Structure of the (Total) Transformation Monoids Under Rank N Generators

Definition 2.1:

¹² A mapping $\sigma$ from a monoid $A$ to a monoid $B$ is named (morphism) if

Definition 2.2:

¹² Let $B$ be a non-empty set and $A$ be a monoid, then $B$ is refed to a (left)- $A$ -act if there is a mapping $\alpha :A\times B⟶B,$ defined by $(a,b)\alpha =\mathit{ab}$ , such the following conditions are satisfied for all $b\in B$ and $a,r\in A$ :

Definition 2.3:

¹² Suppose that $Y$ and $T$ are two (left)- $A$ -acts, then a mapping $\alpha :Y⟶T$ is named an $A$ - acts morphism if $\left(\mathit{ay}\right)\alpha =a\left(\mathit{y\alpha }\right)$ , $\forall a\in A,$ $y\in Y$ .

Definition 2.4:

^12,13 The (total) transformation semigroup on a non-empty set $\mathcal{X}$ is denoted by $T_{\mathcal{X}}=\left\{\sigma \right|\sigma :\mathcal{X}\to \mathcal{X}\}.$ If $n\ge 0$ , such that $n\in Z$ , where $\mathcal{X}={\mathcal{X}}_n=\{1,2,\dots ,n\},$ we write $T_{\mathcal{X}}$ to be $T_n$ . With respect to the semigroup operation of composition that $\mathcal{x}(\delta \circ \gamma )=\mathcal{x}$ ( $\mathit{\delta \gamma }$ ) $=\left(\mathit{x\delta }\right)\gamma ,$ for all $\mathcal{x}\in \mathcal{X}$ and $\delta ,\gamma \in T_x$ . Clearly, $T_n$ is monoid with identity transformation $I_{\mathcal{X}}=\left(\begin{array}{ccc}1& 2\dots & n\\ 1& 2\dots & n\end{array}\right)$ .

Remark 2.5:

^12,13

Recall that, if $S$ is a semigroup and if $t\in S$ , the (principal) left ideal generated by $t$ is the smallest left ideal of $S$ containing $t$ , and it is denoted by $S^{1}t$ . Dually, for ${\mathit{tS}}^1$ .²²

Definition 2.6:

²² Let $S$ be any semigroup, the Green’s preorder binary relations are defined as follows:

Definition 2.7:

²² Let $S$ be any semigroup, the Green’s relations are defined by:

Clearly, if $S$ is finite, then $\mathcal{D}=\mathcal{J}.$ So, the above definition can be translated as follows:

Definition 2.8:

²² Let $S$ be a finite semigroup, the Green’s relations can be defined as:

Definition 3.1:

^12,22 Let $G$ be a finite group and $\mathcal{X}$ be a non-empty set, then ${\left(\mathcal{F}\ell \right)}_{\mathcal{X}}\left(G\right)$ is known as a free (left)- $G$ -acts on $\mathcal{X}$ if:

If $\mathcal{X}={\mathcal{X}}_n=\{1,2,\dots ,n\}$ where $n\ge 0$ and $n\in Z,$ the free (left)- $G$ -act can be written as ${\left(\mathcal{F}\ell \right)}_n\left(G\right)$ .

