Subset of elements in magma that commute with every element cannot have complement of size one

Statement
Suppose $$(S,*)$$ is a magma. Let $$A$$ be the subset of $$S$$ given by:

$$A = \{ a \in S \mid a * b = b * a \ \forall \ b \in S \}$$

Then, the complement of $$A$$ in $$S$$, i.e., the set $$S \setminus A$$, cannot have size one. It must either be empty or have size two or more.

Facts in magmas

 * Nucleus of a commutative magma cannot have complement of size one
 * Unital magma of size two is abelian monoid

Facts in groups

 * Cyclic over central implies abelian: The proof technique is basically the same. However, because of associativity in the group context, we can talk of centralizers and obtain much stronger results involving subgroups generated.

Proof
We will show that if $$S \setminus A$$ has size at most 1, then it is empty.

Given: Magma $$(S,*)$$, $$A = \{ a \in S \mid a * b = b * a \ \forall \ b \in S \}$$. Further, $$S = A \cup \{ x \}$$ for some $$x \in S$$.

To prove: $$x \in A$$, so $$S = A$$.

Proof: To show that $$x \in A$$, we need to show it commutes with everything in $$S$$. We note this in two steps:


 * $$x$$ commutes with everything in $$A$$, because everything in $$A$$ commutes with everything in $$S$$.
 * $$x$$ commutes with itself.

Hence, $$x$$ commutes with everything in $$A \cup \{ x \} = S$$, completing the proof.