# Semilattices of Stratified Extensions

James Renshaw
School of Mathematical Sciences
University of Southampton
Southampton, SO17 1BJ, England
ORCID: 0000-0002-5571-8007
<EMAIL_ADDRESS>

William Warhurst
School of Mathematical Sciences
University of Southampton
Southampton, SO17 1BJ, England
<EMAIL_ADDRESS>

(May 2023)

###### Abstract

In 1995 Grillet introduced the concept of a stratified semigroup as a kind of
generalisation of finite nilsemigroups. We extend these ideas here by allowing
a more general $\operatorname{Base}$ and describe them in terms of extensions
of semigroups by stratified semigroups. We consider semillatices of certain
types of group-bound semigroups and also semillatices of Clifford semigroups
and show how to describe them as semilattices of these stratified extensions
and provide a number of interesting examples.

Keywords Semigroup, stratified, extension, semilattice, group-bound, Clifford
semigroup.
Mathematics Subject Classification 2020: 20M10.

##  1 Introduction and Preliminaries

Grillet [2] defines a semigroup $S$ with zero to be stratified whenever
$\bigcap_{m>0}S^{m}=\\{0\\}$. A semigroup without zero is called stratified if
$S^{0}$ is stratified. He shows that this class of semigroups includes the
class of all free semigroups, free commutative semigroups, homogeneous
semigroups and nilpotent semigroups with finite index. Our aim is to
generalise this concept and consider some semigroups that can be decomposed as
semilattices of some of these more general kinds of stratified semigroups.

After some basic definitions and preliminary results, in section 2, we
introduce the concept of a stratified extension as a generalisation of
Grillet’s stratified semigroups, and we provide a number of interesting
results on the overall structure of such semigroups. The final two sections
then examine two families of semigroups that exhibit this stratified
structure. In section 3 our focus is on semigroups in which every regular
${\cal H}-$class contains an idempotent. We show that group-bound semigroups
with this property are semilattices of stratified extensions of completely
simple semigroups and describe the semilattice structure. Finally in section 4
we look at strict extensions of Clifford semigroups and show amongst other
things that strict stratified extensions of Clifford semigroups are
semilattices of stratified extensions of groups. For all terminology in
semigroups not otherwise defined see Howie ([4]).

Let $S$ and $T$ be semigroups, with $T$ containing a zero. A semigroup
$\Sigma$ is called an ideal extension of $S$ by $T$ if it contains $S$ as an
ideal and the Rees quotient $\Sigma/S$ is isomorphic to $T$. Grillet and
Petrich [3] define an extension as strict if every element of $\Sigma\setminus
S$ has the same action on $S$ as some element of $S$ and pure if no element of
$\Sigma\setminus S$ does. They also showed that any extension of an arbitrary
semigroup $S$ is a pure extension of a strict extension of $S$.

######  Proposition 1.1 ([3, Proposition 2.4])

Every extension of $S$ is strict if and only if $S$ has an identity.

Let $S$ and $T$ be disjoint semigroups. A partial homomorphism [1] from $T$ to
$S$ is a map $f:T\setminus\\{0\\}\to S$ such that for all $x,y\in
S,f(xy)=f(x)f(y)$ whenever $xy\neq 0$.

We adopt the convention used by Clifford and Preston ([1]) that elements of
$T\setminus\\{0\\}$ are denoted by capital letters and elements of $S$ by
lowercase letters. A partial homomorphism from $T\setminus\\{0\\}$ to $S$
given by $A\mapsto\overline{A}$ defines an extension $\Sigma=S\bigcup
T\setminus\\{0\\}$ with multiplication given by

  1. 1.

$A\ast B=\begin{cases}AB&AB\neq 0\\\
\overline{A}\;\overline{B}&AB=0\end{cases}$

  2. 2.

$A\ast s=\overline{A}s$

  3. 3.

$s\ast A=s\overline{A}$

  4. 4.

$s\ast t=st$

where $A,B\in T\setminus\\{0\\}$ and $s,t\in S$. From parts (2) and (3) above,
all extensions defined in this way are strict.

Let $S$ be a semigroup and let $a,b\in S$. We say that $a$ and $b$ are
interchangeable if

$\forall x\in S,ax=bx\text{ and }xa=xb.$

A semigroup is called weakly reductive if it contains no interchangeable
elements. Notice that every monoid is weakly reductive.

######  Theorem 1.2 ([3, Theorem 2.5])

Let $S$ be weakly reductive. Then every strict extension of $S$ is determined
by a partial homomorphism, and conversely.

Recall that a semigroup is said to be $E-$dense (or $E-$inversive) if for all
$s\in S$ there exists $t\in S$ such that $st\in E(S)$. The following is well-
known

###### Lemma 1.3

The following are equivalent

  1. 1.

$S$ is $E-$dense,

  2. 2.

for all $s\in S$ there exists $t\in S$ such that $ts\in E(S)$,

  3. 3.

for all $s\in S$ there exists $t\in S$ such that $st,ts\in E(S)$,

  4. 4.

for all $s\in S$ there exists $s^{\prime}\in S$ such that
$s^{\prime}ss^{\prime}=s^{\prime}$.

Such an element, $s^{\prime}$, in (4) is called a weak inverse of $s$ and the
set of all weak inverses of $s$ is denoted by $W(s)$. The set of all weak
inverses of elements of $S$ is denoted by $W(S)$. Note that
$W(S)=\operatorname{Reg}(S)$. It is easily shown that for all $s\in
S,s^{\prime}\in W(s)$, $ss^{\prime},s^{\prime}s\in E(S)$ and
$ss^{\prime}\mathcal{L}s^{\prime}\mathcal{R}s^{\prime}s$.

###### Lemma 1.4

Let $S$ be a semigroup.

  1. 1.

If $s,t\in S$ then $W(st)\subseteq W(t)W(s)$.

  2. 2.

If $S$ is an $E-$dense semigroup and $s^{\prime}\in W(s)$ then
$J_{s^{\prime}}\leq J_{s}$.

  * Proof.

These are fairly straightforward.

    1. 1.

Let $(st)^{\prime}\in W(st)$. Then
$(st)^{\prime}=(st)^{\prime}st(st)^{\prime}$ and so
$t(st)^{\prime}=t(st)^{\prime}st(st)^{\prime}$ and hence $t(st)^{\prime}\in
W(s)$. Similarly $(st)^{\prime}s\in W(t)$. Then
$(st)^{\prime}=((st)^{\prime}s)(t(st)^{\prime})\in W(t)W(s)$.

    2. 2.

Since $s^{\prime}\in W(s)$, $s^{\prime}=s^{\prime}ss^{\prime}$ and so
$J_{s^{\prime}}=J_{s^{\prime}ss^{\prime}}$. By [4, Equation 2.1.4] we have
$J_{s^{\prime}}=J_{s^{\prime}ss^{\prime}}\leq J_{s}$.

Note that it is well known that if $E(S)$ forms a band then $W(st)=W(t)W(s)$.

An element $s$ in a semigroup $S$ is called eventually regular if there exists
$n\geq 1$ such that $s^{n}$ is regular. A semigroup is eventually regular if
all of its elements are eventually regular. It is clear that eventually
regular semigroups are $E-$dense. A semigroup $S$ is called group-bound if for
every $s\in S$, there exists $n\geq 1$ such that $s^{n}$ lies in a subgroup of
$S$. Clearly group-bound semigroups are eventually regular. If $S$ is
eventually regular and each regular ${\cal H}-$class is a group then $S$ is
group-bound.

A semigroup $S$ is called Archimedean if for any $a,b\in S$ there exists
$n\in\hbox{\Bbb N}$ such that $a^{n}\in SbS$.

######  Theorem 1.5 ([5, Theorem 3])

Let $S$ be a group-bound semigroup. Then $S$ is a semillatice of Archimedean
semigroups if and only if every regular ${\cal H}-$class of $S$ is a group.

Let $S$ be a semigroup with $0$. We say that an element $x\in S$, is nilpotent
if there is $n\in\hbox{\Bbb N}$ such that $x^{n}=0$. The semigroup $S$ is
called a nilsemigroup if every element of $S$ is nilpotent. The semigroup $S$
is called nilpotent with index $n\in\hbox{\Bbb N}$ if $S^{n}=\\{0\\}$.

