Acta Sci. Math. (Szeged) (2021) 87, 367 – 379
doi: 10.14232/actasm-020-558-7
arXiv version: layout, fonts, pagination and numbering of lemmas, theorems and
formulas may differ from those in the ACTASM published paper; moreover,
references [12] and [13] in the published paper were printed with errors which
have been corrected in this version (the author was not responsible for those
errors – according to one of the editors of Acta Sci. Math. (Szeged), they
were caused by a mistake during their copy-editing process)

# On lattice isomorphisms of orthodox semigroups

Simon M. Goberstein

###### Abstract.

Two semigroups are lattice isomorphic if the lattices of their subsemigroups
are isomorphic, and a class of semigroups is lattice closed if it contains
every semigroup which is lattice isomorphic to some semigroup from that class.
An orthodox semigroup is a regular semigroup whose idempotents form a
subsemigroup. We prove that the class of all orthodox semigroups in which
every nonidempotent element has infinite order is lattice closed.
2020 Mathematics Subject Classification: primary 20M15, 20M18, 20M19;
secondary 08A30.

Department of Mathematics and Statistics, California State University, Chico,
CA 95929-0525, U.S.A.
e-mail<EMAIL_ADDRESS>
##  1\. Introduction

Let $S$ be a semigroup. The set of all subsemigroups of $S$ (including, by
convention, the empty one) is a lattice under set-theoretic inclusion, and the
relationship between the properties of this lattice and the properties of $S$
has been studied in numerous publications for over 60 years. We say that two
semigroups are lattice isomorphic whenever their subsemigroup lattices are
isomorphic. If $S$ is isomorphic or antiisomorphic to any semigroup which is
lattice isomorphic to $S$, then $S$ is called lattice determined in the class
of all semigroups, and if for each isomorphism $\Phi$ of the subsemigroup
lattice of $S$ onto that of a semigroup $T$, there exists an isomorphism or an
antiisomorphism $\varphi$ of $S$ onto $T$ such that $U\Phi=U\varphi$ for every
subsemigroup $U$ of $S$, then $S$ is termed strongly lattice determined. A
class of semigroups is said to be lattice closed (in the class of all
semigroups) if it contains every semigroup which is lattice isomorphic to some
semigroup from that class. Of course, a lattice closed class of semigroups may
contain members which are not lattice determined.
Finding lattice closed classes of semigroups and identifying lattice
determined semigroups in such classes are among the most important problems in
this area of research (see [13]). It is well known that the class of all
groups is not lattice closed. However, the classes of all torsion-free groups
and of all completely simple semigroups, which are neither groups nor left or
right zero semigroups, are lattice closed [13, §§ 34, 38], and it is also
easily seen (and well known) that the class of bands is lattice closed. All
semigroups mentioned in the preceding sentence are regular (in the sense of
von Neumann for rings). A regular semigroup is orthodox if the set of its
idempotents is a subsemigroup, and orthodox semigroups form one of the most
important classes of regular semigroups. Adopting for semigroups the standard
group-theoretic term, we will say that a semigroup is torsion-free if all of
its nonidempotent elements have infinite order. Note that according to Lemma
2.5 in Section 2, a monogenic inverse semigroup whose generators have infinite
order is torsion-free. The principal goal of this paper is to prove that the
class of all torsion-free orthodox semigroups is lattice closed (see Theorem
3.3 in Section 3). An important role in that proof is played by the author’s
theorem stating that every bisimple orthodox semigroup generated by a pair of
mutually inverse elements of infinite order is strongly lattice determined [5,
Theorem 4.1]. Using Theorem 3.3 (and its proof) together with Lemma 2.5, we
also show that the class of all monogenic orthodox semigroups whose generators
have infinite order is lattice closed and every semigroup in that class is
torsion-free (Proposition 3.8). In Section 2, we establish a few auxiliary
results and, for the reader’s convenience, review some basic semigroup-
theoretic concepts and include preliminaries on lattice isomorphisms of
semigroups.
We use [1] and [7] as standard references for the algebraic theory of
semigroups and, in general, follow the terminology and notation of these
monographs. We also refer to [13] for information about lattice isomorphisms
of semigroups and to [10] for a detailed description of the structure of
monogenic inverse semigroups.

##  2\. Preliminaries

We use the term “order” instead of “partial order” and refer to a linearly
ordered set as a chain. In this paper, $\mathbb{N}$ stands for the set of all
natural numbers (= positive integers), and
$\mathbb{N}_{0}\,(=\mathbb{N}\cup\\{0\\})$ for the set of all nonnegative
integers. Let $S$ be a semigroup. The equality (or “diagonal”) relation on $S$
will be denoted by $\Delta_{S}$. We say that $x\in S$ is a group element of
$S$ if it belongs to some subgroup of $S$; otherwise $x$ is a nongroup element
of $S$. The set of nongroup elements of $S$ will be denoted by $N_{S}$, and
the set of idempotents of $S$ by $E_{S}$. To indicate that $U$ is a
subsemigroup of $S$, we write $U\leq S$. Under the convention that
$\emptyset\leq S$, the set of all subsemigroups of $S$ ordered by inclusion is
a lattice which we denote by $\operatorname{Sub}(S)$. If
$w=w(x_{1},\ldots,x_{n})$ is a word in the alphabet
$\\{x_{1},\ldots,x_{n}\\}\subseteq S$, we will say shortly that $w$ is a word
in $x_{1},\ldots,x_{n}$, and if no confusion is likely, $w$ will be identified
with its value in $S$. As usual, $\langle X\rangle$ stands for the
subsemigroup of $S$ generated by $X\subseteq S$, and if
$X=\\{x_{1},\ldots,x_{n}\\}$ is a finite subset of $S$, then $\langle
X\rangle$ is written as $\langle x_{1},\ldots,x_{n}\rangle$. Let $x\in S$.
Then $\langle x\rangle$ is the cyclic subsemigroup of $S$ generated by $x$. If
$\langle x\rangle$ is a finite semigroup, the number of its elements is called
the order of $x$ and denoted by $o(x)$. If $\langle x\rangle$ is infinite, we
write $o(x)=\infty$ and say that $x$ has infinite order. If every
nonidempotent element of $S$ has infinite order, we call $S$ a torsion-free
semigroup.
If $S$ and $T$ are semigroups such that
$\operatorname{Sub}(S)\cong\operatorname{Sub}(T)$, then $S$ and $T$ are called
lattice isomorphic, and any isomorphism of $\operatorname{Sub}(S)$ onto
$\operatorname{Sub}(T)$ is referred to as a lattice isomorphism of $S$ onto
$T$. If $\Phi$ is a lattice isomorphism of $S$ onto $T$, then $\Phi$ is said
to be induced by a mapping $\varphi\colon S\rightarrow T$ if $U\Phi=U\varphi$
for all $U\leq S$. If $S$ is isomorphic or antiisomorphic to any semigroup
that is lattice isomorphic to it, then $S$ is called lattice determined, and
if each lattice isomorphism of $S$ onto a semigroup $T$ is induced by an
isomorphism or an antiisomorphism of $S$ onto $T$, we say that $S$ is strongly
lattice determined. A class $\mathcal{K}$ of semigroups is said to be lattice
closed (in the class of all semigroups) if $\mathcal{K}$ contains every
semigroup which is lattice isomorphic to some semigroup that belongs to
$\mathcal{K}$. It is easily seen (and well known) that bands, in general, are
not lattice determined – for instance, any left or right zero semigroup $S$ is
clearly lattice isomorphic to a chain (that is, a semilattice with a linear
natural order) having the same cardinality as $S$. On the other hand, as an
immediate corollary of [11, Proposition 1] (reproduced as [13, Proposition
3.2(a)]) we have:

######  Result 2.1.

(From [11, Proposition 1] or [13, Proposition 3.2(a)]) A semigroup which is
lattice isomorphic to a band is itself a band, that is, the class of all bands
is lattice closed.

For convenience of reference, we also record here the following two results
(which imply, in particular, that the class of torsion-free semigroups is
lattice closed).

######  Result 2.2.

[13, Lemma 34.8] The infinite cyclic group is strongly lattice determined.

######  Result 2.3.

(From [13, Subsections 31.1–31.5]) Let $S$ and $T$ be torsion-free semigroups,
and let $\Phi$ be a lattice isomorphism of $S$ onto $T$. Then $\Phi$ is
induced by a unique bijection $\varphi\colon S\rightarrow T$ defined by the
formula $\langle x\rangle\Phi=\langle x\varphi\rangle$ for all $x\in S$, and
$(x^{n})\varphi=(x\varphi)^{n}$ for all $x\in S$ and $n\in\mathbb{N}$. Thus
the infinite cyclic semigroup is strongly lattice determined.

If $S$ and $T$ are torsion-free semigroups and $\Phi$ is a lattice isomorphism
of $S$ onto $T$, the bijection $\varphi\colon S\rightarrow T$ in Result 2.3
will be called the $\Phi$-associated bijection of $S$ onto $T$.
Let $S$ be a semigroup. An element $x\in S$ is regular if there is
$x^{\prime}\in S$ such that $xx^{\prime}x=x$, and if all elements of $S$ are
regular, $S$ is a regular semigroup. A $\mathcal{D}$-class $D$ of a regular
semigroup $S$ is said to be combinatorial if all subgroups of $S$ contained in
$D$ are trivial (or, equivalently, if $H_{x}=\\{x\\}$ for all $x\in D$), and
$S$ itself is combinatorial if all $\mathcal{D}$-classes of $S$ are
combinatorial, so $S$ is combinatorial precisely when
$\mathcal{H}=\Delta_{S}$. If $x,x^{\prime}\in S$ satisfy $xx^{\prime}x=x$,
then by setting $y=x^{\prime}xx^{\prime}$, we have $xyx=x$ and $yxy=y$, in
which case, $x$ and $y$ are said to be mutually inverse. As in [4], we
indicate that $x$ and $y$ are mutually inverse elements of $S$ by writing
$x\perp y$, and the phrase “$x\perp y$ in $S$” will always mean that $x,y\in
S$ and $x\perp y$. We also denote by $V_{S}(x)$ the set of all inverses of
$x\in S$, so $y\in V_{S}(x)$ if and only if $x\perp y$ in $S$. In general, a
regular element of $S$ may have more than one inverse but if each $x\in S$ has
a unique inverse (usually denoted by $x^{-1}$), then $S$ is an inverse
semigroup. If $S$ is an inverse semigroup and $x\in S$, the inverse
subsemigroup $\langle x,x^{-1}\rangle$ of $S$ is called monogenic, and if
$S=\langle x,x^{-1}\rangle$ then $S$ is a monogenic inverse semigroup.
The bicyclic semigroup $\mathcal{B}(a,b)$ is often defined as a semigroup with
identity $1$ generated by the two-element set $\\{a,b\\}$ and given by one
defining relation $ab=1$ [1, §1.12]. Of course, $\mathcal{B}(a,b)$ can also be
defined without mentioning the identity as a semigroup given by the following
presentation: $\mathcal{B}(a,b)=\langle
a,b\;|\;aba=a,bab=b,a^{2}b=a,ab^{2}=b\rangle$. It is well known that
$\mathcal{B}(a,b)$ is a combinatorial bisimple inverse semigroup with
$b=a^{-1}$, each of its elements has a unique representation in the form
$b^{m}a^{n}$, where $m$ and $n$ are nonnegative integers (and
$a^{0}=b^{0}=ab$), and the semilattice of idempotents of $\mathcal{B}(a,b)$ is
a chain: $ab>ba>b^{2}a^{2}>\cdots$. Observe that if $S=\langle
x,x^{-1}\rangle$ is a monogenic inverse semigroup such that $xx^{-1}>x^{-1}x$,
then $S=\mathcal{B}(x,x^{-1})$, and if $x^{-1}x>xx^{-1}$, then
$S=\mathcal{B}(x^{-1},x)$. Let $\mathcal{C}=\\{(m,n)\colon
m,n\in\mathbb{N}_{0}\\}$ with multiplication defined as follows:
$(m,n)(p,q)=(m+p-r,n+q-r)$ where $r=\min\\{n,p\\}$. Then $\mathcal{C}$ is a
semigroup and the mapping $(m,n)\to b^{m}a^{n}$ is an isomorphism of
$\mathcal{C}$ onto $\mathcal{B}(a,b)$ (see [1, §1.12]). Therefore
$\mathcal{C}$ can be viewed as another copy of the bicyclic semigroup (in
[10], the bicyclic semigroup was, in fact, defined as the semigroup
$\mathcal{C}$ described above).
A comprehensive description of the structure of monogenic inverse semigroups
is given in [10, Chapter IX]. We will recall only a few basic facts about them
(see [10] for more details). Note, in particular, that if $S$ is a monogenic
inverse semigroup, then $\mathcal{D}=\mathcal{J}$, so the order relation on
the set of $\mathcal{J}$-classes of $S$ is actually an order relation on the
set of its $\mathcal{D}$-classes. Let $S=\langle x,x^{-1}\rangle$ be a