It is clear that ${\left(\mathcal{F}\ell \right)}_n\left(G\right)$ consists elements of the form $g\mathcal{x},$ where $g\in G$ and $\mathcal{x}\in \mathcal{X}$ i.e., ${\left(\mathcal{F}\ell \right)}_n\left(G\right)={\dot{\mathbf{\cup }}}_{i=1}^{i=n}G{\mathcal{x}}_i,$ such that $G{\mathcal{x}}_i=\{g{\mathcal{x}}_i:g\in G,1\le i\le n,{\mathcal{x}}_i\in \mathcal{X}\}.$ ^1,12 It is well-known that the set of all the morphisms $\sigma :B⟶B$ is called the endomorphism monoid, and it is denoted by $\mathit{End}\left(B\right).$ Therefore, the set of all morphisms from the free (left)- $G$ -acts into itself is an endomorphism monoid and it is denoted by $\mathit{End}{\left(\mathcal{F}\ell \right)}_n\left(G\right)$ such that $\mathit{End}{\left(\mathcal{F}\ell \right)}_n\left(G\right)\mathit{=}\left\{\sigma \right|\sigma :{\left(\mathcal{F}\ell \right)}_n\left(G\right)⟶{\left(\mathcal{F}\ell \right)}_n\left(G\right)$ ; and $\sigma$ is a $G-$ morphism}. If $\sigma \mathit{\in }\mathit{End}{\left(\mathcal{F}\ell \right)}_{\mathrm{n}}\left(G\right)$ , we define $\sigma$ to be ${\mathcal{x}}_j\sigma =g_j^{\sigma }{\mathcal{x}}_{j\acute{\sigma },}$ where 1 $\le j\le n$ , $\acute{\sigma }\in T_n$ , and $g_j^{\sigma }\in G^n$ . As $\sigma$ is a $G-$ morphism and for $g\in G$ and 1 $\le j\le n$ we have $\left(g{\mathcal{x}}_j\right)\sigma =g\left({\mathcal{x}}_j\sigma \right)=g\left(g_j^{\sigma }{\mathcal{x}}_{j\acute{\sigma }}\right)$ .

Definition 4.1:

^12,13 Let $G$ be a finite group and $T_n$ be the (total) transformation on ${\mathcal{X}}_n$ , where $n\ge 0,n\in Z.$

Then, the (full) wreath product of $G^n$ by $T_n$ is

Theorem 6.1:

For all $\sigma ,\tau \in T_{{\left(\mathcal{F}\mathbf{\ell }\right)}_{\mathit{n}}\left(\mathit{G}\right)}$ , we get the following identities:

Proof:

That means, $(\mathcal{x},\mathcal{y})\in \mathit{Ker}\left\{\sigma \right\},$ and so $\mathit{Ker}\left\{\tau \right\}\subseteq \mathit{Ker}\left\{\sigma \right\}.$

$\Leftarrow$ ) If $\mathit{Ker}\left\{\tau \right\}\subseteq \mathit{Ker}\left\{\sigma \right\},$ let $\varpi \in T_{{\left(\mathcal{F}\mathbf{\ell }\right)}_{\mathbf{n}}\left(\mathit{G}\right)}$ such that $\varpi$ : ${\left(\mathcal{F}\mathbf{\ell }\right)}_{\mathit{n}}\left(\mathit{G}\right)\mathit{⟶}{\left(\mathcal{F}\mathbf{\ell }\right)}_{\mathit{n}}\left(\mathit{G}\right)\mathbf{.}$ Since $\mathit{Im}\left\{\tau \right\}=G{\mathcal{x}}_{j_1}\dot{\cup }\cdots \dot{\cup }{G\mathcal{x}}_{j_n}$ by,¹² [Lemma 5.1], define ${\mathcal{x}}_{j_{ı}}\varpi$ = ${\mathcal{y}}_{ı}\sigma ,$ such that ${\mathcal{y}}_{ı}\tau$ = ${\mathcal{x}}_{j_{ı}}$ , and ${\mathcal{x}}_j\varpi ={\mathcal{x}}_j$ , for all $j\notin \{j_{1,}{j}_{2,}$ …, $j_m\}$ .

Now, let ${\mathcal{y}}_{ı}\tau$ = $\acute{{\mathcal{y}}_i}\tau =x_{j_{ı}}$ then ( ${\mathcal{y}}_{ı}$ , $\acute{{\mathcal{y}}_i}$ ) $\in \mathit{Ker}\left\{\tau \right\}\subseteq \mathit{Ker}\left\{\sigma \right\}$ means ${\mathcal{y}}_{ı}\sigma =$ $\acute{{\mathcal{y}}_i}\sigma$ that implies ${\mathcal{y}}_{ı}\varpi =\acute{{\mathcal{y}}_i}\varpi$ , therefore, $\varpi$ is well –defined.

