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Lagrange's Theorem

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.


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 

- Doctor Paul, The Math Forum   
Associated Topics:
College Modern Algebra

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