Element Commutes with Product of Commuting Elements/General Theorem

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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.

$\blacksquare$


Sources

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