Inverse of Product/Monoid

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Theorem

Let $\left({S, \circ}\right)$ 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:

$\left({a \circ b}\right)^{-1} = b^{-1} \circ a^{-1}$


General Result

Let $\left({S, \circ}\right)$ 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:

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


Proof

\(\displaystyle \left({a \circ b}\right) \circ \left({b^{-1} \circ a^{-1} }\right)\) \(=\) \(\displaystyle \left({\left({a \circ b}\right) \circ b^{-1} }\right) \circ a^{-1}\) Associativity
\(\displaystyle \) \(=\) \(\displaystyle \left({a \circ \left({b \circ b^{-1} }\right)}\right) \circ a^{-1}\) Associativity
\(\displaystyle \) \(=\) \(\displaystyle \left({a \circ e}\right) \circ a^{-1}\) Behaviour of Inverse
\(\displaystyle \) \(=\) \(\displaystyle a \circ a^{-1}\) Behaviour of Identity
\(\displaystyle \) \(=\) \(\displaystyle e\) Behaviour of Inverse


Similarly for $\left({b^{-1} \circ a^{-1}}\right) \circ \left({a \circ b}\right)$.

$\blacksquare$

Sources