# On amalgamation of inverse semigroups over ample semigroups

Nasir Sohail  Institute of Mathematics and Statistics, the University of
Tartu, Tartu, Estonia

###### Abstract

A semigroup amalgam $(S;T_{1},T_{2})$ is known to be non-embeddable if $T_{1}$
and $T_{2}$ are both groups (completely regular semigroups, Clifford
semigroups) but $S$ is not such. We prove some non-embeddability conditions
for semigroup amalgams $(S;T_{1},T_{2})$ in which $T_{1}$ and $T_{2}$ are
inverse semigroups but $S$ is not such. We also introduce some sufficient
conditions that make the inverse hulls of the copies of $S$ in $T_{1}$ and
$T_{2}$ isomorphic; making, in turn, the amalgam $(S;T_{1},T_{2})$ weakly
embeddable.

Keywords— Inverse semigroup, Amalgam, Antiamalgamation pair, Ample, Rich
ample,

AMS subject classification — 20M18, 20M15, 20M10

##  1 Motivation

The amalgamation problem of semigroups has its origins in the early work of J.
M. Howie from the 1960s. (The inpiration came from group amalgams, which were
considered earlier by O. Schreier.) The topic was then extensively studied by
various mathematicians during the second half of the previous century.
References to this work may be found in Howie’s celebrated monograph [4], of
which the last chapter is also dedicated to semigroup amalgams. The main
emphasis, during all these years, had been on determining the embeddability
conditions for semigroup amalgams. Non-embeddable amalgams were discovered
sporadically, usually as by-products. One of Howie’s pioneering articles [5],
however, provided an important class of non-embeddable amalgams, that may
essentially be viewed as groups intersecting in semigroups. Generalizing
Howie’s result, Rahkema and Sohail [6] came up in 2014 with two more classes
of non-embeddable semigroup amalgams. The current article furthers the same
line of research of investigating the (non-embeddability of) amalgams that may
essentially be viewed as inverse semigroups intersecting in a non-inverse
semigroup.

The study of ample semigroups, and their variants, has been an active area of
research for many decades, see for instance [2] and its references. As every
ample semigroup $S$ gives rise to an amalgam $(S;T_{1},T_{2})$, where $T_{1}$
and $T_{2}$ are inverse semigroups, it was natural for us to consider the
amalgams $(S;T_{1},T_{2})$, such that $T_{1}$ and $T_{2}$ are inverse
semigroups and $S$ belongs to certain class of ample semigroups. In fact, we
introduce, in this connection, the classes of rich and ultra-rich ample
semigroups; the intersection of the latter class with the class of inverse
semigroups is precisely the class of all groups.

##  2 Introduction and preliminaries

Given a semigroup $S$, an element $x\in S$ is called invertible if there
exists a unique element $x^{-1}\in S$ such that $xx^{-1}x=x$ and
$x^{-1}xx^{-1}=x^{-1}$. We call $S$ an inverse semigroup if every $x\in S$ is
invertible. Inverse monoids are defined similarly. Let $X$ be a non-empty set.
Then the set $\mathcal{I}_{X}$ of all partial bijections of $X$ is an inverse
semigroup under the usual composition of partial maps. We call
$\mathcal{I}_{X}$ the symmetric inverse semigroup over $X$. By the Wagner-
Preston representation theorem (see, for instance, [4] Theorem 5.1.7) any
inverse semigroup $S$ can be embedded in the symmetric inverse semigroup
$\mathcal{I}_{S}$. If $S$ is a subsemigroup of an inverse semigroup $T$ then
the inverse subsemigroup of $T$ generated by $S$ is called the inverse hull of
$S$ in $T$. Homomorphisms of inverse semigroups (monoids) are precisely the
semigroup homomorphisms. Throughout this article, the maps will be written to
the right of their arguments. Also, we shall omit parentheses around the
arguments if there is no risk of confusion. For further details about inverse
semigroups, as well as standard definitions in semigroup theory, the reader
may refer to the texts [4, Lawson].

A semigroup $S$ is called right ample if it can be embedded in an inverse
semigroup $T$ (typically, in the symmetric inverse semigroup $\mathcal{I}_{X}$
of a non-empty set $X$) such that the image of $S$ is closed under the unary
operation $s\longmapsto s^{-1}s$, where $S$ is identified with its isomorphic
copy in $T$ and $s^{-1}\in T$ denotes the inverse of $s\in S$. We shall call
$T$ an inverse semigroup associated with $S$. Left ample semigroups are
defined analogously. We say that $S$ is ample if it is both right and left
ample. If $S$ is a subsemigroup of an associated inverse semigroup $T$ then we
shall say that $S$ is (right, left) ample in $T$. Every full subsemigroup of
an inverse semigroup $S$ is ample in $S$. The converse is not true; for
example, $\mathbb{N}$ is ample but not full in the multiplicative monoid
$\mathbb{Q}$. It is possible that $S$ is made into a left and a right ample
semigroup by different associated inverse semigroups. In such a case, the
problem of finding a single (associated) inverse semigroup making $S$ into a
left as well as right ample semigroup is, in general, undecidable ([3],
Theorem 3.4 and Corollary 4.3). If $T_{1}$ and $T_{2}$ are inverse semigroups
admitting a homomorphism $\phi:T_{1}\longrightarrow T_{2}$ and $S$ is right
(respectively, left) ample in $T_{1}$, then one can easily verify that $S\phi$
is right (respectively, left) ample in $T_{2}$. For further details about
ample semigroups, the reader is referred to [2] and the references contained
therein.

A semigroup amalgam is a $5$-tuple $\mathcal{A\equiv}$
$(S;T_{1},T_{2};\phi_{1},\phi_{2})$ comprising pair-wise disjoint semigroups
$S$, $T_{1}$, $T_{2}$ and monomorphisms

$\phi_{i}:S\longrightarrow T_{i}\text{, }1\leq i\leq 2\text{.}$

We say that $\mathcal{A}$ is embeddable (or strongly embeddable, for emphasis)
if there exists a semigroup $T$ admitting monomorphisms
$\psi_{i}:T_{i}\longrightarrow T$, $1\leq i\leq 2$, such that

  1. $(i)$

$\phi_{1}\psi_{1}=\phi_{2}\psi_{2}$,

  2. $(ii)$

$\forall$ $t_{1}\in T_{1}$, $\forall$ $t_{2}\in T_{2}$,
$t_{1}\psi_{1}=t_{2}\psi_{2}$ $\Longrightarrow$ $\exists$ $s\in S$ such that
$t_{1}=s\phi_{1}$, $t_{2}=s\phi_{2}$.

