## Exhibit all of the principal ideals of a given semigroup

Let $S = [1,2] \cup [3,4] \cup [5,\infty) \subseteq \mathbb{R}$, and define a binary operator $\star$ on $S$ by $x \star y = x+y$ if $x,y \geq 5$ and $\mathsf{max}(x,y)$ otherwise. Show that $S$ is a semigroup and exhibit all of its principal ideals. Does $S$ have a kernel? Compute the orders of the elements of $S$.

Note first that $\star$ is commutative.

Given a triple $(a,b,c)$ of elements in $S$, each is either less than 5 or not. This leads us to consider 8 cases. Let $\alpha,\beta, \gamma \in S$ be at least 5, and let $a,b,c \in S$ be less than 5.

• $(\alpha \star \beta) \star \gamma = (\alpha + \beta) \star \gamma$ $= (\alpha + \beta) + \gamma$ $= \alpha + (\beta + \gamma)$ $= \alpha \star (\beta + \gamma)$ $= \alpha \star (\beta \star \gamma)$
• $(a \star \beta) \star \gamma = \beta \star \gamma$ $= \beta + \gamma$ $= a \star (\beta + \gamma)$ $= a \star (\beta \star \gamma)$
• $\alpha \star b) \star \gamma = \alpha \star \gamma$ $= \alpha \star (b \star \gamma)$
• $(\alpha \star \beta) \star c = (\alpha + \beta) \star c$ $= \alpha + \beta$ $= \alpha \star \beta$ $= \alpha \star (\beta \star c)$
• $(a \star b) \star \gamma = a \star \gamma$ $= \gamma$ $= a \star \gamma$ $= a \star (b \star \gamma)$ (Without loss of generality, $a \geq b$.)
• $(a \star \beta) \star c = \beta \star c$ $= \beta$ $= a \star \beta$ $= a \star (\beta \star c)$
• $(\alpha \star b) \star c = \alpha \star c$ $= \alpha$ $= \alpha \star (b \star c)$
• $(a \star b) \star c = \mathsf{max}(\mathsf{max}(a,b),c)$ $= \mathsf{max}(a,\mathsf{max}(b,c))$ $= a \star (b \star c)$.

So $\star$ is associative, and $(S,\star)$ is a semigroup. Since $S$ is commutative, the left, right, and two-sided ideals of $S$ coincide. Recall that the principal left ideal generated by $a \in S$ is $L(a) = a \cup Sa$.

If $a < 5$, then $L(a) = a \cup Sa$ $= a \cup \{sa \ |\ s \in S, s < a\} \cup \{sa\ |\ s \in S, a \leq s\}$ $= a \cup ([a,\infty) \cap S)$.

If $a \geq 5$, then $L(a) = a \cup Sa$ $= a \cup \{sa\ |\ s \in S, s < 5\} \cup \{sa\ |\ s \in S, s \geq 5\}$ $= a \cup [a+5,\infty)$

In particular, for every element $a$ in $S$, there is a principal ideal which does not contain $a$. (For example, $L(a+5)$.) So the kernel of $S$ is empty.

If $a < 5$, then $a^2 = a$, so that $a$ has order 1. If $a \geq 5$, then $a$ has infinite order.