y ) - 1 = x - j So, the conjugacy class of x j is { x j ,x - j } . Next, x i ( x j y )( x i ) - 1 = x 2 i ( x j y ) and ( x i y )( x j y )( x i y ) - 1 = x i yx j yy - 1 x - i = = x 2 i - j y = x 2( i - j ) ( x j y ) So, the conjugacy class of x j y is { x 2 i ( x j y ) | i = 0 ,...,n - 1 } . Case 1. n is odd, n = 2 m +1 . Then x j 6 = x - j for any j = 1 ,...,n - 1 . So, the powers of x split into one 1-element conjugacy class { 1 } and m 2-element conjugacy classes { x j ,x - j } , j = 1 ,...,m. Since n is odd, any power of x can be written in the form x 2 i , so all elements of the form x j y form a single conjugacy class. So, the class equation for G looks as follows: 2 n = 1 + 2 + ··· + 2 | {z } m + n Case 2. n is even, n = 2 m. Then x j = x - j iﬀ j = 0 or m, so there are two 1-element conjugacy classes { 1 } and { x m } . The conjugacy class of any other power of x contains 2 elements; these classes are { x j ,x - j } , where j = 1 ,...m - 1 . Since n is even, there are m elements of the form x 2 i . So, the conjugacy class of an element of the form x j y contains m elements, and therefore there are 2 such classes. So, the class equation for G looks as follows: 2 n = 1 + 1 + 2 + ... + 2 | {z } m - 1 + m + m 3. Suppose G/Z is generated by a coset gZ. This means that any coset xZ is of the form g i Z for some integer i. It follows that any x G can be written in the form g i z for z Z. Now, let us take two arbitrary a,b G and write then in the form a = g i z 1 and b = g j z 2 . Then ab = ( g i z 1 )( g j z 2 ) = g i + j z 1 z 2 and ba = ( g j z 2 )( g i z 1 ) = g i + j z 2 z 1 because z 1 and z 2 commute with g. Since z 1 z 2 = z 2 z 1 , we see that ab = ba, so G is abelian. 1

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## This note was uploaded on 01/18/2012 for the course INFORMATIK 2011 taught by Professor Phanthuongcang during the Winter '11 term at Cornell University (Engineering School).

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Problem Set 7 Solutions - Problem Set 7 Solutions Igor...

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