Proof. In order to show that \ker(\phi) is a subgroup of H, it must hold the subgroup criterion. For this proof, let 1_G and 1_H be the identities of G and H, respectively. Firstly, we know that \ker(\phi) is nonempty since it contains the identity element 1_G \in \ker(\phi). Secondly, let x,y \in \ker(\phi). To satisfy the subgroup criterion, we only need to show that xy^{-1} \in \ker(\phi). Since \phi(x) = \phi(y) = 1_H, we get the following:
\begin{align*}
\phi(xy^{-1}) &= \phi(x)\phi(y^{-1}) \\
&= \phi(x)\phi(y)^{-1} \\
&= 1_H 1_H^{-1} \\
&= 1_H.
\end{align*}
