Adv Alegbra HW Solutions 50

Adv Alegbra HW Solutions 50 - d , by Lemma 2.53. But if x...

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50 Its order is 6, its inverse is ( 145 )( 23 ) , and it is odd. (ii) What are the respective orders of the permutations in Exercises 2.22 and 2.28? Solution. 2 and 4. 2.39 (i) How many elements of order 2 are there in S 5 and in S 6 ? Solution. In S 5 , there are 1 2 ( 5 × 4 ) = 10 transpositions and 1 2 [ 1 2 ( 5 × 4 ) × 1 2 ( 3 × 2 ) ]= 15 products of two disjoint transpositions (the extra factor 1 2 so that ( ab )( cd ) = ( cd )( ab ) not be counted twice). In S 6 , there are 1 2 ( 6 × 5 ) = 15 transpositions, 1 2 [ 1 2 ( 6 × 5 ) × 1 2 ( 4 × 3 ) ]= 45 products of two disjoint transpositions, and 1 6 [ 1 2 ( 6 × 5 ) × 1 2 ( 4 × 3 ) × 1 2 ( 2 × 1 ) ]= 15 products of three disjoint transpositions. (ii) How many elements of order 2 are there in S n ? Solution. 1 2 n ( n 1 ) + 1 2 ! [ 1 2 n ( n 1 ) 1 2 ( n 2 )( n 3 ) ] + 1 3 ! [ 1 2 n ( n 1 ) 1 2 ( n 2 )( n 3 ) 1 2 ( n 4 )( n 5 ) ]+··· . 2.40 Let y be a group element of order m ;if m = dt for some d 1, prove that y t has order d . Solution. Let x = y t .Now x d = ( y t ) d = t td = y m = 1, and so the order k of x is a divisor of
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Unformatted text preview: d , by Lemma 2.53. But if x has order 1, then x = 1 and so y t = 1. This contradicts m = pt being the smallest positive integer with y m = 1. Therefore, x has order p . 2.41 Let G be a group and let a ∈ G have order dk , where d , k > 1. Prove that if there is x ∈ G with x d = a , then the order of x is d 2 k . Conclude that the order of x is larger than the order of a . Solution. It is clear that x d 2 k = ( x d ) dk = a dk = 1; thus, the order of x is a divisor of d 2 k . Now x dk = a k ±= 1 because k < dk . Therefore, x has order d 2 k . because p being a prime implies that there is no divisor d of d 2 k with dk < d < d 2 k ....
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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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