# Index Laws/Sum of Indices/Monoid

## Theorem

Let $\left({S, \circ}\right)$ be a monoid whose identity element is $e$.

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

$a^n = \begin{cases} e & : n = 0 \\ a^x \circ a & : n = x + 1 \end{cases}$

That is:

$a^n = \underbrace{a \circ a \circ \cdots \circ a}_{n \text{ copies of } a} = \circ^n \left({a}\right)$

while:

$a^0 = e$

Then:

$\forall m, n \in \N: a^{n + m} = a^n \circ a^m$

## Proof

Because $\left({S, \circ}\right)$ is a monoid, it is a fortiori also a semigroup.

$\forall m, n \in \N_{>0}: \circ^{n + m} a = \left({\circ^n a}\right) \circ \left({\circ^m a}\right)$

That is:

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

It remains to be shown that the result holds for the cases where $m = 0$ and $n = 0$.

Let $n \in \N$:

 $\displaystyle a^{n + 0}$ $=$ $\displaystyle a^n$ Definition of Zero $\displaystyle$ $=$ $\displaystyle a^n \circ e$ Definition of Identity Element $\displaystyle$ $=$ $\displaystyle a^n \circ a^0$ Definition of $a^0$

Similarly, let $m \in \N$:

 $\displaystyle a^{0 + m}$ $=$ $\displaystyle a^m$ Definition of Zero $\displaystyle$ $=$ $\displaystyle e \circ a_m$ Definition of Identity Element $\displaystyle$ $=$ $\displaystyle a^0 \circ a^m$ Definition of $a^0$

and:

 $\displaystyle a^{0 + 0}$ $=$ $\displaystyle a^0$ Definition of Zero $\displaystyle$ $=$ $\displaystyle e$ Definition of $a^0$ $\displaystyle$ $=$ $\displaystyle e \circ e$ Definition of Identity Element $\displaystyle$ $=$ $\displaystyle a^0 \circ a^0$ Definition of $a^0$

Thus:

$a^{n + m} = a^n \circ a^m$

holds for $n = 0$ and $m = 0$.

Thus:

$\forall m, n \in \N: a^{n + m} = a^n \circ a^m$

$\blacksquare$