If Condition $(ii)$ is not necessarily satisfied, then $\mathcal{A}$ is said
to be weakly embeddable. We call $(S;T_{1},T_{2};\phi_{1},\phi_{2})$ a special
amalgam if $T_{1}$ and $T_{2}$ are isomorphic, say, via
$\nu:T_{1}\longrightarrow T_{2}$, such that $s\phi_{1}\nu=s\phi_{2}$ for all
$s\in S$. Any special amalgam is weakly embeddable; for instance, in $T_{1}$.
It is customary to denote a semigroup amalgam by $(S;T_{1},T_{2})$ if no
explicit mention of $\phi_{1}$ and $\phi_{2}$ is needed. We shall also call
$(S;T_{1},T_{2})$ an amalgam over $S$. Every ample semigroup $S$ gives rise to
an amalgam $(S;T_{1},T_{2})$ in which $S$ is right (respectively, left) ample
in the inverse semigroup $T_{i}$ (respectively, $T_{j}$), where
$\\{i,j\\}=\\{1,2\\}$. We shall consider these amalgams in Theorem 3.4.

Let $T_{1}\ast T_{2}$ denote the free product of semigroups $T_{1}$ and
$T_{2}$. Then by the amalgamated coproduct of
$(S;T_{1},T_{2};\phi_{1},\phi_{2})$ we mean the quotient semigroup
$\left(T_{1}\ast T_{2}\right)/\theta_{R}$, where $\theta_{R}$ denotes the
congruence on $T_{1}\ast T_{2}$ generated by the relation

$R=\\{\left(s\phi_{1},s\phi_{2}\right):s\in S\\}.$

We denote $\left(T_{1}\ast T_{2}\right)/\theta_{R}$ by $T_{1}\ast_{S}T_{2}$.
In fact, the following diagram is a pushout in the category of all semigroups,
where the homomorphisms

$\eta_{i}:T_{i}\longrightarrow T_{1}\ast_{S}T_{2},\,1\leq i\leq 2,$

send $x\in T_{i}$ to the congruence class $(x)_{\theta_{R}}\in
T_{1}\ast_{S}T_{2}$.

Figure 1: Amalgamated coproduct

######  Theorem 2.1 ([4], Theorem 8.2.4).

A semigroup amalgam $(S;T_{1},T_{2})$ is (weakly) embeddable if and only if it
is (weakly) embedded in $T_{1}\ast_{S}T_{2}$ via the homomorphisms
$\eta_{i}:T_{i}\longrightarrow T_{1}\ast_{S}T_{2}$, $i\in\\{1,2\\}$, defined
above.

###### Proof.

Follows immediately from the properties of a pushout. ∎

A semigroup $S$ is called an amalgamation base for a class (equivalently,
category) $\mathcal{C}$ of semigroups if every amalgam $(S;T_{1},T_{2})$, with
$T_{1},T_{2}\in\mathcal{C}$, is embeddable in some $T\in\mathcal{C}$. Given a
semigroup $T_{1}\in\mathcal{C}$ containing an isomorphic copy of a semigroup
$S$, we say that $(S;T_{1})$ is an  amalgamation pair for $\mathcal{C}$ if for
all $T_{2}\in\mathcal{C}$ the amalgam $(S;T_{1},T_{2})$ is embeddable in some
$T\in\mathcal{C}$. Weak amalgamation bases (pairs) are defined similarly.

######  Theorem 2.2 ([4], Theorems 8.6.1 and 8.6.4).

Inverse semigroups are amalgamation bases for the classes of all semigroups
and inverse semigroups.

Let $S$ be an inverse and $T$ be an arbitrary semigroup. Then, by the above
theorem, $(S;T)$ is an amalgamation pair for the class of all semigroups. If
$S$ and $T$ are both inverse then $(S;T)$ is also an amalgamation pair for the
class of inverse semigroups. All of the assertions made in this section about
semigroups are also true for monoids.

##  3 Amalgamation over ample semigroups

It was shown by Howie [5] that a semigroup amalgam $(S;T_{1},T_{2})$ does not
embed if $T_{1}$ and $T_{2}$ are both groups but $S$ is not such. Generalizing
this result, Rahkema and Sohail [6] showed that $(S;T_{1},T_{2})$ is non-
embeddable if $T_{1}$ and $T_{2}$ are both completely regular (respectively,
Clifford) semigroups but $S$ is not completely regular (respectively,
Clifford). In this section we shall consider the amalgams $(S;T_{1},T_{2})$ in
which $T_{1}$ and $T_{2}$ are both inverse semigroups but $S$ is not such—the
non-embeddability of such amalgams was left as an open problem in [6].

######  Definition 3.1.

Let $\mathcal{C}$ be a class of semigroups. Suppose that $T_{1}\in\mathcal{C}$
contains an isomorphic copy of a semigroup $S$ via $\phi_{1}:S\longrightarrow
T_{1}$. Then the pair $(S;T_{1})$ will be called an antiamalgamation pair for
$\mathcal{C}$ if for every $T_{2}\in\mathcal{C}$ and every monomorphism
$\phi_{2}:S\longrightarrow T_{2}$ the amalgam
$(S;T_{1},T_{2};\phi_{1},\phi_{2})$ is non-embeddable (in any semigroup).

######  Theorem 3.2.

Let $T_{1}$ be an inverse semigroup and $\phi_{1}:S\longrightarrow T_{1}$ be a
monomorphism such that $S\phi_{1}$ is right as well as left ample in $T_{1}$.
If $S$ is non-inverse then $(S;T_{1})$ is an antiamalgamation pair for the
class of inverse semigroups.

