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, $$b^{-1}a^{-1}$$.