Index Laws for Monoids/Negative Index

Theorem
Let $\struct {S, \circ}$ be a monoid whose identity is $e_S$.

Let $a \in S$ be invertible for $\circ$.

Let $n \in \N$.

Let $a^n = \map {\circ^n} a$ be defined as the power of an element of a monoid:


 * $a^n = \begin{cases}

e_S : & n = 0 \\ a^x \circ a : & n = x + 1 \end{cases}$

that is:
 * $a^n = \underbrace {a \circ a \circ \cdots \circ a}_{n \text { instances} } = \map {\circ^n} a$

For each $n \in \N$ we define:


 * $a^{-n} = \paren {a^{-1} }^n$

Then:
 * $\forall n \in \Z: \paren {a^n}^{-1} = a^{-n} = \paren {a^{-1} }^n$

Proof
We have $a^0 = e$ so it follows trivially that $a^{-0} = \paren {a^{-1} }^0$.

From the general inverse of product, we have:


 * $\paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = a_n^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$

where $a_1, a_2, \ldots, a_n \in S$ are all invertible for $\circ$.

Hence we have:
 * $a_1, a_2, \ldots, a_n = a$

and we have that:


 * $a^n$ is invertible for all $n \in \N$
 * $\forall n \in \N: \paren {a^n}^{-1} = \paren {a^{-1} }^n$

From the above:
 * $a^{-n} = \paren {a^{-1} }^n$

Thus:

Similarly, if $a$ is invertible then $a^{-1}$ is also invertible.

So we also have:
 * $\circ^{-n} \paren {a^{-1} } = \circ^n \paren {\paren {a^{-1} }^{-1} }$

Thus:

Thus the result holds for all $n \in \Z$.