5 Structure of Fundamental Regular Semigroups

Chapter 5
Structure of Fundamental Regular Semigroups

Definition 5.1. We call a regular semigroup $S$ a fundamental regular semigroup if and only if the greatest idempotent separating congruence on $S$ is the identity congruence. Given a regular semigroup $S$ , it can be easily seen that $S | μ (s)$ is a fundamental regular semigroup.

Throughout this chapter, we denote a biordered set by $E$ and a partially transitive set on $E$ by $T$ . We give below some additional notations needed in the sequel. For $e, f \in E$ define

T (e, f) = {α \in T ∣ e_{α} = e, f_{α} = f} .

(5.1)

Then

T (e, f) \neq 0 \Leftrightarrow e δ (T) f .

If $α \in T (e, f), β \in T (g, h)$ , then by axiom $(T_{3})$ we have

α \circ β \in \{\begin{matrix} T (e, f_{β}) & if f ω g \\ T (g α^{- 1}, h) & if g ω f . \end{matrix}

(5.2)

Also for $e ω^{r} e_{α}$ and $f ω^{l} f_{α} (α \in Ẽ_{1})$ define

e \land α = τ^{r} (e, e τ^{r} (e_{α})) \circ α, α \land f = α \circ τ^{l} (f τ^{l} (f_{2}), f) .

(5.3)

Then if $α \in T, e \land α \in T (e, (e τ^{r} (e_{α})) α)$ and $α \land f \in T ((f τ^{l} (f_{α})) α^{- 1}, f)$ . Also for any $α, β \in T$ and $h \in S (f_{α}, e_{β})$ , it follows from (5.2) and (5.3) that

α \land h \circ h \land β \in T ((h τ^{l} (f_{α})) α^{- 1}, (h τ^{r} (e_{β})) β) .

(5.4)

For $e ω e_{α} (f ω f_{α})$ , $e \land α$ is the domain restriction ( $α \land f$ is the range restriction) of $α$ to $ω (e) (ω (f))$ . Therefore when $f_{α} ω e_{β} (e_{β} ω f_{α})$ , (5.4) reduces to (5.2) if we choose $h = f_{α} (h = e_{β})$ .

In $Ẽ_{1}$ define a releation $σ$ as follows:

α σ β \Leftrightarrow e_{α} \overset{r}{\approx} e_{β}, f_{α} \overset{l}{\approx} f_{β} and e_{β} \land α = β \land f_{α} .

(5.5)

It is not difficult to see that $σ$ is an equivalence relation on $Ẽ_{1}$ . We denote the $σ$ -class of $α \in Ẽ_{1}$ by $\bar{α}$ . If $α \in T$ since $Ẽ_{0} \subseteq T$ it is clear from (5.5) that $β \in T$ for all $β σ α$ or $\bar{α} \subseteq T$ . Hence

\bar{T} = {\bar{α} ∣ α \in T} \subseteq {\bar{Ẽ}}_{1} = Ẽ_{1} ∕ σ

(5.6)

and therefore if $T, T^{'} \in P T (E)$

T \subseteq T^{'} \Leftrightarrow \bar{T} \subseteq {\bar{T}}^{'} .

(5.7)

Following is the main structure theorem of this chapter.

Theorem 5.1. Let $E$ be a biordered set and $T \in P T (E)$ . Define for $α, β \in T$ ,

\bar{α} \cdot \bar{β} = \bar{α \land h \circ h \land β} where h \in S (f_{α}, e_{β}) .

(5.8)

Then $\bar{T}$ is a fundamental regular semigroup under the operation defined by (5.8) and $E (\bar{T})$ is isomorphic to $E$ .

Conversely if $S$ is a fundamental regular semigroup, then $S$ is isomorphic to ${\bar{T}}_{s}$ .

We prove this theorem through a sequence of lemmas.

Lemma 5.2. Let $α \in Ẽ_{1}$ ,

If $f_{1}, f_{2} \in ω^{l} (f_{α}), f_{1} \overset{r}{\approx} f_{2}$ then $τ^{r} (e_{1}^{'}, e_{2}^{'}) \circ (α \land f_{2}) = (α \land f_{1}) \circ τ^{r} (f_{1}, f_{2})$
where $e_{i}^{'} = (f_{i} τ^{l} (f_{α})) α^{- 1}, i = 1, 2 .$
If $e_{1}, e_{2} \in ω^{r} (e_{α}), e_{1} \overset{l}{\approx} e_{2}$ then $τ^{l} (e_{1}, e_{2}) \circ (e_{2} \land α) = (e_{1} \land α) \circ τ^{l} (f_{1}^{'}, f_{2}^{'})$
where $f_{i}^{'} = (e_{i} τ^{r} (e_{α})) α, i = 1, 2$ .

Proof. By hypothesis $e_{1}^{'} \overset{r}{\approx} e'$ and for all $g \in ω (e_{1}^{'})$ ,

(g) τ^{r} (e_{1}^{'}, e_{2}^{'}) \circ α = ((g) τ^{r} (e_{2}^{'})) α = (g α) τ^{r} (e_{2}^{'} α) .

Since $g α \in ω (e_{1}^{'} α) = ω (f_{1} τ^{l} (f_{α}))$ ,

(g) τ^{r} (e_{1}^{'}, e_{2}^{'}) \circ α = (g) α \circ τ^{r} (f_{1} τ^{l} (f_{2}), f_{2} τ^{l} (f_{α})) .

Hence the required result is equivalent to the following:

α \circ τ^{r} (f, τ^{l} (f_{α}), f_{2} τ^{l} (f α)) \circ τ^{l} (f_{2} τ^{l} (f_{α}), f_{2}) = α \circ τ^{l} (f, τ^{l} (f α), f_{1}) \circ τ^{r} (f_{1}, f_{2}) .

Since $α$ is a partial $ω$ -isomorphism, this is equivalent to the equality established in Lemma 3.7. Statement (2), which is the dual of (1), can be proved similarly. □

Lemma 5.3. Let $α, β \in T$ and $h_{1}, h_{2} \in S (f_{α}, e_{β})$ . Then

α \land h_{1} \circ h_{1} \land β σ α \land h_{2} \circ h_{2} \land β .

