# Index Laws for Monoids/Product of Indices It has been suggested that this article or section be renamed: Should have invertible elements referenced in its name One may discuss this suggestion on the talk page. It has been suggested that this page or section be merged into Index Laws/Product of Indices/Monoid. (Discuss)

## 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:

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

## 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:

 $\ds a^{n m}$ $=$ $\ds \map {\paren {g_a \circ h} } n$ $\ds \paren {a^m}^n$ $=$ $\ds \map {g_c} n$

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:

 $\ds a^{n \paren {-m} }$ $=$ $\ds a^{-\paren {n m} }$ $\ds$ $=$ $\ds \paren {a^{-1} }^{n m}$ $\ds$ $=$ $\ds \map {\paren {g_{a^{-1} } \circ h} } n$

and:

 $\ds \paren {a^{-m} }^n$ $=$ $\ds \paren {\paren {a^{-1} }^m}^n$ $\ds$ $=$ $\ds \map {g_d} n$

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.

$\blacksquare$