Every inverse semigroup is a group

Question

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?

Answer 1

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.

Answer 2

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.