Proof. First suppose that $h_{1} \overset{r}{\approx} h_{2}$ . Then $h_{1} τ^{r} (e_{β}) = h_{2} τ^{r} (e_{β})$ and therefore

\begin{array}{rcl} h_{1} \land β & = & τ^{r} (h_{1}, h_{1} τ^{r} (e_{β})) \circ β \\ = & τ^{r} (h_{1}, h_{2} τ^{r} (e_{β})) \circ β \\ = & τ^{r} (h_{1}, h_{2}) \circ h_{2} \land β . \end{array}

If $h_{i}^{'} = (h_{i} τ^{l} (f_{2})) α^{- 1}, i = 1, 2$ then $h_{1}^{'} \overset{r}{\approx} h_{2}^{'}$ , and by Lemma 5.2,

\begin{array}{rcl} τ^{r} (h_{1}^{'}, h_{2}^{'}) \circ (α \land h_{2} \circ h_{2} \land β) & = & α \land h_{1} \circ τ^{r} (h_{1}, h_{2}) \circ h_{2} \land β \\ = & α \land h_{1} \circ h_{1} \land β . \end{array}

If $h^{''} = (h_{1} τ^{r} (e_{β})) β = (h_{2} τ^{r} (e_{β})) β$ , by (5.4),

α \land h_{i} \circ h_{i} \land β \in T (h_{i}^{'}, h^{''}), i = 1, 2,

and hence by (5.5) $α \land h_{1} \circ h_{1} \land β σ α \land h_{2} \circ h_{2} \land β$ .

If $h_{1} \overset{l}{\approx} h_{2}$ , the required relation can be proved analogously. If $h_{1}$ and $h_{2}$ are any two members of $S (f_{α}, e_{β})$ then by Lemma 2.7 there exists $h_{2}^{'} \in S (f_{α}, e_{β})$ such that $h_{1} \overset{r}{\approx} h_{2}^{'} \overset{l}{\approx} h_{2}$ and hence the lemma. □

Lemma 5.4. For $i = 1, 2$ , let $α_{i} \in T (e_{i}, f_{i}), β_{i} \in T (g_{i}, h_{i})$ and $k_{i} \in S (f_{i}, g_{i})$ . Suppose that $α_{1} σ α_{2}$ and $β_{1} σ β_{2}$ . Then

α_{1} \land k_{1} \circ k_{1} \land β_{1} σ α_{2} \land k_{2} \circ k_{2} \land β_{2} .

Proof. The hypotheses imply that

\begin{array}{rcl} e_{1} \overset{r}{\approx} e_{2}, f_{1} \overset{l}{\approx} f_{2} \\ e_{1} \land α_{2} = α_{1} \land f_{α} \end{array}

and

\begin{array}{rcl} g_{1} \overset{r}{\approx} g_{2}, h_{1} \overset{l}{\approx} h_{2} \\ g_{1} \land β_{2} = β_{1} \land h_{2} . \end{array}

Then by Lemma 2.6, $S (f_{1}, g_{1}) = S (f_{2}, g_{2})$ . In view of Lemma 5.3, we need only establish that for any $k \in S (f_{1}, g_{1})$

α_{1} \land k \circ k \land β_{1} σ α_{2} \land k \circ k \land β_{2} .

The statement $α_{1} σ α_{2}$ implies that

α_{2}^{- 1} = τ^{l} (f_{2}, f_{1}) \circ α^{- 1} \circ τ^{r} (e_{1}, e_{2}) .

Hence if $e_{i}^{'} = (k τ^{l} (f_{i})) α_{i}^{- 1}$ and $h_{i}^{'} = (k τ^{r} (g_{i})) β_{i}$

\begin{array}{rcl} e_{2}^{'} & = & (k τ^{l} (f_{2})) τ^{l} (f_{2}, f_{1}) \circ α_{1}^{- 1} \circ τ^{r} (e_{1}, e_{2}) \\ = & ((k τ^{l} (f_{1})) α_{1}^{- 1}) τ^{r} (e_{1}, e_{2}) \\ = & e_{1}^{'} τ^{r} (e_{1}, e_{2}) . \end{array}

Dually $h_{2}^{'} = h_{1}^{'} τ^{l} (h_{1}, h_{2})$ and thus we get $e_{1}^{'} \overset{r}{\approx} e_{2}^{'}$ and $h_{1}^{'} \overset{l}{\approx} h_{2}^{'}$ .

Since $α_{i} \land k \circ k \land β_{i} \in T (e_{i}^{'}, h_{i}^{'})$ we need only establish the following assertion:

τ^{l} (e_{1}^{'}, e_{2}^{'}) \circ α_{2} \land k \circ k \land β_{2} = α_{1} \land k \circ k \land β_{1} \circ τ^{l} (h_{1}^{'}, h_{2}^{'}) .

For Lemma 3.1, taking $c = (f_{1}, f_{2}), h = k τ^{l} (f_{1})$ we get

τ^{l} (k τ^{l} (f_{1}), k τ^{l} (f_{1})) \circ τ^{l} (f_{1}, f_{2}) = τ^{l} (k τ^{l} (f_{1}), k τ^{l} (f_{2})) .

Similarly it can be seen that

\begin{array}{rcl} τ^{r} (k τ^{r} (g_{1}), k τ^{r} (g_{1})) \circ τ^{r} (g_{1}, g_{2}) = τ^{r} (k τ^{r} (g_{1}), k τ^{r} (g_{2})) \\ τ^{r} (e_{1}^{'}, e_{1}^{'}) \circ τ^{r} (e_{1}, e_{2}) = τ^{r} (e_{1}^{'}, e_{2}^{'}) \end{array}

and

τ^{l} (h_{1}^{'}, h_{1}^{'}) \circ τ^{l} (h_{1}, h_{2}) = τ^{l} (h_{1}^{'}, h_{2}^{'}) .

