Index Laws for Semigroup

Theorem
Let $$\left ({S, \odot}\right)$$ be a semigroup.

Let $$a \in S$$.

Let $$n \in \mathbb{N}^*$$.

Let $$\odot^n \left({a}\right)$$ be defined as in Recursive Mapping to Semigroup:

$$ \odot^n \left({a}\right) = \begin{cases} a : & n = 1 \\ \odot^x \left({a}\right) \odot a : & n = x + 1 \end{cases} $$

... that is, $$\odot^n \left({a}\right) = a \odot a \odot \cdots \left({n}\right) \cdots \odot a$$.

Then the following results hold:

Sum of Indices
Let $$a \in S$$. Then:

$$\forall m, n \in \mathbb{N}^*: \odot^{n+m} \left({a}\right) = \left({\odot^n \left({a}\right)}\right) \odot \left({\odot^m \left({a}\right)}\right)$$

Product of Indices
Let $$a \in S$$. Then:

$$\forall m, n \in \mathbb{N}^*: \odot^{n m} \left({a}\right) = \odot^n \left({\odot^m \left({a}\right)}\right) = \odot^m \left({\odot^n \left({a}\right)}\right)$$

Product of Commutative Elements
Let $$a, b \in S$$ commute with each other. Then:

$$\forall n \in \mathbb{N}^*: \odot^n \left({a \odot b}\right) = \left({\odot^n \left({a}\right)}\right) \odot \left({\odot^n \left({b}\right)}\right)$$

Commutation of Powers
Let $$a, b \in S$$ commute with each other. Then:

$$\forall m, n \in \mathbb{N}^*: \left({\odot^m \left({a}\right)}\right) \odot \left({\odot^n \left({b}\right)}\right) = \left({\odot^n \left({b}\right)}\right) \odot \left({\odot^m \left({a}\right)}\right)$$

Sum of Indices
Because $$\left({S, \odot}\right)$$ is a semigroup, $$\odot$$ is associative on $$S$$.

Let $$T$$ be the set of all $$m \in \mathbb{N}$$ such that this result holds.

Let $$n \in \mathbb{N}^*$$.

So $$1 \in T$$.

Now suppose $$m \in T$$. Then we have:

So $$m + 1 \in T$$.

So by the Principle of Finite Induction, $$T = \mathbb{N}^*$$.

Thus this result is true for all $$m, n \in \mathbb{N}^*$$.

Product of Indices
This is proved in Naturally Ordered Semigroup Power Law.

Product of Commutative Elements
Let $$T$$ be the set of all $$n \in \mathbb{N}$$ such that:

$$\odot^n \left({a \odot b}\right) = \left({\odot^n \left({a}\right)}\right) \odot \left({\odot^n \left({b}\right)}\right)$$

Now:

So $$1 \in T$$.

Now suppose $$n \in T$$. Then we have:

So $$n + 1 \in T$$.

Thus by the Principle of Finite Induction, $$T = \mathbb{N}^*$$, and the result holds for all $$n \in \mathbb{N}^*$$.

Commutation of Powers
Let $$a, b \in S: a \odot b = b \odot a$$.

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

Let $$T$$ be the set of all $$n \in \mathbb{N}^*$$ such that:

$$\left({\odot^n \left({a}\right)}\right) \odot b = b \odot \left({\odot^n \left({a}\right)}\right)$$

We have $$a \odot b = b \odot a \Longrightarrow \left({\odot^1 \left({a}\right)}\right) \odot b = b \odot \left({\odot^1 \left({a}\right)}\right)$$.

So $$1 \in T$$.

Now suppose $$n \in T$$. Then we have:

So $$n + 1 \in T$$.

Thus by the Principle of Finite Induction, $$T = \mathbb{N}^*$$. Thus:

$$\forall m \in \mathbb{N}^*: \left({\odot^m \left({a}\right)}\right) \odot b = b \odot \left({\odot^m \left({a}\right)}\right)$$

Thus, from the preceding: $$\forall m, n \in \mathbb{N}^*: \odot^m \left({a}\right)$$ and $$\odot^n \left({b}\right)$$ also commute with each other.

The result follows.