monogenic inverse semigroup. If $xx^{-1}=x^{-1}x$, then $S$ is a cyclic group.
As noted in the preceding paragraph, if $xx^{-1}<x^{-1}x$ or
$x^{-1}x<xx^{-1}$, then $S$ is a bicyclic semigroup. Assume that $S$ is
neither a cyclic group nor a bicyclic semigroup (that is, $xx^{-1}$ and
$x^{-1}x$ are incomparable with respect to the natural order on $E_{S}$). Then
the set of $\mathcal{D}$-classes of $S$ is a nontrivial chain with the largest
element $D_{x}$. Moreover, one of the following holds: (i) $S$ is a free
monogenic inverse semigroup, or (ii) $S$ has a smallest proper ideal $K$ (the
kernel of $S$), which is either a bicyclic semigroup or a cyclic group. In
case (i), all $\mathcal{D}$-classes of $S$ are combinatorial and form an
infinite chain: $D_{x}>D_{x^{2}}>\cdots>D_{x^{m}}>\cdots$, where the
$\mathcal{D}$-class $D_{x^{m}}$ consists of $(m+1)^{2}$ elements for each
$m\in\mathbb{N}$. In case (ii), there is $n\in\mathbb{N}$ such that the
$\mathcal{D}$-classes of $S$ form a finite chain:
$D_{x}>D_{x^{2}}>\cdots>D_{x^{n}}>D_{x^{n+1}}=K$, where for each $1\leq m\leq
n$, the $\mathcal{D}$-class $D_{x^{m}}$ is combinatorial and consists of
$(m+1)^{2}$ elements. If $S$ is either a cyclic group or a bicyclic semigroup,
it is convenient (and common) to assume that $S$ itself is the kernel of $S$,
and with this assumption in mind we can state that if $S$ is an arbitrary
monogenic inverse semigroup, then $S$ is not combinatorial if and only if it
has a kernel which is a nontrivial cyclic group (and is the only nontrivial
$\mathcal{H}$-class of $S$).
If $S=\langle x,x^{-1}\rangle$ is a free monogenic inverse semigroup, each
$s\in S$ can be uniquely written in the form $s=x^{-p}x^{q}x^{-r}$ where $q>0$
and $0\leq p,r\leq q$, and $x^{-p}x^{q}x^{-r}\in E_{S}$ if and only if
$p+r=q$; this copy of the free monogenic inverse semigroup was denoted in [10]
by $C_{3}$. If the set $\\{((m,n),(p,q))\in\mathcal{C}\times\mathcal{C}\colon
m+p=n+q>0\\}$ is equipped with multiplication of the direct square
$\mathcal{C}\times\mathcal{C}$ of the bicyclic semigroup $\mathcal{C}$, one
obtains another copy of the free monogenic inverse semigroup, denoted in [10]
by $C_{2}$. Clearly, $((m,n),(p,q))\in E_{C_{2}}$ if and only if $m=n$ and
$p=q$. By setting $x=((1,0),(0,1))$ in $C_{2}$, it is easily shown that
$C_{2}$ is isomorphic to $C_{3}=\langle x,x^{-1}\rangle$ (and a few other
isomorphic copies of the free monogenic inverse semigroup were described in
[10]). We are going to use several results about monogenic inverse semigroups
expressed in [10] in terms of $C_{2}$ and its congruences, and for the
reader’s convenience we will recall some relevant definitions and notations
from [10].
Let $S=\\{((m,n),(p,q))\in\mathcal{C}\times\mathcal{C}\colon m+p=n+q>0\\}$
with multiplication of the direct product $\mathcal{C}\times\mathcal{C}$. Let
$\rho$ be a congruence on $S$. The least $n\in\mathbb{N}$ such that
$((n,n),(0,0))$ and $((n+1,n+1),(0,0))$ are $\rho$-related is denoted by
$l(\rho)$; if such $n$ does not exist, put $l(\rho)=\infty$. Similarly,
$r(\rho)$ is the least natural number $n$ such that $((0,0),(n,n))$ and
$((0,0),(n+1,n+1))$ are $\rho$-related, and if no such $n$ exists, then
$r(\rho)=\infty$. By [10, Lemma IX.2.6], if $l(\rho)<\infty$ and
$r(\rho)<\infty$, then $l(\rho)=r(\rho)$. Let $k\in\mathbb{N}$. As above, set
$x=((1,0),(0,1))$. If $l(\rho)=r(\rho)=k$ and $\rho|_{\langle x\rangle}$ has
infinitely many classes, then $\rho$ is said to have type $(k,\omega)$. If
$l(\rho)=\infty$ and $r(\rho)=k$, the type of $\rho$ is defined to be
$(k,\infty^{+})$, and if $l(\rho)=k$ and $r(\rho)=\infty$, the type of $\rho$
is $(k,\infty^{-})$. The weight of an element $u=((m,n),(p,q))$ of $S$ is the
number $w(u)=m+p$. Now for all
$u=((m,n),(p,q)),u^{\prime}=((m^{\prime}\\!,n^{\prime}),(p^{\prime}\\!,q^{\prime}))\in
S$, define

$\;u\rho_{(k,\,\omega)}u^{\prime}\;\,\Longleftrightarrow u=u^{\prime}\text{ or
}[w(u),w(u^{\prime})\geq k\text{ and }m-n=m^{\prime}-n^{\prime}],$
$u\rho_{(k,\,\infty^{+})}u^{\prime}\\!\Longleftrightarrow u=u^{\prime}\text{
or }[w(u),w(u^{\prime})\geq k\text{ and }(m,n)=(m^{\prime}\\!,n^{\prime})],$
$\\!\\!\\!\\!\\!u\rho_{(k,\,\infty^{-})}u^{\prime}\\!\Longleftrightarrow
u=u^{\prime}\text{ or }[w(u),w(u^{\prime})\geq k\text{ and
}(p,q)=(p^{\prime}\\!,q^{\prime})].$

