Exercise 3.3

Let \(G\) be a group, and let \(g, h \in G\). We’re going to try multiplying \(g^{-1} h^{-1}\) and \(hg\) and seeing what happens:

\[ \begin{align}\begin{aligned}&g^{-1} h^{-1} hg\\&= g^{-1} (h^{-1}h) g\\&= g^{-1} e g\\&= g^{-1} g\\&= e.\end{aligned}\end{align} \]

Since inverses are unique, we know that \(hg\) must be the unique inverse of \(g^{-1} h^{-1}\). That is, \(g^{-1} h^{-1} = (hg)^{-1}\).