Index Laws/Product of Indices/Monoid

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Theorem

Let $\struct {S, \circ}$ be a monoid whose identity element is $e$.

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


Then:

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


That is:

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


Proof

Because $\struct {S, \circ}$ is a monoid, it is a fortiori a semigroup.


Hence, from Index Laws for Semigroup: Product of Indices:

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

That is:

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


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

\(\displaystyle n \times 0\) \(=\) \(\displaystyle 0\)
\(\displaystyle \) \(=\) \(\displaystyle 0 \times m\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \circ^{n \times 0} a\) \(=\) \(\displaystyle e\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \) \(=\) \(\displaystyle \circ^{0 \times m} a\)

so the condition holds for either $n = 0$ or $m = 0$.


Finally, we also have:

$\circ^n \paren {\circ^0 a} = e = \circ^0 \paren {\circ^m a}$
$\circ^0 \paren {\circ^n a}\= e = \circ^m \paren {\circ^0 a}$

$\blacksquare$


Notation

Let $a^n$ be defined as the power of an element of a magma:

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

... that is:

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


Recall the index law for product of indices:

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


This result can be expressed:

$a^{n m} = \paren {a^n}^m = \paren {a^m}^n$


When additive notation $\struct {S, +}$ is used, the following is a common convention:

$\paren {n m} a = m \paren {n a} = n \paren {m a}$

or:

$\forall m, n \in \N_{>0}: \paren {n m} \cdot a = m \cdot \paren {n \cdot a} = n \cdot \paren {m \cdot a}$


Sources