The maximal subsemigroups of the ideals in a monoid of partial injections

Abstract

We study a submonoid of the well studied monoid $POI_n$ of all order-preserving partial injections on an n-element chain. The set $IOF_n^{par}$ of all partial transformations in $POI_n$ which are fence-preserving as well as parity-preserving forms a submonoid of $POI_n$. We describe Green’s relations and ideals of $IOF_n^{par}$. For each ideal of $IOF_n^{par}$, we characterize the maximal subsemigroups. We observe that there are three different types of maximal subsemigroups.

1 Introduction and preliminaries

Let $\overline{n}$ be a finite chain with n elements (where n is a positive integer), denoted as $\overline{n} = \{1< 2< \cdots < n\}$. A partial transformation $\alpha$ on the set $\overline{n}$ is a mapping from a subset A of $\overline{n}$ into $\overline{n}$. The domain (respectively, image) of $\alpha$ is denoted by . We denote by $PT_n$ the monoid (under composition) of all partial transformations on $\overline{n}$. The empty transformation is symbolized as $\varepsilon$, and it is the transformation with . Let $Id_{\overline{n}}$ be the set of all partial identities on $\overline{n}$, where id is the identity mapping on $\overline{n}$. A transformation $\alpha \in PT_n$ is called order-preserving if $x<y$ implies $x\alpha \le y\alpha$ for all . An injective $\alpha \in PT_n$ is called partial injection. The set $I_n$ (under composition) of all partial injections on $\overline{n}$ forms a monoid, referred to as the symmetric inverse semigroup, which was introduced by Wagner [23]. We denote by $POI_n$ the monoid of all partial order-preserving injections on $\overline{n}$.

Recall, a subsemigroup T of a semigroup S is called a maximal subsemigroup of S if T is contained in no other proper subsemigroup of S. An ideal of S is a subset I of S such that $xs\in I$ and $sx\in I$ for all $x\in I$ and $s\in S$. For more general background on semigroups and standard notations, we refer the reader to [1, 14].

There has been a growing interest in the study of a maximal subsemigroups within transformation semigroups. Notably, several researchers have made significant contributions. In [24], Yang characterized the maximal subsemigroups of the semigroup $O_n$ of all full order-preserving transformations. Dimitrova and Koppitz classified the maximal subsemigroups of the ideals of $O_n$ in [6]. Ganyushkin and Mazorchuk provided a description of the maximal subsemigroups of the semigroup $POI_n$ in [11]. Dimitrova and Koppitz offered a characterization of the maximal subsemigroups of the ideals of the semigroup $POI_n$ [5]. In [8], Dimitrova and Mladenova explored the maximal subsemigroups of the semigroup of all partial order-preserving transformations. Recently, Zhao and Hu have determined both the maximal subsemigroups and the maximal subsemibands of the ideals of the monoid of all orientation-preserving and extensive full transformations on $\overline{n}$ [25]. Additionally, in [12], Graham, Graham, and Rhodes have demonstrated that every maximal subsemigroup of a finite semigroup has certain features, and that every maximal subsemigroup must be one of a small number of types. As is often the case for semigroups, this classification depends on the description of maximal subgroups of certain finite groups. It is worth noting that maximal subsemigroups in many other families of transformation monoids have already been described or quantified, primarily through the work by Dimitrova, East, Fernandes, and other co-authors, as detailed in references such as [3, 4, 9, 13] and the associated literature.

A non-linear order that is close to a linear order in some sense is the so-called zig-zag order. The pair $(\overline{n}, \preceq )$ is called a zig-zag poset or fence if

$1 \prec 2 \succ \cdot \cdot \cdot \prec n - 1 \succ n$ if n is odd and $1 \prec 2 \succ \cdot \cdot \cdot \succ n - 1 \prec n$ if n is even, respectively.

The definition of the partial order $\preceq$ is self-explanatory. The number of order-preserving maps of fences and crowns, as well as transformations on fences, was first considered by Currie and Visentin [2] and Rutkowski [18]. The formula for the number of order-preserving self-mappings of a fence was also introduced by Rutkowski [18]. We observe that every element in a fence is either minimal or maximal, and for all $x,y\in \overline{n}$ with $x \prec y$, it follows $y\in \{x - 1, x + 1\}$. We say that a transformation $\alpha \in I_n$ is fence-preserving if $x \prec y$ implies $x\alpha \prec y\alpha$, for all $x, y \in {{\,\textrm{dom}\,}}(\alpha )$. We denote by $PFI_n$ the submonoid of $I_n$ of all fence-preserving partial injections on $\overline{n}$. Fernandes et al. determined the rank and a minimal generating set of the monoid of all order-preserving transformations on an n-element zig-zag ordered set [10]. It is worth mentioning that several other properties of monoids of fence-preserving full transformations were also studied in [7, 15,16,17, 20,21,22]. We denote by $IF_n$ the inverse subsemigroup of $PFI_n$ of all regular elements in $PFI_n$. It is easy to see that $IF_n$ is the set of all $\alpha \in PFI_n$ with $\alpha ^{-1}\in PFI_n$. In the present paper, we consider a submonoid of both monoids $IF_n$ and $POI_n$, i.e. a submonoid of $IOF_n=IF_n\cap POI_n$. Let $a\in {{\,\textrm{dom}\,}}(\alpha )$ for some $\alpha \in IOF_n$. If $a+1\in {{\,\textrm{dom}\,}}(\alpha )$ or $a-1\in {{\,\textrm{dom}\,}}(\alpha )$ then it is easy to verify that a and $a\alpha$ have the same parity, that is, a is odd if and only if $a\alpha$ is odd. However, if $a-1$ and $a+1$ are not in ${{\,\textrm{dom}\,}}(\alpha )$, then a and $a\alpha$ can have different parity. In order to exclude this case, we require that the image of any $a\in {{\,\textrm{dom}\,}}(\alpha )$ has the same parity as a. In this scenario, we refer to $\alpha$ as parity-preserving. Our focus lies on the set $IOF_n^{par}$ of all parity-preserving transformations in $IOF_n$. Notably, for any $\alpha \in IOF_n^{par}$, the inverse partial injection $\alpha ^{-1}$ exists and possesses order-preserving, fence-preserving, and parity-preserving. This observation implies that $IOF_n^{par}$ can be considered as an inverse submonoid of $I_n$, as explained in [19]. Initial studies showed that the monoid $IOF_n^{par}$ has an interesting structure, which value is to be examined in more detail.

In Sect. 2, we will repeat a characterization of the elements within $IOF_n^{par}$. We introduce a relation, denoted as $\sim$, on the power set $\mathcal {P}(\overline{n})$ of $\overline{n}$. This relation offers an alternative characterization the monoid $IOF_n^{par}$, and furthermore, this characterization leads us to an immediate descriptions of Green’s relation $\mathcal {J}$. Note that Green’s relations $\mathcal {R}, \mathcal {L}$, and $\mathcal {H}$ are already known, given that $IOF_n^{par}$ is an inverse submonoid of $I_n$. However, this paper deals mainly with the ideals of $IOF_n^{par}$. Of course, the sets $IOF_n^{par}$ and $\{\varepsilon \}$ are ideals of $IOF_n^{par}$, which are often referred to as the trivial ideals. We will demonstrate that the ideals of $IOF_n^{par}$ are of the form:

$I_{P^*}=\{\alpha \in IOF_n^{par}: {{\,\textrm{dom}\,}}(\alpha )\in P^*\}$

for particular subsets $P^*\subseteq \mathcal {P}(\overline{n})$. For each ideal $I\ne \{\varepsilon \}$ of $IOF_n^{par}$, we characterize the maximal subsemigroups of I. Our investigation will reveal that there are three distinct types of maximal subsemigroups within an ideal $I\ne \{\varepsilon \}$ of $IOF_n^{par}$. Therefore, the characterization of the ideals of $IOF_n^{par}$ (Sect. 2) and the description of the maximal subsemigroups of these ideals (Sect. 3) constitute the main results of this paper.

