Power of Product of Commuting Elements in Semigroup equals Product of Powers
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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:
- $\forall n \in \N_{>0}: \map {\circ^n} {a \circ b} = \paren {\circ^n a} \circ \paren {\circ^n b}$
That is:
- $\forall n \in \N_{>0}: \paren {a \circ b}^n = a^n \circ b^n$
Proof
The proof proceeds by the Principle of Mathematical Induction:
Let $\map P n$ be the proposition:
- $\map {\circ^n} {a \circ b} = \paren {\circ^n a} \circ \paren {\circ^n b}$
Basis of the Induction
\(\ds \circ^1 \paren {a \circ b}\) | \(=\) | \(\ds a \circ b\) | Definition of $\circ^1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\circ^1 a} \circ \paren {\circ^1 b}\) | Definition of $\circ^1$ |
So $\map P 1$ holds.
This is the basis for the induction.
Induction Hypothesis
Suppose that $\map P k$ holds:
- $\map {\circ^k} {a \circ b} = \paren {\circ^k a} \circ \paren {\circ^k b}$
This is the induction hypothesis.
It remains to be shown that:
- $\map P k \implies \map P {k + 1}$
That is, that:
- $\map {\circ^{k + 1} } {a \circ b} = \paren {\circ^{k + 1} a} \circ \paren {\circ^{k + 1} b}$
Induction Step
This is the induction step:
\(\ds \map {\circ^{k + 1} } {a \circ b}\) | \(=\) | \(\ds \paren {\map {\circ^k} {a \circ b} } \circ \paren {a \circ b}\) | Definition of $\circ^{k + 1}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\paren {\circ^k a} \circ \paren {\circ^k b} } \circ \paren {a \circ b}\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\circ^k a} \circ \paren {\paren {\paren {\circ^k b} \circ a} \circ b}\) | Semigroup Axiom $\text S 1$: Associativity | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\circ^k a} \circ \paren {\paren {a \circ \paren {\circ^k b} } \circ b}\) | Powers of Commuting Elements of Semigroup Commute | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\paren {\circ^k a} \circ a} \circ \paren {\paren {\circ^k b} \circ b}\) | Semigroup Axiom $\text S 1$: Associativity | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\circ^{k + 1} a} \circ \paren {\circ^{k + 1} b}\) | Definition of $\circ^{k + 1}$ |
So $\map P {k + 1}$ holds.
Thus by the Principle of Mathematical Induction, the result holds for all $n \in \N_{>0}$:
- $\forall n \in \N_{>0}: \map {\circ^n} {a \circ b} = \paren {\circ^n a} \circ \paren {\circ^n b}$
$\blacksquare$
Sources
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 4.2$. Commutative and associative operations: Example $70$
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.8 \ (3)$