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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.
 Fall '11
 KeithCornell

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