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ahw3 - Math 5125 Monday September 12 Third Homework...

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Math 5125 Monday, September 12 Third Homework Solutions 1. The action of G on G / H (the left cosets of H in G ) yields a homomorphism θ : G S n , and by hypothesis θ ( G ) A n . Note that θ ( g ) has order 1 or 2, because g has order 2. Since θ ( g ) A n and n 4, we see that θ ( g ) must fix an element of A n (ungraded homework of August 30). Now the permutation θ ( g ) is determined by labeling the left cosets of H 1 , 2 ,..., n , and sending the coset yH to gyH . This means that θ ( g ) must fix one of the left cosets yH ; in other words, there exists x G such that gxH = xH . 2. The elements of order 2 in S 6 have cycle shape (1 2), (1 2)(3 4), (1 2)(3 4)(5 6). The sizes of the conjugacy classes are 6 * 5 / 2 = 15, 6 * 5 * 4 * 3 / ( 2 * 2 * 2 ) = 45 and 6! / ( 2 * 2 * 2 * 3! ) = 15. Since | S 6 | = 720, the sizes of the centralizers are | C G ( 1 2 ) | = 720 / 15 = 48, | C G (( 1 2 )( 3 4 )) | = 720 / 45 = 16, | C G (( 1 2 )( 3 4 )( 5 6 )) | = 48. 3. The sizes of the conjugacy classes divide 21, so must be 1,3,7 or 21. Since Z ( G ) = 1, there is 1 conjugacy class of size 1; also there is no conjugacy class of size 21. Let x
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