Hence if $e^{''} \in ω (e_{1}^{'})$

\begin{array}{rcl} (e^{''}) τ^{r} (e_{1}^{'}, e_{2}^{'}) \circ α_{2} & = & (e^{''}) τ^{r} (e_{1}, e_{2}) \circ α_{2} \\ = & (e^{''}) α_{1} \circ τ^{l} (f_{1}, f_{2}) \\ = & (e^{''}) α_{1} \circ τ^{l} (k τ^{l} (f_{1}), k τ^{l} (f_{2})) . \end{array}

Therefore

\begin{array}{rcl} τ^{r} (e_{1}^{'}, e_{2}^{'}) \circ α_{2} \land k & = & α_{1} \circ τ^{l} (k τ^{l} (f_{1}), k τ^{l} (f_{2})) \circ τ^{l} (k τ^{l} (f_{2}), k) \\ = & α_{1} \circ τ^{l} (k τ^{l} (f_{1}), k) \\ = & α_{1} \land k . \end{array}

The dual statement

k \land β_{2} = k \land β_{1} \circ τ^{l} (h_{1}^{'}, h_{2}^{'}),

can be proved similarly. Thus

α_{1} \land k \circ k \land β_{1} \circ τ^{l} (h_{1}^{'}, h_{2}^{'}) = τ^{r} (e_{1}^{'}, e_{2}^{'}) \circ α_{2} \land k \circ k \land β_{2},

and hence the lemma. □

Lemma 5.5. Let $α, β \in T$ and for $k_{i} \in ω^{l} (f_{α}) \cap ω^{r} (e_{β}), i = 1, 2$ define

e_{i}^{'} = (k_{i} τ^{l} (f_{α})) α^{- 1} and h_{i}^{'} = (k_{i} τ^{r} (e_{β})) β .

(k_{2} τ^{l} (f_{α}), k_{2} τ^{r} (e_{β})) ω^{r} \times ω^{l} (k_{1} τ^{l} (f_{α}), k_{1} τ^{r} (e_{β}))

then

e_{2}^{'} \land (α \land k_{1} \circ k_{1} \land β) σ α \land k_{2} \circ k_{2} \land β σ (α \land k_{1} \circ k_{1} \land β) \land h_{2}^{'} .

Proof. By (5.4), we have for $i = 1, 2$ , $α \land k_{i} \circ k_{i} \land β \in T (e_{i}^{'}, h_{i}^{'})$ . Also $e_{2}^{'} ω^{r} e_{1}^{'}$ and $h_{2}^{'} ω^{l} h_{1}^{'}$ . We obtain the lemma if we establish the following equality:

τ^{r} (e_{2}^{'}, e_{2}^{'} τ^{r} (e_{1}^{'})) \circ α \land k_{1} \circ k_{1} \land β \circ τ^{l} (h_{2}^{'} τ^{l} (h_{1}^{'}), h_{2}^{'}) = α \land k_{2} \circ k_{2} \land β .

(5.9)

From the fact that $α$ is a partial $ω$ -isomorphism we have

τ^{r} (e_{2}^{'}, e_{2}^{'} τ^{r} (e_{1}^{'}) \circ α = α \circ α^{r} (k_{2} τ^{l} (f_{α}), k_{1} τ^{l} (f_{α}))

where $α^{r} (e_{1}, e_{2})$ is defined by (3.26). Similarly

β \circ τ^{l} (h_{2}^{'} τ^{l} (h_{1}^{'}), h_{2}^{'}) = α^{l} {(k_{2} τ^{r} (e_{β}), k_{1} τ^{r} (e_{β}))}^{- 1} \circ β .

Hence (5.9) is equivalent to

α \circ α^{l} {(k_{2}, f_{α})}^{- 1} \circ α^{r} (k_{2}, e_{β}) \circ β = α \circ T \circ β

where

\begin{array}{rcl} T & = & α^{r} (k_{2} τ^{l} (f_{α}), k_{1} τ^{l} (f_{α})) \circ α^{l} {(k_{1}, f_{2})}^{- 1} \\ \circ α^{r} (k_{1}, e_{β}) \circ α^{l} (k_{2} τ^{r} (e_{β}), k_{1} τ^{r} (e_{β})) . \end{array}

Since $α$ and $β$ are partial $ω$ -isomorphisms, this is equivalent to

T = α^{l} {(k_{2}, f_{α})}^{- 1} \circ α^{r} (k_{2}, e_{β}),

which is the same as the equality in Lemma 3.8. □

Lemma 5.6. Let $α, β, γ \in T, h_{1} \in S (f_{α}, e_{β}) h_{2} \in S (f_{β}, e_{r}), h_{1}^{'} = (h_{1} τ^{r} (e_{β})) β$ and $h_{2}^{'} = (h_{2} τ^{l} (f_{β})) β^{- 1}$ . If $h \in S (f_{α}, h_{2}^{'})$ and $h^{'} \in S (h_{1}^{'}, e_{r})$ then

(α \land h_{1} \circ h_{1} \land β) \land h^{'} \circ h^{'} \land r σ α \land h \circ h \land (β \land h_{2} \circ h_{2} \land r) .

Proof. By (5.4)

α \land h_{1} \circ h_{1} \land β \in T (h_{1} τ^{l} (f_{α})) α^{- 1}, h^{'},

and

β \land h_{2} \circ h_{0} \land r \in T (h_{2}^{'}, (h_{2} τ^{r} (e_{r})) r) .

Hence the expressions on either side of $σ$ in the required relation are defined. Further by Lemma 3.9, there exist $h \in S (f_{α}, h_{2}^{'})$ and $h^{'} \in S (h_{1}^{'}, e_{r})$ such that

(h τ^{r} (e_{β})) β = h^{'} τ^{l} (f_{β}) .

By Lemma 5.3 we may without loss of generality choose $h$ and $h'$ so as to satisfy the above equality. □

Since $h \in ω^{l} (f_{α}) \cap ω^{r} (e_{β})$ and $h_{1} \in S (f_{α}, e_{β}), h$ and $h_{1}$ satisfy conditions of Lemma 5.5 and hence we get by the equality $(h τ^{r} (e_{β})) β = h' τ^{l} (f_{β})$ ,

(α \land h_{1} \circ h_{1} \land β) \land h' τ^{l} (f_{β}) σ α \land h \circ h \land β .

It is clear that

(α \land h_{1} \circ h_{1} \land p) \land h' τ^{l} (f_{β}) σ (α \land h_{1} \circ h_{1} \land β) \land h'

and

(α \land h \circ h \land β) \land h' σ α \land h \circ h \land β .

