# Inverse of Group Product/Proof 2

## Theorem

Let $\struct {G, \circ}$ be a group whose identity is $e$.

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

Then:

$\paren {a \circ b}^{-1} = b^{-1} \circ a^{-1}$

## Proof

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

The result follows from Inverse of Product in Monoid.

$\blacksquare$