2 The ideals

In this section, we will describe the ideals of $IOF_n^{par}$. First, we will provide a characterization of the elements within $IOF_n^{par}$.

Proposition 1

[19] Let $p\le n$ and let $\alpha = \bigl ({\begin{matrix} d_1 &{}<&{}d_2&{}<&{} \cdots &{} < &{}d_p \\ m_1 &{} &{}m_2 &{} &{} \cdots &{} &{} m_p \end{matrix}}\bigr ) \in I_n$. Then $\alpha \in IOF_n^{par}$ if and only if the following four conditions hold:

(i) $m_1<m_2< \cdots <m_p$;

(ii) $d_1$ and $m_1$ have the same parity;

(iii) $d_{i+1}-d_i=1$ if and only if $m_{i+1}-m_i=1$ for all $i\in \{1,\dots ,p-1\}$;

(iv) $d_{i+1}-d_i$ is even if and only if $m_{i+1}-m_i$ is even for all $i\in \{1,\dots ,p-1\}$.

A set $X\subseteq \mathcal {P}(\overline{n})$ is called convex if, for all $A,B\in X$ with $A\subseteq B$ and for all $C\in \mathcal {P}(\overline{n})$, the following condition holds: if $A\subseteq C\subseteq B$, then $C\in X$. Here, $\mathcal {P}(A)$ denotes the power set of A, for any $A\subseteq \overline{n}$. The following is easy to verify:

Remark 1

If the empty set is contained in $X\subseteq \mathcal {P}(\overline{n})$, i.e. $\emptyset \in X$, then X is convex if and only if $\mathcal {P}(A)\subseteq X$ for all $A\subseteq X$.

We will observe that the domains of all partial transformations within an ideal form a convex set with an additional requirement. In order to describe this requirement, we define a partial order $\sim$ on $\mathcal {P}(\overline{n})$. Let $k_1, k_2\in \mathcal {P}(\overline{n})$ with $k_1=\{i_1<i_2< \cdots <i_k\}$ and $k_2= \{j_1<j_2< \cdots <j_l\}$ for some positive integers k, l. We put $k_1\sim k_2$, if the following three properties are satisfied:

(i) $k=l$;

(ii) $i_r$ and $j_r$ have the same parity for all $r\in \{1,\dots ,k\}$;

(iii) $i_r-i_{r-1}=1$ if and only if $j_r-j_{r-1}=1$ for all $r\in \{2,\dots ,k\}$.

It is worth mentioning that here (iii) (above) corresponds with (iii) in Proposition 1. In fact, for $A,B\subseteq \overline{n}$, there is with and if and only if $A\sim B$. So any is uniquely determined by domain and image, i.e. by ${{\,\textrm{dom}\,}}(\alpha )$ and . In particular, we have for any $\alpha \in IOF_n^{par}$. Using this description of the elements in $IOF_n^{par}$, we obtain immediately Green’s relations $\mathcal {L}, \mathcal {R}, \mathcal {H}$, and $\mathcal {J}$ (see [14]) for the inverse submonoid $IOF_n^{par}$ of $PT_n$:

Proposition 2

Let $a,b\in IOF_n^{par}$ then

(1) $a\mathcal {L}b$ if and only if ;

(2) $a\mathcal {R}b$ if and only if ;

(3) $a\mathcal {H}b$ if and only if $a=b$;

(4) $a\mathcal {J}b$ if and only if (or ${{\,\textrm{im}\,}}(a)\sim {{\,\textrm{im}\,}}(b)).$

For a set $X\subseteq \mathcal {P}(\overline{n})$ let

$I_X=\{\alpha \in IOF_n^{par}: {{\,\textrm{dom}\,}}(\alpha )\in X\}$.

Now, we are able to characterize the ideals of $IOF_n^{par}$.

Proposition 3

Any ideal I of $IOF_n^{par}$ is of the form $I=I_{P^*}$, where $P^*$ is a convex subset of $\mathcal {P}(\overline{n})$ with $\emptyset \in P^*$ such that $y\in P^*$ implies $z\in P^*$ for all $z\in \mathcal {P}(\overline{n})$ with $z\sim y$.

Proof

Let $P^*$ be a convex subset of $\mathcal {P}(\overline{n})$ as defined in the statement. Additionally, let $a\in I_{P^*}$ and $b\in IOF_n^{par}$. We observe that ${{\,\textrm{im}\,}}(a)\sim {{\,\textrm{dom}\,}}(a)$ by Proposition 1. Consequently, ${{\,\textrm{im}\,}}(a)\in P^*$. Further, we have ${{\,\textrm{dom}\,}}(ab)\subseteq {{\,\textrm{dom}\,}}(a)$, i.e. ${{\,\textrm{dom}\,}}(ab)\in P^*$ by Remark 1. This implies $ab\in I_{P^*}$. Moreover, we observe that ${{\,\textrm{im}\,}}(ba)\in P^*$ since ${{\,\textrm{im}\,}}(ba)\subseteq {{\,\textrm{im}\,}}(a)$ and Remark 1. This gives ${{\,\textrm{dom}\,}}(ba)\in P^*$ since ${{\,\textrm{dom}\,}}(ba)\sim {{\,\textrm{im}\,}}(ba)$. So $ba\in I_{P^*}$, i.e. $I_{P^*}$ is an ideal.

Conversely, let I be an ideal and let $P^*=\{{{\,\textrm{dom}\,}}(a):a\in I\}$. Note that any ideal contains the empty transformation $\varepsilon$. This gives $\emptyset \in P^*$. Now, let $A\in P^*$. This means that there exists $a\in I_{P^*}=I$ such that $A={{\,\textrm{dom}\,}}(\alpha )$. Let $B\subseteq A$ and $c\in Id_{\overline{n}}$ with ${{\,\textrm{dom}\,}}(c)=B$. Then $c\in IOF_n^{par}$, where $B={{\,\textrm{dom}\,}}(ca)$ and we observe that $ca\in I$. This provides $B={{\,\textrm{dom}\,}}(ca)\in P^*$. So, we have shown that $\mathcal {P}(A)\subseteq P^*$. Thus, $P^*$ is a convex set as established in Remark 1.