###### Proof.

Let $S$, $T_{1}$ and $\phi_{1}$ be as described in the statement of the
theorem. Let $T_{2}$ be an inverse semigroup admitting a monomorphism
$\phi_{2}:S\longrightarrow T_{2}$. Given $s\in S$, let us denote $s\phi_{1}$
and $s\phi_{2}$ by $s_{1}$ and $s_{2}$, respectively. Identifying $S$ with its
isomorphic copies $S\phi_{1}$ and $S\phi_{2}$, and using the properties of
inverses, we may calculate in $T_{1}*_{S}T_{2}$:

$\begin{split}ss_{1}^{-1}ss_{2}^{-1}&=ss_{2}^{-1},\hskip
10.0ptss_{2}^{-1}ss_{1}^{-1}=ss_{1}^{-1},\\\
s_{2}^{-1}ss_{1}^{-1}s&=s_{2}^{-1}s,\hskip
10.0pts_{1}^{-1}ss_{2}^{-1}s=s_{1}^{-1}s.\end{split}$ (3.1)

Since $S\phi_{1}$ is right and left ample in $T_{1}$, the identification of
$S$ with $S\phi_{1}$ also gives,

$ss_{1}^{-1},\,s_{1}^{-1}s\in S.$

By the commutativity of idempotents in $T_{2}$, we may write from (3.1):

$ss_{1}^{-1}=ss_{2}^{-1},\hskip 10.0pts_{1}^{-1}s=s_{2}^{-1}s.$ (3.2)

Now, using (3.2), we calculate in $T_{1}*_{S}T_{2}$:

$s_{1}^{-1}=s_{1}^{-1}(ss_{1}^{-1})=s_{1}^{-1}(ss_{2}^{-1})=(s_{1}^{-1}s)s_{2}^{-1}=(s_{2}^{-1}s)s_{2}^{-1}=s_{2}^{-1}.$
(3.3)

Because $S\phi_{1}$ and $S\phi_{2}$ are non-inverse, there exists $s\in S$
such that $s_{i}^{-1}\notin T_{i},\,1\leq i\leq 2$. The amalgam
$(S;T_{1},T_{2})$, therefore, fails to embed by (3.3). ∎

######  Example 3.3.

Let $\mathbb{L}$ denote the lattice of ample submonoids of the symmetric
inverse semigroup $\mathcal{I}_{n}$ over a finite chain $C_{n}:1<2<\cdots<n$,
given in [7], Figure 1. The chain
$\mathbb{L}_{\text{INV}}:\\{\iota\\}\subseteq\mathcal{OI}_{n}\subset\mathcal{RI}_{n}\subset\mathcal{I}_{n}^{\prime}\subset\mathcal{I}_{n}$
constitutes the sublattice of $\mathbb{L}$ comprising the inverse submonoids
of $\mathcal{I}_{n}$. This gives a (finite) set

$\\{(S,T):S\in\mathbb{L\smallsetminus
L}_{\text{INV}}~{},T\in\mathbb{L}_{\text{INV}}\text{ with }S\subseteq
T\mathcal{\\}}$

of antiamalgamation pairs for the class of inverse semigroups.

######  Theorem 3.4.

Let a non-inverse semigroup $S$ be made into a right (respectively, left)
ample semigroup by an associated inverse semigroup $T_{1}$ (respectively,
$T_{2}$). Then the amalgam $(S;T_{1},T_{2})$ is not embeddable (in any
semigroup).

###### Proof.

Let $S$, $T_{1}$ and $T_{2}$ be as given in the statement of the theorem.
Then, as before, the identification of $S$ with its isomorphic copies in
$T_{1}$ and $T_{2}$ gives equations (3.1). Since $S$ is right ample in $T_{1}$
and left ample in $T_{2}$, we have $s_{1}^{-1}s,\,ss_{2}^{-1}\in S$.
Subsequently, $ss_{1}^{-1}$, $ss_{2}^{-1}$ commute in $T_{1}$ and
$s_{1}^{-1}s$, $s_{2}^{-1}s$ commute in $T_{2}$. Using the argument from the
proof of Theorem 3.2, we can once more deduce equations (3.2) from equations
(3.1). However, equations (3.2) give $s_{1}^{-1}=s_{2}^{-1}$ implying (as in
the said proof) that $(S;T_{1},T_{2})$ is non-embeddable. ∎

##  4 Weak amalgamation

Given a subsemigroup $S$ of an inverse semigroup $T$, we define its dual to be
the subsemigroup

$S^{\prime}=\\{s^{-1}\in T:s\in S\\}.$

Defining $\alpha:S\longrightarrow S^{\prime}$ by $s\mapsto s^{-1}$, we have

$(xy)\alpha=\left(xy\right)^{-1}=y^{-1}x^{-1}=(y)\alpha(x)\alpha,\forall\,x,y\in
S,$

implying that $S$ and $S^{\prime}$ are anti-isomorphic. Clearly, if non-empty,
$S\cap S^{\prime}$ is an inverse subsemigroup of $S$ and $S^{\prime}$ with
$E(S)=E(S^{\prime})\subseteq S\cap S^{\prime}$, where $E(U)$ denotes the set
of idempotents of any semigroup $U$. Also, if $S$ is right (respectively,
left) ample in $T$ then $S^{\prime}$ is a left (respectively, right) ample
subsemigroup of $T$.

######  Lemma 4.1.

Let $T_{1}$ and $T_{2}$ be inverse semigroups containing isomorphic copies,
say $S_{1}$ and $S_{2}$, of a semigroup $S$. Then, there exists a bijection
between the sets $S_{1}\cup S^{\prime}_{1}$ and $S_{2}\cup S^{\prime}_{2}$.
Moreover, for any $x\in S_{1}\cup S^{\prime}_{1}$ one has:

$(x^{-1})\psi=(x\psi)^{-1}.$

###### Proof.

Let $\phi$ be the isomorphism from $S_{1}$ to $S_{2}$. Then
$\phi^{\prime}=\alpha_{1}^{-1}\circ\phi\circ\alpha_{2}$ is an isomorphism from
$S^{\prime}_{1}$ to $S^{\prime}_{2}$, where $\alpha_{i}:S_{i}\longrightarrow
S^{\prime}_{i}$, $i=1,2$, are the anti-isomorphisms defined by $s_{i}\mapsto
s_{i}^{-1},\,s_{i}\in S_{i}$. Let $x\in S_{1}\cap S^{\prime}_{1}$. Then
$x^{-1}\in S_{1}\cap S^{\prime}_{1}$, and, in particular, $x,x^{-1}\in S_{1}$.
Using the assumption that $\phi$ is an isomorphism, we have:

$\begin{split}x\phi&=(xx^{-1}x)\phi=(x)\phi(x^{-1})\phi(x)\phi,\vspace{6pt}\\\
(x^{-1})\phi&=(x^{-1}xx^{-1})\phi=(x^{-1})\phi(x)\phi(x^{-1})\phi.\end{split}$

Hence, we have $(x^{-1})\phi=((x)\phi)^{-1}$ whenever $x\in S_{1}\cap
S^{\prime}_{1}$. We can now calculate:

$x\phi=((x^{-1})^{-1})\phi=((x^{-1})\phi)^{-1}=x\phi^{\prime}.$

This implies that $\phi$ and $\phi^{\prime}$ agree on $S_{1}\cap
S^{\prime}_{1}$. Consequently, the map

$\psi=\phi\cup\phi^{\prime}:S_{1}\cup S^{\prime}_{1}\longrightarrow S_{2}\cup
S^{\prime}_{2}$

is well-defined. Using a similar argument one may also construct

$\psi^{-1}:S_{2}\cup S^{\prime}_{2}\longrightarrow S_{2}\cup S^{\prime}_{2},$

such that $\psi\circ\psi^{\prime-1}$ and $\psi^{\prime-1}\circ\psi$ are both
identity functions. This implies that $\psi$ is a bijection, as required.

