If $M$ is a finite monoid and $au=bu$ in $M$ implies that $a=b$, show that $u$ is a unit.
Hint: If $m=\{a_1,\cdots, a_n\}$ show that $a_1u,\dots,a_nu$ are distinct.
A unit is defined as an element of a monoid with an inverse.
The hint confuses me. I don't see why it is necessary that M is finite. Doesn't $au=bu$ implies $a=b$ mean that $u^{-1}$ exists? How else could you cancel $u$ if it did not.
Proof. Let $f:M \to M$ be the map defined by $f(x)= xu$. By hypothesis, if $f(a) = f(b)$, then $a = b$, which means that $f$ is injective. Since $M$ is finite, the map $f$ is actually bijective. In particular, there exists $v \in M$ such that $f(v)= 1$, that is, $vu = 1$. Thus $v$ is a left inverse of $u$.
Let now $g:M \to M$ be the map defined by $g(x) = ux$. If $g(a) = g(b)$, then $ua = ub$, whence $a = vua = vub = b$. Thus $g$ is injective and hence bijective. In particular, there exists $w$ such that $g(w)= 1$, that is $uw = 1$. Finally, $v = v(uw)= (vu)w = w$. Thus $v$ is an inverse of $u$.