Let $y\in P^*$. Then there is $a\in I$ with ${{\,\textrm{dom}\,}}(a)=y$. Further, let $z\in \mathcal {P}(\overline{n})$ with $z\sim y$. Then, there is $b\in IOF_n^{par}$ with ${{\,\textrm{dom}\,}}(b)=z$ amd ${{\,\textrm{im}\,}}(b)=y$. We get that ${{\,\textrm{dom}\,}}(ba)={{\,\textrm{dom}\,}}(b)$ and $ba\in I$ because of $a\in I$. So $z={{\,\textrm{dom}\,}}(b)={{\,\textrm{dom}\,}}(ba)\in P^*$. It clear that $I\subseteq I_{P^*}$ by the definition of the sets $P^*$ and $I_{P^*}$. Let $b\in I_{P^*}$. Then there is $\gamma \in I$ with ${{\,\textrm{dom}\,}}(b)={{\,\textrm{dom}\,}}(\gamma )$ and then ${{\,\textrm{im}\,}}(\gamma )\sim {{\,\textrm{im}\,}}(b)$. Furthermore, there are $\alpha _1,\alpha _2\in IOF_n^{par}$ with ${{\,\textrm{dom}\,}}(\alpha _1)={{\,\textrm{dom}\,}}(b),{{\,\textrm{im}\,}}(\alpha _1)={{\,\textrm{dom}\,}}(\gamma ), {{\,\textrm{dom}\,}}(\alpha _2)={{\,\textrm{im}\,}}(\gamma )$, and ${{\,\textrm{im}\,}}(\alpha _2)={{\,\textrm{im}\,}}(b)$. So, we see that $\alpha _1\gamma \alpha _2=b$, i.e. $b\in I$. Consequently, $I=I_{P^*}$. $\square$

Clearly, $\{\emptyset \}$ is a convex set and $I_{\{\emptyset \}}=\{\varepsilon \}$ is the least ideal of $IOF_n^{par}$. Now, we determine the minimal ideals of $IOF_n^{par}$. An ideal I is called minimal if $I \ne \{\varepsilon \}$ and $M\subseteq I$ implies $M=I$, for all non-trivial ideals M of $IOF_n^{par}$.

Proposition 4

Let I be a non-trivial ideal of $IOF_n^{par}$. Then I is a minimal ideal of $IOF_n^{par}$ if and only if $I=I_{P^*}$, where either $P^*=\{\emptyset , \{1\}, \{3\},\dots ,\{n-1\}\}$ or $P^*=\{\emptyset ,\{2\},\{4\},\dots ,\{n\}\}$ if n is even and either $P^*=\{\emptyset , \{1\}, \{3\},\dots ,\{n\} \}$ or $P^*=\{\emptyset ,\{2\},\{4\},\dots ,\{n-1\}\}$ if n is odd, respectively.

Proof

Without loss of generality, we can assume that n is even. The proof for n is odd is similar.

Note that $\{1\}\sim \{3\}\sim \cdots \sim \{n-1\}$ and $\{2\}\sim \{4\}\sim \cdots \sim \{n\}$. By Proposition 3, we get that $I_{P^*}$ is a minimal ideal, whenever $P^*=P^O=\{\emptyset , \{1\}, \{3\},\dots ,\{n-1\} \}$ or $P^*=P^E=\{\emptyset ,\{2\},\{4\},\dots ,\{n\}\}$.

Conversely, let I be a minimal ideal of $IOF_n^{par}$. Then, there is $P^* \subseteq \mathcal {P}(\overline{n})$, satisfying the conditions in Proposition 3, such that $I=I_{P^*}$. Assume there is $a\in I$ such that ${{\,\textrm{rank}\,}}(a)\ge 2$. This provides that there exists $b\in I$ with ${{\,\textrm{rank}\,}}(b)=1$ and ${{\,\textrm{dom}\,}}(b)\in \mathcal {P}({{\,\textrm{dom}\,}}(a))$. Let $I'=\{b'\in I: {{\,\textrm{rank}\,}}(b')\le 1\}$. It is obvious that $I'$ is an ideal. This gives $\{\varepsilon \}\ne I'\subset I$, a contradiction to I is a minimal ideal. So, ${{\,\textrm{rank}\,}}(a)\le 1$ for all $a\in I$. We have now $P^* = \{{{\,\textrm{dom}\,}}(a): a\in I\}\subseteq \{\emptyset , \{1\}, \{2\},\dots ,\{n\}\}$. Assume $P^*\ne P^O$ and $P^*\ne P^E$. Moreover, assume $P^E\cap P^*\ne \{\emptyset \}$. Then $P^E\subseteq P^*$ since $P^*$ satisfies the conditions in Proposition 3. Since $I_{P^*}$ as well as $I_{P^E}$ are minimal ideals, we obtain $P^*=P^E$, a contradiction to $P^*\ne P^E$. Hence $P^E\cap P^*=\{\emptyset \}$. Similarly, we have $P^O\cap P^*=\{\emptyset \}$. But $P^E\cap P^*=P^O\cap P^*=\{\emptyset \}$ gives $P^*=\{\emptyset \}$, a contradiction to $I_{P^*}\ne \{\varepsilon \}$. Thus, $I=I_{P^*}$ with $P^*=P^O$ or $P^* = P^E$. $\square$

3 The maximal subsemigroups of the ideals on $IOF_n^{par}$

In this section, we determine the maximal subsemigroups of the ideals on $IOF_n^{par}$. First, we need a few technical tools. Let $y=\{i_1<i_2<\cdots < i_k\}\in P^*$ for some positive integer $k\ge 2$ and let $t\in \{1,2,\dots ,k\}$. Then we put $y^{[t]}=y\backslash \{i_t\}$. For $x,y\in P^*$, we write if there is $t\in \{1,\dots ,|y|\}$, such that $x=y^{[t]}$. Otherwise, we write .

Lemma 1

Let $y_1,y_2,z_1,z_2\in P^*$ with $|y_1|=|z_1| \ge 2$ and let $r,s\in \{1,2,\dots ,|y_1|\}$ such that $y_1^{[r]}=z_1^{[s]}= y_1\cap z_1$, $y_2\sim y_1$, $z_2\sim z_1$, and $y_2^{[r]}\sim z_2^{[s]}\sim y_1\cap z_1$. Then there are $\theta ,\delta \in I_{P^*}$ with $\text {rank} (\theta )={{\,\text {rank}\,}}(\delta )=|y_1|$ such that ${{\,\textrm{dom}\,}}(\theta \delta )=y_2^{[r]}$ and ${{\,\textrm{im}\,}}(\theta \delta )=z_2^{[s]}$.

Proof

Let $y_1,y_2,z_1,z_2\in P^*$ with $|y_1|=|z_1| \ge 2$ and let $r,s\in \{1,2,\dots ,|y_1|\}$ such that $y_1^{[r]}=z_1^{[s]}= y_1\cap z_1$, $y_2\sim y_1$, $z_2\sim z_1$, and $y_2^{[r]}\sim z_2^{[s]}\sim y_1\cap z_1$. Then there are $\theta ,\delta \in I_{P^*}$ with ${{\,\textrm{dom}\,}}(\theta )= y_2, {{\,\textrm{im}\,}}(\theta )=y_1, {{\,\textrm{dom}\,}}(\delta )=z_1$, and ${{\,\textrm{im}\,}}(\delta )= z_2$. Because of $y_1^{[r]}=z_1^{[s]} = y_1\cap z_1$, then ${{\,\textrm{rank}\,}}(\theta \delta ) = |y_1|-1$. Because of $y_2\sim y_1$ and $z_2\sim z_1$ with $y_2^{[r]}\sim z_2^{[s]}\sim y_1\cap z_1$, then ${{\,\textrm{dom}\,}}(\theta \delta )=y_2^{[r]}$ and ${{\,\textrm{im}\,}}(\theta \delta )=z_2^{[s]}$. $\square$

Let $I_{P^*}^u$ be the set of all $\alpha \in I_{P^*}$ with ${{\,\textrm{dom}\,}}(\alpha )\ne y_2^{[r]}$ or ${{\,\textrm{im}\,}}(\alpha )\ne z_2^{[s]}$, whenever $y_2, z_2\in P^*$ and $r,s\in \{1,2,\dots ,|y_1|\}$ with $y_2\sim y_1, z_2\sim z_1, y_1^{[r]}=z_1^{[s]}= y_1\cap z_1$, and $y_2^{[r]}\sim z_2^{[s]}\sim y_1\cap z_1$ for some $y_1,z_1\in P^*$ with $|y_1|=|z_1| \ge 2$. Directly from Lemma 1, we obtain: $\alpha \notin I_{P^*}^u$ if and only if there are $\theta , \delta \in I_{P^*}$ with ${{\,\textrm{rank}\,}}(\theta )={{\,\textrm{rank}\,}}(\delta )={{\,\textrm{rank}\,}}(\theta \delta )+1$ such that $\alpha =\theta \delta$. Since $P^*$ is a convex subset of $\mathcal {P}(\overline{n})$, we can conclude:

Corollary 1

Let $\alpha \in I_{P^*}$. Then $\alpha \notin I_{P^*}^u$ if and only if there are $\theta ,\delta \in I_{P^*}$ with ${{\,\textrm{rank}\,}}(\theta ), {{\,\textrm{rank}\,}}(\delta )>{{\,\textrm{rank}\,}}(\theta \delta )$ such that $\alpha =\theta \delta$.

