Index Laws/Sum of Indices/Monoid

Theorem
Let $\left({T, \oplus}\right)$ be a monoid whose identity element is $e$.

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


 * $a^n = \begin{cases}

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

That is:
 * $a^n = \underbrace{a \oplus a \oplus \cdots \oplus a}_{n \text{ copies of } a} = \oplus^n \left({a}\right)$

while:
 * $a^0 = e$

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

Proof
Let $a \in T$.

Because $\left({T, \oplus}\right)$ is a monoid, $\oplus$ is associative on $T$.

By the definition of the power of an element, the mapping $\oplus^n: \N \to T$ is defined as:


 * $\forall n \in \N: \oplus^n a = g_a \left({n}\right)$

where $g_a: \N \to T$ is the recursively defined mapping:


 * $\forall n \in \N: g_a \left({n}\right) = \begin{cases}

e & : n = 0 \\ g_a \left({r}\right) \oplus a & : n = r + 1 \end{cases}$

Consider the recursively defined mapping $f_a: \N_{>0} \to T$ defined as:


 * $\forall n \in \N_{>0}: f_a \left({n}\right) = \begin{cases}

a & : n = 1 \\ f_a \left({r}\right) \oplus a & : n = r + 1 \end{cases}$

From Restriction of Recursive Mapping from Naturally Ordered Semigroup to Algebraic Structure:


 * $f_a$ is the restriction of $g_a$ to $\N_{>0}$.

From Index Laws for Semigroup: Sum of Indices:
 * $\forall m, n \in \N_{>0}: \oplus^{n + m} a = \left({\oplus^n a}\right) \oplus \left({\oplus^m a}\right)$

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

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

Let $n \in \N$:

Similarly, let $m \in \N$:

and:

Thus:
 * $a^{n + m} = a^n \oplus a^m$

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

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

Also see

 * Power of Identity is Identity