Since ${\left(\mathcal{F}\mathbf{\ell }\right)}_{\mathit{n}}\left(\mathit{G}\right)$ is free on ${\mathcal{X}}_n=\{{\mathcal{x}}_1,\dots .,{\mathcal{x}}_n$ }, and as $\varpi \in T_{{\left(\mathcal{F}\mathbf{\ell }\right)}_{\mathit{n}}\left(\mathit{G}\right)}$ then $\varpi$ must be a $G$ -acts morphism. Let $z\in$ ${\left(\mathcal{F}\mathbf{\ell }\right)}_{\mathit{n}}\left(\mathit{G}\right)$ be such that $z=g{\mathcal{x}}_k,$ also let ${\mathcal{x}}_k\tau$ = $h{\mathcal{x}}_{j_{ı}}$ , that implies $\mathit{z\tau }=\left(g{\mathcal{x}}_k\right)\tau =g\left(x_k\tau \right)=g\left(h{\mathcal{x}}_{j_{ı}}\right)=\mathit{gh}\left({\mathcal{x}}_{j_{ı}}\right)=$ $\mathit{gh}\left({\mathcal{y}}_{ı}\tau \right)=\left(\mathit{gh}{\mathcal{y}}_{ı}\right)\tau .$ Now, $\mathit{z\tau \varpi }$ = $\left(\mathit{gh}{\mathcal{x}}_{j_i}\right)\varpi =\left(\mathit{gh}\right)$ ( ${\mathcal{x}}_{j_{ı}}\varpi$ ) = $\left(\mathit{gh}\right)\left({\mathcal{y}}_{ı}\sigma \right)=\left(\mathit{gh}{\mathcal{y}}_{ı}\right)\sigma .$

Because $\mathit{z\tau }=\left(\mathit{gh}{\mathcal{y}}_{ı}\right)\tau ,$ and $\mathit{Ker}\left\{\tau \right\}\subseteq \mathit{Ker}\left\{\sigma \right\}$ we must have z $\sigma$ $=$ $\left(\mathit{gh}{\mathcal{y}}_{ı}\right)\sigma =\mathit{z\tau \varpi }$ . Hence, $\sigma =\mathit{\tau \varpi }.$

Theorem 6.2:

For all $\sigma ,\tau \in T_{{\left(\mathcal{F}\mathbf{\ell }\right)}_{\mathit{n}}\left(\mathit{G}\right)}$ , we have the following:

Proof:

The point (1) and (2) can be easily verified using Theorem 6.1.

3) It is direct result from (1) and (2).

$4)⟹)$ Let $\sigma \mathcal{D}\tau$ that means $\sigma \mathcal{R}\delta \mathcal{L}\tau$ for any $\delta \in T_{{\left(\mathcal{F}\mathbf{\ell }\right)}_{\mathit{n}}\left(\mathit{G}\right)}$ . From (1) and (2) we get $\mathit{Im}\left\{\delta \right\}=\mathit{Im}\left\{\tau \right\}$ and $\mathit{Ker}\left\{\sigma \right\}=\mathit{Ker}\left\{\delta \right\}$ . Now, $\mathit{Im}\left\{\sigma \right\}\cong {\left(\mathcal{F}\mathbf{\ell }\right)}_{\mathit{n}}\left(\mathit{G}\right)/$ $\mathit{Ker}\left\{\sigma \right\}$ (by the Fundamental Theorem of Semigroup), therefore, $m\left\{\sigma \right\}\cong {\left(\mathcal{F}\mathbf{\ell }\right)}_{\mathit{n}}\left(\mathit{G}\right)/\mathit{Ker}\left\{\delta \right\}$ $\cong \mathit{Im}\left\{\delta \right\}$ , and so $\wp \left(\sigma \right)=\wp (\mathit{Im}\left\{\sigma \right\})=\wp (\mathit{Im}\left\{\delta \right\})=\wp \left(\delta \right),$ and then we get $\wp \left(\sigma \right)=\wp \left(\delta \right)$ , furthermore, $\wp \left(\delta \right)=\wp (\mathit{Im}\left\{\delta \right\})=\wp (\mathit{Im}\left\{\tau \right\})=\wp \left(\tau \right).$ Hence, $\wp \left(\sigma \right)=\wp \left(\tau \right).$

