Inverse of Product

Monoid
Let $\left({S, \circ}\right)$ be a monoid whose identity is $e$.

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

Then $x \circ y$ is invertible for $\circ$, and $\left({x \circ y}\right)^{-1} = y^{-1} \circ x^{-1}$.

Generalized 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 $\left({a_1 \circ a_2 \circ \cdots \circ a_n}\right)^{-1} = a_n^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$.

Group
For any $a$ and $b$ in a group $G$, $(ab)^{-1}=b^{-1}a^{-1}$.

In general:
 * $\left({a_1 a_2 \cdots a_{n-1} a_n}\right)^{-1} = a_n^{-1} a_{n-1}^{-1} \cdots a_2^{-1} a_1^{-1}$

Proof for Monoid
Similarly for $\left({y^{-1} \circ x^{-1}}\right) \circ \left({x \circ y}\right)$.

Proof of Generalized Result
Proof by induction:

We have, from above, $\left({a_1 \circ a_2}\right)^{-1} = a_2^{-1} \circ a_1^{-1}$, and (trivially) $\left({a_1}\right)^{-1} = a_1^{-1}$.

Assume that $\left({a_1 \circ a_2 \circ \cdots \circ a_k}\right)^{-1} = a_k^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$.

Then:

So the assumption being true for $n = k$ implies that it is also true for $n = k + 1$.

It is also true for $n = 1$ (trivially) and $n = 2$ (proved above).

So, by the Principle of Mathematical Induction, it is true for all $n \in \N^*$.

Proof for Group
As a group is also a monoid, then the main result applies directly.

The general result follows from the same source.

Comment
This theorem is also known 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. $ab$ would represent putting on socks followed by shoes. In order to take them off, they must be removed in reverse order, that is, $b^{-1}a^{-1}$.