Concerning the second part of the lemma, we have seen above that
$(x^{-1})\phi=((x)\phi)^{-1}$ for all $x\in S_{1}$. Similarly, it can be shown
that $((x^{\prime})^{-1})\phi^{\prime}=((x)\phi^{\prime})^{-1}$ for all
$x^{\prime}\in S^{\prime}_{1}$. This gives $(x^{-1})\psi=(x\psi)^{-1}$, and
the proof is complete. ∎

######  Proposition 4.2.

Let $S$ be any semigroup and $T_{i}$, $1\leq i\leq 2$, be inverse semigroups
admitting monomorphisms $\phi_{i}:S\longrightarrow T_{i}$. Then the amalgam
$(S,T_{1},T_{2};\phi_{1},\phi_{2})$ is weakly embeddable in an inverse
semigroup if and only if $(S;V_{1},V_{2})$ constitutes a special amalgam,
where $V_{i}$ is the inverse hull of $S\phi_{i}$ in $T_{i}$.

###### Proof.

($\implies$) Let $S$, $T_{1}$, $T_{2}$, $V_{1}$, $V_{2}$, and $\phi_{1}$,
$\phi_{2}$ be as described in the statement of the theorem. Assume that
$(S;T_{1},T_{2})$ is weakly embeddable in an inverse semigroup $W$, via
monomorphisms $\mu_{1}:T_{1}\longrightarrow W$ and
$\mu_{2}:T_{2}\longrightarrow W$. We shall denote $S\phi_{i},\,1\leq i\leq 2,$
by $S_{i}$.

Observe that any element of $V_{1}$ may be written in the form
$x_{1}x_{2}\cdots x_{n}$, where $x_{1},x_{2},\dots,x_{n}\in S_{1}\cup
S^{\prime}_{1}$ and for all $1\leq i\leq n-1$ the elements $x_{i},x_{i+1}$ are
not both in $S_{1}$ or $S^{\prime}_{1}\smallsetminus S_{1}$. Similarly, the
elements of $V_{2}$ can be written as $y_{1}y_{2}\cdots y_{m}$, where
$y_{1},y_{2},\dots,y_{m}\in S_{2}\cup S^{\prime}_{2}$ and for all $1\leq i\leq
m-1$ the elements $y_{i},y_{i+1}$ do not both belong to $S_{2}$ or
$S^{\prime}_{2}\smallsetminus S_{2}$. Also, for each $i\in\\{1,2\\}$ and $x\in
S_{i}$ we have:

$(x^{-1})\mu_{i}=(x\mu_{i})^{-1},\hskip 6.0pt\text{where }\,x^{-1}\in
S^{\prime}_{i}.$

We define $\theta:V_{1}\longrightarrow V_{2}$ by

$(x_{1}x_{2}\cdots x_{n})\theta=(x_{1}x_{2}\cdots x_{n})\mu_{1}\mu_{2}^{-1}.$

Then, $\theta$ is clearly an isomorphism from $V_{1}$ to $V_{2}$.

($\impliedby$) Let $(S;V_{1},V_{2};\phi_{1},\phi_{2})$ be made into a special
amalgam by the isomorphism $\nu:V_{1}\longrightarrow V_{2}$. Then

$\phi_{1}\circ\nu=\phi_{2}.$ (4.1)

Consider a semigroup $V$ admitting isomorphisms $\gamma_{i}:V\longrightarrow
V_{i}$, for each $1\leq i\leq 2$, with $V\cap V_{i}=\emptyset$, and

$\gamma_{1}\circ\nu=\gamma_{2};$ (4.2)

that is $(V;V_{1},V_{2})$ is a special amalgam. Then, being an inverse
semigroups amalgam, $(V;T_{1},T_{2};\gamma_{1},\gamma_{2})$ is embeddable in
an inverse semigroup, say $W$, via monomorphisms, say,
$\mu_{i}:T_{i}\longrightarrow W$. This implies that

$\gamma_{1}\circ\mu_{1}=\gamma_{2}\circ\mu_{2}.$ (4.3)

Now, using (4.1) and (4.2), we have:

$\phi_{1}\circ\gamma_{1}^{-1}=\phi_{1}\circ(\nu\circ\gamma_{2}^{-1})=(\phi_{1}\circ\nu)\circ\gamma_{2}^{-1}=\phi_{2}\circ\gamma_{2}^{-1}.$
(4.4)

Finally, using (4.3) and (4.4), we may calculate:

$\phi_{1}\circ\mu_{1}=\phi_{1}\circ\gamma_{1}^{-1}\circ\gamma_{1}\circ\mu_{1}=\phi_{2}\circ\gamma_{2}^{-1}\circ\gamma_{2}\circ\mu_{2}=\phi_{2}\circ\mu_{2}.$

Hence, $(S;T_{1},T_{2})$ is weakly embeddable. ∎

###  4.1 Weak amalgamation over rich ample semigroups

In this subsection, we introduce the notion of rich (right, left) ample
semigroups. Given inverse semigroups (respectively, groups) $T_{1}$ and
$T_{2}$, we show that an amalgam $(S;T_{1},T_{2})$ is weakly embeddable in an
inverse semigroup (respectively, group) if $S$ is rich ample in $T_{1}$ and
$T_{2}$. We begin by recalling that any inverse semigroup $S$ comes equipped
with the natural partial order:

$\forall x,y\in S,\,x\leq y\,\text{ iff }\,x=ey,\text{ for some }\,e\in E(S).$

######  Remark 4.3.

Let $U$ be an inverse semigroup. Then $uu^{-1}$ is the minimum idempotent,
with respect to the natural partial order, such that $(uu^{-1})u=u$. To see
this, let $eu=u$ for some idempotent $e\in U$. Then $u^{-1}e=u^{-1}$, and we
have

$uu^{-1}=euu^{-1}e=uu^{-1}e.$

This implies that $uu^{-1}\leq e$, and hence the assertion.

