Inverse of Product
Theorem
Monoid
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:
- $\forall n \in \N_{> 0}: \paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = {a_n}^{-1} \circ \cdots \circ {a_2}^{-1} \circ {a_1}^{-1}$
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).