##  2 Stratified Extensions

Let $S$ be a semigroup (not necessarily stratified) and define the base of $S$
to be the subset $\operatorname{Base}(S)=\bigcap_{m>0}S^{m}$. We shall say
that a semigroup $S$ is a stratified extension of $\operatorname{Base}(S)$ if
$\operatorname{Base}(S)\neq\emptyset$. The reason for this name will become
apparent later. Clearly $\operatorname{Base}(S)$ is a subsemigroup of $S$.
When $\operatorname{Base}(S)$ is a trivial subgroup then $S$ is a stratified
semigroup. A stratified semigroup $S$ is not in general a stratified extension
as we may have $\operatorname{Base}(S)=\emptyset$, however if $S$ is a
stratified semigroup then $S^{0}$ is also stratified and is a stratified
extension with trivial base. Further, $S$ is called a finitely stratified
extension if there exists $m\in\mathbb{N}$ such that
$S^{m}=S^{m+1}=\operatorname{Base}(S)$. The smallest such $m$ is called the
height of $S$ and where necessary we shall refer to $S$ as a finitely
stratified extension with height $m$. If for every $s$ in $S$ there is an
$m\in\hbox{\Bbb N}$ such that $s^{m}\in\operatorname{Base}(S)$ then $S$ is a
nil-stratified extension. All finitely stratified extensions are nil-
stratified extensions, but it is easy to demonstrate that not all nil-
stratified extensions are finitely stratified extensions.

A finitely stratified extension is a stratified extension over the same base,
since $S^{m}=S^{m+1}$ implies $S^{n}=S^{m}$ for all $n\geq m$ and so
$\bigcap_{k>0}S^{k}=S^{m}$, where $m$ is the height of $S$. The converse is
not true since, for example, if $S$ is a free semigroup with a zero adjoined,
then $S$ is a stratified extension with trivial base but not a finitely
stratified extension. It is clear that a (finitely) stratified extension has a
unique base.

Clearly, for all $m\geq 1$, $S^{m+1}\subseteq S^{m}$ and so we define the
layers of $S$ as the sets $S_{m}=S^{m}\setminus S^{m+1}$, $m\geq 1$. Every
element of $S\setminus\operatorname{Base}(S)$ lies in exactly one layer, and
if $s\in S_{m}$ then $m$ is the depth of $s$. The layer $S_{1}$ generates
every element of $S\setminus\operatorname{Base}(S)$ and is contained in any
generating set of $S$. However $\operatorname{Base}(S)\not\subseteq\langle
S_{1}\rangle$ in general. For example, let $S$ be a semigroup with $0$, with
no zero divisors. Then $0\in\operatorname{Base}(S)$ but $0\not\in\langle
S_{1}\rangle$.

Since $\operatorname{Base}(S)\subseteq S^{m}$ for any $m\in\mathbb{N}$, we
have an alternative characterisation for the elements of
$\operatorname{Base}(S)$. Any $s\in S$ lies in $\operatorname{Base}(S)$ if and
only if $s$ can be factored into a product of $m$ elements for any
$m\in\mathbb{N}$, i.e. $s=a_{1}a_{2}\dots a_{m}$ for some $a_{i}\in S$. This
characterisation gives us some immediate properties of
$\operatorname{Base}(S)$ as a subsemigroup of $S$.

###### Lemma 2.1

Let $S$ be a semigroup and let $s\in S$. If $s\in Ss\cup sS\cup SsS$ then
$s\in\operatorname{Base}(S)$.

  * Proof.

It follows that for $m\geq 1$, $s=x^{m}s$ or $s=sy^{m}$ or $s=x^{m}sy^{m}$ and
so the result follows from the previous observation.

###### Corollary 2.2

Suppose that $S$ is a semigroup.

  1. 1.

Every monoid subsemigroup of $S$ is a submonoid of $\operatorname{Base}(S)$.

  2. 2.

$\operatorname{Reg}(S)\subseteq\operatorname{Base}(S)$. Hence if $S$ is
regular, $\operatorname{Base}(S)=S$.

  3. 3.

$E(S)=E(\operatorname{Base}(S))$.

  4. 4.

If $s\in S\setminus\operatorname{Base}(S)$ then $|J_{s}|=1$, where $J_{s}$ is
the ${\mathcal{J}}-$class of $s$.

To see (4) notice that if $a\mathcal{J}b$ and $a\neq b$ then we have $a=ubv$
for some $u,v\in S^{1}$ and since $a\neq b$ we have $u$ and $v$ not both equal
to $1$. Similarly $b=sat$ with $s,t\in S^{1}$ not both equal to $1$ and hence
$a\in Sa\cup aS\cup SaS$. The converse is not true, since for example in a
semigroup with zero we have $J_{0}=\\{0\\}$ but $0\in\operatorname{Base}(S)$.

If follows immediately that the class of stratified extensions contains the
class of semigroups with regular elements and hence in particular the classes
of monoids, finite semigroups and regular semigroups. However, not every
semigroup is a stratified extension. Consider for example a semigroup with a
length function (i.e. a function $l:S\to\hbox{\Bbb N}$ such that for all
$x,y\in S,l(xy)=l(x)+l(y)$). If $T$ is the subsemigroup of elements with non-
zero length, then the elements of $T^{m}$ each have length at least $m$. Hence
the elements of length exactly $m$ lie in
$T_{m}\not\subseteq\operatorname{Base}(T)$ and so the base is empty. In
particular, a free semigroup is not a stratified extension, nor is the
semigroup of polynomials of degree $\geq 1$ over any ring, under
multiplication.

###### Proposition 2.3

Let $S$ be a stratified extension. Then

  1. 1.

$\operatorname{Base}(S)$ is an ideal of $S$ and $S$ is an ideal extension of
$\operatorname{Base}(S)$ by a stratified semigroup with $0$.

  2. 2.

If $S$ is a nil-stratified extension then it is an ideal extension of
$\operatorname{Base}(S)$ by a nilsemigroup.

  3. 3.

If $S$ is a finitely stratified extension then it is an ideal extension of
$\operatorname{Base}(S)$ by a nilpotent semigroup of finite index.

  * Proof.

    1. 1.

For any $u,v\in S^{1}$, $t\in\operatorname{Base}(S)$ and $m>3$, we have $t\in
S^{m-2}$ so $utv\in S^{m}$ and hence $utv\in\operatorname{Base}(S)$. Hence we
can regard $S$ as being an ideal extension of $\operatorname{Base}(S)$ by
$S/\operatorname{Base}(S)$ and note that $S/\operatorname{Base}(S)$ is a
stratified semigroup with $0$.

    2. 2.

If $S$ is a nil-stratified extension then it follows that for every $s\in S$
there exists $m\in\hbox{\Bbb N}$ such that $s^{m}\in\operatorname{Base}(S)$.
Hence in the Rees quotient $S/\operatorname{Base}(S)$, $s^{m}=0$ and so
$S/\operatorname{Base}(S)$ is a nilsemigroup and $S$ is an ideal extension by
a stratified nilsemigroup.

    3. 3.

Recall that the nilpotency index of a semigroup is the smallest value $m$ such
that every product of $m$ elements is zero. It is easy to see that if the
nilpotency index of $S/\operatorname{Base}(S)$ is $m$ then the height of $S$
is $m$ and so $S$ is a finitely stratified extension. Conversely, any
nilpotent semigroup $S$ of finite nilpotency index $m$ is a stratified
semigroup with $S^{m}=\\{0\\}$. We have hence proved the following.

The converses of these results do not hold. To see this, let $S$ be a free
semigroup and $T$ be the two element nilsemigroup. Then $T$ is a stratified
semigroup with $0$ but an extension of $S$ by $T$ is not a stratified
extension. Further, $T$ is a nilpotent semigroup of finite index and an
extension of $S^{0}$ by $T$ is a stratified extension, but is not a finitely
stratified nor nil-stratified extension.

###### Proposition 2.4

Let $S$ be a stratified extension.

  1. 1.

If $S$ is a nil-stratified extension then $\operatorname{Base}(S)$ is periodic
if and only if $S$ is periodic;

  2. 2.

If $S$ is a nil-stratified extension then $\operatorname{Base}(S)$ is
eventually regular if and only if $S$ is eventually regular;

  3. 3.

$\operatorname{Base}(S)$ is $E-$dense if and only if $S$ is $E-$dense.

So a stratified extension with a periodic base is $E-$dense.

  * Proof.

If $S$ is either periodic or eventually regular, then so is any subsemigroup
and in particular $\operatorname{Base}(S)$. Conversely, if
$\operatorname{Base}(S)$ is periodic (resp. eventually regular) and $s\in S$,
then since $S$ is nil-stratified it follows that there exists $m\in\hbox{\Bbb
N}$ such that $s^{m}\in\operatorname{Base}(S)$ and so there exists
$n\in\hbox{\Bbb N}$ such that $s^{mn}$ is idempotent (resp. regular) and so
$\operatorname{Base}(S)$ is periodic (resp. eventually regular).