Then $\rho_{(k,\,\omega)}$ is the unique congruence on $S$ of type
$(k,\,\omega)$ [10, Proposition IX.2.11], and $\rho_{(k,\,\infty^{+})}$ is the
unique congruence on $S$ of type $(k,\,\infty^{+})$ [10, Proposition IX.2.10];
of course, by symmetry, $\rho_{(k,\,\infty^{-})}$ is the unique congruence on
$S$ of type $(k,\,\infty^{-})$.
Let $Q$ be a semigroup with zero $0$. Denote $Q^{\ast}=Q\setminus\\{0\\}$. Let
$T$ be a semigroup disjoint from $Q$, and let $\varphi\colon Q^{\ast}\to T$ be
a partial homomorphism. For all $a,b\in T\cup Q^{\ast}$ define $a\circ b$ as
follows: $a\circ b=a(b\varphi)$ if $a\in T$ and $b\in Q^{\ast}$; $a\circ
b=(a\varphi)b$ if $a\in Q^{\ast}$ and $b\in T$; $a\circ
b=(a\varphi)(b\varphi)$ if $a,b\in Q^{\ast}$ and $ab=0$ in $Q$; $a\circ b=ab$
if either $a,b\in T$ or $a,b\in Q^{\ast}$ and $ab\neq 0$ in $Q$. Then $T\cup
Q^{\ast}$ is a semigroup called a retract ideal extension (or simply an ideal
extension) of $T$ by $Q$ whose operation $\circ$ is determined by the partial
homomorphism $\varphi$ (see [10, Section I.9]).
Consider again $S=\\{((m,n),(p,q))\in\mathcal{C}\times\mathcal{C}\colon
m+p=n+q>0\\}$ with multiplication of the direct product
$\mathcal{C}\times\mathcal{C}$, so $S$ is the free monogenic inverse semigroup
$C_{2}$. For any $n\in\mathbb{N}$, denote by $I_{n}$ the set $\\{u\in
C_{2}\colon w(u)\geq n\\}$. Then $I_{n}$ is an ideal of $C_{2}$, and we let
$M_{n}=C_{2}/I_{n}$. In the notation of this paragraph, we can state the
following result.

######  Result 2.4.

(From [10, Theorem IX.3.4 (i), (ii)]) Let $k\in\mathbb{N}$. Then
(i) $C_{2}/\rho_{(k,\,\infty^{+})}$ [respectively,
$C_{2}/\rho_{(k,\,\infty^{-})}$] is an ideal extension of the bicyclic
semigroup $\mathcal{C}$ by $M_{k}$ with the operation in $\mathcal{C}\cup
M^{\ast}_{k}$ determined by the partial homomorphism
$\theta:M^{\ast}_{k}\to\mathcal{C}$ where
$\theta:((m,n),(p,q))\mapsto(m,n)\;{\rm[respectively,}\;\theta:((m,n),(p,q))\mapsto(p,q){\rm]}$;
(ii) $C_{2}/\rho_{(k,\,\omega)}$ is an ideal extension of the cyclic group
$\mathbb{Z}$ by $M_{k}$ with the operation in $\mathbb{Z}\cup M^{\ast}_{k}$
determined by the partial homomorphism $\eta:((m,n),(p,q))\mapsto m-n$ of
$M^{\ast}_{k}$ to $\mathbb{Z}$.

Of course, in Case (ii) of Result 2.4, instead of $\mathbb{Z}$ one can use a
multiplicative infinite cyclic group $G=\langle g\rangle$, so
$C_{2}/\rho_{(k,\,\omega)}$ will be an ideal extension of $G$ by $M_{k}$ with
the operation in $G\cup M^{\ast}_{k}$ determined by the partial homomorphism
$\eta:((m,n),(p,q))\mapsto g^{m-n}$ of $M^{\ast}_{k}$ to $G$. Take any
nonidempotent element $((m,n),(p,q))$ of $M^{\ast}_{k}$. According to Result
2.4, $((m,n),(p,q))\theta\notin E_{\mathcal{C}}$ in both situations considered
in Case (i), and if in Case (ii) instead of $\mathbb{Z}$ we use a
multiplicative infinite cyclic group $G=\langle g\rangle$ with the identity
$1_{G}$, then $((m,n),(p,q))\eta\neq 1_{G}$.

######  Lemma 2.5.

Let $S=\langle x,x^{-1}\rangle$ be a monogenic inverse semigroup such that
$o(x)=\infty$. Then every nonidempotent element of $S$ has infinite order,
that is, $S$ is torsion-free.

