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 http://mathworld.wolfram.com/Group.html 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 more. - Doctor Paul, The Math Forum http://mathforum.org/dr.math/
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