$\Leftarrow )$ Let $\wp \left(\sigma \right)=\wp \left(\tau \right).$ As $\mathit{Im}\left\{\sigma \right\}={\dot{\cup }}_{y\in Y}\mathit{Gy},\mathit{Im}\left\{\tau \right\}={\dot{\cup }}_{z\in Z}G\mathrm{z}$ for some $Y,Z\subseteq {\mathcal{X}}_n$ . With | $Y$ |=| $Z$ | = $\wp \left(\sigma \right)=\wp \left(\tau \right).$

If we let $\phi$ :Y $\mathit{⟶}{\rm Z}$ be a bijection and by defining $\varphi :\mathit{Im}\left\{\sigma \right\}\mathit{⟶}\mathit{Im}\left\{\tau \right\}$ by $(g\mathcal{y}$ ) $\varphi =g\left(\mathcal{y}\phi \right),$ for all $g\in G$ and $\mathcal{y}\in Y.$ Clearly, $\varphi$ is one to one as where $(g\mathcal{y}$ ) $\varphi =\left({h\mathcal{y}}^{\prime }\right)\varphi$ that implies $g\left(\mathcal{y}\phi \right)$ = $h\left({\mathcal{y}}^{\prime }\phi \right)$ , for all $g,h\in G$ and $\mathcal{y},\acute{\mathcal{y}}\in Y$ . Then we must have $g=h$ and $\left(\mathcal{y}\phi \right)$ = $\left(\acute{\mathcal{y}}\phi \right),$ this is because they are in $\mathit{Im}\left\{\tau \right\}$ and as $\mathit{Im}\left\{\tau \right\}$ is free (left)- $G$ -act and $\phi$ is one-one, then $\mathcal{y}=\acute{\mathcal{y}}$ . Furthermore, $\varphi$ is onto since from definition of $\varphi$ we get, $\left(g\mathcal{y}\right)\varphi =g\left(\mathcal{y}\phi \right)$ for all $g\in G,\mathcal{y}\in Y$ and because $\phi$ is bijection hence for all $\mathit{gz}\in \mathit{Im}\left\{\tau \right\}$ , choose $\mathcal{y}\in Y$ with $\mathcal{y}\phi =z,\text{then}g\mathcal{y}\in \mathit{Im}\left\{\sigma \right\}\text{and}\mathit{gz}=g\left(\mathcal{y}\right)\phi =(g\mathcal{y}$ ) $\varphi$ .

Suppose $\xi =\mathit{\sigma \varphi },\text{where}\xi \in T_{{\left(\mathcal{F}\mathbf{\ell }\right)}_{\mathit{n}}\left(\mathit{G}\right)}$ . We have $\mathit{Im}\left\{\xi \right\}$ = $\mathit{Im}\left\{\mathit{\sigma \varphi }\right\}$ = $\mathit{Im}\left\{\sigma \right\}\varphi =\mathit{Im}\left\{\tau \right\}$ , therefore, $\tau \mathcal{L}\xi .$

Let $\mathcal{c},\mathcal{d}\in {\left(\mathcal{F}\mathbf{\ell }\right)}_{\mathit{n}}\left(\mathit{G}\right).$ Clearly, $\mathcal{c}\sigma =\mathcal{d}\sigma$ if and only if $\left(\mathcal{c}\sigma \right)\varphi =\left(\mathcal{d}\sigma \right)\varphi .$ Since $\varphi$ is one to one, then $\mathit{Ker}\left\{\sigma \right\}=\mathit{Ker}\left\{\mathit{\sigma \varphi }\right\}=\mathit{Ker}\left\{\xi \right\}$ , therefore, $\sigma \mathcal{R}\xi$ . Hence, $\sigma \mathcal{D}\tau .$

5) $⟹)$ Let $\sigma {\le }_{\mathcal{I}}\tau$ that means $\sigma =\mathit{\zeta \tau \eta }$ , for any $\zeta ,\eta \in T_{{\left(\mathcal{F}\mathbf{\ell }\right)}_{\mathit{n}}\left(\mathit{G}\right)}$ . From Theorem 6.1, (III), we have $\wp \left(\sigma \right)=\wp \left(\mathit{\zeta \tau \eta }\right)\le \wp \left(\mathit{\zeta \tau }\right)\le \wp \left(\tau \right)$ .

