Index Laws for Monoids/Product of Indices

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Theorem

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

Let $a \in S$ be invertible for $\circ$.

Let $n \in \N$.

Let $a^n = \map {\circ^n} a$ be defined as the power of an element of a monoid:

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

that is:

$a^n = \underbrace {a \circ a \circ \cdots \circ a}_{n \text { instances} } = \map {\circ^n} a$


Also, for each $n \in \N$ we can define:

$a^{-n} = \paren {a^{-1} }^n$

Then:

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


Proof

Let $m \in \N, c = a^m, d = \left({a^{-1}}\right)^m$.

We define the mapping $g_c: \Z \to S$ as:

$\forall n \in \Z: g_c \left({n}\right) = \circ^n \left({c}\right)$

as defined in the proof of the Index Law for Sum of Indices.

Let $h: \Z \to \Z$ be the mapping defined as:

$\forall z \in \Z: h \left({z}\right) = z m$

Then:

\(\displaystyle a^{n m}\) \(=\) \(\displaystyle \left({g_a \circ h}\right) \left({n}\right)\)
\(\displaystyle \left({a^m}\right)^n\) \(=\) \(\displaystyle g_c \left({n}\right)\)


By Index Law for Sum of Indices and Index Laws for Semigroup: Product of Indices, $g_a \circ h$ and $g_c$ are homomorphisms from $\Z$ to $S$ which coincide on $\N$.

So by the Extension Theorem for Homomorphisms, $g_a \circ h = g_c$.

Therefore:

$\forall n \in \Z, m \in \N: a^{n m} = \left({a^m}\right)^n$


Also:

\(\displaystyle a^{n \left({-m}\right)}\) \(=\) \(\displaystyle a^{-\left({n m}\right)}\)
\(\displaystyle \) \(=\) \(\displaystyle \left({a^{-1} }\right)^{n m}\)
\(\displaystyle \) \(=\) \(\displaystyle \left({g_{a^{-1} } \circ h}\right) \left({n}\right)\)

and

\(\displaystyle \left({a^{-m} }\right)^n\) \(=\) \(\displaystyle \left({\left({a^{-1} }\right)^m}\right)^n\)
\(\displaystyle \) \(=\) \(\displaystyle g_d \left({n}\right)\)


So, by the same reasoning as before, $g_{a^{-1}} \circ h = g_d$.

Therefore:

$\forall n \in \Z, m \in \N: a^{n \left({-m}\right)} = \left({a^{-m}}\right)^n$


Thus:

$\forall n, m \in \Z: a^{n m} = \left({a^m}\right)^n$


As $n m = m n$, the result follows.

$\blacksquare$


Sources