Element Commutes with Product of Commuting Elements/General Theorem

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
The proof proceeds by induction on $n$, the length of $\left \langle {a_k} \right \rangle_{1 \mathop \le k \mathop \le n}$.

Basis for the Induction
The case $n = 1$ states that:
 * $b \circ a_1 = a_1 \circ b$

That is, $b$ commutes with $a_1$, which holds by hypothesis.

This constitutes the basis for the induction.

Induction Hypothesis
Fix $n \in \N$ with $n \ge 1$.

Assume that the theorem holds for sequences of length $n$.

This is our induction hypothesis.

Induction Step
This is our induction step:

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

Suppose that $b$ commutes with $a_k$ for all $k \in \left[{0 \,.\,.\, n+1}\right]$.

Then (suppressing brackets as $\circ$ is associative):

We conclude that the theorem holds for sequences of length $n+1$.

The result follows by the Principle of Mathematical Induction.