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
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.8$