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The Wikipedia page about inverse semigroups defines them as follows:

In mathematics, an inverse semigroup (occasionally called an inversion semigroup) $S$ is a semigroup in which every element $x$ in $S$ has a unique inverse $y$ in $S$ in the sense that $x = xyx$ and $y = yxy$.

There is a question on this site (A semigroup $X$ is a group iff for every $g\in X$, $\exists! x\in X$ such that $gxg = g$) whose answer is "A nonempty semigroup $S$ is a group iff for every $x\in S$ there is a unique $y\in S$ such that $xyx=x$."

So it follows that every nonempty inverse semigroup is a group. It seems weird that this fact is not listed on Wikipedia, and that some literature seems to exist about inverse semigroups. Am I not understanding correctly the definitions?

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    $\begingroup$ I am not 100% on this, but I suspect the difference comes from requiring a unique $y$ such that $xyx=x$ vs. requiring a unique $y$ such that $xyx = x$ and $yxy=y$. The second condition is weaker than the first, since there could be many $y'$ which satisfy $xy'x = x$ while not simultaneously satisfying $y'xy'=y'$. Very subtle difference. $\endgroup$ – EuYu Feb 29 '16 at 10:38
  • $\begingroup$ @EuYu: But the condition on Wikipedia is (at least) stronger than the condition in the MSE question, so that the answer by Bill Dubuque still applies. $\endgroup$ – slowlearner Feb 29 '16 at 12:21
  • $\begingroup$ @EuYu: Also, if $xyx=x$ then by uniqueness $yxy=y$, so that I believe the two conditions are in fact equivalent. $\endgroup$ – slowlearner Feb 29 '16 at 12:22
  • $\begingroup$ I don't think the wikipedia condition ($A$) is (at least) stronger than the MSE condition ($B$). Bill Dubuque proved that $B\implies A$. But there can be semigroups which satisfy $A$ but not $B$. You can imagine a semigroup where every element has multiple $y'$ such that $y'xy' = y'$ and multiple $y''$ such that $xy''x=x$, but exactly one $y$ for which both $yxy=y$ and $xyx=x$ holds. Condition $A$ should be strictly weaker than $B$. $\endgroup$ – EuYu Feb 29 '16 at 19:56
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Let's consider the canonical example of an inverse semigroup that J.-E. Pin mentioned in his answer, the semigroup of partial bijections on a set. For concreteness, let's take our set to be $[3]=\{1,2,3\}$ and call our resulting semigroup $S$.

Consider the element $f:\{1,2\}\rightarrow\{2,3\}$ defined by $f(1)=2$ and $f(2)=3$. Then $f$ has a unique inverse $f^*:\{2,3\}\rightarrow\{1,3\}$ defined by $f^*(2)=1$ and $f^*(3)=2$. You can easily verify that $$f\circ f^*\circ f = f,\ \ \ \ \ \text{and}\ \ \ \ f^*\circ f \circ f^* = f^*,$$ and indeed, you can verify that $f^*$ is the unique element of $S$ which satisfy both these conditions. But there are others which satisfy the each of the two above conditions individually. For example, $f^{**}:[3]\rightarrow [3]$ defined by $f^{**}(1)=3$, $f^{**}(2)=1$, and $f^{**}(3)=2$. Then we also have $$f\circ f^{**} \circ f = f,$$ but we no longer have $$f^{**} \circ f \circ f^{**} = f^{**}.$$ In fact, $f^{**} \circ f \circ f^{**}$ is the restriction of $f^{**}$ to $\{2,3\}$.

So there is a distinct difference between:

i. A semigroup $S$ such that for every $x\in S$, there exists a unique $y\in S$ such that $xyx=x$.

ii. A semigroup $S$ such that for every $x\in S$, there exists a unique $y\in S$ such that both $xyx=x$ and $yxy=y$.

And the condition ii is strictly weaker than condition i. You can prove that any semigroup which satisfies condition i is a full group. You cannot do the same for condition ii, which defines inverse semigroups. The semigroup of partial bijections is an explicit example of this.

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  • $\begingroup$ Thanks so much for your answer. It clearly sealed the deal for me. $\endgroup$ – slowlearner Mar 1 '16 at 10:00
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There are indeed many inverse monoids that are not groups. Just consider for instance the set of all partial injective functions from a set $E$ into itself, under composition.

An element $x$ of a semigroup is called a weak inverse of $y$ if $xyx = x$. It is called an inverse of $y$ if $xyx = x$ and $yxy = x$, that is, if $x$ is a weak inverse of $y$ and $y$ is a weak inverse of $x$.

The result you mention states that

A nonempty semigroup is a group if and only if every element is a weak inverse of exactly one element.

Edit, following the comment of Tristan Brice.

As observed by Tristan Brice, the following dual result does not hold:

A nonempty semigroup is a group if and only if every element has exactly one weak inverse.

Only a weaker (and simpler) result holds

A monoid is a group if and only if every element has exactly one weak inverse.

Let $M$ be the monoid and let $s$ be an element of $M$. Let $\bar s$ be the unique weak inverse of $s$. Then $\bar s s \bar s = \bar s$ and hence $s\bar s$ and $\bar ss$ are idempotent. I claim that $1$ is the unique idempotent of $M$ contains a unique idempotent. Indeed, let $e$ be an idempotent. Since $1e1 = eee = e$, $e$ and $1$ are both weak inverses of $e$ and thus $e = 1$. Coming back to $s$, we get $s\bar s = \bar ss = 1$ and thus $S$ is a group.

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  • $\begingroup$ In your example (partial injective functions), an element can have multiple inverses. This does not match the definition given on Wikipedia, but rather seems to concern regular semigroups. $\endgroup$ – slowlearner Feb 29 '16 at 14:19
  • $\begingroup$ @slowlearner Not only it matches the definition of wikipedia, but this is the first example given in this article. "Inverse semigroups were introduced independently by Viktor Vladimirovich Wagner in the Soviet Union in 1952, and by Gordon Preston in Great Britain in 1954.] Both authors arrived at inverse semigroups via the study of partial one-one transformations of a set (...)" $\endgroup$ – J.-E. Pin Feb 29 '16 at 15:11
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    $\begingroup$ Your proof doesn't quite show that S is a group, as the unique idempotent is only an identity for weak inverses. Indeed, in the semigroup with $ab=0$, for all $a$ and $b$, $0$ is the unique weak inverse of every element but $0$ is not an identity (as long as the semigroup contains at least 2 elements). $\endgroup$ – Tristan Bice Mar 1 '16 at 13:05
  • $\begingroup$ @tristan-bice You are absolutely right, the result is only true for monoids, and it is much simpler in this case. I will edit my answer. $\endgroup$ – J.-E. Pin Mar 1 '16 at 13:26

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