Powers of Commuting Elements of Semigroup Commute
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Theorem
Let $\struct {S, \circ}$ be a semigroup.
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:
- $\forall m, n \in \N_{>0}: \paren {\circ^m a} \circ \paren {\circ^n b} = \paren {\circ^n b} \circ \paren {\circ^m a}$
That is:
- $\forall m, n \in \N_{>0}: a^m \circ b^n = b^n \circ a^m$
Proof
The proof proceeds by the Principle of Mathematical Induction:
Let $\map P n$ be the proposition:
- $\paren {\circ^n a} \circ b = b \circ \paren {\circ^n a}$
Basis of the Induction
\(\ds a \circ b\) | \(=\) | \(\ds b \circ a\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {\circ^1 a} \circ b\) | \(=\) | \(\ds b \circ \paren {\circ^1 a}\) | Definition of $\circ^1$ |
demonstrating that $\map P 1$ holds.
This is the basis for the induction.
Induction Hypothesis
Suppose that $\map P k$ holds:
- $\paren {\circ^k a} \circ b = b \circ \paren {\circ^k a}$
This is the induction hypothesis.
It remains to be shown that:
- $\map P k \implies \map P {k + 1}$
That is, that:
- $\paren {\circ^{k + 1} a} \circ b = b \circ \paren {\circ^{k + 1} a}$
Induction Step
This is the induction step:
Thus:
\(\ds \paren {\circ^{k + 1} a} \circ b\) | \(=\) | \(\ds \paren {\paren {\circ^k a} \circ a} \circ b\) | Definition of $\circ^n a$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\circ^k a} \circ \paren {a \circ b}\) | Semigroup Axiom $\text S 1$: Associativity | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\circ^k a} \circ \paren {b \circ a}\) | $b$ commutes with $a$ under $\circ$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\paren {\circ^k a} \circ b} \circ a\) | Semigroup Axiom $\text S 1$: Associativity | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {b \circ \paren {\circ^k a} } \circ a\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds b \circ \paren {\paren {\circ^k a} \circ a}\) | Semigroup Axiom $\text S 1$: Associativity | |||||||||||
\(\ds \) | \(=\) | \(\ds b \circ \paren {\circ^{k + 1} a}\) | Definition of $\circ^n a$ |
So $\map P {k + 1}$ is true.
Thus:
- $\forall m \in \N_{>0}: \paren {\circ^m a} \circ b = b \circ \paren {\circ^m a}$
$\Box$
By repeating the argument above, replacing $a$ with $b$ and $b$ with $\circ^m a$, we have:
\(\ds b \circ \paren {\circ^m a}\) | \(=\) | \(\ds \paren {\circ^m a} \circ b\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {\circ^n b} \circ \paren {\circ^m a}\) | \(=\) | \(\ds \paren {\circ^m a} \circ \paren {\circ^n b}\) |
Hence the result:
- $\forall m, n \in \N_{>0}: \paren {\circ^m a} \circ \paren {\circ^n b} = \paren {\circ^n b} \circ \paren {\circ^m a}$
That is:
- $\forall m, n \in \N_{>0}: a^m \circ b^n = b^n \circ a^m$
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.8 \ (2)$