This is Exercise 1.9.13 of Howie's "Fundamentals of Semigroup Theory".

The Details:

Definition: The partial transformation semigroup $\mathscr{P}_n$ is the set of all partial maps from $\{1, 2, \dots , n\}$ to itself, together with composition of partial transformations.

Let $\xi$ be the partial map $$\begin{pmatrix} 2 & 3 & \dots & n \\ 2 & 3 & \dots & n \end{pmatrix}$$ and let $\zeta$, $\tau$, $\pi$ be as defined in this question.

Definition: For each subset $Y$ of $X=\{1, 2, \dots , n\}$, define $\xi_{Y}$ by $1_{X\setminus Y}$; thus $\xi=\xi_{\{1\}}$.

Some Lemmas:

Lemma 1: We have $(1i)\xi(1i)=\xi_{\{i\}}$ for $i\in X$ and $$\xi_Y=\prod_{i\in Y}\xi_{\{i\}}.$$

Proof: This is basic computation and the recognition that $$1_{X\setminus Y}=\prod_{i\in Y}1_{X\setminus \{i\}}.$$ $\square$

Lemma 2: For each $\alpha$ in $\mathscr{P}_n$, let $\hat{\alpha}$, the completion of $\alpha$, be given by $$x\hat{\alpha}=\begin{cases} x\alpha &: x\in \operatorname{dom} \alpha \\ x &: \text{ otherwise.}\end{cases}$$ Then, for $Y=X\setminus \operatorname{dom} \alpha$, $$\alpha=\xi_Y\hat{\alpha}.$$

Proof: Let $x\in X$. Then $$\begin{align} x(\xi_Y\hat{\alpha})&=(x 1_{X\setminus Y})\hat{\alpha} \\ &=(x 1_{X\setminus (X\setminus \operatorname{dom} \alpha)})\hat{\alpha} \\ &=(x 1_{\operatorname{dom} \alpha})\hat{\alpha} \\ &=x\alpha. \end{align}$$ $\square$

The Question:

Deduce (from Lemma 1 and Lemma 2) that $\mathscr{P}_n=\langle\zeta, \tau, \pi, \xi\rangle$.


I'm stuck.

Please help :)


Each $\alpha\in\mathscr{P}_n$ can be written as $$\alpha=\left(\prod_{i\in X\setminus \operatorname{dom}\alpha}(1i)\xi(1i)\right)\hat{\alpha}$$ and $\hat{\alpha}\in\mathscr{T}_n$.

Hence $\mathscr{P}_n=\langle \zeta, \tau, \pi, \xi\rangle$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.