\begin{equation}\label{eq:idG} \phi (x \ast x^{-1}) = \phi (1_G) = \phi (x) \star \phi (x^{-1}) \end{equation} Recall that, for any group homomorphism $\phi\colon G \to K$

\begin{equation}\label{eq:idK} \phi (1_G) = 1_K \end{equation} In other words, homomorphisms preserve identity. ¹ It follows from () and () that

\begin{equation} \phi (x) \star \phi (x^{-1}) = 1_K \end{equation} Because the inverse of any group is unique, the only value of $\phi (x^{-1})$ whose product with $\phi (x)$ is $1_K$ is, of course, $\phi (x)^{-1}$ Therefore, all group homomorphisms preserve the inverse.