Index Laws/Sum of Indices/Semigroup

Theorem
Let $\struct {S, \circ}$ be a semigroup.

For $a \in S$, let $\circ^n a = a^n$ be defined as the $n$th power of $a$:


 * $a^n = \begin{cases}

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

That is:
 * $a^n = \underbrace {a \circ a \circ \cdots \circ a}_{n \text{ copies of } a} = \circ^n \paren a$

Then:


 * $\forall m, n \in \N_{>0}: a^{n + m} = a^n \circ a^m$

Proof
Let $a \in S$.

Because $\struct {S, \circ}$ is a semigroup, $\circ$ is associative on $S$.

The proof proceeds by the Principle of Mathematical Induction.

Let $\map P m$ be the proposition:
 * $\forall n \in \N_{>0}: a^{n + m} = a^n \circ a^m$

that is:
 * $\forall n \in \N_{>0}: \circ^{n + m} a = \paren {\circ^n a} \circ \paren {\circ^m a}$

Basis for the Induction
So $\map P 1$ holds.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:
 * $\circ^{n + k} a = \paren {\circ^n a} \circ \paren {\circ^k a}$

It is then to be shown that:
 * $\circ^{n + \paren {k + 1} } a = \paren {\circ^n a} \circ \paren {\circ^{k + 1} a}$

Induction Step
This is our induction step:

So $\map P {k + 1}$ is true.

So by the Principle of Mathematical Induction, this result is true for all $m, n \in \N_{>0}$:


 * $\forall m, n \in \N_{>0}: \circ^{n + m} a = \paren {\circ^n a} \circ \paren {\circ^m a}$

or:


 * $\forall m, n \in \N_{>0}: a^{n + m} = a^n \circ a^m$