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Show That G is a Group

Date: 02/28/2002 at 21:41:33
From: Stephanie Worthington
Subject: Modern/Abstract Algebra

Two questions:

1. Let G be a finite group.  Show that there exists a positive 
   integer "m" such that a^m=e for all a in G.  (I'm pretty much stuck 
   on this, I'm not sure what I can assume 'm' to be.)

2. Suppose that G is a set closed under an associative operation 
   such that: 
   a) for every a,y in G, there exists an x in G such that ax = y;   
   b) for every a,w in G, there exsits a u in G such that ua = w. 
   Show that G is a group.

Thanks for your help!

Date: 03/01/2002 at 19:24:50
From: Doctor Paul
Subject: Re: Modern/Abstract Algebra

For your first question, it follows as a pretty easy corollary of 
Lagrange's Theorem that picking 

   m = |G| 

will work for all a in G.

I'll leave the details for you to mull over. If you need a hint, 
please write back.

For your second question, according to the definition of a group, we 
need to show four things:

   Group - MathWorld, Eric Weisstein

     1. Closure
     2. Associativity
     3. Identity
     4. Inverse

1 and 2 are given. To see 3, use condition (b). Pick w to be an 
element of the group (do we know that G is nonempty? If not, then this 
step might break down. We need to be sure that such an element exists) 
and pick a = w. Then there exists u in G such that u*a = w = a.

So u*a = a, which forces u to be the identity element.

To see that every element is invertible, use condition (a):

Let a be any element of G. Pick y = the identity element 
(which we now know exists). Then there exists x in G such that 
a*x = 1 ==> x = a^(-1).

This completes the proof.

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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