For the third, let $\operatorname{Base}(S)$ be $E-$dense and let $s\in S$.
Then for any $t\in\operatorname{Base}(S),ts\in\operatorname{Base}(S)$ and so
there exists $u\in\operatorname{Base}(S)$ such that $uts\in
E(\operatorname{Base}(S))=E(S)$ and so $S$ is $E-$dense. Conversely suppose
that $S$ is $E-$dense. Since
$W(S)=\operatorname{Reg}(S)\subseteq\operatorname{Base}(S)$, then
$\operatorname{Base}(S)$ is $E-$dense.

Notice that periodic $\Rightarrow$ eventually regular $\Rightarrow$ $E-$dense
$\Rightarrow$ stratified extension.

There is in general little control over the base as a stratified extension can
be constructed with any given semigroup as its base.

###### Proposition 2.5

Let $T$ and $R$ be any semigroups.Then there exists a stratified extension $S$
such that $T\subseteq\operatorname{Base}(S)$ and $S/T\cong R$. Moreover, if
$R$ is stratified without a zero then $T=\operatorname{Base}(S)$.

  * Proof.

Let $S=R\dot{\cup}T$ and define a binary operation $\ast$ on $S$ by $r_{1}\ast
r_{2}=r_{1}r_{2}$ for $r_{1},r_{2}\in R$, $t_{1}\ast t_{2}=t_{1}t_{2}$ for
$t_{1},t_{2}\in T$, and $r\ast t=t\ast r=t$ for $r\in R$ and $t\in T$. It is
easy to verify that this operation is associative and so $(S,\ast)$ is a
semigroup. Then $T\subseteq\bigcap_{m>0}(R\dot{\cup}T)^{m}$ and so $S$ is a
stratified extension. Moreover, if we choose $R$ such that
$\bigcap_{m>0}R^{m}=\emptyset$, for example $R=A^{+}$, a free semigroup, we
see that $\bigcap_{m>0}(R\dot{\cup}T)^{m}=T$ and so we can obtain a stratified
extension with base $T$.

In contrast, the possible bases for a finitely stratified extension are much
more restricted. Let $S$ be a finitely stratified extension with
$T=\operatorname{Base}(S)$ and consider $T^{2}$. There exists $m\in\mathbb{N}$
such that $T=S^{m}$, so $T^{2}=S^{2m}$. But by definition
$S^{m}=S^{m+1}=S^{m+2}=\dots=S^{2m}$ and so $T^{2}=T$. A semigroup $T$
satisfying $T^{2}=T$ is said to be globally idempotent and so the base of a
finitely stratified extension is globally idempotent. Note also that if $S$ is
globally idempotent, then $S$ is a finitely stratified extension in a trivial
sense, with base $S$ and height 1.

###### Proposition 2.6

A semigroup $\Sigma$ is a finitely stratified extension if and only if it is
an ideal extension of a globally idempotent semigroup by a nilpotent semigroup
of finite index.

  * Proof.

We need only justify the converse. Let $\Sigma$ be an ideal extension of a
globally idempotent semigroup $S$ by a nilpotent semigroup $T$ of finite index
$m$. Then $\Sigma^{m}=S$ and $S=S^{2}$ so $\Sigma^{m}=\Sigma^{2m}$ and as each
$\Sigma^{i}\subseteq\Sigma^{i+1}$ it follows that $\Sigma^{m}=\Sigma^{m+1}$.
Hence $\Sigma$ is a finitely stratified extension with base $S$ and height
$m$.

This is still a very broad class of semigroup, including among its members
every monoid and every regular semigroup. It should also be noted that a
globally idempotent semigroup need not contain idempotents, the Baer-Levi
semigroup being one such example.

###### Proposition 2.7

There exists a finitely stratified extension of height $h$, for any
$h\in\hbox{\Bbb N}$.

  * Proof.

Let $S=\langle a\rangle$ be the monogenic semigroup of index $h$ and period
$r$ and let $G=\\{a^{h},\ldots,a^{h+r-1}\\}$ be the kernel of $S$, a finite
cyclic group of order $r$. For any $x=a^{n_{1}}\ldots a^{n_{h}}\in S^{h}$ it
follows that $x=a^{n_{1}+\ldots+n_{h}}$ and $n_{1}+\ldots+n_{h}\geq h$ so that
$S^{h}\subseteq G$. However, since $a^{l}\in S^{l}$ then $S^{l}\not\subseteq
G$ for any $l<h$. Since $S^{h+r-1}\subseteq S^{h}$, $S^{h+1}\subseteq S^{h}$
and $G\subseteq S^{h+r-1}$ then $S^{h}=S^{h+1}=G$ and $S$ is a finitely
stratified extension with base $G$ and height $h$.

###### Theorem 2.8

Let $S$ be a (finitely, nil-) stratified extension and let $S_{i}$ for $i\in
I$ be a family of stratified extensions.

  1. 1.

If $\rho$ is a congruence on $S$ then $S/\rho$ is a (finitely, nil-)
stratified extension with base $\operatorname{Base}(S)/\rho$;

  2. 2.

the direct product $\prod_{i\in I}S_{i}$ is a stratified extension with base
$\prod_{i\in I}\operatorname{Base}(S_{i})$;

  3. 3.

if $|I|<\infty$ and each $S_{i}$ is a (finitely, nil-) stratified extension
then the direct product $\prod_{i\in I}S_{i}$ is a (finitely, nil-) stratified
extension with base $\prod_{i\in I}\operatorname{Base}(S_{i})$.

  * Proof.

    1. 1.

Let $S$ be a semigroup and let $\rho$ be a congruence on $S$. It is easy to
see that for any $m\in\hbox{\Bbb N}$ we have $(S/\rho)^{m}=S^{m}/\rho$. Hence
if $S$ is a stratified extension then $S/\rho$ is also a stratified extension,
with $\operatorname{Base}(S/\rho)=\operatorname{Base}(S)/\rho$. Further, if
$S$ is a finitely stratified or nil-stratified extension then so is $S/\rho$.

    2. 2.

Let $S_{i}$ be a family of semigroups. Then $(\prod_{i\in
I}S_{i})^{m}=\prod_{i\in I}{S_{i}}^{m}$. Hence if each $S_{i}$ is a stratified
extension, the product $\prod_{i\in I}S_{i}$ is also a stratified extension
with $\operatorname{Base}(\prod_{i\in I}S_{i})=\prod_{i\in
I}\operatorname{Base}(S_{i})$.

    3. 3.

If $I$ is a finite set and each $S_{i}$ is a nil-stratified extension then so
is $\prod_{i\in I}S_{i}$. Similarly if each $S_{i}$ is a finitely stratified
extension then so is $\prod_{i\in I}S_{i}$. To see that we cannot remove the
condition $|I|<\infty$, let $I=\hbox{\Bbb N}$ and for each $i\in I$ let
$S_{i}$ be a finitely stratified (and hence nil-stratified) extension of
height $i$. Then $\prod_{i\in I}S_{i}$ is a stratified extension but is
neither a finitely stratified extension nor a nil-stratified extension.

Subsemigroups of (finitely, nil-) stratified extensions are not necessarily
(finitely, nil-) stratified extensions. For example the bicyclic semigroup is
a finitely stratified extension (in fact globally idempotent) but contains
$(\mathbb{N},+)$ as a subsemigroup which is free and hence not even a
stratified extension. The class of (finitely, nil-) stratified semigroups
therefore does not form a variety.

Having considered direct products of stratified extensions, and in
anticipation of the families of semigroups in the final two sections, we
observe

###### Proposition 2.9

Let $S=\bigcup_{\alpha\in Y}S_{\alpha}$ be a semilattice of stratified
extensions. Then $S$ is a stratified extension.

  * Proof.

This follows easily on observing that $\bigcup_{\alpha\in
Y}\operatorname{Base}(S_{\alpha})\subseteq\bigcap_{m>0}S^{m}$.