######  Definitions 4.4.

A subsemigroup $S$ of an inverse semigroup $T$ is called rich right ample in
$T$, if

$\forall x,y\in S,\,x^{-1}y\in S\cup S^{\prime},$

where $S^{\prime}=\\{z^{-1}\in T:z\in S\\}$ is the dual of $S$. We say that
$S$ is rich left ample in $T$ if

$\forall x,y\in S,\,xy^{-1}\in S\cup S^{\prime}.$

A subsemigroup of $T$ is called rich ample in $T$ if it is both rich right and
rich left ample in $T$. A submonoid $S$ of an inverse semigroup $T$ is rich
(right, left) ample in $T$ if it is such as a subsemigroup. By saying that $S$
is a rich (left, right) ample subsemigroup of $T$ we shall mean that $S$ is
rich (right, left) ample in $T$.

######  Lemma 4.5.

Let $S,\,S^{\prime}$ and $T$ be as defined above. Then

  1. 1.

$S$ is rich right (respectively, left) ample in $T$ if and only if (its dual)
$S^{\prime}$ is a rich left (respectively, right) ample in $T$,

  2. 2.

$S$ is (rich) ample in $T$ if and only if $S^{\prime}$ is (rich) ample in $T$.

###### Proof.

The proof is straightforward. ∎

######  Proposition 4.6.

Let $S$ be a subsemigroup of an inverse semigroup $T$. Then $S$ is rich ample
in $T$ if and only if the inverse hull of $S$ in $T$ equals $S\cup
S^{\prime}$.

###### Proof.

The proof is straightforward. ∎

######  Lemma 4.7.

Let $S$ be a rich ample subsemigroup of an inverse semigroup $T$. Then $S$ and
$S^{\prime}$ are the down closed subsemigroups of $S\cup S^{\prime}$ with
respect to the natural partial order.

###### Proof.

Let $s\in S$ and $s^{\prime}\in S^{\prime}$ be such that $s^{\prime}\leq s$,
where $\leq$ denotes the natural partial order on the inverse semigroup $S\cup
S^{\prime}$. Then $s^{\prime}=es$ for some idempotent $e\in E(S\cup
S^{\prime})$. Now, because $e\in S\cap S^{\prime}$, it follows (in particular)
from $e\in S$ that $s^{\prime}=es\in S$. Similarly, one can show that $s\leq
s^{\prime}$ implies that $s\in S^{\prime}$. ∎

######  Proposition 4.8.

A subsemigroup $S$ of an inverse semigroup $T$ is rich right ample if and only
if

  1. 1.

$S$ is right ample in $T$, and

  2. 2.

for all $x,y\in S$, $x^{-1}y=x^{-1}xa$, where $a\in S\cup S^{\prime}$.

###### Proof.

($\implies$) Let $S$ be rich right ample in $T$. Then for all $x\in U$, the
element $x^{-1}x$ belongs to $S\cup S^{\prime}$. Because $x^{-1}x$ is an
idempotent, we in fact have $x^{-1}x\in S\cap S^{\prime}\subseteq S$, meaning
that $S$ is right ample in $T$. Next, let $x,y\in S$ and observe that

$x^{-1}y=x^{-1}xx^{-1}y=x^{-1}xa,$

where $a=x^{-1}y\in S\cup S^{\prime}$. Hence the second condition is also
satisfied.

($\impliedby$) Let $S$ be a subsemigroup of an inverse semigroup $T$
satisfying Conditions (1) and (2) of the proposition. Let $x,y\in S$. Then by
the second condition, $x^{-1}y=x^{-1}xa$, where $a\in S\cup S^{\prime}$. Now,
$x^{-1}x\in S\cap S^{\prime}$, because, by the first condition, $S$ is right
ample. This implies that

$x^{-1}y=x^{-1}xa\in S\cup S^{\prime}\text{,}$

whence $S$ is rich right ample in $T$. ∎

######  Proposition 4.9.

A subsemigroup $S$ of an inverse semigroup $T$ is rich left ample if and only
if

  1. 1.

$S$ is left ample in $T$, and

  2. 2.

for all $x,y\in S$, $xy^{-1}=byy^{-1}$, where $b\in S\cup S^{\prime}$ .

###### Proof.

Similar to the proof of the above proposition. ∎

######  Remark 4.10.

If $\phi:T_{1}\longrightarrow T_{2}$ is a homomorphism of inverse semigroups
and $S$ is rich right (left) ample in $T_{1}$ then it can be easily verified
that such is $S\phi$ in $T_{2}$.

######  Theorem 4.11.

Let $T_{1}$ and $T_{2}$ be inverse semigroups admitting monomorphisms
$\phi_{1}$ and $\phi_{2}$ from a semigroup $S$, respectively, such that
$S\phi_{1}$ is rich right ample in $T_{1}$ but $S\phi_{2}$ is not such in
$T_{2}$. Then the amalgam $(S;T_{1},T_{2})$ fails to embed weakly in any
inverse semigroup.

###### Proof.

Let $S,T_{1},T_{2},\phi_{1}$ and $\phi_{2}$ be as described in the statement
of the Theorem. Assume, on the contrary, that $(S;T_{1},T_{2})$ is weakly
embeddable in an inverse semigroup. Let $(S;V_{1},V_{2})$ be the special
amalgam considered in Proposition 4.2. Clearly, $S_{1}=S\phi_{1}$ is rich
right ample in $V_{1}$. But then its image $S_{1}\mu=S\phi_{2}$ must be such
in $V_{2}$ and hence in $T_{2}$, a contradiction. ∎

######  Remark 4.12.

The dual statement, obtained by replacing ’rich right ample’ with ’rich left
ample’ in Theorem 4.11, can be proved on similar lines.

######  Lemma 4.13.

Let $T_{1}$ and $T_{2}$ be inverse semigroups containing isomorphic copies
$S_{1}$ and $S_{2}$ of a semigroup $S$, respectively, that are rich ample in
the respective oversemigroups. Then the postes $S_{1}\cup S^{\prime}_{1}$ and
$S_{2}\cup S^{\prime}_{2}$ are order-isomorphic.

###### Proof.

Let the map

$\psi=\phi\cup\phi^{\prime}:S_{1}\cup S^{\prime}_{1}\longrightarrow S_{2}\cup
S^{\prime}_{2}$

be as defined in Lemma 4.1. Then, it follows from the Lemma 4.7 that $\psi$ is
indeed an order-embedding. ∎

######  Theorem 4.14.

Let $T_{1}$ and $T_{2}$ be inverse semigroups containing isomorphic copies,
say $S\phi_{1}$ and $S\phi_{2}$, respectively, of a semigroup $S$. Assume also
that $S\phi_{i}$ is rich ample in $T_{i}$ for each $i\in\\{1,2\\}$. If $S$ is
not inverse then the amalgam $(S;T_{1},T_{2};\phi_{1},\phi_{2})$ is weakly
(but not strongly) embeddable in an inverse semigroup.

