In a finite monoid, every left cancellable element is right invertible?

Question

In class, my professor told us that in a finite monoid, if an element is left cancellable, then it is also right invertible and that we should think about why this is. I've thought about it, but I'm actually unsure about how to go showing this is true. I've tried to see if my book has some proof about it, but it doesn't. Any guidance or proof would be helpful. Thank you.

Answer 1

An element $m$ is left cancellable if its corresponding function $f_m: x \mapsto mx$ is left cancellable, which is if and only if $f_m$ is injective, which since the monoid in question is finite is if and only if $f_m$ is surjective, i.e. there exists $n$ such that $mn=e$ where $e$ is the identity.

Consider $g_m:x \mapsto xm$. Let $g_m(x)=g_m(y)$, i.e. $xm=ym$. Multiply both sides by $n$ to the right and we get $xmn=ymn$, whence $xe=ye$, whence $x=y$.