For any Group homomorphism $~$f: G \to H$~$, we have $~$f(e_G) = e_H$~$ where $~$e_G$~$ is the identity of $~$G$~$ and $~$e_H$~$ the identity of $~$H$~$.

Indeed, $~$f(e_G) f(e_G) = f(e_G e_G) = f(e_G)$~$, so premultiplying by $~$f(e_G)^{-1}$~$ we obtain $~$f(e_G) = e_H$~$.