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This question is a follow-up to this one. I tried to check whether the same statement as discussed for rings there is true for monoids too, but without success.

Let $M$ be a monoid and $u\in M$. Suppose there exists a unique $z\in M$ such that $uzu=u$. Does it imply that $u$ is an invertible element of $M$, with $u^{-1}=z?$

The only thing I see is that it implies that $z=zuz,$ since $$u(zuz)u=(uzu)zu=uzu=u.$$

So $z$ must be a (unique) von Neumann generalized inverse of $u$. But this is far from enough...

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Having the addition operation figured prominently in that proof, so I am interested in seeing the result of this :) – rschwieb Jun 11 '12 at 14:40
Hmmm... If you require that for every $u$ there exist a unique $z$ such that $uzu=u$, then the answer is "no", with the bicyclic semigroup (actually a monoid) as an example; but in that example, for each $u$ there may be several $z$ that work (though only one in which both equalities hold). Interestingly, your condition implies that $u$ is the unique element such that $zuz=z$, so that this condition (universally quantified) implies you have an inverse monoid, but you can have an inverse monoid in which this condition fails. – Arturo Magidin Jun 11 '12 at 16:59
Small correction: above it should be "for every $u$ there exists a unique $z$ such that $uzu=u$ and $zuz=z$". – Arturo Magidin Jun 11 '12 at 17:15
@ArturoMagidin I think the answer is actually "yes" with the (later corrected) requirements from your first comment. Suppose the for every $u$ there is a unique $z$ such that $uzu=u$. Let $e=e^2$. The equation $e=exe$ has a unique solution $e=1$. Let $u=uzu$. $uz$ and $zu$ are idempotent, and therefore $uz=zu=1.$ – user23211 Jun 12 '12 at 0:17
@ymar: Careful: the uniqueness is not about $e$, it's about $x$. So the unique solution is that $exe=e$ implies $x=1$, not that it implies $e=1$. Also the condition given is not that every $u$ has a unique $z$, rather that a given $u$ happens to have a unique $z$, so there may be no unique $x$ for $e$. – Arturo Magidin Jun 12 '12 at 3:16
up vote 4 down vote accepted

EDIT: this has been reworked multiple times due to discussion below. I believe it is now correct.

Let $S$ be the set of sequences of natural numbers, almost all of which are zero. It is endowed with a function $\Sigma$ which returns the sum in a sequence. Let $M$ be the monoid of endomorphisms $T$ of $S$ such that $\Sigma(Ts) \leq \Sigma(s)$ for all $s \in S$.

Let $z$ be the "right shift operator" which inserts a 0 in the first slot of the sequence and shifts everything else to the right, and let $u$ be the left shift operator which shifts everything to the left (eliminating the number in the first slot).

Clearly $uzu = u$, and $u$ can't be invertible. I claim that $z$ is unique with this property; clearly any other element with this property must be a right shift operator inserting something in the first spot (one gets this immediately upon writing $uzu(a_0, \ldots) = (a_1, \ldots)$), and that something has to be 0 or the sum will increase.

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Could you please say a word about the uniqueness of $z$? – user23211 Jun 11 '12 at 14:26
Oh, good point - it's clearly not unique since you could insert anything you like into the first slot of the sequence, not just 0. – user29743 Jun 11 '12 at 14:28
working on a repair. – user29743 Jun 11 '12 at 14:29
I hope it's fixed now – user29743 Jun 11 '12 at 14:38
Couldn't $z$ insert different numbers in the first spot for different inputs? For example, let $z$ be exactly as you say, except for $z(1,1,\ldots)=(1,1,\ldots).$ – user23211 Jun 11 '12 at 15:00

Let $M = \{1, a, b, ab, ba, 0\}$ be the monoid defined by the relations $aba = a$ and $bab = b$ and $0x = 0 = x0$ for all $x \in M$. Then the equation $axa = a$ has $b$ as unique solution. However, $a$ is not invertible.

By the way, the nickname of this monoid is the universal counterexample and once again, it did its job!

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