Proof. If $S$ is a free monogenic inverse semigroup, [10, Lemma IX.3.8] shows
that every nonidempotent element of $S$ has infinite order (this also follows
from [10, Exercise IX.2.14(iii)], which asserts that $\langle s,s^{-1}\rangle$
is a free monogenic inverse subsemigroup of $S$ for every $s\in S\setminus
E_{S}$). Suppose that $S$ is not free. Then $S$ has a kernel $K$ (perhaps
coinciding with $S$), which is either a bicyclic semigroup or a cyclic group
(the latter must be infinite since $o(x)=\infty$).
Take any $s\in S\setminus E_{S}$. If $s\in K$, it is immediate that
$o(s)=\infty$. Now assume that $K$ is a proper ideal of $S$ and $s\notin K$.
Thus there is $k\geq 2$ such that $D_{x}>\cdots>D_{x^{k-1}}>D_{x^{k}}=K$, and
$s\in D_{x}\cup\cdots\cup D_{x^{k-1}}$. In the notation of the paragraph
preceding Result 2.4, $K=I_{k}$ and $D_{x}\cup\cdots\cup
D_{x^{k-1}}=M^{\ast}_{k}$. Moreover, if $K$ is an infinite cyclic group, then
$S=C_{2}/\rho_{(k,\,\omega)}$, and if $K$ is a bicyclic semigroup, then
$S=C_{2}/\rho_{(k,\,\infty^{+})}$ or $S=C_{2}/\rho_{(k,\,\infty^{-})}$. Since
all nonidempotent elements of a free monogenic inverse semigroup have infinite
order, $\langle s\rangle\nsubseteq M^{\ast}_{k}$ and hence $\langle
s\rangle\cap K\neq\emptyset$. Let $l$ be the smallest natural number such that
$s^{l}\in K$. Then $s^{l-1}\in M^{\ast}_{k}$. Using Result 2.4, the remark
that follows it, and the formulas for the powers of a nonidempotent element
$((m,n),(p,q))$ of $C_{2}$ given in [10, Lemma IX.3.8], one can deduce that if
$K$ is a bicyclic semigroup, then $s^{l}=(s\theta)(s^{l-1}\theta)\notin
E_{S}$, and if $K$ is an infinite cyclic group, then
$s^{l}=(s\eta)(s^{l-1}\eta)\notin E_{S}$. We have shown that in both cases
$s^{l}\in K\setminus E_{S}$. If $o(s)<\infty$, the cyclic semigroup $\langle
s\rangle$ is finite and since $s^{l}\in\langle s\rangle$, we would have
$o(s^{l})<\infty$, contradicting the fact that all nonidempotent elements of
$K$ have infinite order. Therefore $o(s)=\infty$. The proof is complete.
$\Box$
An orthodox semigroup is a regular semigroup in which the idempotents form a
subsemigroup. By [7, Theorem VI.1.1], if $S$ is a regular semigroup, the
following conditions are equivalent: (a) $S$ is orthodox; (b)
$V_{S}(e)\subseteq E_{S}$ for all $e\in E_{S}$; (c) $V_{S}(b)V_{S}(a)\subseteq
V_{S}(ab)$ for all $a,b\in S$. It follows that if $a\perp b$ in an orthodox
semigroup $S$, then $a\in N_{S}$ if and only if $b\in N_{S}$, and $a^{n}\perp
b^{n}$ for all $n\in\mathbb{N}$, which implies that $o(a)=o(b)$. According to
the terminology introduced by the author in [4] (by analogy with the inverse
semigroup case), a monogenic orthodox semigroup is an orthodox semigroup
generated by a pair of mutually inverse elements. In what follows, the phrase
“let $A=\langle a,b\rangle$ be a monogenic orthodox semigroup” will always
mean that $A$ is an orthodox semigroup with $a\perp b$ in $A$.
It was shown by Hall (see [6, Theorem 2] or [7, Theorem VI.1.10]) that a
regular semigroup $S$ is orthodox if and only if for all $x,y\in S$, if
$V_{S}(x)\cap V_{S}(y)\neq\emptyset$ then $V_{S}(x)=V_{S}(y)$. Let $S$ be an
arbitrary orthodox semigroup. Since $E_{S}$ is a band, it is a semilattice $Y$
of rectangular bands $E_{\alpha}\,({\alpha}\in Y)$, that is,
$E_{S}=\bigcup_{\alpha\in Y}E_{\alpha}$ where $Y$ is a semilattice and
$E_{\alpha}\,(\alpha\in Y)$ are pairwise disjoint rectangular bands satisfying
$E_{\alpha}E_{\beta}\subseteq E_{\alpha\beta}$ for all $\alpha,\beta\in Y$
(see [6, §3] or [7, Section VI.1]); as in [6] and [7], we may denote
$E_{\alpha}$ by $E(e)$ if $e\in E_{\alpha}$. Now let $\gamma_{S}=\\{(x,y)\in
S\times S\colon V_{S}(x)=V_{S}(y)\\}$. If no confusion is likely, we will omit
the subscript $S$ in $\gamma_{S}$ and in $V_{S}(x)$ for all $x\in S$. It was
proved by Hall that $\gamma$ is the smallest inverse semigroup congruence on
$S$ (see [6, Theorem 3] or [7, Theorem VI.1.12]), so
$S/\gamma\,(=S\gamma^{\natural})$ is the maximum inverse semigroup homomorphic
image of $S$. According to [6, Remark 1], $e\gamma=V(e)=E(e)$ for all $e\in
E_{S}$. Thus $e\gamma$ is a rectangular band for every $e\in E_{S}$, the
semilattices $E_{S/\gamma}$ and $Y$ are isomorphic, and $s\gamma^{\natural}\in
E_{S/\gamma}$ if and only if $s\in E_{S}$ for all $s\in S$. It follows that
for any $x\in S$, we have $o(x)=\infty$ if and only if $o(x\gamma)=\infty$.
Therefore $S$ is torsion-free if and only if $S/\gamma$ is torsion-free. The
results reviewed in this paragraph will be used below without any further
reference or explanation.
Let $S$ be an orthodox semigroup. Since $\gamma\cap\mathcal{H}=\Delta_{S}$
(see [7, Chapter VI, formula (1.16)]), if $H$ is any $\mathcal{H}$-class of
$S$, the restriction of $\gamma^{\natural}$ to $H$ is an injection. As noted
in [2], from this observation and the fact that a monogenic inverse semigroup
contains a nontrivial $\mathcal{H}$-class if and only if it has a kernel which
is a nontrivial cyclic group (and is the only nontrivial $\mathcal{H}$-class
of that semigroup), one can deduce the following assertion:

######  Result 2.6.

(From [2, Proof of Theorem 2.1] Let $A=\langle a,b\rangle$ be a monogenic
orthodox semigroup. If $H$ is a nontrivial $\mathcal{H}$-class of $A$, then
$A$ has a kernel $K$ and $H\subseteq K$.

Let us show that the assertion stated in Result 2.6 can be made more precise.

######  Proposition 2.7.

Let $A=\langle a,b\rangle$ be a monogenic orthodox semigroup. Suppose that $H$
is a nontrivial $\mathcal{H}$-class of $A$. Then $H$ is a cyclic group
isomorphic to the kernel of $A/\gamma$ and $A$ has a completely simple kernel
$K$, which contains $H$ and is isomorphic to the direct product of $H$ and a
rectangular band.