Our initial observation is that all maximal subsemigroups of an ideal I have the form $I\backslash T$, where all transformations in T have the same rank.

Lemma 2

Let J be a maximal subsemigroup of $I_{P^*}$ and let $\alpha \notin J$. Then $\beta \in J$ for all $\beta \in I_{P^*}$ with $\text {rank} (\beta )\ne \text {rank} (\alpha )$.

Proof

Assume there is $\beta \in I_{P^*}$ such that $\beta \notin J$ with $\text {rank} (\beta )\ne \text {rank} (\alpha )$. Suppose $\text {rank} (\beta ) >\text {rank} (\alpha )$. We observe that $\langle J,\alpha \rangle$ is semigroup, where ${{\,\textrm{rank}\,}}(a\alpha ), {{\,\textrm{rank}\,}}(\alpha a) \le {{\,\textrm{rank}\,}}(\alpha )$ for all $a\in J$ and we see that $\langle J,\alpha \rangle \ne I_{P^*}$ since $\beta \in I_{P^*}$ but $\beta \notin \langle J,\alpha \rangle$. Moreover, $J\subset \langle J,\alpha \rangle \ne I_{P^*}$, a contradiction to J is a maximal subsemigroup of $I_{P^*}$. Suppose $\text {rank} (\beta )< \text {rank} (\alpha )$. Then we can show by contradiction that $J\subset \langle J,\beta \rangle \ne I_{P^*}$ in the same way. $\square$

For the remainder of this section, let $P^*$ be a convex subset of X, satisfying the conditions in Proposition 3, with $I_{P^*}\ne \{ \varepsilon \}$. Now, we determine several subsemigroups of $I_{P^*}$ and will show that they are exactly the maximal ones.

Let .

For $\Gamma _1\in P^*$ with $|\Gamma _1|\ge 2$, we define $A_{\Gamma _1}=\{\Gamma \in P^*: \Gamma \sim \Gamma _1\}$ and $B_{\Gamma _1}=\{\Gamma ^t: \Gamma \in A_{\Gamma _1}, t\in \{1,2,\dots ,|\Gamma |\}$.

We put $A=\{A_{\Gamma }: \Gamma \in P^*, |\Gamma |\ge 2, |A_{\Gamma }|\ge 2 \}$ and $B=\{B_{\Gamma }: \Gamma \in P^*, |\Gamma |\ge 2, |A_{\Gamma }|\ge 2 \}$.

Let $\Delta _1\in B\cup C$. We define

.

For $\Delta \in BC_{\Delta _1}$, we also define

.

And for a partition $Q=(Q_1,Q_2)$ of $BC_{\Delta _1}$, let

$T_{Q}=\{c\in I_{P^*} : {{\,\textrm{dom}\,}}(c)\in \tilde{\Delta },{{\,\textrm{im}\,}}(c)\in \hat{\Delta } \ \hbox {with} \ \tilde{\Delta } \in Q_1, \hat{\Delta }\in Q_2 \}$.

Lemma 3

Let $g\in P^*$ with $\{g\}\in C$ and $|BC_{\{g\}}|=1$. Then $I_{P^*}\backslash T_{\{g\}}$ with $T_{\{g\}}\subseteq I_{P^*}^u$ is a semigroup.

Proof

Let $\alpha \in T_{\{g\}}$. Because of $\{g\}\in C$ and $|BC_{\{g\}}|=1$, we get that $\alpha \in Id_{\overline{n}}$ and $T_{\{g\}}=\{\alpha \}$. Let $a,b\in I_{P^*}\backslash \{\alpha \}$. If ${{\,\textrm{rank}\,}}(a)\le {{\,\textrm{rank}\,}}(\alpha )$ or ${{\,\textrm{rank}\,}}(b)\le {{\,\textrm{rank}\,}}(\alpha )$ then $ab\ne \alpha$ because ${{\,\textrm{dom}\,}}(a)\not \sim {{\,\textrm{dom}\,}}(\alpha )$ and ${{\,\textrm{dom}\,}}(b)\not \sim {{\,\textrm{im}\,}}(\alpha )$. If ${{\,\textrm{rank}\,}}(a),{{\,\textrm{rank}\,}}(b)>{{\,\textrm{rank}\,}}(\alpha )$ and ${{\,\textrm{rank}\,}}(ab)={{\,\textrm{rank}\,}}(\alpha )$ then by Corollary 1, we have $ab \ne \alpha$ since $\alpha \in I_{P^*}^u$. This shows $ab\in I_{P^*}\backslash \{\alpha \}$. Consequently, $I_{P^*}\backslash \{\alpha \}$ is a semigroup. $\square$

Lemma 4

Let $y_1\in P^*$ with $|y_1|\ge 2$, $B_{y_1}\in B$, and $|BC_{B_{y_1}}|=1$. Then $I_{P^*}\backslash T_{B_{y_1}}$ with $T_{B_{y_1}}\subseteq I_{P^*}^u$ is a semigroup.

Proof

Let $\alpha \in T_{B_{y_1}}$ and let $a, b\in I_{P^*}\backslash T_{B_{y_1}}$. Note that if ${{\,\textrm{rank}\,}}(a)\le {{\,\textrm{rank}\,}}(\alpha )$ or ${{\,\textrm{rank}\,}}(b)\le {{\,\textrm{rank}\,}}(\alpha )$ then we can immediately deduce that ${{\,\textrm{dom}\,}}(ab)\not \sim {{\,\textrm{dom}\,}}(\alpha )$. Thus, $ab\notin B_{y_1}$ and $ab\in I_{P^*}\backslash T_{B_{y_1}}$. Suppose ${{\,\textrm{rank}\,}}(a),{{\,\textrm{rank}\,}}(b)>{{\,\textrm{rank}\,}}(\alpha )$ and ${{\,\textrm{rank}\,}}(ab)={{\,\textrm{rank}\,}}(\alpha )$. Then by Corollary 1, we have $ab\notin I_{P^*}^u$. This shows $ab\in I_{P^*}\backslash T_{B_{y_1}}$. Consequently, $I_{P^*}\backslash T_{B_{y_1}}$ is a semigroup. $\square$

Lemma 5

Let $Q=( Q_1, Q_2 )$ be a partition of $BC_{\Delta _1}$, where $\Delta _1\in B\cup C$. Then $I_{P^*}\backslash T_{Q}$ with $T_Q\subseteq I_{P^*}^u$ is a semigroup.

