Index Laws/Sum of Indices/Monoid

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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.

From Index Laws for Semigroup: Sum of Indices:

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

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:

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

$\blacksquare$


Also see


Sources