Hence

(α \land h_{1} \circ h_{1} \land β) \land h' \circ h^{'} \land γ σ (α \land h \circ h \land β) \land h' \circ h^{'} \land γ .

But

(α \land h \circ h \land β) \land h' = (α \land h \circ τ^{r} (h_{1} h τ^{r} (e_{β})) \circ β) \circ τ^{l} (h' τ^{l} (f_{β}), h') .

Since $(h τ^{r} (e_{β})) β = h' τ^{l} (f_{β})$ , the range of $h \land β$ is $ω (h' τ^{l} (f_{β}))$ and the domain of $β \land h'$ is $ω (h τ^{r} (e_{β}))$ , and therefore,

(h \land β) \land h' = h \land (β \land h') .

Hence

(α \land h_{1} \circ h_{1} \land β) \land h' \circ h^{'} \land γ σ α \land h \circ (h \land β) \land h' \circ h^{'} \land γ

and the dual statement

α \land h \circ h \land (β \land h_{2} \circ h_{2} \land γ) σ α \land h \circ h \land (β \land h') \circ h^{'} \land γ

can be obtained in the same way. Since $(h \land β) \land h' = h \land (β \land h')$ , the lemma is proved.

Lemma 5.7. If $T \in P ℐ (E)$ , $\bar{T} = T ∕ σ$ is a regular semigroup under the operation defined by Lemma 5.8.

Proof. If $x, y \in \bar{T}$ , for $α \in x, β \in y$ and $h \in S (f_{α}, e_{β})$ , by Lemma 5.8

x \cdot y = \bar{α \land h \circ h \land β} .

□

By Lemma 5.4, the $σ$ -class $α \land h \circ h \land β$ does not depend on the choice of $α$ in $x$ , $β$ in $y$ and $h \in S (f_{α}, e_{β})$ . Hence $\bar{α \land h \circ h \land β}$ is a $σ$ -class of $T$ depending only on the classes $x$ and $y$ , and therefore Lemma 5.8 defines a binary operation on $\bar{T}$ .

Let $x, y, z \in \bar{T}$ , $α \in x$ , $β \in y$ and $γ \in z$ . Then if $h_{1} \in S (f_{α}, e_{β})$ , by (5.4)

α \land h_{1} \circ h_{1} \land β \in T (h_{1} τ^{l} (f_{α}) α^{- 1}, (h_{1} τ^{r} (e_{β})) β) .

Now for $h' \in S (h'_{1}, e_{r})$ , $h'_{1} = (h_{1} τ^{r} (e_{β})) β$

(x \cdot y) \cdot z = \bar{(α \land h_{1} \circ h_{1} \land β) \land h' \circ h^{'} \land γ} .

Similarly, if $h_{2} \in S (f_{β}, e_{γ})$ , $h \in S (f_{2}, h'_{2})$ where $h'_{2} = (h_{2} τ^{2} (f_{β})) β^{- 1}$ then

(x \cdot (y \cdot z)) = \bar{α \land h \circ h \land (β \land h_{2} \circ h_{2} \land γ)} .

Thus by Lemma 5.6,

(x \cdot y) \cdot z = x \cdot (y \cdot z) .

If $x \in \bar{T}$ , $α \in x$ then $α^{- 1} \in Ẽ, (f_{α}, e_{α})$ and by axiom ( $T_{2}$ ) of Definition 3.1, $α^{- 1} \in T$ . Also if $x' = \bar{α^{- 1}}$

x \cdot x' \cdot x = \bar{α \circ α^{- 1} \circ α} = \bar{α} = x,

x' \cdot x \cdot x' = \bar{α^{- 1} \circ α \circ α^{- 1}} = \bar{α^{- 1}} = x' .

Hence $\bar{T}$ is a regular semigroup.

Lemma 5.8. For every $e \in E$ . $\bar{ℰ} (e, e) \in E (\bar{T})$ and

e ω^{r} f \Leftrightarrow \bar{ϵ} (e, e) ω^{r} \bar{ϵ} (f, f) .

(5.10)

Moreover, the mapping

α \to \bar{α} : T (e, f) \to \bar{T}

(5.11)

is a bijection of $T (e, f)$ onto the $ℋ$ -class $H_{\bar{ϵ} (e, e), \bar{ϵ} (f, f)}$ of $\bar{T}$ . Consequently,

e δ (T) f \Leftrightarrow \bar{ϵ} (e, e) δ (\bar{T}) \bar{ϵ} (f, f) .

(5.12)

Proof. That $\bar{ϵ} (e, e) \in E (\bar{T})$ for all $e \in E$ is clear from the definition of operation in $\bar{T}$ . □

Now let $e ω^{r} f$ . Then $e τ^{r} (f) \in S (e, f)$ and $\bar{ϵ} (f, f) \cdot \bar{ϵ} (e, e)$ = $\bar{ϵ (e τ^{r} (f), e τ^{r} (f)) \circ ϵ (e τ^{r} (f), e)} = \bar{ϵ} (e τ^{r} (f), e)$ . But clearly $ϵ (e τ^{r} (f), e) σ ϵ (e, e)$ and hence

\bar{ϵ} (f, f) \cdot \bar{ϵ} (e, e) = \bar{ϵ} (e, e)

which implies that $\bar{ϵ} (e, e) ω^{r} \bar{ϵ} (f, f)$ in $E (\bar{T})$ , by (1.10)

On the other hand if $\bar{ϵ} (e, e) ω^{r} \bar{ϵ} (f, f)$ and if $h \in S (f, e)$ then

\bar{ϵ} (e, e) = \bar{ϵ} (f, f) \cdot \bar{ϵ} (e, e) = \bar{ϵ (h τ^{l} (f)), h \circ ϵ (h, h τ^{r} (e))} .

Hence

h τ^{l} (f) \overset{r}{\approx} e \overset{l}{\approx} h τ^{r} (e) .

Since $h τ^{r} (e) ω e, e \overset{l}{\approx} h τ^{r} (e)$ implies $e = h τ^{r} (e)$ or $h \overset{r}{\approx} e$ . Hence $h = h τ^{l} (f)$ .

e \overset{r}{\approx} h ω f .

