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:

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


For each $n \in \N$ we 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 $m \in \N, c = a^m, d = \paren {a^{-1}}^m$.

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

$\forall n \in \Z: \map {g_c} n = \map {\circ^n} c$

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

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

$\forall z \in \Z: \map h z = z m$

Then:

\(\ds a^{n m}\) \(=\) \(\ds \map {\paren {g_a \circ h} } n\)
\(\ds \paren {a^m}^n\) \(=\) \(\ds \map {g_c} n\)


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:

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


Also:

\(\ds a^{n \paren {-m} }\) \(=\) \(\ds a^{-\paren {n m} }\)
\(\ds \) \(=\) \(\ds \paren {a^{-1} }^{n m}\)
\(\ds \) \(=\) \(\ds \map {\paren {g_{a^{-1} } \circ h} } n\)

and:

\(\ds \paren {a^{-m} }^n\) \(=\) \(\ds \paren {\paren {a^{-1} }^m}^n\)
\(\ds \) \(=\) \(\ds \map {g_d} n\)


So, by the same reasoning as before:

$g_{a^{-1}} \circ h = g_d$

Therefore:

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


Thus:

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


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

$\blacksquare$


Sources