Simple proof involving groups

JohnDuck

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.

matt grime

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).

JohnDuck

I'm still not making any progress. :(

NateTG

Here's a start on one of them:
Let's say we have some [itex]i[/itex] (not necessarily [itex]e[/itex]) with
[tex]i \times x = x[/tex]
then
[tex]i \times i = i[/tex]
[tex]i^{-1} \times (i \times i) =i^{-1} \times i[/tex]
.
.
.