Inverse of Product/Monoid/General Result

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Theorem

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

Let $a_1, a_2, \ldots, a_n \in S$ be invertible for $\circ$, with inverses ${a_1}^{-1}, {a_2}^{-1}, \ldots, {a_n}^{-1}$.


Then $a_1 \circ a_2 \circ \cdots \circ a_n$ is invertible for $\circ$, and:

$\forall n \in \N_{> 0}: \paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = {a_n}^{-1} \circ \cdots \circ {a_2}^{-1} \circ {a_1}^{-1}$


Proof

Proof by induction:

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

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


$\map P 1$ is (trivially) true, as this just says:

$\paren {a_1}^{-1} = {a_1}^{-1}$


Basis for the Induction

$\map P 2$ is the case:

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

which has been proved in Inverse of Product in Monoid.

This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.


So this is our induction hypothesis:

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


Then we need to show:

$\paren {a_1 \circ a_2 \circ \cdots \circ a_{k + 1} }^{-1} = {a_{k + 1} }^{-1} \circ \cdots \circ {a_2}^{-1} \circ {a_1}^{-1}$


Induction Step

This is our induction step:

\(\ds \paren {a_1 \circ a_2 \circ \cdots \circ a_k \circ a_{k + 1} }^{-1}\) \(=\) \(\ds \paren {\paren {a_1 \circ a_2 \circ \cdots \circ a_k} \circ a_{k + 1} }^{-1}\) General Associativity Theorem
\(\ds \) \(=\) \(\ds {a_{k + 1} }^{-1} \circ \paren {a_1 \circ a_2 \circ \cdots \circ a_k}^{-1}\) Basis for the Induction
\(\ds \) \(=\) \(\ds {a_{k + 1} }^{-1} \circ \paren { {a_k}^{-1} \circ \cdots \circ {a_2}^{-1} \circ {a_1}^{-1} }\) Induction Hypothesis
\(\ds \) \(=\) \(\ds {a_{k + 1} }^{-1} \circ {a_k}^{-1} \circ \cdots \circ {a_2}^{-1} \circ {a_1}^{-1}\) General Associativity Theorem


So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \N_{> 0}: \paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = {a_n}^{-1} \circ \cdots \circ {a_2}^{-1} \circ {a_1}^{-1}$

$\blacksquare$

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