###### Proof.

Let $S_{1}=S\phi_{1}$ and $S_{2}=S\phi_{2}$. Then, by Proposition 4.6, the
inverse hull of $S_{i}$ in $T_{i}$, $1\leq i\leq 2$, is $S_{i}\cup
S^{\prime}_{i}$. The main objective will be proving that $S_{1}\cup
S^{\prime}_{1}$ and $S_{2}\cup S^{\prime}_{2}$ are isomorphic. We shall prove,
to this end, that the poset order-isomorphism $\psi:S_{1}\cup
S^{\prime}_{1}\longrightarrow S_{2}\cup S^{\prime}_{2}$, considered in Lemma
4.13, is a homomorphism of semigroups.

Clearly, if $x,y\in S_{1}$ (equivalently, $S^{\prime}_{1}$) then
$\psi(xy)=\psi(x)\psi(y)$. We prove that $(xz)\psi=(x)\psi(z)\psi$, for all
$x\in S_{1}$ and $z\in S^{\prime}_{1}$—that $(zx)\psi=(z)\psi(x)\psi$ will
follow from the symmetry. If $x\in S_{1}$ and $z\in S^{\prime}_{1}$ then
$z=y^{-1}$ for some $y\in S_{1}$ and we observe, by Proposition 4.9, that

$(xz)\psi=(xy^{-1})\psi=(byy^{-1})\psi=(b)\psi(yy^{-1})\psi,\,\,\text{where
}\,b\in S\cup S^{\prime}.$

We first show that $(yy^{-1})\psi=(y)\psi(y^{-1})\psi$. To this end, let us
first recall from Remark 4.3 that

$yy^{-1}=\text{min}\,\\{e\in E_{1}:ey=y\\},$ (4.5)

where $E_{1}=E(S_{1}\cup S^{\prime}_{1})$. Then, we note that the restriction
of $\psi$ to $E_{1}$ is identity and $ey=y$ for $e\in E_{1}$ if and only if
$(ey)\psi=(e\psi)(y\psi)=y\psi$. Thus

$(\\{e\in E_{1}:ey=y\\})\psi=\\{e\psi\in E_{1}\psi:(e)\psi(y)\psi=y\psi\\}.$
(4.6)

Now, recall from Lemma 4.13 the map $\psi:S_{1}\cup
S^{\prime}_{1}\longrightarrow S_{2}\cup S^{\prime}_{2}$ is an order-
isomorphism of posets, whence we may write from (4.5) and (4.6)

$\begin{split}(yy^{-1})\psi&=(\text{min}\\{e\in
E_{1}:ey=y\\})\psi\vspace{6pt}\\\ &=\text{min}(\\{e\in
E_{1}:ey=y\\})\psi\vspace{6pt}\\\ &=\text{min}\,\\{e\psi\in
E_{1}\psi:(e)\psi(y)\psi=y\psi\\}\vspace{6pt}\\\
&=(y)\psi(y\psi)^{-1}\text{,}\hskip 15.0pt\text{since }E_{1}\psi=E(S_{2}\cup
S^{\prime}_{2})\text{,}\\\ &=(y)\psi(y^{-1})\psi\text{,}\hskip 15.0pt\text{by
Lemma }\ref{Rmrk3.1}.\end{split}$

Coming back to proving that $(xz)\psi=(x)\psi(z)\psi$, we consider, in view of
Definitions 4.4, two cases.

If $xy^{-1}=b\in S_{1}$, then we have:

$\begin{split}(xz)\psi&=(xy^{-1})\psi=(byy^{-1})\psi=(b)\psi(yy^{-1})\psi\\\
&=(b)\psi(y)\psi(y^{-1})\psi=(by)\psi(y^{-1})\psi\\\
&=(byy^{-1}y)\psi(y^{-1})\psi=(xy^{-1}y)\psi(y^{-1})\psi\\\
&=(x)\psi(y^{-1}yy^{-1})\psi=(x)\psi(y^{-1})\psi=(x)\psi(z)\psi.\end{split}$

On the other hand, if $xy^{-1}\in S^{\prime}_{1}\smallsetminus S_{1}$, then,
using the right rich ampleness of $S^{\prime}_{1}$, we first write

$xz=xy^{-1}=(x^{-1})^{-1}y^{-1}=(x^{-1})^{-1}x^{-1}a.$

Note that $a\in S^{\prime}_{1}\smallsetminus S_{1}$, for otherwise we get
$xy^{-1}\in S_{1}$, a contradiction. Now, one may calculate:

$\begin{split}(xz)\psi&=(xy^{-1})\psi=((x^{-1})^{-1}y^{-1})\psi\\\
&=((x^{-1})^{-1}x^{-1}a)\psi=((x^{-1})^{-1}x^{-1})\psi(a)\psi\\\
&=(xx^{-1})\psi(a)\psi=(x)\psi(x^{-1})\psi(a)\psi\\\
&=(x)\psi(x^{-1}a)\psi=(x)\psi(x^{-1}xx^{-1}a)\psi\\\
&=(x)\psi(x^{-1}xy^{-1})\psi=(x)\psi(x^{-1}x)\psi(y^{-1})\psi\\\
&=(xx^{-1}x)\psi(y^{-1})\psi=(x)\psi(y^{-1})\psi\\\
&=(x)\psi(z)\psi\end{split}$

This completes the proof that $\psi:S_{1}\cup S^{\prime}_{1}\longrightarrow
S_{2}\cup S^{\prime}_{2}$ is an isomorphism of semigroups. That
$(S;T_{1},T_{2})$ is weakly embeddable follows from Proposition 4.2. The
amalgam $(S;T_{1},T_{2})$, however, fails to embed strongly by Propositions
4.8 and 4.9, and Theorem 3.2 (alternatively, Theorem 3.4). ∎

######  Corollary 4.15.

Consider an amalgam $\mathcal{A}=(S;G_{1},G_{2};\phi_{1},\phi_{2})$, in which
$G_{1}$ and $G_{2}$ are groups and $S$ is not a group. Let $S\phi_{i}$ be rich
ample in $G_{i}$ for each $i\in\\{1,2\\}$. Then $\mathcal{A}$ is weakly
embeddable in a group.