Proof

Let $a,b\in I_{P^*}\backslash T_{Q}$ and let $c\in T_Q$. If ${{\,\textrm{rank}\,}}(a)<{{\,\textrm{rank}\,}}(c)$ or ${{\,\textrm{rank}\,}}(b)<{{\,\textrm{rank}\,}}(c)$ then ${{\,\textrm{rank}\,}}(ab)<{{\,\textrm{rank}\,}}(c)$. So, $ab\in I_{P^*}\backslash T_{Q}$. If ${{\,\textrm{rank}\,}}(a)={{\,\textrm{rank}\,}}(b)={{\,\textrm{rank}\,}}(c)$ then we need to consider only the case that ${{\,\textrm{dom}\,}}(a)\in \Delta '$ for some $\Delta '\in Q_1$ and ${{\,\textrm{im}\,}}(b)\in \Delta ''$ for some $\Delta ''\in Q_2$. We get that ${{\,\textrm{im}\,}}(a)\in Y$ for some $Y\in Q_1$ and ${{\,\textrm{dom}\,}}(b)\in Y'$ for some $Y'\in Q_2$ because $a,b\in I_{P^*}\backslash T_Q$. We see that ${{\,\textrm{rank}\,}}(ab)<{{\,\textrm{rank}\,}}(c)$ because $Q_1 \cap Q_2= \emptyset$. Suppose that ${{\,\textrm{rank}\,}}(a)>{{\,\textrm{rank}\,}}(b)={{\,\textrm{rank}\,}}(c)$ and ${{\,\textrm{rank}\,}}(ab)= {{\,\textrm{rank}\,}}(c)$. Note if ${{\,\textrm{im}\,}}(b)\in \Delta \in Q_1$ then ${{\,\textrm{im}\,}}(ab)\in \Delta \in Q_1$. So, $ab\in I_{P^*}\backslash T_{Q}$. Suppose ${{\,\textrm{im}\,}}(b)\in \Delta$ for some $\Delta \in Q_2$. Then ${{\,\textrm{dom}\,}}(b) \in \Delta '$ for some $\Delta '\in Q_2$ because of $b\in I_{P^*}\backslash T_{Q}$ and we have ${{\,\textrm{dom}\,}}(b)\subset {{\,\textrm{im}\,}}(a)$. We put ${{\,\textrm{dom}\,}}(a)=y_1, {{\,\textrm{im}\,}}(a)=y_2$, and ${{\,\textrm{dom}\,}}(b)=y_2^{[t]}$ for some $t\in \{1,2,\dots ,|y_2|\}$. This gives ${{\,\textrm{dom}\,}}(ab)=y_1^{[t]}$. So, we see that $y_1^{[t]}\in \Delta '\in Q_2$ because of $y_1\sim y_2$. Then $ab\in I_{P^*}\backslash T_{Q}$. If ${{\,\textrm{rank}\,}}(b)>{{\,\textrm{rank}\,}}(a)={{\,\textrm{rank}\,}}(\alpha )$ then we obtain $ab\in I_{P^*}\backslash T_Q$ dually. If ${{\,\textrm{rank}\,}}(a),{{\,\textrm{rank}\,}}(b)>{{\,\textrm{rank}\,}}(\alpha )$ with ${{\,\textrm{rank}\,}}(ab)={{\,\textrm{rank}\,}}(\alpha )$ then by Corollary 1, we have $ab\notin I_{P^*}^u$. This shows $ab\in I_{P^*}\backslash T_{Q}$. Consequently, $I_{P^*}\backslash T_{Q}$ is a semigroup. $\square$

Now, we are able to characterize the maximal subsemigroups of $IOF_n^{par}$, which is the main result of this section.

Theorem 1

Let J be a subsemigroup of $I_{P^*}$. Then J is a maximal subsemigroup of $I_{P^*}$ if and only if J has one of the following forms.

(1) $J=I_{P^*}\backslash T_{\{g\}}$ with $|BC_{\{g\}}|=1$ for some $g\in P^*$ such that $\{g\}\in C$ and $T_{\{g\}}\subseteq I_{P^*}^u$;

(2) $J=I_{P^*}\backslash T_{B_{y_1}}$ with $|BC_{B_{y_1}}|=1$ for some $y_1\in P^*$ such that $|y_1|\ge 2$, $B_{y_1}\in B$, and $T_{B_{y_1}}\subseteq I_{P^*}^u$;

(3) $J=I_{P^*}\backslash T_{Q}$ for some partition $Q=( Q_1, Q_2 )$ of $BC_{\Delta _1}$, where $\Delta _1\in B \cup C$ and $T_Q\subseteq I_{P^*}^u$.

Proof

Let J be a maximal subsemigroup of $I_{P^*}$ and let $\alpha \in I_{P^*}\backslash J$.

Assume $\alpha \notin I_{P^*}^u$. Then by the definition of $I_{P^*}^u$, we get that ${{\,\textrm{dom}\,}}(\alpha )= y_2^{[r]}$ and ${{\,\textrm{im}\,}}(\alpha )= z_2^{[s]}$, where $y_2, z_2\in P^*$ and $r,s\in \{1,2,\dots ,|y_1|\}$ with $y_2\sim y_1, z_2\sim z_1, y_1^{[r]}=z_1^{[s]}= y_1\cap z_1$, and $y_2^{[r]}\sim z_2^{[s]}\sim y_1\cap z_1$ for some $y_1,z_1\in P^*$ with $|y_1|=|z_1| \ge 2$. By Lemma 1, we get that there are $\theta , \delta \in I_{P^*}$ with ${{\,\textrm{rank}\,}}(\theta )={{\,\textrm{rank}\,}}(\delta )=|y_2|$ such that $\theta \delta =\alpha$. Then $\theta , \delta \in J$ by Lemma 2, a contradiction to $\alpha \notin J$. So, $\alpha \in I_{P^*}^u$.

Suppose that $m={{\,\textrm{dom}\,}}(\alpha )$ and $m={{\,\textrm{im}\,}}(\alpha )$ for all $m\in P^*$ with $m \sim {{\,\textrm{dom}\,}}(\alpha )$. So, we can put $m={{\,\textrm{dom}\,}}(\alpha )={{\,\textrm{im}\,}}(\alpha )$. This provides $\{m\}\in C$, i.e. $BC_{\{m\}}=\{\{m\}\}$. So, $\alpha \in Id_{\overline{n}}$ and by the definition of $T_{\{m\}}$, we have that $T_{\{m\}}=\{\alpha \}$. It is easy to see that $J\cap \{\alpha \}=\emptyset$, this means $J\subseteq I_{P^*}\backslash \{\alpha \}$ and we have $I_{P^*}\backslash \{\alpha \}$ is a semigroup by Lemma 3. Together with J is a maximal subsemigroup of $I_{P^*}$, we have $J=I_{P^*}\backslash \{\alpha \}$.

Suppose there is $m \in P^*$ with $m\sim {{\,\textrm{dom}\,}}(\alpha )$ such that $m \ne {{\,\textrm{dom}\,}}(\alpha )$ or $m \ne {{\,\textrm{im}\,}}(\alpha )$ and there are $y_1\in P^*$ with $|y_1|\ge 2$ and $t\in \{1,2,\dots ,|y_1|\}$ such that for all $k\in P^*$ with $k\sim {{\,\textrm{dom}\,}}(\alpha )$ there is $y_3\in P^*$ with $y_3\sim y_1$ and $k=y_3^{[t]}$. This implies that there exists $y_2\in P^*$ with $y_2\sim y_1\ne y_1$. Then $A_{y_1}\in A$ and thus, $B_{y_1}\in B$. We can conclude that $\{B_{y_1}\}=BC_{B_{y_1}}$ and $|B_{y_1}|\ge 2$. So, we get ${{\,\textrm{dom}\,}}(\alpha ), {{\,\textrm{im}\,}}(\alpha )\in B_{y_1}$. Then there are $y_4,y_5\in A_{y_1}$ such that ${{\,\textrm{dom}\,}}(\alpha )=y_4^{[t]}$ and ${{\,\textrm{im}\,}}(\alpha )= y_5^{[t]}$.

