Index Laws for Monoids/Sum 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:

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


For each $n \in \N$ we define:

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

Then:

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


Proof

For each $c \in S$ which is invertible for $\circ$, we define the mapping $g_c: \Z \to S$ as:

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

By the Index Law for Monoids: Negative Index, $\map {g_a} n$ is invertible for all $n \in \Z$.

By definition of Power of Element of Monoid, the restriction of $g_a$ to $\N$ is a homomorphism from $\struct {\N, +}$ to $\struct {S, \circ}$.

From the definition of Power of Element of Monoid, $\map {g_a} 0$ is the identity for $\circ$.

Hence, by the Extension Theorem for Homomorphisms, there is a unique homomorphism $h_a: \paren {\N, +} \to \paren {S, \circ}$ which coincides in $\N$ with $g_c$.

But by Homomorphism with Identity Preserves Inverses:

\(\, \ds \forall n > 0: \, \) \(\ds \map {h_a} {-n}\) \(=\) \(\ds \paren {\map {h_a} n}^{-1}\)
\(\ds \) \(=\) \(\ds \paren {a^n}^{-1}\)
\(\ds \) \(=\) \(\ds a^{-n}\)
\(\ds \) \(=\) \(\ds \map {g_a} {-n}\)


Hence $h_a = g_a$ and so $g_a$ is a homomorphism and so the result follows.

$\blacksquare$


Sources