Index Laws/Product of Indices/Semigroup

Theorem

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

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

Then:

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

That is:

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

Proof

Let $b = \circ^m a$.

Let $h: \N_{>0} \to S$ be the mapping defined as:

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

Let the mapping $f_b: \N_{>0} \to S$ be recursively defined as:

$\forall n \in \N_{>0}: f_b \paren n = \circ^n b$

From the Principle of Recursive Definition:

$f_b$ is the unique mapping which satisfies:
$\forall n \in \N_{>0}: f_b \paren n = \begin{cases} b & : n = 1 \\ f_b \paren r \circ b & : n = r \circ 1 \end{cases}$

But $h \paren 1 = \circ^{1 \times m} a = \circ^m a = b$.

So:

 $\, \displaystyle \forall n \in \N_{>0}: \,$ $\displaystyle h \paren {n + 1}$ $=$ $\displaystyle \circ^{\paren {n + 1} m} a$ $\displaystyle$ $=$ $\displaystyle \circ^{\paren {n m} + m} a$ Natural Number Multiplication Distributes over Addition $\displaystyle$ $=$ $\displaystyle \paren {\circ ^{n \ast m} } \circ \paren {\circ^m a}$ Index Laws for Semigroup: Sum of Indices $\displaystyle$ $=$ $\displaystyle h \paren n) \circ b$

Thus $h = f_b$, and so:

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

$\blacksquare$

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:

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