Index Laws for Monoids/Negative Index

Theorem
Let $\left ({S, \odot}\right)$ be a monoid whose identity is $e_S$.

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

Let $n \in \N$.

Let $a^n = \odot^n \left({a}\right)$ extend the definition in Power of an Element to include the identity as an index:


 * $a^n = \begin{cases}

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

... that is, $a^n = a \odot a \odot \cdots \left({n}\right) \cdots \odot a = \odot^n \left({a}\right)$.

Also, for each $n \in \N$ we can define:


 * $a^{-n} = \left({a^{-1}}\right)^n$

Then:
 * $\forall n \in \Z: \left({a^n}\right)^{-1} = a^{-n} = \left({a^{-1}}\right)^n$

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


 * From the generalized inverse of product, we have:


 * $\left({a_1 \odot a_2 \odot \cdots \odot a_n}\right)^{-1} = a_n^{-1} \odot \cdots \odot a_2^{-1} \odot a_1^{-1}$

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

So we can put $a_1, a_2, \ldots, a_n = a$ and we have that


 * $a^n$ is invertible for all $n \in \N$;
 * $\forall n \in \N: \left({a^n}\right)^{-1} = \left({a^{-1}}\right)^n$.


 * From the above, we have $a^{-n} = \left({a^{-1}}\right)^n$. Thus:

Similarly, if $a$ is invertible then $a^{-1}$ is also invertible, so we also have:
 * $\odot^{-n} \left({a^{-1}}\right) = \odot^n \left({\left({a^{-1}}\right)^{-1}}\right)$

Thus:

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