S an inverse semigroup with semilattice of idempotents E and $\sigma$ the minimum group congruence on S.

Question

Let S be an inverse semigroup with semilattice of idempotents E, and let $\sigma$ be the minimum group congruence on S. Show that the following statements are equivalent:

(a) $x\sigma y$;

(b) $(\exists e \in E) \; xe=ye $;

(c) $(\exists a \in S) \; ax=ay $;

(d) $(\exists a \in S) \; xa=ya $;

(e) $(\exists e, f \in E) \; ex=fy $;

(f) $(\exists e, f \in E) \; xe=yf $;

(g) $x^{-1}y\in E\omega$;

(h) $(Ex)\omega = (Ey)\omega $;

(i) $y\in (Ex)\omega$;

(j) $x\in (Ey)\omega$

I can show that $(a)\Rightarrow (e)$. Suppose: $x\sigma y$

by definition $\sigma = \lbrace (a, b)\in S\times S: ab^{-1}\in E\omega\rbrace$

where $E\omega = \lbrace s\in S: (\exists e\in E) h\leq s\rbrace$ is a closure $E\omega$ of $E$.

Now $x\sigma y \iff \exists c\in S \;\;s.t.\;\; c\leq x,y$

$c \leq x \iff \exists e\in S \;\; c=ex$ similarly for $c \leq y \iff \exists f\in S \;\; c=fy$ Hence, we have shown that $((\exists e, f \in E) \; ex=c=fy)$

$(e)\Rightarrow (f)$

Given that $x\leq y \Rightarrow x^{-1}\leq y^{-1}$, and the fact that $e=e^{-1}, \forall e\in E$ where $E$ is a semilattice of idempotents for $S$; we have

$c\leq x \Rightarrow c^{-1}\leq x^{-1}, \text{ so } c^{-1}= ex^{-1} =e^{-1}x^{-1} = (xe)^{-1}$ the last step is by (Prop 5.1.2 Howie's book), then

$c=(xe)$ and similarly we can show that $c=yf$ hence $(\exists e, f \in E) \; xe=yf $;

If I suppose that (g)i.e. $x^{-1}y\in E\omega$ is true then by (a) and definition of $\sigma$ I have

$xy^{-1}\in E\omega$ since $\sigma = \lbrace (x, y)\in S\times S: xy^{-1}\in E\omega\rbrace$ .

However I can't find the link between (f) and (g).

The fact that $(g)\Rightarrow (h)\Rightarrow (i)\Rightarrow (j)$ follows from the fact that E is an inverse subsemigroup of S.

Thanks in advance

Answer 1

we have $ E\omega = \{s \in S : (\exists e \in E)\; e \leq s\} $

where $ c \leq s \iff \exists e \in E\quad c = es \iff \exists f \in E \quad c=sf$

we start with $ (\exists e,f \in E)\; xe=yf $

$x^{-1}xe = x^{-1}yf$

$x^{-1}x$ is idempotent, therefore, so is $x^{-1}xe$, call this idempotent $g$, giving us

$ g = (x^{-1}y)f$

$g \leq x^{-1}y $

and so, as $g \in E$, $(x^{-1}y) \in E\omega$