Definition 1: Let $X$ be a countably infinite set and let $B$ be the set of one-to-one maps $\alpha: X\to X$ with the property that $X\setminus X\alpha$ is infinite. Then $B$ with composition is the Baer-Levi semigroup.
Definition 2: We call $S$ right simple if $\mathcal R=S\times S$, and left simple if $\mathcal L=S\times S$, where $\mathcal R$ and $\mathcal L$ are Green's $R$- and $L$-relation, respectively.
Definition 3: We call $S$ right cancellative if $(\forall a, b, c\in S) ac=bc\implies a=b$, and left cancellative if $(\forall a, b, c\in S) ca=cb\implies a=b$.
The Question:
Show that the Baer-Levi semigroup $B$ is right simple and right cancellative, but is neither left simple nor left cancellative.
Let $\alpha, \beta \in B$. Then $\alpha \mathcal R \beta$ iff there exist $\gamma, \delta \in B$ such that $$\alpha = \beta \gamma \; , \; \beta = \alpha \delta$$
we need to show that $ \mathcal R = B \times B$. Let $\alpha, \beta \in B$, then define $\gamma : X \rightarrow X$ such that
$$ y \gamma = \begin{cases} x \alpha & \text{if} \; y\; \in \text{im}\beta ; x\beta = y \\ a_y & \text{otherwise} \end{cases}$$ where $a_y \notin$ im$\beta$ is fixed such that $a_y = a_{y'}$ iff $y = y'$.
Cleary $\gamma$ is a one-one mapping. How to prove $X\backslash X\gamma$ is infinite.
Let $a, b \in B$. Since the sets $X \setminus Xa$ and $X \setminus Xb$ are countable infinite, there exists an injective function $h$ from $X \setminus Xa$ into $X \setminus Xb$ such that the set $F = (X \setminus Xb) \setminus (X \setminus Xa)h$ is infinite. We now define $c$ as follows: for each $q \in X$, $$ q c = \begin{cases} p b & \text{if $q = pa$ for some $p \in X$,} \\ q h & \text{if $q \in X \setminus Xa$} \end{cases} $$ By construction, $c$ is injective and $ac = b$. Furthermore, the set $X \setminus Xc$ contains $F$ and hence is infinite. Thus $c \in B$ and $\mathcal{R}$ is the universal relation on $B$.