Let $(M,\circ)$ be a finite monoid. Suppose the identity element $e\in M$ is the only idempotent element. Then prove that each element in $M$ has inverse.
How can I prove this?
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MAC
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Let $a \in M$. Consider the set ${e, a, a^2, \cdots }$. Can this set be infinite? – SMG Dec 02 '14 at 18:32
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This set is finite. From this we can say $a^{mn}$ is idempotent for some integers $m,n$, i.e., $a^{mn}\circ a^{mn}=a^{mn}$. Can I show the result from this? – MAC Dec 02 '14 at 18:41
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@Mac If you can show what you say in your comment then you are ready. You are allowed to conclude that $a^{mn}=e$ since $e$ is unique as idempotent. Then $a^{mn-1}$ will serve as inverse of $a$. You could edit the proof of what you say in your comment so that it can be checked by others. – drhab Dec 02 '14 at 19:17
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Duplicate of A finite Monoid $M$ is a group if and only if it has only one idempotent element – J.-E. Pin Dec 15 '14 at 12:22
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Let $a$ be an element of a finite monoid.
Then $n>0$ and $k>0$ exist with $a^{n+k}=a^{n}$, leading to $a^{m+kr}=a^{m}$ for $m\geq n$ and $r\geq0$.
Choose some $r$ such that $kr\geq n$ and note that $a^{kr}$ is idempotent.
If identity $e$ is unique as idempotent then $a^{kr}=e$, showing that $a^{kr-1}$ serves as inverse of $a$.
drhab
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