Let $α \in T (e, f)$ . Then $α^{- 1} \in T (f, e)$ and we have

\bar{α} \cdot \bar{α^{- 1}} = \bar{ϵ} (e, e), \bar{α^{- 1}} \cdot \bar{α} = \bar{ϵ} (f, f)

\bar{ϵ} (e, e) \cdot \bar{α} = \bar{α} = \bar{α} \cdot \bar{ϵ} (f, f) .

Hence $\bar{α} \in H_{\bar{ϵ} (e, e), \bar{ϵ} (f, f)}$ and so the mapping $α \to \bar{α}$ maps $T (e, f)$ into $H_{\bar{ϵ} (e, e), \bar{ϵ} (f, f)}$ . Let $x \in H_{\bar{E} (e, e), \bar{ϵ} (f, f)}$ . Then if $α \in x$ as observed earlier,

x = \bar{α} \in H_{\bar{ϵ} (e_{α}, e_{α}), \bar{ϵ} (f_{α}, f_{α})} .

Hence by (5.10) and (5.10)*, $e \overset{r}{\approx} e_{α}$ , $f \overset{l}{\approx} f_{α}$ . Then if $α' = τ^{r} (e, e_{2}) \circ α \circ τ^{l} (f_{α}, f), α' σ α$ and $α' \in T (e, f)$ . Thus the mapping defined by (5.11) is onto $H_{\bar{ϵ} (e, e), \bar{ϵ} (f, f)}$ . From the definition of $σ$ it is clear that

α, α' \in T (e, f), α σ α' \Rightarrow α = α';

which shows that the mapping defined by (5.11) is one-to-one.

Further

\begin{array}{rcl} e δ (T) f & \Leftrightarrow & T (e, f) \neq \emptyset \\ \Leftrightarrow & H_{\bar{ϵ} (e, e), \bar{ϵ} (f, f)} \neq \emptyset \\ \Leftrightarrow & \bar{ϵ} (e, e) δ (\bar{T}) \bar{ϵ} (f, f), \end{array}

which is statement (5.12).

Lemma 5.9. Let $ψ$ denote the mapping of $E$ into $E (T)$ defined by

e ψ = \bar{ϵ} (e, e)

(5.13)

for every $e \in E$ . Then $ψ$ is an isomorphism of $E$ onto the biordered set of idempotents $E (\bar{T})$ of $\bar{T}$ . Moreover,

α \circ ψ = ψ \circ a (\bar{α}, \bar{α^{- 1}})

(5.14)

for every $α \in T$ .

Proof. We have already noted that $ψ$ maps $E$ into $E (\bar{T})$ and that $ψ$ and $ψ^{- 1}$ preserve both $ω^{r}$ and $ω^{l}$ (by (5.10) and (5.10)*).

Let $\bar{α} \in E (\bar{T})$ where $α \in T (e, f)$ . Then if $h \in S (f, e)$ we have

\bar{α} \cdot \bar{α} = \bar{α \land h \circ h \land α} = \bar{α} .

Hence

e \overset{r}{\approx} (h τ^{l} (f)) α^{- 1} a n d f \overset{l}{\approx} (h τ^{r} (e)) α .

Since $α$ is a partial $ω$ -isomorphism we have

h τ^{l} (f) \overset{r}{\approx} e α a n d h τ^{r} (e) \overset{l}{\approx} f α^{- 1} .

Hence

h τ^{l} (f) = f a n d h τ^{r} (e) = e .

Therefore

f \overset{l}{\approx} h \overset{r}{\approx} e .

By Lemma 5.8, we see that $\bar{ϵ} (h, h)$ and $\bar{α}$ are $ℋ$ -equivalent idempotents and consequently

\bar{α} = \bar{ϵ} (h, h) .

This proves that $ψ$ is onto. Being one-to-one, it is a bijection.

Let $e ω^{r} f$ . Then $e \in S (e, f)$ and by (4.1),

\begin{array}{rcl} e ψ τ^{r} (f ψ) & = & e ψ \cdot f ψ \\ = & \bar{ϵ (e, e τ^{r} (f)) \circ ϵ (f, f)} \\ = & \bar{ϵ} (e, e τ^{r} (f)) . \end{array}

But

ϵ (e, e τ^{r} (f)) σ ϵ (e τ^{r} (f), e τ^{r} (f))

and therefore

e ψ τ^{r} (f ψ) = \bar{ϵ} (e τ^{r} (f), e τ^{r} (f)) = (e τ^{r} (f)) ψ .

Let $e, f \in E$ and $h \in S (e, f)$ . Then $h \in S (e, h) \cap S (h, f)$ and hence

\begin{array}{rcl} e ψ \cdot h ψ \cdot f ψ & = & \bar{ϵ (e, e) \circ ϵ (h τ^{l} (e), \circ ϵ (h, h τ^{r} (f))) \circ ϵ (f, f)} \\ = & ϵ (e, e) \cdot \bar{ϵ} (f, f) . \end{array}

Then by (4.4) $h ψ \in S (e ψ, f ψ)$ .

Hence $ψ$ is a bimorphism and similarly $ψ^{- 1}$ is also a bimorphism. Thus $ψ$ is an isomorphism.

Let $α \in T$ and $e' \in ω (e_{α})$ . Then

\begin{array}{rcl} (e' ψ) a (\bar{α}, \bar{α^{- 1}}) & = & \bar{α^{- 1}} \cdot e' ψ \cdot \bar{α} \\ = & \bar{α^{- 1} \circ ϵ (e', e') \circ α} . \end{array}

Now

α^{- 1} \circ ϵ (e', e') \circ α = ϵ (e' α, e' α)

and therefore

(e' ψ) a (\bar{α}, \bar{α^{- 1}}) = (e' α) ψ .

This proves (5.14). □

Remark 5.10: We have already noted that given any regular semigroup $S$ we can associate with $S$ a partially transitive set $T_{s}$ on $E (S)$ (of Theorem 4.12). For any $T \in P T (E), \bar{T}$ is a regular semigroup and hence we can define a partially transitive set $T_{\bar{T}}$ on $E (\bar{T})$ by (4.13). It will be of interest to find out the relation between $T$ and $T_{\bar{T}}$ . In fact the mapping

α \to a (\bar{α}, \bar{α^{- 1}})

(5.15)

is a bijection of $T$ onto $T_{\bar{T}}$ which reduces to identity mapping when $E$ and $E (\bar{T})$ are identified by $ψ$ .