Proof. As noted prior to Result 2.6, the restriction of $\gamma^{\natural}$ to
$H$ is an injection from $H$ to the kernel of $A/\gamma$, which is a
nontrivial cyclic group. Denote the kernel of $A/\gamma$ by $G$, and let
$K=G\left(\gamma^{\natural}\right)^{-1}$. Then $K$ is an ideal of $A$ and
$H\subseteq K$. Clearly, $E_{K}=\\{e\in K\colon e\gamma^{\natural}=1_{G}\\}$
where $1_{G}$ is the identity of $G$. If $e,f\in E_{K}$, then
$e\gamma^{\natural}=f\gamma^{\natural}$, so $e\perp f$ and hence
$(e,f)\in\mathcal{D}$. Therefore $K$ is a regular $\mathcal{D}$-class (=
$\mathcal{J}$-class) of $A$, which shows that $K$ is the kernel of $A$. If $e$
is any element of $E_{K}$, then $E_{K}=\\{f\in K\colon f\in
e\gamma\\}=e\gamma$. Thus $E_{K}$ is a rectangular band. By [9, Corollary
IV.3.5] (or [7, Exercise III.12]), $K\cong H\times E_{K}$. Since each $e\in
E_{K}$ is primitive, $K$ is completely simple. Clearly, the maximum inverse
semigroup homomorphic image of $H\times E_{K}$ is isomorphic to $H$, and
$K\gamma^{\natural}=G$, so $H\cong G$. The proof is complete. $\Box$

##  3\. The class of torsion-free orthodox semigroups is lattice closed

It is well known that cyclic groups and the bicyclic semigroup are the only
bisimple monogenic inverse semigroups. However, as shown in [4], the class of
bisimple monogenic orthodox semigroups is substantially more diverse. In
particular, the author constructed in [4] a family of pairwise nonisomorphic
bisimple orthodox semigroups $\mathcal{O}_{(\nu,\,\mu)}(a,b)$ indexed by
ordered pairs $(\nu,\mu)\in\mathbb{N}^{\ast}\times\mathbb{N}^{\ast}$ (where
$\mathbb{N}^{\ast}=\mathbb{N}\cup\\{\infty\\}$), each being generated by a
pair of mutually inverse elements $a$ and $b$ satisfying $ab=a^{2}b^{2}$ and
$ba\neq b^{2}a^{2}$, and proved that if $S$ is an arbitrary bisimple monogenic
orthodox semigroup with nongroup generators, then $S$ or its dual is
isomorphic to one of the semigroups of that two-parameter family.

######  Theorem 3.1.

(From [4, Lemma 2.7 and Theorem 2.9]) Let $S$ be an orthodox semigroup, let
$a$ be an arbitrary nongroup element of $S$, and let $b\in V(a)$. Then either
$\\{a,b,ab,ba\\}$ is a $\mathcal{D}$-class of $\langle a,b\rangle$ such that
$\langle a,b\rangle\setminus\\{a,b,ab,ba\\}$ is an ideal of $\langle
a,b\rangle$, or $\langle a,b\rangle$ is a bisimple monogenic orthodox
semigroup and $o(a)=o(b)=\infty$, in which case $\langle a,b\rangle$ or its
dual is isomorphic to $\mathcal{O}_{(\nu,\,\mu)}(a,b)$ for some
$\mu,\nu\in\mathbb{N}^{\ast}$.

As shown in [12] (or [13, Theorem 41.8]), the bicyclic semigroup is strongly
lattice determined. A much more general result was established in [5] where it
was proved that for all $\mu,\nu\in\mathbb{N}^{\ast}$, the semigroups
$\mathcal{O}_{(\nu,\,\mu)}(a,b)$ are strongly lattice determined (the bicyclic
semigroup is just one member of that infinite two-parameter family – namely,
$\mathcal{O}_{(1,1)}(a,b)$ is bicyclic); this is the main part of the
following theorem, from which it is obtained when the generators $a$ and $b$
of a bisimple monogenic orthodox semigroup $S=\langle a,b\rangle$ are assumed
to be nongroup.

######  Theorem 3.2.

[5, Theorem 4.1] Let $S=\langle a,b\rangle$ be an arbitrary bisimple monogenic
orthodox semigroup such that $a$ (and hence $b$) has infinite order. Then $S$
is strongly lattice determined.

We are ready to establish the main result of the paper:

######  Theorem 3.3.

Let $S$ be a torsion-free orthodox semigroup, and let $\Phi$ be a lattice
isomorphism of $S$ onto a semigroup $T$. Then $T$ is also a torsion-free
orthodox semigroup. Thus the class of all torsion-free orthodox semigroups is
lattice closed.

Proof. By Result 2.1, $E_{T}\neq\emptyset$ and $E_{T}$ is a subsemigroup of
$T$ such that $E_{T}=E_{S}\Phi$. Since the class of torsion-free semigroups is
lattice closed, we can apply Result 2.3 and denote by $\varphi$ the
$\Phi$-associated bijection of $S$ onto $T$. Group elements of $T$ are
regular. To prove that $T$ is an orthodox semigroup, it remains to show that
all nongroup elements of $T$ are regular.
Take an arbitrary $x\in N_{T}$, and let $a=x\varphi^{-1}$. By Result 2.2, the
infinite cyclic group is strictly lattice determined. Since $S$ is torsion-
free, if $a$ were a group element of $S$, it would be contained in some
infinite cyclic subgroup $G$ of $S$, so $x$ would belong to the infinite
cyclic subgroup $G\varphi$ of $T$, contradicting the assumption that $x\in
N_{T}$. Therefore $a\in N_{S}$. Choose an arbitrary $b\in V(a)$. Then $\langle
a,b\rangle$ is a monogenic orthodox semigroup such that $a,b\in N_{S}$ and
$o(a)=o(b)=\infty$. Let $y=b\varphi$. By Result 2.3, $o(x)=o(y)=\infty$. Note
that

$\langle a,b\rangle\Phi=\left(\langle a\rangle\vee\langle
b\rangle\right)\Phi=\langle a\rangle\Phi\vee\langle b\rangle\Phi=\langle
x\rangle\vee\langle y\rangle=\langle x,y\rangle.$

