Commutative connected components

A number of different types of subsets of graphs provide example of commutatively necessary subsemigroups. A special case worth examining is when the connected components of a commuting graph are subsemigroups. A further special case is when commutative connectivity forms a congruence then its quotient will be a rectangular band.

In the other direction, if a semigroup has a rectangular band quotient, then its commuting graph is necessarily disconnected. The congruence classes of the congruence with rectangular band quotient are all disjoint unions of connected components of the commuting graph.

In order for the commutative connectivity partition of a semigroup to be non-trivial, the commuting graph of the semirgoup must not be connected. This already rules out any finite semigroup with idempotent commutative, such as any finite inverse semigroup.

Theorem. let $S$ be a semigroup and let $C$ be a commutativity connected subsemigroup of $S$ then $C$ is a radical subsemigroup.

Proof. let $y$ be an element in $C$ and suppose that there exists $x \in S$ and $n \in \mathbb{Z}_+$ such that $x^n = y$. Then $x * x^n = x^{n+1} = x^n * x$ which implies that $xy=yx$. Now since $x$ and $y$ commute we have that $x$ is in the same commutativity connected component as $y$, therefore $y \in C$. It follows that $C$ is a radical subsemigroup. $\square$

An example is the free semigroup on $n$ generators, in that case two lists only commute if they share a common generator. The commutativity connected components of the free semigroup are all principal radical subsemigroups, and so the commuting graph of a free semigroup is a cluster graph. A cluster graph is a special case of a class of graphs whose connected components all form subsemigroups.

Theorem. let $G$ be a trivially perfect commuting graph of a semigroup. Then all connected components of $G$ are radical subsemigroups.

Proof. trivially perfect graphs are precisely the graphs for which every connected component has a universal vertex. Therefore, let $C$ be a connected component of $G$, then let $x$ be a universal vertex of $C$. Then since $x$ is a universal vertex of $C$, $C$ is equal to the centralizer of $x$. Therefore, $C$ is a subsemigroup of $G$, and by the previous theorem it is a radical subsemigroup as well. $\square$

Every single graph of order four or less is either trivially perfect or connected. It is therefore an immediate consequence that commutative connected components are subsemigroups for all semigroups that are of order four or less. It seems that in at least the simplest cases, the connected components of the commuting graph are subsemigroups but this doesn't appear to be the case in general.