Power of Identity is Identity

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 $16.8 \ (4)$