In a finite monoid (M, $\circ$) if the identity element $e$ is the only idempotent element, prove that each element of the monoid is invertible.
As the set $M$ is finite, $\exists$ $y \in M$, s.t $y \circ a = e$ $\forall a \in M$.
Similarly, we have $x \in M$, s.t $a \circ x = e$ $\forall a \in M$.
Now, we consider:
$a \circ x=e \implies y \circ (a \circ x) =y \circ e \implies (y \circ a)\circ x=y \implies x=y $ [by associative property of monoids]
Hence, for every $a \in M$, $\exists x \in M$ s.t $x \circ a=a \circ x = e$; i.e, every element is invertible.
Now, my question is, is this at all the correct approach towards the solution?
Evidently, if $a \circ a=a$, then $y \circ a= (y \circ a) \circ a= e \circ a= a$ which is ultimately $a=e$ But does this imply that every $ y \circ a=e $ and $ a \circ x=e$ has a solution?
[ Alternative form of the question: If a Monoid is a Quasigroup, then it is a Group (also kindly justify the alternative form of the question)]