# Element Commutes with Product of Commuting Elements/General Theorem

## 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

The proof proceeds by induction on $n$, the length of $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$.

### 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 $\sequence {a_k}_{1 \mathop \le k \mathop \le n + 1}$ be a sequence of length $n + 1$ in $S$.

Suppose that $b$ commutes with $a_k$ for all $k \in \closedint 0 {n + 1}$.

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

 $\ds b \circ a_1 \circ \cdots \circ a_n \circ a_{n + 1}$ $=$ $\ds a_1 \circ \cdots \circ a_n \circ b \circ a_{n + 1}$ Induction Hypothesis $\ds$ $=$ $\ds a_1 \circ \cdots a_n \circ a_{n + 1} \circ b$ $a_{n + 1}$ commutes with $b$

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

The result follows by the Principle of Mathematical Induction.

$\blacksquare$