###### Proof.

By the above theorem $(S;G_{1},G_{2})$ weakly embeds in an inverse semigroup,
say $W$. Let $\psi_{1}:G_{1}\longrightarrow W$ and
$\psi_{1}:G_{1}\longrightarrow W$ be the embedding monomorphisms. Then, by the
properties of a pushout, there exists a unique homomorphism

$\psi:G_{1}*_{S}G_{2}\longrightarrow W,$

such that $\eta_{i}\psi=\psi_{i}$ for $1\leq i\leq 2$, where $\eta_{i}$ are as
given in Figure 1. Clearly, $(S;G_{1},G_{2})$ embeds in
$(G_{1}*_{S}G_{2})\psi$, which is a subgroup of $W$ by the uniqueness of
inverses in an inverse semigroup. ∎

###  4.2 Connection with dominions

Let $U$ be a subsemigroup of a semigroup $S$. Then, recall, for instance from
[4], that an element $d\in S$ is said to be dominated by $U$ if for all
homomorphisms $f,g:S\longrightarrow T$ with $f|_{U}=g|_{U}$ we have
$(d)f=(d)g$. The set Dom${}_{S}U$ of all elements of $S$ dominated by $U$ is a
subsemigroup of $S$, called the dominion of $U$ in $S$.

######  Proposition 4.16.

An ample subsemigroup $S$ of an inverse semigroup $T$ is rich ample in $T$ if
and only if Dom${}_{T}S=S\cup S^{\prime}$.

###### Proof.

($\implies$) Let $S$ be rich ample in $T$. Then by Proposition 4.6, the
inverse hull of $S$ in $T$ is $S\cup S^{\prime}$. Now, it follows from
Proposition 1 of [7] that Dom${}_{T}S=S\cup S^{\prime}$.

($\impliedby$) Assume that Dom${}_{T}S=S\cup S^{\prime}$, with $S$ being an
ample in $T$. Then by Proposition 1 of [7] the inverse hull of $S$ in $T$ is
$S\cup S^{\prime}$. Consequently, $S$ is rich ample in $T$ by Proposition 4.6.
∎

######  Remark 4.17.

It follows from the above proposition and zigzag theorem (see, for instance,
[4], Theorem 8.3.3) that $S$ is rich ample in $T$ if and only if for all $t\in
T$, the equality $t\otimes 1=1\otimes t$ in the tensor product $T\otimes_{S}T$
implies that $t\in S\cup S^{\prime}$.

##  5 Ultra-rich ample semigroups

In this section we introduce a special class of rich ample semigroups, namely
the ultra-rich ample semigroups, that stems from our earlier considerations.
The reader may refer to Section 6 for some natural examples of rich and ultra-
rich ample semigroups.

######  Definitions 5.1.

Let a subsemigroup $S$ be rich right (respectively, left) ample in an inverse
semigroup $T$ such that the elements $a$ (respectively, $b$) given in
Proposition 4.8 (respectively, 4.9) are uniquely determined. Then we say that
$S$ is ultra-rich right (respectively, left) ample in $T$. We call $S$ ultra-
rich ample in $T$ if it is both ultra-rich right and ultra-rich left ample in
$T$. We also recall [1, page 21] that a semigroup is called unipotent if it
contains precisely one idempotent.

######  Proposition 5.2.

If $S$ is ultra-rich right (left) ample subsemigroup of an inverse semigroup
$T$ then $S$ is unipotent.

###### Proof.

Let $S$ be an ultra-rich right ample subsemigroup of an inverse semigroup $T$.
Let $e_{1},e_{2}\in E(S)$. Then

$e_{1}^{-1}e_{2}=e_{1}^{-1}e_{1}a$

for a unique $a\in S\cup S^{\prime}$. Because

$e_{1}^{-1}e_{2}=e_{1}^{-1}e_{1}(e_{1}^{-1}e_{2}),\hskip 4.0pt\text{and}\hskip
6.0pte_{1}^{-1}e_{2}=e_{1}^{-1}e_{1}(e_{2}),$

we get from the uniqueness of $a$:

$e_{1}e_{2}=e_{1}^{-1}e_{2}=e_{2}.$

Similarly, we may calculate $e_{2}e_{1}=e_{1}$, whence by the commutativity of
idempotents, $e_{1}=e_{2}$, and we conclude that $E(S)$ is a singleton.

Similarly, one can show that any ultra-rich left ample subsemigroup of an
inverse semigroup is unipotent. ∎

######  Corollary 5.3.

Let $S$ be an inverse semigroup. Then $S$ is ultra-rich right (left) ample if
and only if it is a group.

###### Proof.

Straightforward. ∎

######  Proposition 5.4.

Let $S$ be a subsemigroup of an inverse semigroup $T$. Then $S$ is ultra-rich
ample in $T$ if and only if it is unipotent and rich ample in $T$.

###### Proof.

($\implies$) A proof of the direct part follows from Propositions 5.2 and the
fact that any ultra-rich ample subsemigroup of $T$ is rich ample in $T$.

($\impliedby$) Let $S$ be a rich ample subsemigroup of an inverse semigroup
$T$ such that $E(S)=\\{e\\}$. This implies that $x^{-1}x=xx^{-1}=e$ for all
$x\in S\cup S^{\prime}$. Now, for any $y\in S$, there exists, by Condition (2)
of Proposition 4.8, an element $a\in S\cup S^{\prime}$ such that

$x^{-1}y=x^{-1}xa\;(=ea).$

To prove the uniqueness of $a$ let

$x^{-1}y=x^{-1}xc\;(=ec),$

for some $c\in S\cup S^{\prime}$. Then

$a=(aa^{-1})a=ea=x^{-1}xa=x^{-1}y=x^{-1}xc=ec=(cc^{-1})c=c.$

This proves that $U$ is ultra-rich right ample. Similarly, one can show that
$U$ is ultra-rich left ample. ∎

######  Lemma 5.5.

A subsemigroup $S$ of an inverse semigroup $T$ is ultra-rich right
(respectively, left) ample if and only if $S^{\prime}$ is an ultra-rich left
(respectively, right) ample subsemigroup of $T$.

###### Proof.

Straightforward. ∎

######  Corollary 5.6.

Let $S$ be an ultra-rich ample subsemigroup of an inverse semigroup $T$. Then
the inverse hull $S\cup S^{\prime}$ of $S$ is a subgroup of $T$.

###### Proof.

It follows from Proposition 4.6 that $S\cup S^{\prime}$ is an inverse
subsemigroup of $T$. Because $S$ and $S^{\prime}$ are ultra-rich ample, such
is $S\cup S^{\prime}$. The proof now follows from Corollary 5.3. ∎

######  Corollary 5.7.

Let $U$ be an ultra-rich ample subsemigroup of an inverse semigroup $S$ and
$\phi:S\longrightarrow T$ be a homomorphism of (inverse) semigroups. Then
$U\phi$ is an ultra-rich ample subsemigroup of $T$.

