If every element in a finite monoid $(M, \star)$ is uniquely-regular then $M$ is a group.

Question

Suppose that $(M, \star)$ is a finite Monoid and for every $m \in M$ there exists a unique element $n \in M$ such that $m\star n \star m=m$ or in other words every element in $M$ is (uniquely!) regular. Prove that $M$ is a group.

I don't know where I should start from. I think $y$ must be the inverse of $x$ but I have no idea how to get there.

Answer 1

From $mnm=m$ it follows $m(nmn)m=m$. Since $n$ is unique we get $nmn=n$. Therefore $M$ is an inverse semigroup. An example such a semigroup, not a group, is a semilattice, i.e. an commutative idempotent semigroup (http://en.wikipedia.org/wiki/Semilattice#Algebraic_definition).

Remark: If you want to have a monoid, you can join an extra identity (http://en.wikipedia.org/wiki/Monoid#Examples) to the given semigroup.