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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)]

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    $\begingroup$ This question might help. $\endgroup$ – drhab Jan 17 '18 at 10:45
  • $\begingroup$ I want someone to verify this approach... $\endgroup$ – Subhasis Biswas Jan 17 '18 at 13:36
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    $\begingroup$ How do you come to the statement that $y\in M$ must exist with $ya=e$? $\endgroup$ – drhab Jan 17 '18 at 13:49
  • $\begingroup$ This is what I need to show. If I can show, that $e$ has a pre image (y,a) then it is done $\endgroup$ – Subhasis Biswas Jan 17 '18 at 13:51
  • $\begingroup$ I would like to see proof that if a monoid is a quasigroup, then it is a group, please. $\endgroup$ – Shaun Jan 17 '18 at 18:20

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