Inverse of Group Product

Theorem
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
We have that a group is also a monoid

The result follows from Inverse of Product/Monoid.

The general result follows from Inverse of Product/General Result.