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An inverse semigroup $S$ is a semigroup in which for each $x\in S$ there exists a unique $y\in S$ such that $xyx=x$ and $yxy=y$. I'm trying to find an explicit example(which is not a group) of such semigroups. is there a way to construct an inverse semigroup? Thank you for your help.

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4 Answers 4

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Let $E$ be a set. Then the set $Sym(E)$ of all permutations on $E$ is a group (the symmetric group on $E$) under function composition. Similarly, the set $Inv(E)$ of all partial one-one transformations is an inverse semigroup (the symmetric inverse semigroup on $E$) under function composition.

Cayley's theorem states that every group can be embedded into a symmetric group. Similarly, the Wagner-Preston theorem states that every inverse semigroup can be embedded into an symmetric inverse semigroup.

For this reason, symmetric inverse semigroups are considered the standard examples of inverse semigroups.

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  • $\begingroup$ How useful is this example, at last, for me. $\endgroup$
    – Ned Dabby
    Commented Jan 12, 2017 at 18:54
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What about a 2-element semilattice? If the elements are $x$ and $y$, then the multiplication could be: $x^2=x$, $xy=y=yx=y^2$. This is an inverse semigroup ($x^{-1}=x$ and $y^{-1}=y$) but not a group (it has 2 idempotents).

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Well, you can take any semilattice and it will be an inverse semigroup.

And if you want to construct an inverse semigroup, I believe with the Green's relation it is possible, since a semigroup is inverse iff every $\mathcal{R}$ and $\mathcal{L}$ classes have precesely one idempotent.

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The bicyclic Semigroup $B=\mathbb{N}_0 \times \mathbb{N}_0$ is an inverse semigroup but it is not a group. for every element $(m, n)\in B$, the element $(n, m)$ is unique inverse for it and $\forall n\in \mathbb{N}$ the element $(n, n)$ is idempotent.

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