That (5.15) defines a mapping of $T$ into $T_{\bar{T}}$ , is clear. Now suppose that $x \in \bar{T}$ and $x' \in i (x)$ . Let $e ψ = x \cdot x', f ψ = x' \cdot x$ so that $x \in H_{e ψ, f ψ}$ and $x' \in H_{f ψ, e ψ}$ . By Lemma 5.8, there exists $α \in T (e, f)$ such that $\bar{α} = x$ . Then $\bar{α^{- 1}}$ is an inverse of $x$ in $H_{f ψ, e ψ}$ and so $\bar{α^{- 1}} = x'$ . Thus

a (\bar{α}, \bar{α^{- 1}}) = a (x, x')

and this proves that the mapping defined by (5.15) is onto. Now if for some $x \in \bar{T}$ and $x' \in i (x)$ , we have

a ({\bar{α}}_{1}, α_{1}^{- 1}) = a (x, x')

where $α_{1} \in T (e_{1}, f_{1})$ , then $\bar{α}, ℋ x, \bar{α_{1}^{- 1}}, ℋ x'$ and therefore there exists $α' \in T (e_{1}, f_{1})$ such that

\bar{α'} = x, \bar{α_{1}^{- 1}} = x' .

However, for all $e' \in ω (e_{1})$ , by (5.14)

\begin{array}{rcl} (e' α_{1}) ψ & = & (e' ψ) a (\bar{α_{1}}, \bar{α_{1}^{- 1}}) \\ = & (e' ψ) a (\bar{α'}, \bar{α_{1}^{- 1}}) \\ = & (e' α') ψ . \end{array}

Thus $e' α_{1} = e' α'$ and this being true for all $e' \in ω (e_{1})$ we conclude that $α' = α_{1}$ . Hence the mapping defined by (5.15) is also one-to-one.

Lemma 5.11. $\bar{T}$ is a fundamental regular semigroup.

Proof. For $x, y \in \bar{T}$ , by Theorem 4.13,

x μ (\bar{T}) y \Leftrightarrow a (x, x') = a (y, y')

for some $x' \in i (x)$ and $y' \in i (y)$ . $\bar{T}$ is fundamental if we establish that

x, y \in \bar{T}, a (x, x') = a (y, y') \Rightarrow x = y .

(5.16)

Suppose that $x, y \in \bar{T}, x' \in i (x), y' \in i (y)$ and $a (x, x') = a (y, y')$ . Choose $α, β \in T$ so that $\bar{α} = x, \bar{α^{- 1}} = x', \bar{β} = y$ and $\bar{β^{- 1}} = y'$ .

Then

a (\bar{α}, \bar{α^{- 1}}) = a (\bar{β}, \bar{β^{- 1}}) .

Hence by Remark 5.10 $α = β$ and consequently

x = \bar{α} = \bar{β} = y .

Thus we have established the first part of Theorem 5.1 (i.e., for all $T \in P T (E), \bar{T}$ under the operation defined by (5.8), is a fundamental regular semigroup and $E$ is isomorphic to $E (\bar{T})$ ). We obtain the converse part of Theorem 5.1 from the following theorem. □

Theorem 5.12. Let $S$ be a regular semigroup and $T_{S}$ denote the partially transitive set on $E = E (S)$ defined by (4.13). Then for $x \in S$ and $x' \in i (x)$

x ϕ_{S} = \bar{a (x, x')}

(5.17)

defines a homomorphism of $S$ onto $T_{S}$ such that

ϕ_{S} \circ ϕ_{S}^{- 1} = μ (S) .

(5.18)

Proof. It follows from the definition of $σ$ and Lemma 4.11 (c), that for all $x \in S$ and $x', x'' \in i (x)$ ,

a (x, x') σ a (x, x'')

and thus $ϕ_{s}$ defines a mapping of $S$ into $T_{S}$ . For $x, y \in S$ , $h \in S (x, y), x' \in i (x)$ and $y' \in i (y)$ we have by Theorem (4.8) $y' h x' \in i (x y)$ . Therefore by Lemma 4.11 (b) and (d),

\begin{array}{rcl} a (x y, y' h x') & = & a (x h, h x') \circ a (h y, y' h) \\ = & a (x, x') \circ τ^{l} (h τ^{l} (f), h) \circ τ^{r} (h, h τ^{r} (e)) \circ a (y y') \\ = & a (x, x') \land h \circ h \land a (y, y') \end{array}

where $e = x' x$ and $f = y y'$ . By (5.8)

\bar{a (x, x') \land h \circ h \land a (y, y')} = \bar{a (x, x')} \cdot \bar{a (y, y')}

and so

\begin{array}{rcl} (x \cdot y) ϕ_{s} & = & \bar{a (x y, y' h x')} \\ = & \bar{a (x, x')} \cdot \bar{a (y, y')} \\ = & x ϕ_{s} \cdot y ϕ_{s} . \end{array}

Now suppose that $x μ (s) y$ . Then by the definition of $μ (s)$ for some $x' \in i (x)$ , $y' \in i (y)$

a (x, x') = a (y, y')

and thus

x ϕ_{s} = y ϕ_{s} .

On the other hand if $x ϕ_{s} = y ϕ_{s}$ , then for some $x' \in i (x), y' \in i (y)$

a (x, x') σ a (y, y') .

Thus by (5.5)

\begin{array}{rcl} x x' & = & e \overset{r}{\approx} e_{1} = y y' \\ x' x & = & f \overset{l}{\approx} f_{1} = y' y . \end{array}

In other words $x ℋ y$ and for $y'' = i (y, f, e)$ ,

a (x, x'), a (y, y'') \in T_{s} (e, f)

and

a (x, x') σ a (y, y'') .

We now conclude that

a (x, x') = a (y, y'');

and thus $x μ (s) y$ .

Proof of the theorem is now complete. □

For any biordered set $E$ , since $P T (E) \neq \emptyset$ , we have the following corollary:

Corollary 5.13. Every biordered set is isomorphic to the biordered set of idempotents of some regular semigroup.

