## Exhibit a semigroup of transformations having no idempotents

Give an example of a semigroup of transformations having no idempotents.

We saw in this previous exercise that every finite semigroup contains idempotent elements. So any example we find will have to be transfinite.

Given $k \in \mathbb{N}^+$, define $\varphi_k : \mathbb{N}^+ \rightarrow \mathbb{N}^+$ by $\varphi_k(a) = a+k$. Let $S \subseteq \mathsf{T}(\mathbb{N}^+)$ be the subset $S = \{\varphi_k\ |\ k \in \mathbb{N}^+\}$. We claim that $S$ is a subsemigroup of the full semigroup of transformations on $\mathbb{N}^+$; indeed, for all $k,\ell \in \mathbb{N}^+$ and $a \in \mathbb{N}^+$, we have $(\varphi_k \circ \varphi_\ell)(a) = \varphi_k(\varphi_\ell(a))$ $= \varphi_k(a + \ell)$ $= a+\ell+k$ $= \varphi(\ell+k)(a)$, so that $S$ is closed under composition. So $S$ is a semigroup of transformations.

If $\varphi_k$ is idempotent, then $a+k = \varphi_k(a) = (\varphi_k \circ \varphi_k)(a)$ $= a+2k$ for all $a \in \mathbb{N}^+$, and so $k = 2k$. This equation has no solutions in $\mathbb{N}^+$, so that $S$ does not contain an idempotent.