# Finite group

1. Oct 2, 2006

### asp

Given a finite set G is closed under an associative product and that both cancellation laws hold in G,Then G must be a group.

I need to prove that G must be a group, I understand that for this
I only need to show that :

1) There exist the identity
2) There exist the inverse.

But while trying to do this problem , i am not able to understand how to use the fact that G is finite ?

2. Oct 2, 2006

### matt grime

Hint: suppose X is a finite set, then any injective map from X to itself is a bijection (this is false for infinite sets).

3. Oct 11, 2006

### GreenApple

If you still can not figure it out, here is a more detailed hint:
the "product" you mentioned is but a fuction G*G -> G. For any a belonging to G, you can get a fuction G -> G f(x), which is a*x, and which is a one to one and onto(from the fact that it is one to one and and G is finite and the product follow the cancellation law), from which we can get that for any x belonging to G, there is a p such that f(p)=x.
Here we have draw that for any a belonging to G and any x belonging to G, there always a p, a*p=x (1)
similarly, we can get that for any a belonging to G and any x belong to G, there always a p, p*a=x (2)
from (1) and (2) and a theorem whick you must have learned, we conclude that G is a group.

Last edited: Oct 11, 2006
4. Oct 11, 2006

### gonzo

It may be easier to start like this. If you have:

a*b = a*c

Using the cancellation laws, what does that tell you? What can you conclude happens if you then multiply every element in your group by "a"?

And then, finally, what conclusions can you draw from this?