Index Laws for Monoids/Product of Indices

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

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


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

Then:
 * $\forall m, n \in \Z: a^{n m} = \paren {a^m}^n = \paren {a^n}^m$

Proof
Let $m \in \N, c = a^m, d = \paren {a^{-1}}^m$.

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


 * $\forall n \in \Z: \map {g_c} n = \map {\circ^n} c$

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

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


 * $\forall z \in \Z: \map h z = z m$

Then:

By Index Law for Sum of Indices and Index Laws for Semigroup: Product of Indices, $g_a \circ h$ 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:


 * $\forall n \in \Z, m \in \N: a^{n m} = \paren {a^m}^n$

Also:

and

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

Therefore:
 * $\forall n \in \Z, m \in \N: a^{n \paren {-m}} = \paren {a^{-m}}^n$

Thus:
 * $\forall n, m \in \Z: a^{n m} = \paren {a^m}^n$

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