Index Laws for Monoids/Product of Indices

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:
 * $\forall m, n \in \Z: a^{n m} = \left({a^m}\right)^n = \left({a^n}\right)^m$

Proof
Let $m \in \N, c = a^m, d = \left({a^{-1}}\right)^m$.

We define the mapping $g_c: \Z \to S$ as:


 * $\forall n \in \Z: g_c \left({n}\right) = \odot^n \left({c}\right)$

as defined in the proof of the Index Law for Sum of Indices.

Let $h: \Z \to \Z$ be the mapping defined as:


 * $\forall z \in \Z: h \left({z}\right) = z m$

Then:

By Index Law for Sum of Indices and Naturally Ordered Semigroup Power Law, $g_a \circ h$ and $g_c$ are homomorphisms from $\Z$ to $S$ which coincide on $\N$.

So by the Extension Theorem for Homomorphisms, $g_a \circ h = g_c$.

Therefore:


 * $\forall n \in \Z, m \in \N: a^{n m} = \left({a^m}\right)^n$

Also:

and

So, by the same reasoning as before, $g_{a^{-1}} \circ h = g_d$.

Therefore:
 * $\forall n \in \Z, m \in \N: a^{n \left({-m}\right)} = \left({a^{-m}}\right)^n$

Thus:
 * $\forall n, m \in \Z: a^{n m} = \left({a^m}\right)^n$

As $n m = m n$, the result follows.