# 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 $16.8 \ (4)$