Element Commutes with Product of Commuting Elements

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Theorem

Let $\struct {S, \circ}$ be a semigroup.

Let $x, y, z \in S$.


Let $x$ commute with both $y$ and $z$.

Then $x$ commutes with $y \circ z$.


General Theorem

Let $\struct {S, \circ}$ be a semigroup.

Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of terms of $S$.


Let $b \in S$ such that $b$ commutes with $a_k$ for each $k \in \closedint 1 n$.

Then $b$ commutes with $a_1 \circ \cdots \circ a_n$.


Proof

\(\ds x \circ \paren {y \circ z}\) \(=\) \(\ds \paren {x \circ y} \circ z\) Semigroup Axiom $\text S 1$: Associativity
\(\ds \) \(=\) \(\ds \paren {y \circ x} \circ z\) $x$ commutes with $y$
\(\ds \) \(=\) \(\ds y \circ \paren {x \circ z}\) Semigroup Axiom $\text S 1$: Associativity
\(\ds \) \(=\) \(\ds y \circ \paren {z \circ x}\) $x$ commutes with $z$
\(\ds \) \(=\) \(\ds \paren {y \circ z} \circ x\) Semigroup Axiom $\text S 1$: Associativity

$\blacksquare$


Sources

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