Index Laws for Monoids/Sum 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:
 * $\forall m, n \in \Z: a^{n+m} = a^n \odot a^m$

Proof
For each $c \in S$ which is invertible for $\odot$, we define the mapping $g_c: \Z \to S$ as:


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

By the Index Law for Negative Index above, $g_a \left({n}\right)$ is invertible for all $n \in \Z$.

By Recursive Mapping to Semigroup, the restriction of $g_a$ to $\N$ is a homomorphism from $\left({\N, +}\right)$ to $\left({S, \odot}\right)$.

From Recursive Mapping to Semigroup, $g_a \left({0}\right)$ is the identity for $\odot$ by definition.

Hence, by the Extension Theorem for Homomorphisms, there is a unique homomorphism $h_a: \left({\N, +}\right) \to \left({S, \odot}\right)$ which coincides in $\N$ with $g_c$.

But by Homomorphism with Identity Preserves Inverses:

Hence $h_a = g_a$ and so $g_a$ is a homomorphism and so the result follows.