###### Proof.

Note that $U\phi$ is rich ample by Remark 4.10. On the other hand, being the
image of a group, $(U\cup U^{\prime})\phi$ is a subgroup of $T$. So, $(U\cup
U^{\prime})\phi$, and consequently $U\phi$, must contain a unique idempotent.
The corollary now follows by Proposition 5.4. ∎

It is an easy exercise to show that every subsemigroup of a finite group $G$
is a subgroup of $G$. Also, recall from Corollary 5.6 that the inverse hull of
an ultra-rich ample subsemigroup of an inverse semigroup $T$ is a subgroup of
$T$. Consequently, we make the following observation.

######  Remark 5.8.

Assume that $S$ is an ultra-rich ample subsemigroup of an inverse semigroup
$T$. If $T$ (or $S$) is finite then $S$ is a subgroup of $T$.

##  6 Examples

By an (ultra-) rich (right, left) ample semigroup we shall mean a semigroup
that is such in some inverse oversemigroup. Clearly every group is an inverse
as well as an ultra-rich ample semigroup, whereas every inverse semigroup is a
rich ample and hence an ample semigroup. In view of Remark 5.8, we shall
consider, in the first five examples, the infinite multiplicative inverse
monoid $\mathbb{Q}$ of rational numbers. Examples 6–9 will concern the finite
case.

  1. 1.

It is straightforward to observe that the (non-inverse) multiplicative monoid
$\mathbb{N}$ of natural numbers is ample but not rich (right, left) ample in
$\mathbb{Q}$.

  2. 2.

On the other hand, if $\mathbb{Q}^{+}_{1}$ denotes the submonoid of
$\mathbb{Q}$ containing the unit $1$ and the elements of the form $p/q$ such
that $p$ and $q$ are relatively prime positive integers with $p>q$. Then
$\mathbb{Q}^{+}_{1}$ is a non-inverse ultra-rich ample submonoid of
$\mathbb{Q}$.

  3. 3.

The non-inverse monoid $\mathbb{Q}^{+}_{1}\cup\\{0\\}$ is rich ample but not
ultra-rich (right, left) ample in $\mathbb{Q}$.

  4. 4.

The inverse monoid $\mathbb{Q}$ is not ultra-rich ample in itself.

  5. 5.

The subgroup $\mathbb{Q}\smallsetminus\\{0\\}$ is inverse, a well as ultra-
rich ample submonoid of $\mathbb{Q}$.

  6. 6.

Let $\mathcal{I}_{n}$, $n\geq 2$, denote the symmetric inverse monoid over a
finite chain ${C}_{n}$: $1<\cdots<n$ of natural numbers. Let
$\mathcal{DI}_{n}$ and $\mathcal{DI}^{+}_{n}$ denote the (non-inverse) ample
submonoids of $\mathcal{I}_{n}$ comprising, respectively, the order decreasing
and order increasing partial bijections [7]. To see $\mathcal{DI}_{n}$ and
$\mathcal{DI}^{+}_{n}$ are rich (right, left) ample in $\mathcal{I}_{n}$ for
$n\leq 3$ one simply needs to effectuate some computations. We show that
$\mathcal{DI}_{n}$ and $\mathcal{DI}^{+}_{n}$ are not rich (right, left) ample
in $\mathcal{I}_{n}$ for all $n\geq 4$. To this end, consider the following
elements of $\mathcal{DI}_{4}$:

$\begin{split}u&=\\{(4,3),(3,2)\\}\text{,}\\\
v&=\\{(4,4),(3,1)\\}.\end{split}$

Then

$u^{-1}v=\\{(3,4),(2,1)\\}\notin U\cup U^{\prime},$

implying that $\mathcal{DI}_{4}$ is not rich right ample in $\mathcal{I}_{n}$.
Similarly, one can show that $\mathcal{DI}_{5}$ is not rich left ample in
$\mathcal{I}_{n}$. That $\mathcal{DI}^{+}_{n}$ is not rich (right, left) ample
in $\mathcal{I}_{n}$ follows from Lemma 4.5. Because $\mathcal{DI}_{m}$
(respectively, $\mathcal{DI}^{+}_{m}$) is contained in $\mathcal{DI}_{n}$
(respectively, $\mathcal{DI}^{+}_{n}$) for $m\leq n$, it follows that
$\mathcal{DI}_{n}$ and $\mathcal{DI}^{+}_{n}$ are not rich (right, left) ample
in $\mathcal{I}_{n}$ for all $n\geq 4$.

  7. 7.

Let $\mathcal{I}_{n}$ ($n\geq 2$) denote the inverse monoid considered in the
previous example. As every inverse subsemigroup of an inverse semigroup is
rich ample, the members of the chain

$\mathbb{L}_{\text{INV}}:\\{\iota\\}\subseteq\mathcal{OI}_{n}\subset\mathcal{RI}_{n}\subset\mathcal{I}_{n}^{\prime}\subset\mathcal{I}_{n}$

from Example 3.3 are rich ample in $\mathcal{I}_{n}$. However, because
$\mathcal{OI}_{n}$, $\mathcal{RI}_{n}$, $\mathcal{I}^{\prime}_{n}$ and
$\mathcal{I}_{n}$ are not unipotent, by Proposition 5.2 none of them is ultra-
rich (right, left) ample in $\mathcal{I}_{n}$.

  8. 8.

The symmetric group $S_{n}$ is ultra-rich ample in $\mathcal{I}_{n}$.

  9. 9.

If $S$ is a group then it is straightforward to observe that every monogenic
subsemigroup of $S$ is ultra-rich ample. Thus, the monogenic subsemigroups of
$S_{\mathbb{N}}$ are ultra-rich ample in $\mathcal{I}_{\mathbb{N}}$.

Based on the above examples, we have the following containment diagram for
various classes of right ample subsemigroups of an inverse semigroup,
considered in earlier sections. There exist similar diagrams for the
corresponding classes of left ample and (two-sided) ample subsemigroups of an
inverse semigroup. (The oval representing ultra-rich ample semigroups
disappears in the finite case.)

######  Acknowledgement 1.

The author is thankful to Professor Victoria Gould for her comments that were
received during his visit to the University of York, UK. I am also grateful to
Professor Valdis Laan and Dr. Ülo Reimaa for their valuable comments on the
earlier drafts of this article. This research has been supported by Estonian
Science Foundation’s grants PRG1204.

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