In the case of finitely stratified extensions, we can construct a semilattice
of finitely stratified extensions which is not a finitely stratified
extension. Let $Y=\mathbb{N}\cup\\{0\\}$ be a semilattice under the
multiplication $ij=0$ for all $i,j\in Y$ with $i\neq j$. For each
$i\in\mathbb{N}$ let $S_{i}$ be a finitely stratified extension with height
$i$ and let $S_{0}$ be globally idempotent. Let $S$ be the union of each
$S_{i}$ as a semilattice of semigroups over $Y$. Then
$S^{m}=S_{0}\cup\bigcup_{i\in\mathbb{N}}{S_{i}}^{m}$. If $i>m$ then there are
elements in ${S_{i}}^{m}$ which are not in ${S_{i}}^{m+1}$ and so $S^{m}\neq
S^{m+1}$ for any $m\in\mathbb{N}$.

##  3 Semilattices of group bound semigroups

In this and the following section we explore two families of examples of
semigroups that can be decomposed as semilattices of stratified extensions. In
this section we introduce the definition of a conditionally completely regular
semigroup and, after some preliminary results, show that if such a semigroup
is also group-bound then it is a semilattice of stratified extensions.

A semigroup in which every regular $\cal H$-class contains an idempotent is
called a conditionally completely regular semigroup. This is clearly
equivalent to every regular ${\cal H}-$class being a group.

Let $S$ be a conditionally completely regular semigroup. We define a relation
$\rho$ on $S$ by $s\rho t$ if and only if for every $\cal D$-class $D$ of $S$
we have

$W(s)\cap D\neq\emptyset\iff W(t)\cap D\neq\emptyset.$

Clearly $\rho$ is an equivalence relation. We will show that $\rho$ is in fact
a congruence, and moreover that $S/\rho$ is a semilattice.

We begin by establishing some properties of such semigroups.

###### Lemma 3.1

Let $S$ be a conditionally completely regular semigroup.

  1. 1.

Every regular $\cal D$-class of $S$ is a completely simple subsemigroup of
$S$.

  2. 2.

Let $s\in S$ and $s^{\prime}\in W(s)$. Every $\cal H$-class of
$D_{s^{\prime}}$ contains a weak inverse of $s$.

  * Proof.

These are fairly straightforward.

    1. 1.

Let $D$ be a regular ${\cal D}-$class and let $a,b\in D$ and let $e$ be the
idempotent lying in $L_{a}\cap R_{b}$. Then $ab\mathcal{L}eb\mathcal{R}ee=e$.
Hence $ab\in D$ and $D$ is a completely simple subsemigroup of $S$.

    2. 2.

Let $D$ be the (regular) ${\cal D}-$class containing $s^{\prime}$ and let $r$
be an idempotent such that $r\mathcal{R}ss^{\prime}$. Then $ss^{\prime}r=r$
and $rss^{\prime}=ss^{\prime}$, and it follows that $s^{\prime}r\in W(s)$ and
$s^{\prime}\mathcal{R}s^{\prime}r\mathcal{L}r$.

$s^{\prime}$$ss^{\prime}$$r$$s^{\prime}r$

Let $I=S/{\cal R}$ and $\Lambda=S/{\cal L}$ and as is normal denote the ${\cal
R}-$classes as $R_{i}$ ($i\in I$), the ${\cal L}-$classes as $L_{\lambda}$
($\lambda\in\Lambda$) and the ${\cal H}-$class $R_{i}\cap L_{\lambda}$ as
$H_{i\lambda}$. Suppose $s^{\prime}\in R_{j}$ and $ss^{\prime}\in R_{i}$. For
each $\lambda\in\Lambda$, let $r_{i\lambda}$ be the idempotent in
$H_{i\lambda}$ so that we produce a weak inverse of $s$,
$s^{\prime}_{j\lambda}$, in $H_{j\lambda}$. Let $s^{\prime}_{j\lambda}s\in
L_{\mu}$, and for each $k\in I$ let $l_{k\mu}$ be the idempotent in $H_{k\mu}$
and note that, using a similar argument to above,
$l_{k\mu}s^{\prime}_{j\lambda}\in W(s)$ and $l_{k\mu}s^{\prime}_{j\lambda}\in
H_{k\lambda}$.

$s^{\prime}_{j\lambda}s$$r_{i\lambda}$$l_{k\mu}$$s^{\prime}_{j\lambda}$$l_{k\mu}s^{\prime}_{j\lambda}$

Notice that the converse of the first point is true as well. The second point
allows us to give an equivalent definition of $\rho$: $s\rho t$ if and only if
for every $\cal H$-class $H$ of $S$ we have

$W(s)\cap H\neq\emptyset\iff W(t)\cap H\neq\emptyset.$

The next result is key in what follows.

###### Lemma 3.2

Let $S$ be a conditionally completely regular semigroup and let $s,t\in S$.
For any $\cal D$-class $D$ of $S$ we have

$W(st)\cap D\neq\emptyset\text{ if and only if }W(s)\cap D\neq\emptyset\text{
and }W(t)\cap D\neq\emptyset.$

  * Proof.

Let $s^{\prime}\in W(s)\cap D$ and suppose $W(t)\cap D\neq\emptyset$. Then
$s^{\prime}s$ is an idempotent lying in $D$. By Lemma 3.1(2) $t$ has a weak
inverse in every $\cal H$-class of $D$, so let $t^{\prime}$ be the weak
inverse of $t$ lying in the $\cal H$-class of $s^{\prime}s$. Then
$t^{\prime}s^{\prime}stt^{\prime}s^{\prime}=t^{\prime}tt^{\prime}s^{\prime}=t^{\prime}s^{\prime}$.
By Lemma 3.1(1) $t^{\prime}s^{\prime}\in D$ and so $W(st)\cap D\neq\emptyset$.

Conversely let $(st)^{\prime}\in W(st)\cap D$. Then
$t(st)^{\prime}st(st)^{\prime}=t(st)^{\prime}$ and so $t(st)^{\prime}$ is a
weak inverse of $s$. As $t(st)^{\prime}\mathcal{L}(st)^{\prime}$ we have
$W(s)\cap D\neq\emptyset$. Similarly we have $(st)^{\prime}s\in W(t)\cap
D\neq\emptyset$.

###### Corollary 3.3

Let $S$ be a conditionally completely regular semigroup and let $s,t\in S$.
Then $s\rho s^{2}$ and $st\rho ts$.

###### Corollary 3.4

Let $S$ be a conditionally completely regular semigroup. Either $S$ is
$E-$dense or the set $\\{s\in S|W(s)=\emptyset\\}$ is an ideal of $S$.

We can now prove the following theorem.

###### Theorem 3.5

Let $S$ be a conditionally completely regular semigroup. Then the relation
$\rho$ is a congruence and $S/\rho$ is a semilattice.

  * Proof.

Let $a,b,c,d\in S$ such that $a\rho b$ and $c\rho d$ and let $D$ be a $\cal
D$-class of $S$. By Lemma 3.2 $W(ac)\cap D\neq\emptyset$ if and only if
$W(a)\cap D\neq\emptyset$ and $W(c)\cap D\neq\emptyset$. As $a\rho b$ and
$c\rho d$ this latter condition is equivalent to $W(b)\cap D\neq\emptyset$ and
$W(d)\cap D\neq\emptyset$ which is in turn equivalent to $W(bd)\cap
D\neq\emptyset$ by Lemma 3.2. It follows that $ac\rho bd$ and so $\rho$ is a
congruence. That $S/\rho$ is a semilattice follows from Corollary 3.3.

We can now prove some results about the structure of $S$.

###### Lemma 3.6

Let $S$ be a conditionally completely regular semigroup and let
$s,t\in\operatorname{Reg}(S)$. Then $s\rho t$ if and only if $s\mathcal{D}t$.

  * Proof.

From Lemma 3.2 it follows that all of Green’s relations are contained in
$\rho$. To see this suppose that $(s,t)\in{\cal J}$. Then there exists $u,v\in
S^{1}$ such that $s=utv$. So for every ${\cal D-}$class $D$, if $W(s)\cap
D\neq\emptyset$ then $W(utv)\cap D\neq\emptyset$. Hence by Lemma 3.2 $W(t)\cap
D\neq\emptyset$. By a dual argument we then deduce that $W(s)\cap
D\neq\emptyset$ if and only if $W(t)\cap D\neq\emptyset$ and so
$(s,t)\in\rho$.

As $s$ is regular it has an inverse which lies in the same $\cal D$-class and
so $W(s)\cap D_{s}\neq\emptyset$. Hence $W(t)\cap D_{s}\neq\emptyset$ and by
Lemma 3.1 there exists $t^{\prime}\in W(t)$ such that
$t^{\prime}\mathcal{L}s$. By a similar argument there exists $s^{\prime}\in
W(s)$ such that $s^{\prime}\mathcal{R}t$. Then

$s\mathcal{L}t^{\prime}\mathcal{R}t^{\prime}t\mathcal{L}st\mathcal{R}ss^{\prime}\mathcal{L}s^{\prime}\mathcal{R}t$

and so $s\mathcal{D}t$ as required.