Suppose that $\langle a,b\rangle$ is a bisimple monogenic orthodox semigroup.
Then, by Theorem 3.2, it is strongly lattice determined. Therefore $\langle
x,y\rangle$ is also a bisimple monogenic orthodox semigroup which is
isomorphic or antiisomorphic to $\langle a,b\rangle$. In particular, $x\perp
y$ in $T$ and hence $x$ is a regular element of $T$.
Now assume that $\langle a,b\rangle$ is not bisimple. Then, according to
Theorem 3.1, $\\{a,b,ab,ba\\}$ is a $\mathcal{D}$-class of $\langle
a,b\rangle$ such that $\langle a,b\rangle\setminus\\{a,b,ab,ba\\}$ is an ideal
of $\langle a,b\rangle$. Let $I=\langle a,b\rangle\setminus\\{a,b,ab,ba\\}$
and $J=I\Phi$. Take an arbitrary $s\in\langle a,b\rangle$. Since $\langle
s,I\rangle=\langle s\rangle\vee I$ and $\langle s\varphi,J\rangle=\langle
s\varphi\rangle\vee J$, we conclude that

$\langle s,I\rangle\Phi=\left(\langle s\rangle\vee I\right)\Phi=\langle
s\rangle\Phi\vee I\Phi=\langle s\varphi\rangle\vee J=\langle
s\varphi,J\rangle.$

If $s\in I$ then $\langle s,I\rangle=I=\\{s\\}\cup I$. If $s\in\\{ab,ba\\}$,
it is also clear that $\langle s,I\rangle=\\{s\\}\cup I$. Finally, if
$s\in\\{a,b\\}$ then again $\langle s,I\rangle=\\{s\\}\cup I$ because
$s^{k}\in I$ for all $k\geq 2$. We have shown that $\langle
s,I\rangle=\\{s\\}\cup I$, and hence $\langle s\varphi,J\rangle=\langle
s,I\rangle\varphi=\left(\\{s\\}\cup I\right)\varphi=\\{s\\}\varphi\cup
I\varphi=\\{s\varphi\\}\cup J$. Since $a,b\notin I$, we have $x,y\notin J$.
Let $e=(ab)\varphi$ and $f=(ba)\varphi$. Then $e,f\in E_{T}$ and $e,f\notin J$
because $ab,ba\notin I$. Note that $\langle x,y\rangle=\langle
a,b\rangle\Phi=\left(\\{a,b,ab,ba\\}\cup
I\right)\varphi=\\{a\varphi,b\varphi,(ab)\varphi,(ba)\varphi\\}\cup I\varphi$.
Therefore we have

(3.4) $\langle x,y\rangle=\\{x,y,e,f\\}\cup J\text{ and }\\{x,y,e,f\\}\cap
J=\emptyset.$

Since $\langle a\rangle\cap I=\langle a\rangle\setminus\\{a\\}$, it follows
that

$\langle x\rangle\cap J=\langle a\rangle\Phi\cap I\Phi=\left(\langle
a\rangle\cap I\right)\Phi=\left(\langle
a\rangle\setminus\\{a\\}\right)\Phi=\langle
a\rangle\varphi\setminus\\{a\\}\varphi=\langle x\rangle\setminus\\{x\\},$

which shows that $x^{k}\in J$ for all $k\geq 2$. By symmetry, we also have
$y^{k}\in J$ for all $k\geq 2$.

######  Lemma 3.5.

One of the following two statements is true:
(i) $e=xy$ and $f=yx$, or
(ii) $e=yx$ and $f=xy$.

Proof. Since $e\in\langle x,y\rangle$, we can choose a shortest possible word
$u$ in $x,y$ representing $e$. Thus $e=u$ and no word in $x,y$ which is
shorter than $u$ has value $e$.
Case 1: The first letter of $u$ is $x$.
Since $ab\neq a$, it follows that $e\neq x$. Therefore $u=xv$ for some
nonempty word $v$ in $x,y$, and $e\neq v$ because $v$ is shorter than $u$.
Since $e=xv$, we conclude that $e\in\langle x,v\rangle=\langle
x\rangle\vee\langle v\rangle$. Let $c=v\varphi^{-1}$. Then

$ab=e\varphi^{-1}\in\left(\langle x\rangle\vee\langle
v\rangle\right)\Phi^{-1}=\langle x\rangle\Phi^{-1}\vee\langle
v\rangle\Phi^{-1}=\langle x\varphi^{-1}\rangle\vee\langle
v\varphi^{-1}\rangle=\langle a\rangle\vee\langle c\rangle=\langle a,c\rangle.$

If $c\in I$, then $ab\in\langle a,I\rangle=\\{a\\}\cup I$, which is not true.
Hence $c\in\\{a,b,ab,ba\\}$. If $c=a$, then $v=a\varphi=x$, so that
$e=x^{2}\in J$; a contradiction. Since $v\neq e$, it is clear that $c\neq ab$.
If $c=ba$, then $ab\in\langle a,ba\rangle$, contradicting the easily verified
fact that $\langle a,ba\rangle\setminus I=\\{a,ba\\}$. Since
$c\notin\\{a,ab,ba\\}$, it follows that $c=b$, so $v=b\varphi=y$. Therefore
$e=xy$.
Case 2: The first letter of $u$ is $y$.
By a symmetric argument to that used in Case 1, we deduce that $e=yx$.
Similarly to the above, starting with $f\in\langle x,y\rangle$, we prove that
either $f=xy$, or $f=yx$. Since $e\neq f$, note that if $e=xy$ then $f\neq xy$
and hence $f=yx$, whereas if $e=yx$ then $f\neq yx$ and so $f=xy$. Therefore
either (i) $e=xy$ and $f=yx$, or (ii) $e=yx$ and $f=xy$. This completes the
proof of Lemma 3.5. $\Box$
By Lemma 3.5, $xy$ and $yx$ are distinct idempotents of $T$ such that
$\\{xy,yx\\}=\\{e,f\\}$. Therefore, using (3.4), we conclude that

(3.6) $\langle x,y\rangle=\\{x,y,xy,yx\\}\cup J\text{ and }\\{x,y,xy,yx\\}\cap
J=\emptyset.$

Our goal is to prove that $x$ is a regular element. In fact, we are going to
show that $x\perp y$.
Since $xyx\in\langle x,y\rangle$, according to (3.6), we have

