Inverse of Group Product

Theorem
Let $\left({G, \circ}\right)$ be a group whose identity is $e$.

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

Then:
 * $\left({a \circ b}\right)^{-1} = b^{-1} \circ a^{-1}$

Proof
We have that a group is a monoid, all of whose elements are invertible.

The result follows from Inverse of Product in Monoid.