Assume there is $\theta \in J$ with ${{\,\textrm{dom}\,}}(\theta ), {{\,\textrm{im}\,}}(\theta ) \in B_{y_1}$. There are $y_6, y_7 \in A_{y_1}$ such that ${{\,\textrm{dom}\,}}(\theta )=y_6^{[t]}, {{\,\textrm{im}\,}}(\theta )=y_7^{[t]}$. We have $\gamma _1,\gamma _2\in I_{P^*}$ with ${{\,\textrm{dom}\,}}(\gamma _1)=y_4, {{\,\textrm{im}\,}}(\gamma _1)=y_6, {{\,\textrm{dom}\,}}(\gamma _2)=y_7$, and ${{\,\textrm{im}\,}}(\gamma _2)=y_5$. This gives $\gamma _1, \gamma _2\in J$ because of ${{\,\textrm{rank}\,}}(\gamma _1), {{\,\textrm{rank}\,}}(\gamma _2)>{{\,\textrm{rank}\,}}(\alpha )$ together with Lemma 2. We get that $\gamma _1\theta \gamma _2=\alpha$, a contradiction to $\alpha \notin J$ and J is semigroup. Thus, $\theta \notin J$ for all $\theta \in I_{P^*}$ with ${{\,\textrm{dom}\,}}(\theta ), {{\,\textrm{im}\,}}(\theta ) \in B_{y_1}$, i.e. we have $\theta \notin J$ for all $\theta \in T_{B_{y_1}}$. This means $J\cap T_{B_{y_1}}=\emptyset$. So $J\subseteq I_{P^*}\backslash T_{B_{y_1}}$ and by Lemma 4, we have that $I_{P^*}\backslash T_{B_{y_1}}$ is a semigroup. Together with J is a maximal subsemigroup of $I_{P^*}$, we have $J=I_{P^*}\backslash T_{B_{y_1}}$.

Assume there is $\theta \in T_{B_{y_1}}$ with $\theta \notin I_{P^*}^u$. By Corollary 1, there are $a_1,a_2\in I_{P^*}$ with ${{\,\textrm{rank}\,}}(a_1),{{\,\textrm{rank}\,}}(a_2)>{{\,\textrm{rank}\,}}(a_1a_2)$ such that $\theta =a_1a_2$. This provides, $a_1,a_2\in J$ by Lemma 2, i.e. $\theta \in J$, a contradiction to $T_{B_{y_1}}\cap J = \emptyset$. Thus, $T_{B_{y_1}}\subseteq I_{P^*}^u$.

Suppose for all $y_1\in P^*$ with $|y_1|\ge 2$ and for all $t\in \{1,2,\dots ,|y_1|\}$, there is $k\in P^*$ with $k\sim {{\,\textrm{dom}\,}}(\alpha )$ such that $k\ne y_3^{[t]}$ for all $y_3\in P^*$ with $y_3\sim y_1$, and there is $m \in P^*$ with $m\sim {{\,\textrm{dom}\,}}(\alpha )$ such that $m \ne {{\,\textrm{dom}\,}}(\alpha )$ or $m \ne {{\,\textrm{im}\,}}(\alpha )$. Let $\lambda \in I_{P^*}\backslash J$ with ${{\,\textrm{dom}\,}}(\lambda )\sim {{\,\textrm{dom}\,}}(\alpha )$. Then we have the following four cases:

1. ${{\,\textrm{dom}\,}}(\lambda )=y_3^{[t]}$ for some $t\in \{1,2,\dots ,|y_3|\}$ and ${{\,\textrm{im}\,}}(\lambda )=z_3^{[s]}$ for some $s\in \{1,2,\dots ,|z_3|\}$ and $y_3,z_3\in P^*$ and there are $y_1,y_2,z_1,z_2\in P^*$ with $y_3\sim y_1\sim y_2\ne y_1$ and $z_3\sim z_1\sim z_2\ne z_1$.

Then there is $k\in P^*$ with $k\sim {{\,\textrm{dom}\,}}(\alpha )$ such that $k\ne y_3^{[t]}$. Assume $y_1\sim z_1$ and $t=s$. For $\theta \in I_{P^*}$, with ${{\,\textrm{dom}\,}}(\theta )=y_4^{[t]}, {{\,\textrm{im}\,}}(\theta )=z_4^{[s]}$, and $y_4\sim z_4\sim y_1$, we have $\theta \notin J$. Otherwise, let $\gamma _1,\gamma _2\in I_{P^*}$ with ${{\,\textrm{dom}\,}}(\gamma _1)=y_3, {{\,\textrm{im}\,}}(\gamma _1)=y_4, {{\,\textrm{dom}\,}}(\gamma _2)=z_4$, and ${{\,\textrm{im}\,}}(\gamma _2)=z_3$. This gives $\gamma _1, \gamma _2\in J$ because of Lemma 2. We get that $\gamma _1\theta \gamma _2=\lambda$, a contradiction to $\lambda \notin J$ because J is semigroup. Moreover, there are $\beta _1, \beta _2\in I_{P^*}$ with ${{\,\textrm{dom}\,}}(\beta _1)=y_3^{[t]}, {{\,\textrm{im}\,}}(\beta _1)={{\,\textrm{dom}\,}}(\beta _2)=k$, and ${{\,\textrm{im}\,}}(\beta _2)=z_3^{[s]}$. So, we observe that $\beta _1\beta _2=\lambda$. This show $\beta _1\notin J$ or $\beta _2\notin J$. Suppose $\beta _1\notin J$. Let $\theta \in I_{P^*}$ with ${{\,\textrm{dom}\,}}(\theta )=z_3^{[s]}, {{\,\textrm{im}\,}}(\theta )=y_3^{[t]}$, where $y_3,z_3\in P^*$ with $y_3\sim z_3\sim y_1$. As we have shown above, we have $\theta \notin J$. Note that $id\in J$ by Lemma 2 and ${{\,\textrm{rank}\,}}(\theta )<n$. So $\beta _1\in \langle J \cup \{\theta \}\rangle$ and we observe that $\beta _1=\alpha ' \theta \rho$ with $\alpha ' \in \langle J \cup \{\theta \} \rangle$ and $\rho \in J$. Since $y_3^{[t]}\ne k$, we can conclude that $\rho =\beta _1$, a contradiction to $\beta _1\notin J$. Dually, we can prove that $\beta _2 \notin J$ is not possible. Therefore, if $y_1\sim z_1$ then $t\ne s$.

Furthermore, for all $\beta \in I_{P^*}$ with ${{\,\textrm{dom}\,}}(\beta )=y_4^{[t]}$ and ${{\,\textrm{im}\,}}(\beta )=z_4^{[s]}$ for some $y_4,z_4\in P^*$ such that $y_4\sim y_1$ and $z_4\sim z_1$, we have $\beta \notin J$. Otherwise, $\theta _1\beta \theta '=\lambda$, where ${{\,\textrm{dom}\,}}(\theta _1)=y_3, {{\,\textrm{im}\,}}(\theta _1)=y_4, {{\,\textrm{dom}\,}}(\theta ')=z_4$, and ${{\,\textrm{im}\,}}(\theta ')=z_3$, i.e. $\theta _1, \theta '\in J$ by Lemma 2, a contradiction to $\lambda \notin J$.

