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).

  • 1
    $\begingroup$ What is a one-time map? $\endgroup$ – HeinrichD Apr 15 '17 at 16:31
  • $\begingroup$ @HeinrichD It's predictive text on my phone taking over. $\endgroup$ – Shaun Apr 15 '17 at 16:35
  • $\begingroup$ Do you mean $X \setminus \alpha(X)$, the complement of the image? In (a), you have to give some evidence for "has at least as many elements as ...". $\endgroup$ – HeinrichD Apr 15 '17 at 16:37
  • $\begingroup$ @HeinrichD Yes. Howie uses $y=xm$ for $y=m(x)$. $\endgroup$ – Shaun Apr 15 '17 at 16:39

(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.

  • $\begingroup$ It doesn't matter whether $X\setminus X\alpha $ is infinite, or countable, or anything. Since $\alpha $ is injective, you always get a bijection. $\endgroup$ – Andrés E. Caicedo Apr 15 '17 at 17:26
  • $\begingroup$ Yeah that was silly. I'll edit to make it simpler. $\endgroup$ – Nathaniel Mayer Apr 15 '17 at 17:40

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