Element Commutes with Product of Commuting Elements

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Theorem

Let $(S, \circ)$ be a semigroup.

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


If $x$ commutes with both $y$ and $z$, then $x$ commutes with $y \circ z$.


General Theorem

Let $(S,\circ)$ be a semigroup.

Let $\left \langle {a_k} \right \rangle_{1 \mathop \le k \mathop \le n}$ be a sequence of terms of $S$.

Let $b \in S$.


If $b$ commutes with $a_k$ for each $k \in \left[{1 \,.\,.\, n}\right]$, then $b$ commutes with $a_1 \circ \cdots \circ a_n$.


Proof

\(\displaystyle x \circ \left({y \circ z}\right)\) \(=\) \(\displaystyle \left({x \circ y}\right) \circ z\) $\circ$ is associative
\(\displaystyle \) \(=\) \(\displaystyle \left({y \circ x}\right) \circ z\) $x$ commutes with $y$
\(\displaystyle \) \(=\) \(\displaystyle y \circ \left({x \circ z}\right)\) $\circ$ is associative
\(\displaystyle \) \(=\) \(\displaystyle y \circ \left({z \circ x}\right)\) $x$ commutes with $z$
\(\displaystyle \) \(=\) \(\displaystyle \left({y \circ z}\right) \circ x\) $\circ$ is associative

$\blacksquare$


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