2. ${{\,\textrm{dom}\,}}(\lambda )=y_3^{[t]}$ for some $t\in \{1,2,\dots ,|y_3|\}$ with $y_3\in P^*$ and there are $y_1,y_2 \in P^*$ with $y_3\sim y_1\sim y_2\ne y_1$ and for all $z\in P^*$ (or ${{\,\textrm{im}\,}}(\lambda )\sqsubset y\in P^*$ with $y\not \sim z$ for all $z\in P^*\backslash \{y\}$).

Then for all $\beta \in I_{P^*}$ with ${{\,\textrm{dom}\,}}(\beta )=y_4^{[t]}$ for some $y_4\in P^*$ with $y_4 \sim y_1$ and ${{\,\textrm{im}\,}}(\beta )={{\,\textrm{im}\,}}(\lambda )$, we have $\beta \notin J$. Otherwise, $\theta \beta =\lambda$, where ${{\,\textrm{dom}\,}}(\theta )=y_3$ and ${{\,\textrm{im}\,}}(\theta )=y_4$ for some $\theta \in J$ by Lemma 2, a contradiction to $\lambda \notin J$.

3. for all $y\in P^*$ (or ${{\,\textrm{dom}\,}}(\lambda )\sqsubset z\in P^*$ with $z\not \sim y$ for all $y\in P^*\backslash \{z\}$) and ${{\,\textrm{im}\,}}(\lambda )=z_3^{[s]}$ for some $s\in \{1,2,...|z_3|\}$, $z_3\in P^*$, and there are $z_1,z_2\in P^*$ with $z_3\sim z_1\sim z_2\ne z_1$.

Then for all $\beta \in I_{P^*}$with ${{\,\textrm{dom}\,}}(\beta )={{\,\textrm{dom}\,}}(\lambda )$ and ${{\,\textrm{im}\,}}(\beta )=z_4^{[s]}$ for some $z_4\in P^*, z_4\sim z_1$, we have $\beta \notin J$. Otherwise, $\beta \theta '=\lambda$ by Lemma 2, where ${{\,\textrm{dom}\,}}(\theta ')=z_4$ and ${{\,\textrm{im}\,}}(\theta ')=z_3$ for some $\theta '\in J$, a contradiction to $\lambda \notin J$.

4. for all $y\in P^*$ (or ${{\,\textrm{dom}\,}}(\lambda )\sqsubset z\in P^*$ with $z\not \sim y$ for all $y\in P^*\backslash \{z\}$) and for all $y\in P^*$ (or ${{\,\textrm{im}\,}}(\lambda )\sqsubset z\in P^*$ with $z\not \sim y$ for all $y\in P^*\backslash \{z\}$).

Note that: $B_{y_3}, B_{z_3}\in B$ with ${{\,\textrm{dom}\,}}(\alpha )\in B_{y_3}$ and ${{\,\textrm{im}\,}}(\alpha )\in B_{z_3}$, if $\alpha =\lambda$ is of form 1;

$B_{y_3}\in B$ with ${{\,\textrm{dom}\,}}(\alpha )\in B_{y_3}$ and $\{{{\,\textrm{im}\,}}(\alpha )\}\in C$, if $\alpha =\lambda$ is of form 2;

$B_{z_3}\in B$ with ${{\,\textrm{im}\,}}(\alpha )\in B_{z_3}$ and $\{{{\,\textrm{dom}\,}}(\alpha )\}\in C$, if $\alpha =\lambda$ is of form 3;

$\{{{\,\textrm{dom}\,}}(\alpha )\}, \{{{\,\textrm{im}\,}}(\alpha )\}\in C$, if $\alpha =\lambda$ is of form 4.

Hence, there is $\Delta _1\in B\cup C$ with ${{\,\textrm{dom}\,}}(\alpha )\in \Delta _1$. We define,

$$\begin{aligned} \tilde{B}\tilde{C}&=\{\Delta \in BC_{\Delta _1} : \hbox {there is} \ \beta \in I_{P^*}\backslash J \ \hbox {with} \ {{\,\textrm{dom}\,}}(\beta )\in \Delta \};\\ \hat{B}\hat{C}&=\{\Delta \in BC_{\Delta _1} : \hbox {there is} \ \beta \in I_{P^*}\backslash J \ \hbox {with} \ {{\,\textrm{im}\,}}(\beta )\in \Delta \};\\ T_{BC}&=\{ c \in I_{P^*} : {{\,\textrm{dom}\,}}(c)\in \tilde{\Delta },{{\,\textrm{im}\,}}(c)\in \hat{\Delta } \ \hbox {with} \ \tilde{\Delta } \in \tilde{B}\tilde{C}, \hat{\Delta }\in \hat{B}\hat{C}\}. \end{aligned}$$

Assume there are $\theta _1,\theta _2\notin J$ with $\theta _1\ne \theta _2$ and ${{\,\textrm{im}\,}}(\theta _1), {{\,\textrm{dom}\,}}(\theta _2)\in \Delta$ for some $\Delta \in BC_{\Delta _1}$. So, we observe that $\gamma \theta _1\rho =\theta _2$ with $\gamma \in \langle J \cup \{\theta _1\} \rangle$ and $\rho \in J$. Then we have ${{\,\textrm{im}\,}}(\theta _2)\subset {{\,\textrm{im}\,}}(\rho )$ or ${{\,\textrm{im}\,}}(\theta _2)={{\,\textrm{im}\,}}(\rho )$. Because of ${{\,\textrm{im}\,}}(\theta _1), {{\,\textrm{dom}\,}}(\theta _2)\in \Delta$ and 1–3, then ${{\,\textrm{im}\,}}(\theta _2)={{\,\textrm{im}\,}}(\rho )$. This gives ${{\,\textrm{dom}\,}}(\rho )={{\,\textrm{im}\,}}(\theta _1)$. Then $\rho \notin J$ because of 2, a contradiction. Thus, $\tilde{B}\tilde{C} \cap \hat{B}\hat{C}= \emptyset$.

Clearly, by the definition of $\tilde{B}\tilde{C}$ and $\hat{B}\hat{C}$, we have $\tilde{B}\tilde{C}\ \cup \hat{B}\hat{C} \subseteq BC_{\Delta _1}$. Let $\Delta \in BC_{\Delta _1}$. Further, let $h\in \Delta$, i.e. $h\sim {{\,\textrm{dom}\,}}(\alpha )$, and let $\gamma _1,\gamma _2 \in I_{P^*}$ with $h={{\,\textrm{im}\,}}(\gamma _1)={{\,\textrm{dom}\,}}(\gamma _2), {{\,\textrm{dom}\,}}(\gamma _1)={{\,\textrm{dom}\,}}(\alpha )$, and ${{\,\textrm{im}\,}}(\gamma _2)={{\,\textrm{im}\,}}(\alpha )$. We have that $\gamma _1\gamma _2=\alpha$. This gives $\gamma _1\notin J$ or $\gamma _2\notin J$. So by the definition of $\hat{B}\hat{C}$($\tilde{B}\tilde{C}$), we see that if $\gamma _1\notin J$( $\gamma _2\notin J$), then $\Delta \in \hat{B}\hat{C}$($\Delta \in \tilde{B}\tilde{C})$. This means $\tilde{B}\tilde{C}\ \cup \hat{B}\hat{C} \supseteq BC_{\Delta _1}$ and together with $\tilde{B}\tilde{C}\ \cup \hat{B}\hat{C} \subseteq BC_{\Delta _1}$, we obtain $\tilde{B}\tilde{C}\ \cup \hat{B}\hat{C} = BC_{\Delta _1}$. Consequently, $BC=(\tilde{B}\tilde{C}, \hat{B}\hat{C})$ is a partition of $BC_{\Delta _1}$.

