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

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In the context of monoids, if uv = 1 for elements u and v in a monoid M, it does not necessarily imply that vu = 1. The discussion highlights the failure to find a counterexample using 2x2 matrices, as the invertible matrices form a group where each element has a unique inverse. The canonical examples provided involve left and right shifts of sequences or countable dimensional vector spaces, illustrating that LR = id while RL ≠ id.

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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.
 
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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.
 
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.
 

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