Theorem 1   A group homomorphism preserves identity elements.

Proof. Let $\phi:G\to K$ be a group homomorphism. For clarity we use $\ast$ and $\star$ for the group operations of $G$ and $K$ respectively. Also, denote the identities by $1_G$ and $1_H$ respectively.

By the definition of identity, \begin{equation}\label{eq:idG} 1_G\ast 1_G=1_G. \end{equation}Applying the homomorphism $\phi$ to () produces: \begin{equation}\label{eq:idPhi} \phi(1_G)\star \phi(1_G)= \phi(1_G\ast 1_G) = \phi (1_G). \end{equation}Multiply both sides of () by the inverse of $\phi(1_G)$ in $K$ and use the associativity of $\star$ to produce: \begin{equation} \phi(1_G)=(\phi(1_G))^{-1}\star \phi(1_G)\star \phi(1_G) = (\phi(1_G))^{-1}\star \phi(1_G)=1_K. \end{equation}