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.

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