For a monoid, if uv = 1, do we know vu = 1?

[AxiomOfChoice]

If $M$ is a monoid and $u,v\in M$, and $uv = 1$, do we know $vu = 1$? Can someone prove this or provide a counterexample? I tried to come up with one (a counterexample, that is) using 2 x 2 matrices but was unsuccessful.

 

[jambaugh]

Assume:
$$uv=1$$
If there exists a w such that:
$$vw=1$$
Then by associativity:
$$w=1w=(uv)w=u(vw)=u1=u$$
Thus if such a w exists it must be u.

This isn't quite enough but I can't find a short proof either. I'll think on it.

 

[matt grime]

The canonical example is left and right shift of a sequence or countable dimensional vector space.

L(a,b,c,d,...) = (b,c,d,...)

R(a,b,c,..)=(0,a,b,c...)

LR=id, and RL=/=id.

You won't find one in 2x2 matrices - the invertible ones form a group, so there's no point looking.