Date: 01/24/2002 at 01:15:30 From: santhosh kumar.L Subject: Modern algebra Let G be a finite group of order p, where p is a prime number and G is a cyclic group. I need the proof of the theorem.
Date: 01/24/2002 at 10:09:27 From: Doctor Paul Subject: Re: Modern algebra Let |G| = p. Pick an element from G which is not the identity element and call it x. Compute: x, x^2, x^3, ..., x^p = e This list of p powers of x must include every element of G. Restating, it must be the case that G = <x>. To see this, suppose that this is not the case. Then <x> is some proper subgroup of G. Lagrange's Theorem says that the order of this proper subgroup must divide the order of G (which is the prime number p). Clearly this cannot happen since the only divisors of the prime p are 1 and p. Thus G = <x> and G is hence cyclic. I hope this helps. Please write back if you'd like to talk about this more. - Doctor Paul, The Math Forum http://mathforum.org/dr.math/
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