Inverse of Product

From ProofWiki
Jump to: navigation, search

Theorem

Monoid

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:

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


Group

Let $\struct {G, \circ}$ be a group whose identity is $e$.

Let $a, b \in G$, with inverses $a^{-1}, b^{-1}$.


Then:

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


General Result

Let $\struct {G, \circ}$ be a group whose identity is $e$.

Let $a_1, a_2, \ldots, a_n \in G$, with inverses $a_1^{-1}, a_2^{-1}, \ldots, a_n^{-1}$.


Then:

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


Also known as

When first introducing this concept in a course on group theory, some tutors refer to this theorem as the Socks-Shoes Property.

If one thinks of $a$ as putting on socks, $b$ as putting on shoes, $a^{-1}$ as taking off socks, and $b^{-1}$ as taking off shoes, the theorem demonstrates the order in which one must perform these actions. $a b$ would represent putting on socks followed by shoes. In order to take them off, they must be removed in reverse order, that is, $b^{-1} a^{-1}$.


Less informally it is known as the reversal rule (for group inverses).


Also see