Let $X$ be a countably infinite set and let $S$ be the set of one-to-one maps $\alpha: X\to X$ with the property that $X\setminus X\alpha$ is infinite.
NB: Howie uses $y=xf$ for $y=f (x)$.
Definition: Let Y be a set. Then $(\mathcal{T}_Y, \circ)$ is the full transformation semigroup consisting of all maps from $Y$ to $Y$ under composition.
(a) Show that S is a subsemigroup of $\mathcal{T}_X$.
My Attempt:
Let $\alpha, \beta\in S$. Then $\alpha\circ\beta$ is injective since $\alpha$ and $\beta$ are. Also $X\setminus X(\alpha\circ\beta)$ has at least as many elements as $X\setminus X\beta$, which is infinite. Hence $\alpha\circ\beta\in S$, so S is subsemigroup of $\mathcal{T}_X$.
The rest of the question:
(b) Show that for all $\alpha\in S$ there exists a bijection between $X\setminus X\alpha$ and $X\alpha\setminus X\alpha^2$.
(c) Deduce that $S$ has no idempotents.
I'm stuck on parts (b) & (c).
(b) In general, $X\alpha \setminus X\alpha^2 \subset (X \setminus X\alpha)\alpha$. The reverse inclusion holds because $\alpha$ is injective. $\alpha$ restricts to a bijection $X \setminus X\alpha \to X\alpha \setminus X\alpha^2$.
(c) If $\alpha$ were idempotent, then $X\alpha \setminus X\alpha^2$ would be empty.