It follows that for each $\rho$-class $S_{\alpha}$ either $S_{\alpha}$ has no
regular elements or the regular elements in $S_{\alpha}$ are contained within
a single $\cal D$-class and hence by Lemma 3.1 form a completely simple
subsemigroup of $S_{\alpha}$. In the latter case $S_{\alpha}$ is an $E-$dense
semigroup as by definition of $\rho$ each element has a weak inverse lying in
the regular $\cal D$-class. Since each $\cal J$-class is contained within a
$\rho$-class, it also follows that the regular $\cal J$-classes of $S$ are
exactly the regular $\cal D$-classes.

###### Lemma 3.7

Let $S$ be a conditionally completely regular semigroup and let $x\in S$. Then
$x\rho$ is an $E-$dense subsemigroup of $S$ if and only if $x\rho$ contains a
regular element.

  * Proof.

One way round is obvious. That $x\rho$ is a subsemigroup of $S$ follows from
the fact that $\rho$ is a semilattice. Let $y\in x\rho$ be regular. Then there
exists $y^{\prime}\in W(y)\cap D_{y}$ and so for any $z\in x\rho$ there exists
$z^{\prime}\in W(z)\cap D_{y}$. Since $D_{y}\subseteq x\rho$ then $x\rho$ is
$E-$dense.

###### Lemma 3.8

Let $S$ be an $E-$dense semigroup such that $\operatorname{Reg}(S)$ is a
completely simple semigroup. Then $\operatorname{Reg}(S)$ is an ideal of $S$.

  * Proof.

Let $s\in\operatorname{Reg}(S)$ and $t\in S$. Let $t^{\prime}\in W(t)$ and let
${\cal H}$ be Green’s ${\cal H}-$relation on $\operatorname{Reg}(S)$. As
$\operatorname{Reg}(S)$ is completely simple every regular $\cal H$-class
contains an inverse of $s$ so we may choose $s^{\prime}\in V(s)$ such that
$s^{\prime}\mathcal{R}tt^{\prime}$. Then
$t^{\prime}s^{\prime}stt^{\prime}s^{\prime}=t^{\prime}s^{\prime}ss^{\prime}=t^{\prime}s^{\prime}$
and $stt^{\prime}s^{\prime}st=ss^{\prime}st=st$. Hence $st$ is regular and so
$\operatorname{Reg}(S)$ is a right ideal. A dual argument shows
$\operatorname{Reg}(S)$ is a left ideal and hence an ideal.

###### Lemma 3.9

Let $S$ be a conditionally completely regular semigroup, let $\alpha\in
S/\rho$, let $S_{\alpha}={\rho^{\natural}}^{-1}(\alpha)$ and let $s\in
S_{\alpha}$. Then for all $\beta\in S/\rho$, $S_{\beta}$ contains a weak
inverse of $s$ if and only if $S_{\beta}$ contains a regular element and
$\beta\leq\alpha$.

  * Proof.

Suppose $\beta\leq\alpha$ and $S_{\beta}$ contains regular elements. Let $t\in
S_{\beta}$. Then $st\in S_{\alpha\beta}=S_{\beta}$. As $S_{\beta}$ contains a
regular element it is $E-$dense by Lemma 3.7, and so there exists
$(st)^{\prime}\in W(st)\cap S_{\beta}$. Then
$t(st)^{\prime}st(st)^{\prime}=t(st)^{\prime}$ so $t(st)^{\prime}\in W(s)\cap
S_{\beta}$ as required. Conversely, let $s^{\prime}\in W(s)\cap S_{\beta}$.
Clearly $s^{\prime}$ is regular, and $s^{\prime}=s^{\prime}ss^{\prime}\in
S_{\beta\alpha\beta}=S_{\alpha\beta}$ so $\alpha\beta=\beta$ and hence
$\beta\leq\alpha$ as required.

From the perspective of stratified extensions, we cannot say anything about
these semigroups in general. For example, a free semigroup $S$ and a group $G$
both satisfy the property that every regular ${\cal H}$-class contains an
idempotent, but $\operatorname{Base}(S)=\emptyset$ and
$\operatorname{Base}(G)=G$. One condition that allows us to make more precise
statements is to require that $S$ is a group-bound semigroup. Note that group-
bound implies eventually regular, and when every regular ${\cal H}$-class
contains an idempotent the two concepts are equivalent.

We will show that applying our results to a semigroup which is also group-
bound gives the same decomposition as that in Theorem 1.5.

If $S$ is a group-bound semigroup and $e\in E(S)$ then let $H_{e}$ denote the
largest subgroup of $S$ containing $e$. The set of elements $s$ such that
$s^{n}\in H_{e}$ for some $n\in\mathbb{N}$ is denoted by $K_{e}$. This is well
defined in the sense that if $s^{n}\in H_{e}$ we have $s^{m}\in H_{e}$ for all
$m>n$ ([5, Lemma 1]). It also follows that the sets $K_{e}$ partition $S$. In
general $K_{e}$ is not a subsemigroup of $S$ ([5, Proposition 7]) and in
addition in a group bound semigroup ${\cal D}={\cal J}$ ([5, Lemma 4]). As is
usual, $J_{s}$ will denote the $\cal J-$class of $s$.

The following result is important in what follows.

###### Lemma 3.10

Let $S$ be an eventually regular conditionally completely regular semigroup.
If $s\in K_{e}$ then $J_{e}$ is the greatest $\mathcal{J}$-class containing a
weak inverse of $s$. Moreover, if $e\mathcal{J}f$ and $s\in K_{e}$ and $t\in
K_{f}$ then $(s,t)\in\rho$.

  * Proof.

Let $S$ be a semigroup satisfying the conditions stated. As $S$ is eventually
regular and every regular element lies in a group then $S$ is group-bound. Let
$s\in K_{e}$ for some idempotent $e$ so that there exists $n\in\hbox{\Bbb N}$
such that $s^{n}\in H_{e}$. Then

$(s^{n}(s^{n+1})^{-1})s(s^{n}(s^{n+1})^{-1})=s^{n}(s^{n+1})^{-1}e=s^{n}(s^{n+1})^{-1}$

where $(s^{n+1})^{-1}$ is the inverse of $s^{n+1}$ in $H_{e}$. Therefore $s$
has a weak inverse in $H_{e}$ and hence in $J_{e}$.

Now let $s_{1}\in W(s)$ and notice that $s_{1}$ is regular and so lies in a
group $H_{f}$, say. By Lemma 3.1 every $\mathcal{H}$-class of $J_{f}$ contains
a weak inverse of $s$. Let $s_{2}$ be a weak inverse of $s$ such that
$s_{2}\mathcal{L}s_{1}s$ and note that $s_{2}\in D_{f}=J_{f}$. Then as
$s_{2}s_{1}s=s_{2}$ we have

$s_{2}s_{1}s^{2}s_{2}s_{1}=s_{2}ss_{2}s_{1}=s_{2}s_{1},$

so $s_{2}s_{1}\in W(s^{2})$ and by Lemma 3.1, $s_{2}s_{1}\in J_{f}$. Let us
denote $w_{2}=s_{2}s_{1}$ so that we proceed inductively as follows. Suppose
that for $n\geq 2$ we have $w_{n-1}\in W(s^{n-1})\cap J_{f}$. Let $s_{n}\in
W(s)\cap L_{w_{n-1}s^{n-1}}$ and let $w_{n}=s_{n}w_{n-1}$, so that
$w_{n}s^{n-1}=s_{n}$ and

$w_{n}s^{n}w_{n}=s_{n}sw_{n}=s_{n}ss_{n}w_{n-1}=s_{n}w_{n-1}=w_{n}.$

Hence $w_{n}\in W(s^{n})\cap J_{f}$.

We see then that there is a weak inverse of $s^{n}$ in $J_{f}$ for any
$n\in\hbox{\Bbb N}$. In particular, since $s\in K_{e}$, we can choose $n$
large enough such that $s^{n}\in H_{e}\subseteq J_{e}$. Let $s^{\ast}$ be the
associated weak inverse of $s^{n}$ in $J_{f}$. Then by Lemma 1.4 we have
$J_{f}=J_{s^{\ast}}\leq J_{s^{n}}=J_{e}$. Consequently if $s\in K_{e}$ then
$J_{e}$ is the greatest $\mathcal{J}$-class containing a weak inverse of $s$.