Theorem 5.14. If

ℱ (E) = {\bar{T} | T \in P T (E)}

(5.19)

then $ℱ (E)$ is a complete lattice under inclusion relation and the mapping

T \to \bar{T}

(5.20)

is a lattice isomorphism of $P T (E)$ onto $ℱ (E)$ .

Proof. The mapping defined by (5.20) is an order isomorphism of $P T (E)$ onto $ℱ (E)$ by (5.7) and hence the result follows from this and Theorem 3.5. □

Remark 5.15: A different construction for the greatest and the least members of the lattice $ℱ (E)$ is given by Hall [5].

Remark 5.16: If $S$ is a regular semigroup, the operation of $S$ induces a biordered structure on $E (S)$ (cf. Theorem 4.5). By 5.13, the structure of any biordered set $E$ may be registered as having been induced by the operation of a regular semigroup $S$ that contains $E$ isomorphically as its idempotent set. It is clear that the operation of $S$ induces a partial operation of $E$ and the partial groupoid so formed determines the structure of $E$ (as a biordered set). But there are in general more than one such partial groupoids on $E$ that determine the same biorder structure on $E$ .

For example let $G$ be any group, $a_{i j} (i, j = 1, 2)$ be distinct elements of $G$ and e be its identity. Suppose that

P = [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}], P' = [\begin{matrix} e & e \\ e & e \end{matrix}] .

Then

S_{1} = M (G, 2, 2; P), S'_{1} = M (G, 2, 2; P')

(where 2 denotes the index set {1, 2}) are two completely simple semigroups such that $E (S_{1})$ and $E (S'_{1})$ are isomorphic as biordered sets but are different as partial groupoids. In fact, as partial groupoids $E (S'_{1})$ is a band, but $E (S_{1})$ may not.

It is natural to define a biordered set E as a band if ${\bar{Ẽ}}_{0}$ is a band semigroup. Also $E$ is a band in $S$ if the partial groupoid $E$ is a band semigroup. Clearly if $E$ is a band in $S$ it is a band (as a biordered set). But as the previous example shows, the converse of this is not true.

The structure of band semigroup was determined by McLean [8], and of those semigroups whose biordered set of idempotents is a band in that semigroup was determined by Yamada [13] (Yamada [13] calls such semigroups as strictly regular semigroups). The following theorem characterizes bands in the class of all biordered sets and strictly regular semigroups in terms of generalised Brandt groupoids (cf [14]).

Theorem 5.17. Let $E$ be a biordered set. Then $E$ is a band if and only if the following conditions are satisfied.

e, f ϵ E, e^{r} \cap f^{l} \neq \emptyset \Rightarrow e^{l} \cap f^{r} \neq \emptyset

(i)

where $e^{r} (e^{l})$ denotes the $\overset{r}{\approx}$ ( $\overset{l}{\approx}$ )-class of $E$ containing e.

ϵ, ϵ' \in Ẽ_{0} (e, f) \Rightarrow ϵ = ϵ' .

(ii)

If $S$ is a regular semigroup, $E (S)$ is a band in $S$ if and only if the trace (cf. p. 92 of [2]) of each $D$ -class of $S$ is isomorphic to a generalized Brandt semigroup.

Proof. Suppose that the biordered set $E$ satisfies the hypothesis and let $e δ_{0} f$ . Then by Theorem 3.3, there exists a $δ_{0}$ -chain $c = (e_{0}, e_{1}, \dots, e_{n})$ such that $e_{0} = e, f_{n} = f$ . We may assume that for $i = 1, 2, \dots, n - 1$ , the elementary factors ( $e_{i - 1}, e_{i}$ ) and ( $e_{i}, e_{i + 1}$ ) of are not similar. Let $e_{0} \overset{r}{\approx} e_{1}$ . Then by hypothesis

e_{0} \overset{r}{\approx} e_{1} \overset{l}{\approx} e_{2} \overset{r}{\approx} e_{3} .

Hence there exists $e_{1} \in e_{0}^{l} \cap e_{2}^{r}$ such that

e_{0} \overset{l}{\approx} e'_{1} \overset{r}{\approx} e_{3}

and again by hypothesis we conclude that there exists $e {''}_{1}$ such that

e_{0} \overset{r}{\approx} e {''}_{1} \overset{l}{\approx} e_{3} .

Continuing this process, we obtain by induction that there exist $g, g' \in E$ such that

e = e_{0} \overset{r}{\approx} g \overset{l}{\approx} e_{n} = f

and

e \overset{l}{\approx} g' \overset{r}{\approx} f .

Now suppose that $ϵ (c) \in Ẽ_{0} (e, f)$ . Then by the definition of $σ$ , and (ii),

ϵ (c) σ ϵ (g, g), ϵ (c^{- 1}) σ ϵ (g', g') .

Thus

\bar{ϵ (c)} = \bar{ϵ (g, g)}, \bar{ϵ (c^{- 1})} = \bar{ϵ (g'), g'} .

Hence by Lemma 5.9, $\bar{ϵ (c)}, \bar{ϵ (c^{- 1})} \in E ({\bar{Ẽ}}_{0})$ . It follows that ${\bar{Ẽ}}_{0}$ is a band semigroup, and hence $E$ is a band.

Conversely let $E$ be a band. Then ${\bar{Ẽ}}_{0}$ is a band semigroup. Let $e, f \in E, e^{r} \cap f^{l} \neq \emptyset$ . Then there exists $g \in E$ such that $e \overset{r}{\approx} g \overset{l}{\approx} f$ and therefore $x = \bar{τ^{r} (e, g) \circ τ^{l} (g, f)} \in {\bar{Ẽ}}_{0}$ . Clearly

{\tilde{E}}_{0} (e, f) = {τ^{r} (e, g) \circ τ^{l} (g, f)} .

