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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.

```

```
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,

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.

more.

- Doctor Paul, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Modern Algebra

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