This is Exercise 2.6.5(b) of Howie's "Fundamentals of Semigroup Theory".
The Details:
(Here I write maps on the right of their arguments, so $X\alpha$ means $\alpha(X)$.)
Let $S$ be a semigroup.
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.
My Attempt:
I think it all follows from the maps being injective but working with the infinite set $X\setminus X\alpha$ for $\alpha\in B$ is tricky.