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 defined by means of $$g_a$$ in Recursive Mapping with Identity:


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



\forall n \in S: g_a \left({n}\right) = \begin{cases} e & : n = 0 \\ g_a \left({r}\right) * a & : n = r \circ 1 \end{cases} $$

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

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

Proof
Proof by finite induction:

Let $$a \in T$$.

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

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

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:
 * $$g_a \left({0}\right) = \ast^0 a = e$$

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 $$k \in S'$$ is true, where $$k \ge 1$$, then it logically follows that $$k \circ 1 \in S'$$ is true.

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

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

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

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

This can be proved by the Principle of Finite Induction.

Let $$Q = \left\{{n \in S: \ast^n e = e}\right\}$$, 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 $$\ast^0 a = e$$, where $$0 \in S$$ is the zero of $$S$$, we have $$\ast^0 e = e$$ and so $$0 \in Q$$.

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

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