Power of Identity is Identity
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Theorem
Let $\struct {M, \circ}$ be a monoid whose identity element is $e$.
Then:
- $\forall n \in \Z: e^n = e$
Proof
Since $e$ is invertible, the power of $e$ is defined for all $n \in \Z$.
We prove the case $n \ge 0$ by induction.
Basis for the Induction
By definition of power of monoid element:
- $e^0 = e$
so the theorem holds for $n = 0$.
This is our basis for the induction.
Induction Hypothesis
Our induction hypothesis is that the theorem is true for $n = k$:
- $e^k = e$
Induction Step
In the induction step, we prove that the theorem is true for $n = k + 1$.
We have:
\(\ds e^{k + 1}\) | \(=\) | \(\ds e^k \circ e\) | Definition of Power of Element of Monoid | |||||||||||
\(\ds \) | \(=\) | \(\ds e^k\) | Definition of Identity Element | |||||||||||
\(\ds \) | \(=\) | \(\ds e\) | Induction Hypothesis |
Therefore, by Principle of Mathematical Induction:
- $\forall n \in \Z_{\ge 0} : e^n = e$
$\Box$
Now we prove the case $n < 0$.
We have:
\(\ds e^n\) | \(=\) | \(\ds \paren {e^{-n} }^{-1}\) | Definition of Power of Element of Monoid | |||||||||||
\(\ds \) | \(=\) | \(\ds e^{-1}\) | since $-n > 0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds e\) | Inverse of Identity Element is Itself |
Thus:
- $\forall n \in \Z : e^n = e$
$\blacksquare$
Sources
- 1964: Walter Ledermann: Introduction to the Theory of Finite Groups (5th ed.) ... (previous) ... (next): Chapter $\text {I}$: The Group Concept: $\S 2$: The Axioms of Group Theory: $(1.8)$
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.8 \ (4)$