$\Leftarrow )\text{If}\wp \left(\sigma \right)\le \wp \left(\tau \right),\mathit{Im}\left\{\sigma \right\}={\dot{\cup }}_{\mathcal{y}\in Y}\mathit{Gy}$ and $\mathit{Im}\left\{\tau \right\}={\dot{\cup }}_{z\in Z}G\mathrm{z}$ for some $Y,{\rm Z}\subseteq {\mathcal{X}}_n$ , with $\wp \left(\sigma \right)=$ | $Y$ | and $\wp \left(\tau \right)=\left|Z\right|.$ Because $\wp \left(\sigma \right)\le \wp \left(\tau \right)$ then by,¹² [Remark 5.7] there is a one to one map $\varphi :Y\mathit{⟶}{\rm Z}$ . Let $P=\mathit{Im}\left\{\varphi \right\}$ , that means $P\subseteq Z$ and $\left|Y\right|=\left|P\right|$ .

Fix $p_0\in P$ and define $\pi :Z\mathit{⟶}P$ by z $\pi =z$ , for all z $\in P$ , and $\mathit{z\pi }=p_0$ for all z $\in Z\backslash P,$ then $\mathit{Im}\left\{\pi \right\}=P$ . Define $\beta :{\dot{\cup }}_{z\in Z}G\mathrm{z}⟶{\dot{\cup }}_{p\in P}{G}_p$ by $\mathit{z\beta }=\mathit{z\pi }.$ Obviously, $\beta$ extends to a $G$ -acts morphism therefore, $\mathit{\beta \pi }\in T_{{\left(\mathcal{F}\mathbf{\ell }\right)}_{\mathit{n}}\left(\mathit{G}\right)}$ . Since $\mathit{Im}\left\{\mathit{\beta \pi }\right\}$ = $(\mathit{Im}\left\{\beta \right\})\pi$ = $\left({\dot{\cup }}_{z\in Z}\mathit{Gz}\right)\pi ={\dot{\cup }}_{z\in Z}\mathit{Gz\pi }$ = ${\dot{\cup }}_{z\in Z}\mathit{Gz\beta }$ = ${\dot{\cup }}_{p\in P}\mathit{Gp}$ , we obtain $\wp \left(\mathit{\beta \pi }\right)=\left|P\right|=\left|Y\right|=\wp \left(\sigma \right)$ so, by (4) we get $\mathit{\beta \pi }\mathcal{D}\sigma$ and then $\mathit{\beta \pi }\mathcal{J}\sigma$ because $\mathcal{D}\subseteq \mathcal{J}$ . Therefore, $\sigma {\le }_{\mathcal{I}}\tau .$

6) $⟹)$ If $\sigma \mathcal{J}\tau ,$ then $\sigma =\mathit{\delta \tau \lambda },\tau =\mathit{\alpha \sigma \upsilon },\text{for some}\delta ,\lambda ,\alpha ,\upsilon \in T_{{\left(\mathcal{F}\mathbf{\ell }\right)}_{\mathit{n}}\left(\mathit{G}\right)}.$ By Theorem 6.1, (III), $\wp \left(\sigma \right)$ = $\wp \left(\mathit{\delta \tau \lambda }\right)\le \wp \left(\mathit{\delta \tau }\right)\le \wp \left(\tau \right)$ , as well as $\wp \left(\tau \right)=\wp \left(\mathit{\alpha \sigma \upsilon }\right)\le \wp \left(\mathit{\alpha \sigma }\right)\le \wp \left(\sigma \right)$ , that gives $\wp \left(\sigma \right)=$ $\wp \left(\tau \right).$

$\Leftarrow )$ Let $\wp \left(\sigma \right)=$ $\wp \left(\tau \right)$ , then $\sigma \mathcal{D}\tau$ by using (4), so that $\sigma \mathcal{J}\tau$ as $\mathcal{D}$ $\subseteq \mathcal{J}$ .

7) This is an immediate consequence of (4) and (6).□