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

AxiomOfChoice

If [itex]M[/itex] is a monoid and [itex]u,v\in M[/itex], and [itex]uv = 1[/itex], do we know [itex]vu = 1[/itex]? 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:
[tex] uv=1[/tex]
If there exists a w such that:
[tex]vw=1[/tex]
Then by associativity:
[tex]w=1w=(uv)w=u(vw)=u1=u[/tex]
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.

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AxiomOfChoice	For a monoid, if uv = 1, do we know vu = 1? Tweet If [itex]M[/itex] is a monoid and [itex]u,v\in M[/itex], and [itex]uv = 1[/itex], do we know [itex]vu = 1[/itex]? 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 Blog Entries: 6 Recognitions: Gold Member Science Advisor	Assume: [tex] uv=1[/tex] If there exists a w such that: [tex]vw=1[/tex] Then by associativity: [tex]w=1w=(uv)w=u(vw)=u1=u[/tex] 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 Recognitions: Homework Help Science Advisor	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.