Further,

i (x; f, e) = \bar{τ^{l} (f, g) \circ τ^{r} (g, e)}

and this is an idempotent. Hence by Lemma 5.9, there exists $g' \in E$ such that

τ^{l} (f, g) \circ τ^{r} (g, e) σ \in (y', y'),

which implies that $f \overset{r}{\approx} g' \overset{l}{\approx} e$

Let $S$ be a regular semigroup such that $E (S)$ is a band in $S$ and $D$ be a $D$ -class of $S$ . Fix a maximal subgroup $H_{e, e}$ of $S$ contained in $D$ . We shall show that it is possible to choose elements

r_{e f} \in H_{e, f}, r_{f e} \in H_{f, e}, f \in E (D),

such that $r_{e f} = r_{e f'}$ if and only if $f \overset{l}{\approx} f' r_{f e} = r_{f' e}$ if and only if $f \overset{r}{\approx} f'$ and $r_{e f} \in i (r_{f e})$

For this let ${e_{α} | α \in E (D) | δ_{0}}$ be a subset of $E (D)$ such that for any $f \in E (D), e \in δ_{0} (f)$ . Choose $r_{e, e_{α}} \in H_{e, e_{α}}$ and let $r_{e_{α}, e}$ be its inverse in $H_{e_{α}, e}$ . For each $f \in δ (e_{α})$ define

r_{e f} = r_{e e_{α}} f .

Clearly $r_{e f} \in H_{e, f}$ and $r_{e f} = r_{e e_{α}}$ if and only if $f \overset{l}{\approx} e_{α}$ . Also if $f \overset{l}{\approx} f'$

r_{e f e_{α}} = r_{e e_{α}} = r_{e f' e_{α}}

(since $f e_{α}$ is an idempotent $ℒ$ -equivalent to $e_{α}$ , $r_{e f} e_{α} = r_{e e_{α}}$ ). Since the mapping $x \to x e_{α}$ of $L_{f}$ onto $L_{e_{α}}$ is one-to-one, we have $r_{e f} = r_{e f'}$ . Thus we have chosen $r_{e f} \in H_{e, f}$ for all $f \in E (D)$ . In a similar way we can choose $r_{f e} \in H_{f, e}$ for all $f \in E (D)$ also by defining $r_{f e} = f r_{e_{α} e}$ if $f \in δ_{0} (e_{α})$ . Then we can show that $r_{f e} = r_{f' e}$ if and only if $f \overset{r}{\approx} f'$ .

Now let $f \in E (D)$ . Then

\begin{array}{rcl} r_{g e} r_{e f} r_{f e} & = & f r_{e_{α} e} r_{e e_{α}} f r_{e_{α} e} \\ = & f e_{α} f r_{e_{α} e} . \end{array}

Now $f e_{α} \overset{r}{\approx} f$ and hence $f e_{α} f = f$ . Therefore

r_{f e} r_{e f} r_{f e} = r_{f e} .

Similarly

r_{e f} r_{f e} r_{e f} = r_{e f} .

Hence

r_{e f} \in i (r_{f e}) .

It is clear that if we define the matrix

P^{e} - (P_{λ i}^{e} | λ \in E (D) ∕ \overset{l}{\approx}, i \in E (D) ∕ \overset{r}{\approx})

P_{λ i}^{e} = \{\begin{matrix} r_{f e} \cdot r_{e f} & if r_{f e} \cdot r_{e f'} \in H_{e, e} \\ 0 & otherwise \end{matrix}

where $λ$ is the $\overset{l}{\approx}$ -class of $E (D)$ containing $f$ and $i$ is the $\overset{r}{\approx}$ -class of $E (D)$ containing $f'$ , then every nonzero entry of $P^{e}$ is the identity of $H_{e, e}$ . Using the fact that $f δ_{0} f'$ if and only if there exist $f'', f'' ϵ E (D)$ such that

f \overset{r}{\approx} f'' \overset{l}{\approx} f', f \overset{l}{\approx} f {''}_{1} \overset{r}{\approx} f .

We can prove that the matrix constructed above is the $(E (D) ∕ \overset{l}{\approx}) \times (E (D) ∕ \overset{r}{\approx})$ semi-identity matrix of type ${(α ∕ \overset{l}{\approx}) \times (α ∕ \overset{r}{\approx})}$ (cf. Yamada [14]).

Now by Theorem 3.4 of [2], the trace of $D$ is isomorphic to

M^{\circ} (H_{e, e}, E (D) ∕ \overset{r}{\approx}, E (D) ∕ \overset{l}{\approx}; P^{e})

and this is a generalized Brandt semigroup by lemma 5 of [14].

Suppose now that trace of each $D$ -class of $S$ is a generalized Brandt semigroup. Let $e, f \in E (S)$ . Then for any $h \in S (e, f), e h, h, h f \in E (S)$ and $e h \overset{l}{\approx} h \overset{r}{\approx} h f$ . Hence the product $(e h) (h f) = e f$ is not zero in the trace of $D_{h}$ . Since the set of idempotents of the trace of $D_{h}$ is a band, it follows that $e f$ is a nonzero idempotent of the trace of $D_{h}$ and hence $e f \in E (S)$ . Hence $E (S)$ is a band. This completes the proof of Theorem 5.17.

If $E (S)$ is a semilattice (i.e., $ω^{r} = ω^{l}$ ) it can be easily seen that it is a semilattice in $S$ (i.e., a commutative subsemigroup of $S$ ). Hence in this case the statement that $E (S)$ is a semilattice is equivalent to the statement that $E (S)$ is a semilattice in $S$ . □

Remark 5.18:

When $E$ is a semilattice, it is clear that $σ$ defined in $Ẽ_{1}$ by (5.5) reduces to the identity relation. Moreover, the operation defined by (5.8) is then equivalent to the usual composition of relations. Therefore $Ẽ_{1}$ is an inverse semigroup whose semilattice is isomorphic to $E$ . Also, any partially transitive set $T$ on $E$ is a subsemigroup of $Ẽ_{1}$ since axiom ( $T_{3}$ ) in this case is equivalent to the fact that $α, β \in T$ implies $α \circ β \in T$ . Hence every fundamental inverse semigroup whose semilattice is isomorphic to $E$ is isomorphic to a subsemigroup of $Ẽ_{1}$ .

The analogous results for bisimple inverse semigroup are due to W. D. Munn (cf. [11]).

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Chapter 5Structure of Fundamental Regular Semigroups

Chapter 5
Structure of Fundamental Regular Semigroups