(3.7) $xyx\in\\{x,y,xy,yx\\}\cup J.$

If $xyx\in J$, then $xy=(xy)^{2}=(xyx)y\in\langle y,J\rangle=\\{y\\}\cup J$; a
contradiction because $xy\neq y$ and $xy\notin J$. Therefore $xyx\notin J$. If
$xyx=y$, then $xy=(xyx)y=y^{2}\in J$, which is not true. Thus $xyx\neq y$.
Assume that $xyx=xy$. Then $xy=(xyx)y=xy^{2}\in\langle x,J\rangle=\\{x\\}\cup
J$; a contradiction since $xy\neq x$ and $xy\notin J$. Hence $xyx\neq xy$. By
a dual argument, $xyx\neq yx$. We have shown that $xyx\notin\\{y,xy,yx\\}\cup
J$. In view of (3.7), it follows that $xyx=x$, that is, $x$ is a regular
element of $T$. By symmetry, we also have $yxy=y$, so $x\perp y$. The proof of
Theorem 3.3 is complete. $\Box$
Remark. By [8, Theorem] and comments after [3, Result 4.5], if
$A\\!=\\!\langle a,a^{-1}\rangle$ is a monogenic inverse semigroup, which is
neither bicyclic nor a group, and $\Psi$ is an isomorphism of the partial
automorphism monoid of $A$ onto that of a semigroup $B$, then $B\\!\cong\\!A$.
One step in the proof of this is [8, Lemma 1] according to which
$B\\!=\\!\langle x,y\rangle$ for mutually inverse $x$ and $y$ defined by
$\Delta_{\langle x\rangle}\\!=\\!\Delta_{\langle a\rangle}\Psi$ and
$\Delta_{\langle y\rangle}\\!=\\!\Delta_{\langle a^{-1}\rangle}\Psi$. The
assumption that $A$ is a monogenic inverse semigroup (neither bicyclic nor a
group) and partial automorphism monoids of $A$ and $B$ are isomorphic, is
crucial for all results of [8]. However, due to Theorem 3.1, certain arguments
in the proof of [8, Lemma 1] have natural analogues in the more general
setting of this paper. For instance, the argument in the last paragraph of the
proof of Theorem 3.3 is similar to the one in the final paragraph of the proof
of [8, Lemma 1].
Using Lemma 2.5 and Theorem 3.3 (and its proof), we can establish the
following result.

######  Proposition 3.8.

Let $A=\langle a,b\rangle$ be a monogenic orthodox semigroup such that $a$
(and hence $b$) has infinite order. Then $A$ is torsion-free and any semigroup
lattice isomorphic to $A$ is a torsion-free monogenic orthodox semigroup. Thus
the class of all monogenic orthodox semigroups whose generators have infinite
order is lattice closed and every semigroup in that class is torsion-free.

Proof. Since $A/\gamma=\langle a\gamma,b\gamma\rangle$ is a monogenic inverse
semigroup with $b\gamma=(a\gamma)^{-1}$ such that $o(a\gamma)=\infty$,
according to Lemma 2.5, $A/\gamma$ is torsion-free. Therefore $A$ is torsion-
free.
By Result 2.3, $\langle a\rangle$ and $\langle b\rangle$ are strongly lattice
determined. Thus $\langle a\rangle\Phi=\langle x\rangle$ and $\langle
b\rangle\Phi=\langle y\rangle$ for some $x,y\in X$ such that
$o(x)=o(y)=\infty$. Then

$X=A\Phi=\langle a,b\rangle\Phi=\left(\langle a\rangle\vee\langle
b\rangle\right)\Phi=\langle a\rangle\Phi\vee\langle b\rangle\Phi=\langle
x\rangle\vee\langle y\rangle=\langle x,y\rangle.$

If $A$ is bisimple, according to Theorem 3.2, $\langle x,y\rangle$ is a
bisimple monogenic orthodox semigroup which is isomorphic or antiisomorphic to
$\langle a,b\rangle$. Suppose that $A$ is not bisimple. Then, as shown in the
proof of Theorem 3.3, $x\perp y$. Moreover, since $A$ is torsion-free,
applying Theorem 3.3, we conclude that $X$ is a torsion-free orthodox
semigroup. Therefore in all cases $X=\langle x,y\rangle$ is a torsion-free
monogenic orthodox semigroup. This completes the proof. $\Box$

Acknowledgement

The author would like to thank a careful referee for useful comments, which
have helped shorten the proofs of Propositions 2.7 and 3.8, resulting in an
improved version of the paper.

## References

  * [1] A. H. Clifford and G. B. Preston, The Algebraic Theory of Semigroups, Math. Surveys No. 7, Amer. Math. Soc., Providence, R.I.; Vol. I, 1961; Vol. II, 1967.
  * [2] C. Eberhart and W. Williams, Elementary orthodox semigroups, Semigroup Forum, $\mathbf{29}$ (1984), 351–364.
  * [3] S. M. Goberstein, $\mathcal{P}\\!\mathcal{A}$-isomorphisms of inverse semigroups, Algebra Universalis, $\mathbf{53}$ (2005), 407–432.
  * [4] S. M. Goberstein, Bisimple monogenic orthodox semigroups, Semigroup Forum, $\mathbf{78}$ (2009), 310–325.
  * [5] S. M. Goberstein, Lattice isomorphisms of bisimple monogenic orthodox semigroups, Semigroup Forum, $\mathbf{83}$ (2011), 250–280.
  * [6] T. E. Hall, On regular semigroups whose idempotents form a subsemigroup, Bull. Australian Math. Soc., $\mathbf{1}$ (1969), 195–208.
  * [7] J. M. Howie, An Introduction to Semigroup Theory, Academic Press, London, 1976.
  * [8] A. L. Libih, Local automorphisms of monogenic inverse semigroups, Theory of Semigroups and Its Applications, No. 4, Saratov Univ. Press, 1978, 54–59 (in Russian).
  * [9] M. Petrich, Introduction to Semigroups, Merrill, Columbus, Ohio, 1973.
  * [10] M. Petrich, Inverse Semigroups, Wiley, New York, 1984.
  * [11] L. N. Shevrin, Lattice properties of idempotent semigroups. I, Sib. Mat. J., $\mathbf{6}$ (1965), 446–470 (in Russian).
  * [12] L. N. Shevrin, The bicyclic semigroup is determined by its subsemigroup lattice, Simon Stevin, $\mathbf{67}$ (1993) Supplement (December 1993), 49–53.
  * [13] L. N. Shevrin and A. J. Ovsyannikov, Semigroups and Their Subsemigroup Lattices, Kluwer Academic Publishers, Dordrecht, 1996.