# Simple proof involving groups

1. ### JohnDuck

74
1. The problem statement, all variables and given/known data
Let G be a set with an operation * such that:
1. G is closed under *.
2. * is associative.
3. There exists an element e in G such that e*x = x.
4. Given x in G, there exists a y in G such that y*x = e.

Prove that G is a group.

2. Relevant equations
I need to prove that x*e = x and x*y = e, with x, y, e as given above.

3. The attempt at a solution
I dunno, I'm stumped. I've tried finding some sort of clever identity without any success.

2. ### matt grime

9,395
I'm just playing around here to reduce the number of things we have to prove. We know x has a left inverse y. What is xyx? It is x(yx)=x, and it is (xy)x, thus if we could show that the e in 3 was unique, we'd have xy=e, and x would have a right inverse.

So we just have to show that fx=x implies f=e, to obtain the existence of right inverses.

Now, supposing that we have inverses on both sides, what can we say? Well, x*e=x(x^-1x)=(xx^-1)x=e*x, so we get a unique two sided identity.

Putting that together, all I need to show is that fx=x implies f=e. Is that any easier? (it is implied by the existence of right inverses, but that is circular logic, so careful how you try to prove it).

Last edited: Jun 4, 2007
3. ### JohnDuck

74
I'm still not making any progress. :(

4. ### NateTG

2,537
Here's a start on one of them:
Let's say we have some $i$ (not necessarily $e$) with
$$i \times x = x$$
then
$$i \times i = i$$
$$i^{-1} \times (i \times i) =i^{-1} \times i$$
.
.
.

Last edited: Jun 5, 2007