Now let $s\in K_{e}$ and $t\in K_{f}$ as in the statement of the lemma. We can
assume that $s$ and $t$ are regular. To see this, let $n\in\hbox{\Bbb N}$ be
the minimum value such that $s^{n}\in H_{e}$ and note that if
$(s^{n})^{\prime}$ is a weak inverse of $s^{n}$ then $s^{n-1}(s^{n})^{\prime}$
is a weak inverse of $s$ with
$s^{n-1}(s^{n})^{\prime}\mathcal{L}(s^{n})^{\prime}$. This, along with the
previous argument, shows that $s$ has a weak inverse in a ${\cal J}-$class $J$
if and only if the regular element $s^{n}$ has a weak inverse in $J$.

Let $J$ be a ${\cal J}-$class containing a weak inverse $s^{\prime}$ of $s$.
If $t\mathcal{L}s$ then $ts^{\prime}\mathcal{L}ss^{\prime}$ and so
$ts^{\prime}\in J$. Then, since $J$ is regular, there exists $r\in J$ such
that $ts^{\prime}r\in J$ is an idempotent, and so $s^{\prime}rts^{\prime}r\in
J$ is a weak inverse of $t$. By a similar argument if $t\mathcal{R}s$ there is
a weak inverse of $t$ in $J$ and so if $s\mathcal{J}t$ there is a weak inverse
of $t$ in $J$. A dual argument then gives the opposite direction, and the
result follows from the definition of $\rho$.

Note that each $\mathcal{H}$-class of $S$ contains at most one weak inverse of
$s$: if $s^{\prime},s^{\ast}\in W(s)$ with $s^{\prime}\mathcal{H}s^{\ast}$
then
$s^{\prime}s\mathcal{R}s^{\prime}\mathcal{R}s^{\ast}\mathcal{R}s^{\ast}s$. As
$\mathcal{L}$ is a right congruence we also have
$s^{\prime}s\mathcal{L}s^{\ast}s$. Since $s^{\prime}s$ and $s^{\ast}s$ are
idempotents it follows that $s^{\prime}s=s^{\ast}s$ and by a similar argument
$ss^{\prime}=ss^{\ast}$. Then
$s^{\prime}=s^{\prime}ss^{\prime}=s^{\ast}ss^{\prime}=s^{\ast}ss^{\ast}=s^{\ast}$.

###### Theorem 3.11

Let $S$ be a conditionally completely regular semigroup. If $S$ is group-bound
then $S$ is a semilattice of Archimedean semigroups of the form
$K_{J_{e}}=\bigcup_{f\in E(J_{e})}K_{f}$ for $e\in E(S)$.

  * Proof.

Let $e\in E(S)$ and define $K_{J_{e}}=\bigcup_{f\in E(J_{e})}K_{f}$. Let
$s,t\in K_{J_{e}}$ and notice that $s\in K_{f},t\in K_{g}$ for some $f,g\in
J_{e}$, so that by Lemma 3.10, $(s,t)\in\rho$. Conversely, if $(s,t)\in\rho$
then there exists $e,f\in E(S)$ such that $s\in K_{e}\subseteq K_{J_{e}},t\in
K_{f}\subseteq K_{J_{f}}$. By Lemma 3.10, $J_{e}$ is the greatest ${\cal
J}-$class containing a weak inverse of $s$ and $J_{f}$ is the greatest ${\cal
J}-$class containing a weak inverse of $t$. Since $(s,t)\in\rho$ it easily
follows that $J_{e}=J_{f}$ and so $s,t\in K_{J_{e}}=K_{J_{f}}$. Hence the sets
$K_{J_{e}}$ are the $\rho-$classes and so partition $S$ and since $S/\rho$ is
a semilattice then the result follows.

For each $e,f\in E(S)$ it follows that there exists $g\in E(S)$ such that
$K_{J_{e}}K_{J_{f}}\subseteq K_{J_{g}}$. Since $e\in K_{J_{e}}$ and $f\in
K_{J_{f}}$ then $ef\in K_{J_{g}}$. In addition there exists a uniquely
determined $h\in E(S)$ such that $ef\in K_{h}\subseteq K_{J_{h}}$ and so
$K_{J_{g}}=K_{J_{h}}$.

To see that $K_{J_{e}}$ is an Archimedean semigroup, let $a,b\in K_{J_{e}}$.
Then there exist $m,n\in\hbox{\Bbb N}$ such that $a^{m},b^{n}\in J_{e}$ and so
$a^{m}\in K_{J_{e}}b^{n}K_{J_{e}}\subseteq K_{J_{e}}bK_{J_{e}}$ as required.

Note that a decomposition into a semilattice of Archimedean semigroups is
necessarily unique: Let $S=[Y;S_{\alpha}]=[Y^{\prime};S_{a}]$ be two
Archimedean semilattice decompositions of the semigroup $S$. If $s,t\in S$ lie
in the same subsemigroup $S_{\alpha}$ where $\alpha\in Y$ and $s\in S_{a},t\in
S_{b}$ where $a,b\in Y^{\prime}$, then there exist $n\in\hbox{\Bbb N}$ and
$u,v\in S$ such that $s^{n}=utv$ and $a\leq b$. Similarly $b\leq a$ and so
$s,t\in S_{a}$ and the two semilattices, $Y$ and $Y^{\prime}$, are isomorphic.
We have hence recovered the same decomposition as Shevrin (Theorem 1.5) in
this case.

It is clear from the above structure that these semigroups are group-bound and
since it is straightforward to check that
$\operatorname{Reg}(K_{J_{e}})=J_{e}$, then the regular elements form a
completely simple subsemigroup.

The converse of Theorem 3.11 does not hold in general as an Archimedean
semigroup need not contain regular elements and hence a semilattice of
Archimedean semigroups may not be group-bound. It is enough, however, to
require that each Archimedean semigroup contains a regular element.

###### Corollary 3.12

Let $S$ be a conditionally completely regular semigroup. Then $S$ is group-
bound if and only if $S=[Y;S_{\alpha}]$ is a semilattice of Archimedean
semigroups $S_{\alpha}$ with $\operatorname{Reg}(S_{\alpha})\neq\emptyset$.

  * Proof.

Clearly if $S$ is group-bound then every subsemigroup contains a regular
element. Conversely, let $s\in S$. Then $s\in S_{\alpha}$ for some $\alpha$
and let $t\in\operatorname{Reg}(S_{\alpha})$. Since $S_{\alpha}$ is an
Archimedean semigroup, there exists $n\in\hbox{\Bbb N}$ such that $s^{n}\in
S_{\alpha}tS_{\alpha}$ and hence
$s^{n}\in\operatorname{Reg}(S_{\alpha})\subseteq\operatorname{Reg}(S)$ by
Lemmas 3.7 and 3.8.

###### Proposition 3.13

Let $S$ be a semigroup. Any two of the following implies the third.

  1. 1.

$S$ is group-bound;

  2. 2.

$S$ is conditionally completely regular;

  3. 3.

$S$ is a semilattice of Archimedean semigroups $S_{\alpha}$ with
$\operatorname{Reg}(S_{\alpha})\neq\emptyset$.

  * Proof.

By Corollary 3.12 we have (1) and (2) imply (3) and (2) and (3) imply (1). The
remaining implication follows from Theorem 1.5.

We now turn our attention to describing the subsemigroups $K_{J_{e}}$ at each
vertex of the semilattice. Since each semigroup contains regular elements,
they are all stratified extensions with a base consisting of at least the
regular elements. From [5, Proposition 3] each $K_{J_{e}}$ is an ideal
extension of the completely simple semigroup $J_{e}$ by a nilsemigroup. If
this nilsemigroup is stratified then $K_{J_{e}}$ is a nil-stratified extension
with base $J_{e}$. However not every nilsemigroup is stratified as
demonstrated by the following example.

Let $S={\cal P}(\hbox{\Bbb N})$ be the set of subsets of ℕ and define a
multiplication on $S$ by

$A\circ B=\begin{cases}A\cup B&\text{if }A\text{ and }B\text{ are non-empty
and }A\cap B=\emptyset\\\ \emptyset&\text{otherwise.}\end{cases}$

Then it is clear that each element is nilpotent of index 2 and so $S$ is a
nilsemigroup, but all infinite elements are in the base. Hence $S$ is not
stratified.