Let $\gamma \in T_{BC}$. Then there are $\Delta \in \tilde{B}\tilde{C}$ and $\Delta ' \in \hat{B}\hat{C}$ such that ${{\,\textrm{dom}\,}}(\gamma )\in \Delta$ and ${{\,\textrm{im}\,}}(\gamma )\in \Delta '$. By the definition of $\tilde{B}\tilde{C}$ and $\hat{B}\hat{C}$, there are $\delta _1,\delta _2\in I_{P^*}\backslash J$ with ${{\,\textrm{dom}\,}}(\delta _1)\in \Delta$ and ${{\,\textrm{im}\,}}(\delta _2)\in \Delta '$. If $\gamma =\delta _1$ or $\gamma =\delta _2$ then we have $\gamma \notin J$. Suppose $\gamma \ne \delta _1$ and $\gamma \ne \delta _2$. Recall, we have ${{\,\textrm{dom}\,}}(\gamma ), {{\,\textrm{dom}\,}}(\delta _1)\in \Delta$. By 1–4, we get that there is $\theta _1\notin J$ with ${{\,\textrm{dom}\,}}(\theta _1)={{\,\textrm{dom}\,}}(\gamma )$. There is $\gamma '\in I_{P^*}$ with ${{\,\textrm{dom}\,}}(\gamma ')={{\,\textrm{im}\,}}(\gamma )$ and ${{\,\textrm{im}\,}}(\gamma ')={{\,\textrm{im}\,}}(\theta _1)$. We observe that $\gamma \gamma '=\theta _1$. This means, $\gamma \notin J$ or $\gamma '\notin J$. Assume that $\gamma '\notin J$. We have ${{\,\textrm{dom}\,}}(\gamma ')={{\,\textrm{im}\,}}(\gamma )\in \Delta '$. Then $\Delta '\in \tilde{B}\tilde{C}$ and we have $\tilde{B}\tilde{C} \cap \hat{B}\hat{C}\ne \emptyset$, a contradiction. Thus, $\gamma \notin J$. We can conclude that $T_{BC}\cap J = \emptyset$, this means $J\subseteq I_{P^*}\backslash T_{BC}$ and by Lemma 5, we have $I_{P^*}\backslash T_{BC}$ is a semigroup. Together with J is a maximal subsemigroup of $I_{P^*}$, we have $J=I_{P^*}\backslash T_{BC}$.

Assume there is $\theta \in T_{BC}$ with $\theta \notin I_{P^*}^u$. By Corollary 1, there are $a_1,a_2\in I_{P^*}$ with ${{\,\textrm{rank}\,}}(a_1),{{\,\textrm{rank}\,}}(a_2)>{{\,\textrm{rank}\,}}(a_1a_2)$ such that $\theta =a_1a_2$. This provides, $a_1,a_2\in J$ by Lemma 2, i.e. $\theta \in J$, a contradiction to $T_{BC}\cap J = \emptyset$. Thus, $T_{BC}\subseteq I_{P^*}^u$.

Conversely, let $J=I_{P^*}\backslash T_{\{g\}}$ with $|BC_{\{g\}}|=1$ for some $g\in P^*$ such that $\{g\}\in C$ and $T_{\{g\}}\subseteq I_{P^*}^u$. Then $I_{P^*}\backslash T_{\{g\}}$ is a semigroup by Lemma 3. Since $| T_{\{g\}} |=1$, we can conclude that J is a maximal subsemigroup of $I_{P^*}$.

Let $J=I_{P^*}\backslash T_{B_{y_1}}$ with $|BC_{B_{y_1}}|=1$ for some $y_1\in P^*$ such that $|y_1|\ge 2$, $B_{y_1}\in B$, and $T_{B_{y_1}}\subseteq I_{P^*}^u$. Then $I_{P^*}\backslash T_{B_{y_1}}$ is a semigroup by Lemma 4. Moreover, we can conclude that $\{B_{y_1}\}=BC_{B_{y_1}}$. Let $\alpha ,\beta \in T_{B_{y_1}}$. There are $y_3,y_4,y_5,y_6\in A_{y_1}$ and some $t\in \{1,2,\dots ,|y_1|\}$ such that ${{\,\textrm{dom}\,}}(\alpha )=y_3^{[t]}, {{\,\textrm{im}\,}}(\alpha )=y_4^{[t]}, {{\,\textrm{dom}\,}}(\beta )=y_5^{[t]}$, and ${{\,\textrm{im}\,}}(\beta )=y_6^{[t]}$. Further, there are $\theta _1,\theta _2\in I_{P^*}$ with ${{\,\textrm{dom}\,}}(\theta _1)=y_5, {{\,\textrm{im}\,}}(\theta _1)=y_3, {{\,\textrm{dom}\,}}(\theta _2)= y_4$, and ${{\,\textrm{im}\,}}(\theta _2)=y_6$. This shows $\theta _1,\theta _2\in J$ because of ${{\,\textrm{rank}\,}}(\theta _1)={{\,\textrm{rank}\,}}(\theta _2)>{{\,\textrm{rank}\,}}(\alpha )$. So, we have $\theta _1\alpha \theta _2=\beta$. Thus, we get $\beta \in \langle J \cup \{\alpha \} \rangle$. Consequently, J is a maximal subsemigroup of $I_{P^*}$.

Let $J=I_{P^*}\backslash T_{Q}$ for some partition $Q=( Q_1, Q_2 )$ of $BC_{\Delta _1}$, where $\Delta _1\in B \cup C$ and $T_Q\subseteq I_{P^*}^u$. We have that $I_{P^*}\backslash T_{Q}$ is a semigroup by Lemma 5. Let $\alpha , \beta \in T_{Q}$. Then ${{\,\textrm{dom}\,}}(\alpha )\in \Delta _2, {{\,\textrm{im}\,}}(\alpha )\in \Delta _3, {{\,\textrm{dom}\,}}(\beta )\in \Delta _4$, and ${{\,\textrm{im}\,}}(\beta )\in \Delta _5$, where $\Delta _2, \Delta _4\in Q_1$ and $\Delta _3,\Delta _5\in Q_2$. There are $\theta _1, \theta _2\in I_{P^*}$ with ${{\,\textrm{dom}\,}}(\theta _1)={{\,\textrm{dom}\,}}(\beta ), {{\,\textrm{im}\,}}(\theta _1)={{\,\textrm{dom}\,}}(\alpha )$, ${{\,\textrm{dom}\,}}(\theta _2)={{\,\textrm{im}\,}}(\alpha )$, and ${{\,\textrm{im}\,}}(\theta _2)={{\,\textrm{im}\,}}(\beta )$. We get that $\theta _1, \theta _2\in J$ because of ${{\,\textrm{im}\,}}(\theta _1)={{\,\textrm{dom}\,}}(\alpha )\in \Delta _2\in Q_1$ and ${{\,\textrm{dom}\,}}(\theta _2)={{\,\textrm{im}\,}}(\alpha )\in \Delta _3\in Q_2$. So, we have $\theta _1\alpha \theta _2=\beta$. Thus, we get $\beta \in \langle J \cup \{\alpha \} \rangle$. Consequently, we have that J is a maximal subsemigroup of $I_{P^*}$. $\square$