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

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