###### Lemma 3.14

Let $S$ be an eventually regular semigroup such that $\operatorname{Reg}(S)$
is completely simple and suppose $S$ is a finitely stratified extension. Then
$\operatorname{Base}(S)\setminus\operatorname{Reg}(S)$ is either empty or
infinite.

  * Proof.

Suppose
$s_{0}\in\operatorname{Base}(S)\setminus\operatorname{Reg}(S)\neq\emptyset$.
Since $S$ is a finitely stratified extension, $\operatorname{Base}(S)$ is a
globally idempotent subsemigroup so $s_{0}=s_{1}t_{1}$ for some
$s_{1},t_{1}\in\operatorname{Base}(S)$. If $s_{1}$ is regular then as
$\operatorname{Reg}(S)$ is an ideal, $s_{0}$ is regular giving a
contradiction. Further, if $s_{1}=s_{0}$ then
$s_{0}=s_{0}t_{1}=s_{0}{t_{1}}^{n}$ for any $n\in\hbox{\Bbb N}$. We can choose
$n$ such that ${t_{1}}^{n}$ is regular, so $s_{0}$ is again regular giving a
contradiction. Hence $s_{1}$ is an element of
$\operatorname{Base}(S)\setminus\operatorname{Reg}(S)$ not equal to $s_{0}$.
By a similar argument, $s_{1}=s_{2}t_{2}$ where
$s_{2}\in\operatorname{Base}(S)\setminus\operatorname{Reg}(S)$ and $s_{2}$ is
not equal to $s_{0}$ nor $s_{1}$. Proceeding inductively we deduce that the
set $\\{s_{0},s_{1},s_{2},\dots\\}$ is an infinite subset of
$\operatorname{Base}(S)\setminus\operatorname{Reg}(S)$.

It follows that any finite semigroup in which every regular
$\mathcal{H}$-class contains an idempotent is a semilattice of finitely
stratified extensions with completely simple bases.

###### Theorem 3.15

A semigroup $S$ is a finite conditionally completely regular semigroup if and
only if $S=[Y;S_{\alpha}]$ is a finite semilattice of finite semigroups
$S_{\alpha}$ where each $S_{\alpha}$ is a finitely stratified extension of a
completely simple semigroup.

  * Proof.

To see that the converse is true, let $s\in S$ be a regular element, so that
there exists $\alpha$ such that $s\in S_{\alpha}$. Let $s^{\prime}$ be an
inverse of $s$ (within $S$) with $s^{\prime}\in S_{\beta}$ for some $\beta$.
Then $s=ss^{\prime}s\in S_{\alpha}S_{\beta}S_{\alpha}\subseteq
S_{\alpha\beta}\cap S_{\alpha}$, and so $S_{\alpha}=S_{\alpha\beta}$.
Similarly $s^{\prime}=s^{\prime}ss^{\prime}\in S_{\alpha\beta}\cap S_{\beta}$
and so $S_{\alpha}=S_{\beta}$. It follows that $s$ is regular within
$S_{\alpha}$ and so $s\in\operatorname{Base}(S_{\alpha})$ and is therefore
${\cal H}-$related to an idempotent as required.

##  4 Strict extensions of Clifford Semigroups

In this section we continue our exploration of examples by looking at Clifford
semigroups. In particular, we show that every strict extension of a Clifford
semigroup can be decomposed as a semilattice of stratified extensions of
groups.

We make use of the notation of Clifford and Preston [1, Section 4.4], and in
particular that relating to ideal extensions determined by partial
homomorphisms. A Clifford semigroup is a completely regular inverse semigroup.
It is well known that a Clifford semigroup $S$ decomposes as a semilattice of
groups $S={\cal S}[Y;G_{\alpha}]$. We begin by showing that a strict extension
$\Sigma$ of a Clifford semigroup $S$ has a semilattice structure isomorphic to
that of the Clifford semigroup itself.

###### Lemma 4.1

Let $S={\cal S}[Y;G_{\alpha}]$ be a Clifford semigroup. An ideal extension of
$S$ is strict if and only if it is determined by a partial homomorphism.

  * Proof.

Let $a,b\in S$ be such that $ax=bx$ and $xa=xb$ for all $x\in S$. As $S$ is a
Clifford semigroup $a\in G_{\alpha}$ and $b\in G_{\beta}$ for some
$\alpha,\beta\in Y$. Let $e,f$ be the identities of $G_{\alpha},G_{\beta}$
respectively. Then $a=ea=eb$ and so $\alpha\leq\beta$. Similarly, $b=fb=fa$ so
$\beta\leq\alpha$ and so $\alpha=\beta$ and $e=f$. Then $a=ea=eb=b$ and hence
$S$ is weakly reductive. The result then follows from Theorem 1.2.

###### Lemma 4.2

Let $\Sigma$ be a strict extension of a Clifford semigroup $S={\cal
S}[Y;G_{\alpha}]$ by a semigroup $T$ defined by a partial homomorphism
$A\mapsto\overline{A}$ and let $\Sigma_{\alpha}=G_{\alpha}\cup\\{A\in
T\setminus\\{0\\}|\overline{A}\in G_{\alpha}\\}$ for each $\alpha\in Y$.
Define a relation $\sim$ on $\Sigma$ by $s\sim t$ if and only if
$s,t\in\Sigma_{\alpha}$ for some $\alpha\in Y$. Then $\sim$ is a congruence
and $\Sigma/\\!\\!\sim$ is a semilattice isomorphic to $Y$.

  * Proof.

Clearly $\sim$ is an equivalence relation. To prove $\sim$ is a congruence and
that $\Sigma/\\!\\!\sim\;\cong Y$ we show that $\sim$ is the kernel of the
homomorphism $\theta:\Sigma\rightarrow Y$ where if $s\in\Sigma_{\alpha}$ then
$\theta(s)=\alpha$. Note that if $A\in T\setminus\\{0\\}$ then
$\theta(A)=\theta(\overline{A})$. We have four cases to consider:

    1. 1.

If $s,t\in S$ then $\theta(s)\theta(t)=\theta(st)$ follows from the
semilattice structure of $S$.

    2. 2.

If $s\in S$ and $A\in T\setminus\\{0\\}$ then
$\theta(s)\theta(A)=\theta(s)\theta(\overline{A})=\theta(s\overline{A})=\theta(sA)$,
where the last two equalities follow from the first case and multiplication in
a strict extension respectively. The case for $\theta(A)\theta(s)$ follows
similarly.

    3. 3.

If $A,B\in T\setminus\\{0\\}$ then
$\theta(A)\theta(B)=\theta(\overline{A})\theta(\overline{B})=\theta(\overline{A}\;\overline{B})$
by the first case. Then if $AB=0$ in $T$ we have
$\theta(AB)=\theta(\overline{A}\;\overline{B})$ and if $AB\neq 0$ in $T$ we
have $\theta(AB)=\theta(\overline{AB})=\theta(\overline{A}\;\overline{B})$. In
either case $\theta(A)\theta(B)=\theta(AB)$.

Hence $\theta$ is a homomorphism as required and $\sim$ is clearly its kernel.

###### Theorem 4.3

Every strict extension $\Sigma$ of a Clifford semigroup $S$ by a semigroup $T$
is a semilattice of extensions of groups. Conversely, if $\Sigma$ is a
semilattice of extensions $\Sigma_{\alpha}$ of groups $G_{\alpha}$ and
$S=\bigcup_{\alpha\in Y}G_{\alpha}$ is an ideal of $\Sigma$ then $\Sigma$ is a
strict extension of the Clifford semigroup $S$.

  * Proof.

By Lemma 4.2, $\Sigma$ is a semilattice of semigroups $\Sigma_{\alpha}$
defined via a partial homomorphism $A\mapsto{\overline{A}}$ from
$T\setminus\\{0\\}\to S$. The restriction of this map to
$\Sigma_{\alpha}\setminus G_{\alpha}$ gives a partial homomorphism defining
the ideal extension $\Sigma_{\alpha}$ of the group $G_{\alpha}$.

