Index Laws for Semigroup

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

Let $$a \in S$$.

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

Let $$\odot^n \left({a}\right) = a^n$$ be defined as in Power of an Element:

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

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

(Note that the notation $$a^n$$ and $$\odot^n \left({a}\right)$$ mean the same thing. The former is more compact and readable, but the latter is more explicit and can be more useful when more than one structure is under consideration.)

Then the following results hold:

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

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

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

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

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.