Inverse of Product/Monoid

Theorem

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

Let $a, b \in S$ be invertible for $\circ$, with inverses $a^{-1}, b^{-1}$.

Then $a \circ b$ is invertible for $\circ$, and:

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

General Result

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

 $\ds \paren {a \circ b} \circ \paren {b^{-1} \circ a^{-1} }$ $=$ $\ds \paren {\paren {a \circ b} \circ b^{-1} } \circ a^{-1}$ Definition of Associative Operation $\ds$ $=$ $\ds \paren {a \circ \paren {b \circ b^{-1} } } \circ a^{-1}$ Definition of Associative Operation $\ds$ $=$ $\ds \paren {a \circ e} \circ a^{-1}$ Behaviour of Inverse $\ds$ $=$ $\ds a \circ a^{-1}$ Behaviour of Identity $\ds$ $=$ $\ds e$ Behaviour of Inverse

Similarly for $\paren {b^{-1} \circ a^{-1} } \circ \paren {a \circ b}$.

$\blacksquare$