Conversely, let $\Sigma$ be a semilattice of semigroups $\Sigma_{\alpha}$
where each $\Sigma_{\alpha}$ is an ideal extension of a group $G_{\alpha}$ by
a stratified semigroup $T_{\alpha}$ and $S=\bigcup_{\alpha\in Y}G_{\alpha}$ is
an ideal of $\Sigma$. It follows that $S$ is a Clifford semigroup and $\Sigma$
is an ideal extension of $S$ by $T=\Sigma/S$, where $T$ can equivalently be
viewed as $\\{0\\}\cup\bigcup_{\alpha\in Y}T_{\alpha}\setminus\\{0\\}$. As
$G_{\alpha}$ has identity $e_{\alpha}$ the extension $\Sigma_{\alpha}$ is
determined by the partial homomorphism $A\mapsto Ae_{\alpha}\;(=e_{\alpha}A)$
(Proposition 1.1 and Theorem 1.2). The union of these maps is then a map
$\varphi:T\setminus\\{0\\}\rightarrow S$ such that $\varphi(A)=Ae_{\alpha}$
for each $A\in T_{\alpha}\setminus\\{0\\}$. We will show that $\varphi$ is a
partial homomorphism and that it defines the ideal extension $\Sigma$. For
clarity, the multiplication determined by $\varphi$ will be denoted by
$\circ$, multiplication within $T$ by $\ast$, and the original multiplication
of the semilattice $\Sigma$ by juxtaposition.

Let $A,B\in T\setminus\\{0\\}$ such that $A\ast B\neq 0$ and assume $A\in
T_{\alpha}$, $B\in T_{\beta}$ so that $A\ast B\in T_{\alpha\beta}$. Then
$\varphi(A)\varphi(B)=Ae_{\alpha}(Be_{\beta})=A(Be_{\beta})e_{\alpha}=ABe_{\alpha\beta}=\varphi(AB)$
as required.

This partial homomorphism determines an ideal extension of $S$ consisting of
the same set $\Sigma$ under the multiplication $\circ$ defined by

  1. 1.

$s\circ t=st$

  2. 2.

$A\circ B=\begin{cases}AB&\text{if }A\ast B\neq 0\\\
\varphi(A)\;\varphi(B)&\text{otherwise}\end{cases}$

  3. 3.

$A\circ s=\varphi(A)s$

  4. 4.

$s\circ A=s\varphi(A)$

where $A,B\in T\setminus\\{0\\}$ and $s,t\in S$. We show that in all cases,
this multiplication is equivalent to the original multiplication on $\Sigma$.
The first condition and the first part of the second condition do not require
proof. For the second part of the second condition, let $A\in
T_{\alpha}\setminus\\{0\\}$ and $B\in T_{\beta}\setminus\\{0\\}$ with $A\ast
B=0$ so $AB\in G_{\alpha\beta}$. Then

$A\circ
B=\varphi(A)\varphi(B)=Ae_{\alpha}(Be_{\beta})=A(Be_{\beta})e_{\alpha}=ABe_{\alpha\beta}=AB$

as required. For the third condition, let $A\in T_{\alpha}\setminus\\{0\\}$
and $s\in G_{\beta}$ with $As\in G_{\alpha\beta}$. Then

$A\circ
s=\varphi(A)s=Ae_{\alpha}(se_{\beta})=A(se_{\beta})e_{\alpha}=Ase_{\alpha\beta}=As$

as required. The fourth condition follows a dual argument. Hence $\varphi$
determines the extension $\Sigma$ and so it is a strict extension of $S$.

###### Corollary 4.4

Let $\Sigma$ be a strict stratified extension of a Clifford semigroup $S$.
Then $\Sigma$ is a semilattice of stratified extensions of groups.

  * Proof.

Let $\Sigma$ be a strict extension of a Clifford semigroup $S$ by a stratified
semigroup $T$. By Theorem 4.3, $\Sigma$ is a semilattice of semigroups
$\Sigma_{\alpha}$, each of which is an ideal extension of a group $G_{\alpha}$
by a subsemigroup of $T$ containing zero. It can be easily verified that such
a subsemigroup is also stratified, and hence $\Sigma$ is a semilattice of
stratified extensions of groups.

The converse of Corollary 4.4 does not hold in general as each $T_{\alpha}$
being a stratified semigroup does not guarantee that $T$ is itself a
stratified semigroup. For example, let $Y=\\{a,b\\}$ with $a\leq b$. For each
$\alpha\in Y$ let $G_{\alpha}$ be a group, $T_{\alpha}$ a free semigroup with
adjoined zero, and $\Sigma_{\alpha}$ an ideal extension of $G_{\alpha}$ by
$T_{\alpha}$. For $s\in T_{a}$ and $t\in T_{b}$ let $st=ts=s$. Along with the
fact that $S=G_{a}\cup G_{b}$ is an ideal of $\Sigma$, this defines a
multiplication on the semilattice $\Sigma=\Sigma_{a}\cup\Sigma_{b}$. Each
$T_{\alpha}$ is a stratified semigroup so each $\Sigma_{\alpha}$ is a
stratified extension of a group, however $T=\Sigma/S$ is not stratified, as
$\bigcap_{i\geq 1}T^{i}\cong T_{a}$. A sufficient, but clearly not necessary,
condition under which $T$ will always be stratified is if $T$ is finite.

As an example of the above construction, consider the following. Let
$n\in\hbox{\Bbb N}$ and let $N=\\{1,\dots,n\\}$. Let
$S=G_{1}^{0}\times\ldots\times G_{n}^{0}$ be a direct product of $0-$groups
$G^{0}_{i}$, $i\in N$. For $s=(a_{1},\ldots,a_{n})\in S$ define
$\hbox{dom}(s)=\\{i\in N|a_{i}\neq 0\\}$.

Let $m\in\hbox{\Bbb N}$ and define a relation $\rho_{m}$ on $(\hbox{\Bbb
N},+)$ by

$\rho_{m}=1_{\hbox{\Bbb N}}\cup\\{(x,y)\in\hbox{\Bbb N}\times\hbox{\Bbb
N}|x,y\geq m\\}.$

Then it is easy to check that $S$ is a Clifford semigroup (and hence a strong
semilattice of groups), $\rho_{m}$ is a congruence on ℕ and $\hbox{\Bbb
N}/\rho_{m}$ is a finite monogenic semigroup with trivial kernel. For
simplicity, we shall identify $\hbox{\Bbb N}/\rho_{m}$ with
$\\{1,\ldots,m\\}$, in the obvious way. Let $T^{\prime}$ be the semigroup of
all partial maps from $N$ to $\hbox{\Bbb N}/\rho_{m}$ with binary operation
$\ast$ given by $(f\ast g)(x)=f(x)+g(x)$ when both are defined and undefined
otherwise. Let $I\subseteq T^{\prime}$ be the set of maps whose image is
$\\{m\\}$. It can be readily seen that $I$ is an ideal of $T^{\prime}$ and
$T=T^{\prime}/I$ is a nilsemigroup.

For each $i\in N$ pick an element $g_{i}\in G_{i}$ and let
$\alpha_{i}:T\setminus\\{0\\}\rightarrow G^{0}_{i}$ be the partial
homomorphism given by

$\alpha_{i}(f)=\begin{cases}g_{i}^{f(i)}&f(i)\text{ is defined}\\\
0&\text{otherwise.}\end{cases}$

Then $\alpha:T\setminus\\{0\\}\rightarrow S$ given by
$\alpha(f)=(\alpha_{1}(f),\dots,\alpha_{n}(f))$ is a partial homomorphism
defining an ideal extension $\Sigma$ of $S$ by $T$.

Notice that $s{\cal J}t$ if and only if $\hbox{dom}(s)=\hbox{dom}(t)$. It
follows that the semilattice structure of $S$ is defined in terms of the power
set of $N$ (i.e. $\hbox{dom}(st)=\hbox{dom}(s)\cap\hbox{dom}(t)$). Let $S_{M}$
be the $\cal J$-class of $S$ with $\hbox{dom}(s)=M$ for $s\in S_{M}$. Then
$T_{M}=\alpha^{-1}(S_{M})$ is the set of maps in $T\setminus\\{0\\}$ whose
domain is exactly $M$. The set $T_{M}^{0}=T_{M}\cup\\{0\\}$ is a subsemigroup
of $T$ and is a nilsemigroup. The restriction of $\alpha$ to $T_{M}$ then
gives a partial homomorphism from $T_{M}^{0}$ to $S_{M}$ which defines an
ideal extension $\Sigma_{M}$ of the group $S_{M}$ by $T_{M}^{0}$. It can then
be shown that $\Sigma$ is a semilattice of these semigroups $\Sigma_{M}$.

The authors thank the anonymous referee for pointing out the example after
Proposition 3.13 and for a number of useful comments which have enhanced the
overall exposition of the paper.

## References

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  * [3] P. A. Grillet and Mario Petrich, Ideal extensions of semigroups, Pacific Journal of Mathematics, Vol. 26, No. 3, 1968.
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