Index Laws for Monoids/Negative Index

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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:

$\forall n \in \Z: \paren {a^n}^{-1} = a^{-n} = \paren {a^{-1} }^n$


Proof

We have $a^0 = e$ so it follows trivially that $a^{-0} = \paren {a^{-1} }^0$.

From the general inverse of product, we have:

$\paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = a_n^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$

where $a_1, a_2, \ldots, a_n \in S$ are all invertible for $\circ$.


So we can put $a_1, a_2, \ldots, a_n = a$ and we have that

$a^n$ is invertible for all $n \in \N$
$\forall n \in \N: \paren {a^n}^{-1} = \paren {a^{-1 }}^n$


From the above:

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

Thus:

\(\displaystyle \paren {a^{-n} }^{-1}\) \(=\) \(\displaystyle \paren {\paren {a^{-1} }^n}^{-1}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {\paren {a^{-1} }^{-1} }^n\)
\(\displaystyle \) \(=\) \(\displaystyle a^{-\paren {-n} }\)


Similarly, if $a$ is invertible then $a^{-1}$ is also invertible.

So we also have:

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


Thus:

\(\displaystyle a^{-\paren {-n} }\) \(=\) \(\displaystyle \paren {\paren {a^{-1} }^{-1} }^n\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {a^{-1} }^{-n}\)


Thus the result holds for all $n \in \Z$.

$\blacksquare$


Sources