Index Laws/Sum of Indices/Monoid

Theorem
Let $\left({S, \circ, \preceq}\right)$ be a naturally ordered semigroup.

Let $\left({T, *}\right)$ be a monoid whose identity is $e$, and let $a \in T$.

Let the mapping $*^n: S \to T$ be recursively defined by means of the mapping $g_a: S \to T$:


 * $\forall n \in \left({S, \circ, \preceq}\right): *^n a = g_a \left({n}\right)$:


 * $\forall n \in S: g_a \left({n}\right) = \begin{cases}

e & : n = 0 \\ a & : n = 1 \\ \end{cases}$
 * ^r * a & : n = r \circ 1

Then:
 * $\forall n \in S: *^{n \circ m} a = \left({*^n a}\right) * \left({*^m a}\right)$

In particular, the following result holds:
 * $\forall m \in S: *^m e = e$

Proof
From Uniqueness of Recursive Mapping from Naturally Ordered Semigroup to Algebraic Structure with Identity:
 * $*^n: S \to T$ is the only mapping with that definition.

Because $\left({T, *}\right)$ is a semigroup, $*$ is associative on $T$.

The proof proceeds by finite induction on $S$.

Let $a \in T$.

Because $\left({T, *}\right)$ is a semigroup, $*$ is associative on $T$.

Let $S'$ be the set of all $m \in S$ such that:
 * $*^{n \circ m} a = \left({*^n a}\right) * \left({*^m a}\right)$

Let $0$ be understood in this immediate context as the zero of the naturally ordered semigroup.

Let $1$ be understood in this immediate context as the one of the naturally ordered semigroup.

First note that from the definition we have immediately:
 * $g_a \left({0}\right) = *^0 a = e$

Basis for the Induction
Let $n \in \left({S, \circ, \preceq}\right)$.

So $0 \in S'$.

So $1 \in S'$.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $k \in S'$, where $k \ge 1$, then it logically follows that $k \circ 1 \in S'$.

So this is our induction hypothesis:
 * $*^{n \circ k} a = \left({*^n a}\right) * \left({*^k a}\right)$

Then we need to show:
 * $*^{n \circ \left({k \circ 1}\right)} a = \left({*^n a}\right) * \left({*^{k \circ 1} a}\right)$

Induction Step
This is our induction step:

So $k \circ 1 \in S'$.

So by the Principle of Finite Induction:
 * $S' = S$

Thus this result is true for all $m, n \in S$:
 * $\forall m, n \in S^*: *^{n \circ m} a = \left({*^n a}\right) * \left({*^m a}\right)$

Powers of Identity
To show that $\forall m \in S: *^m e = e$:

This can be proved by the Principle of Finite Induction.

Let $Q = \left\{{n \in S: *^n e = e}\right\}$, that is, the set of all elements of $S$ for which the result holds.

We need to show that $Q$ is the same as $S$, that is, that the result holds for all $n \in S$.

By $*^0 a = e$, where $0 \in S$ is the zero of $S$, we have $*^0 e = e$ and so $0 \in Q$.

We also have $*^1 e = e$, where $1 \in S$ is the one of $S$, and so $1 \in Q$.

Now suppose $k \in Q$ where $1 \preceq k$.

We have $*^{k \circ 1} e = *^k e * e = e * e = e$ and so $k \circ 1 \in Q$.

Thus by the Principle of Finite Induction, $Q = S$, and so $\forall n \in S: *^n e = e$.