# Element Commutes with Product of Commuting Elements

## 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

 $\ds x \circ \paren {y \circ z}$ $=$ $\ds \paren {x \circ y} \circ z$ $\circ$ is associative $\ds$ $=$ $\ds \paren {y \circ x} \circ z$ $x$ commutes with $y$ $\ds$ $=$ $\ds y \circ \paren {x \circ z}$ $\circ$ is associative $\ds$ $=$ $\ds y \circ \paren {z \circ x}$ $x$ commutes with $z$ $\ds$ $=$ $\ds \paren {y \circ z} \circ x$ $\circ$ is associative

$\blacksquare$