Adv Alegbra HW Solutions 76

Adv Alegbra HW - 76 Solution The following permutations commute with(1 2(3 4 5(1(1 2(3 4 5(1 2(3 4 5(1 2(3 5 4(3 5 4(6 7(1 2(3 4 5(6 7(1 2(6 7(1

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76 Solution. The following permutations commute with ( 12 )( 345 ) : ( 1 )( 67 ) ( )( )( )( ) ( )( ) ( )( )( )( ) ( )( 354 )( )( ) ( )( ) Since there are only 12 permutations commuting with α , this must be all of them. 2.125 (i) Show that there are two conjugacy classes of 5-cycles in A 5 , each of which has 12 elements. Solution. The hint shows that | C S 5 (α) |= 5. Since |h α i| = 5 and h α i≤ C S 5 (α) ,wehave h α i= C S 5 (α) .By( i) , C A 5 (α) = A 5 C S 5 (α) = A 5 ∩h α i=h α i , so that | C A 5 (α) 5. Therefore, the number of conjugates of α in A 5 is 60 / | C A 5 (α) 60 / 5 = 12. (ii) Prove that the conjugacy classes in A 5 have sizes 1, 12, 12, 15, and 20. Solution. There are exactly 4 cycle structures in A 5 : ( 1 ) ; ( 123 ) ; ( 12345 ) ; ( )( 34 ) . Using Example 2.30, these determine conjugacy classes in S 5 of sizes 1, 20, 24, and 15, respectively. In part (ii), we saw that the class of 5-cycles splits, in A 5 , into two conjugacy classes of size 12. The centralizer C S 5 ( ) consists of ( 1 ), ( ), ( 132 ), ( 45 ), ( )( ), ( )( ) ; Only the f rst 3 of these are even, and so
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

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