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:

$\forall n \in \Z : e^n = e$

$\blacksquare$