Power of Product of Commutative Elements in Semigroup

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Theorem

Let $\left ({S, \circ}\right)$ be a semigroup.

Let $a, b \in S$ both be cancellable elements of $S$.

Then:

$\forall n \in \N, n > 1: \left({x \circ y}\right)^n = x^n \circ y^n \iff x \circ y = y \circ x$

Let $P \left({n}\right)$ be the proposition that $\left({x \circ y}\right)^n = x^n \circ y^n$ iff $x$ and $y$ commute.

We note in passing that $P \left({1}\right)$ does not hold: it says $x \circ y = x \circ y \iff x \circ y = y \circ x$ which is just wrong.

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle \left({x \circ y}\right) \circ \left({x \circ y}\right)$	$=$	$\displaystyle \left({x \circ x}\right) \circ \left({y \circ y}\right)$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle \iff$	$\displaystyle $	$\displaystyle x \circ \left({y \circ x}\right) \circ y$	$=$	$\displaystyle x \circ \left({x \circ y}\right) \circ y$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle \iff$	$\displaystyle $	$\displaystyle x \circ y$	$=$	$\displaystyle y \circ x$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Cancellability of $x$ and $y$

Now suppose $P \left({n}\right)$ holds. We need to show that $P \left({n+1}\right)$ holds as a result of this.

That is, that $\left({x \circ y}\right)^{n+1} = x^{n+1} \circ y^{n+1} \iff x \circ y = y \circ x$.

Suppose $x \circ y = y \circ x$ commute. Then:

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle \left({x \circ y}\right)^{n+1}$	$=$	$\displaystyle \left({x \circ y}\right)^n \circ \left({x \circ y}\right)$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle x^n \circ y^n \circ x \circ y$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle x^n \circ x \circ y^n \circ y$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Commutativity of Powers in Semigroup
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle x^{n+1} \circ y^{n+1}$	$\displaystyle $	$\displaystyle $	$\displaystyle $

As $x \circ y^n = y^n \circ x$ if and only if $x$ and $y$ commute, the result follows.

$\blacksquare$

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({x \circ y}\right) \circ \left({x \circ y}\right)\)	\(=\)	\(\displaystyle \left({x \circ x}\right) \circ \left({y \circ y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \iff\)	\(\displaystyle \)	\(\displaystyle x \circ \left({y \circ x}\right) \circ y\)	\(=\)	\(\displaystyle x \circ \left({x \circ y}\right) \circ y\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \iff\)	\(\displaystyle \)	\(\displaystyle x \circ y\)	\(=\)	\(\displaystyle y \circ x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Cancellability of $x$ and $y$

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({x \circ y}\right)^{n+1}\)	\(=\)	\(\displaystyle \left({x \circ y}\right)^n \circ \left({x \circ y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle x^n \circ y^n \circ x \circ y\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle x^n \circ x \circ y^n \circ y\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Commutativity of Powers in Semigroup
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle x^{n+1} \circ y^{n+1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)