Power of Product of Commuting Elements in Monoid equals Product of Powers

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Theorem

Let $\struct {S, \circ}$ be a monoid whose identity element is $e$.

For $a \in S$, let $\circ^n a = a^n$ denote the $n$th power of $a$.

Let $a, b \in S$ such that $a$ commutes with $b$:

$a \circ b = b \circ a$


Then:

$\forall n \in \N: \circ^n \paren {a \circ b} = \paren {\circ^n a} \circ \paren {\circ^n b}$

That is:

$\forall n \in \N: \paren {a \circ b}^n = a^n \circ b^n$


Proof

Because $\struct {S, \circ}$ is a monoid, it is a fortiori also a semigroup.

From Power of Product of Commuting Elements in Semigroup equals Product of Powers:

$\forall n \in \N_{>0}: \circ^n \paren {a \circ b} = \paren {\circ^n a} \circ \paren {\circ^n b}$

That is:

$\forall n \in \N_{>0}: \paren {a \circ b}^n = a^n \circ b^n$


It remains to be shown that the result holds for the cases where $n = 0$.

Thus:

\(\ds \paren {a \circ b}^0\) \(=\) \(\ds \circ^0 \paren {a \circ b}\) Definition of $\circ$
\(\ds \) \(=\) \(\ds e\) Definition of $\circ^0$
\(\ds \) \(=\) \(\ds e \circ e\) Definition of Identity Element
\(\ds \) \(=\) \(\ds \paren {\circ^0 a} \circ \paren {\circ^0 b}\) Definition of $\circ^0$
\(\ds \) \(=\) \(\ds a^0 \circ b^0\) Definition of $\circ$

Thus:

$\paren {a \circ b}^n = a^n \circ b^n$

holds for $n = 0$.

Thus:

$\forall n \in \N: \paren {a \circ b}^n = a^n \circ b^n$

$\blacksquare$


Sources

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