\section{Power of Identity is Identity}
Tags: Monoids, Powers (Abstract Algebra), Proofs by Induction

\begin{theorem}
Let $\struct {M, \circ}$ be a [[Definition:Monoid|monoid]] whose [[Definition:Identity Element|identity element]] is $e$.
Then:
:$\forall n \in \Z: e^n = e$
\end{theorem}

\begin{proof}
Since [[Inverse of Identity Element is Itself|$e$ is invertible]], the [[Definition:Power of Element of Monoid|power of $e$]] is defined for all $n \in \Z$.
We prove the case $n \ge 0$ by [[Principle of Mathematical Induction|induction]].
=== Basis for the Induction ===
By definition of [[Definition:Power of Element of Monoid|power of monoid element]]:
:$e^0 = e$
so the theorem holds for $n = 0$. 
This is our [[Definition:Basis for the Induction|basis for the induction]].
=== Induction Hypothesis ===
Our [[Definition:Induction Hypothesis|induction hypothesis]] is that the theorem is true for $n = k$:
:$e^k = e$
=== Induction Step ===
In the [[Definition:Induction Step|induction step]], we prove that the theorem is true for $n = k + 1$.
We have:
{{begin-eqn}}
{{eqn | l = e^{k + 1}
      | r = e^k \circ e
      | c = {{Defof|Power of Element of Monoid}}
}}
{{eqn | r = e^k
      | c = {{Defof|Identity Element}}
}}
{{eqn | r = e
      | c = [[Power of Identity is Identity#Induction Hypothesis|Induction Hypothesis]]
}}
{{end-eqn}}
Therefore, by [[Principle of Mathematical Induction]]:
:$\forall n \in \Z_{\ge 0} : e^n = e$
{{qed|lemma}}
Now we prove the case $n < 0$.
We have:
{{begin-eqn}}
{{eqn | l = e^n
      | r = \paren {e^{-n} }^{-1}
      | c = {{Defof|Power of Element of Monoid}}
}}
{{eqn | r = e^{-1}
      | c = since $-n > 0$
}}
{{eqn | r = e
      | c = [[Inverse of Identity Element is Itself]]
}}
{{end-eqn}}
Thus:
:$\forall n \in \Z : e^n = e$
{{qed}}
\end{proof}


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