Suppose $G$ is a groups with order $56$ and $H$ is a normal Sylow $7$-subgroup of $H$. Let $K=Z_4\times Z_2$. Then $G\cong H\rtimes K$ for some homomorphism $\varphi:K\rightarrow \text{Aut}(H)\cong Z_6 =\langle\gamma\rangle$ for some generator $\gamma$. Now, acknowledging that the order of an image of an element must divide the order of the corresponding element, it follows that $\varphi(K)\in\{1,\gamma^3\}$, which implies by the First Isomorphism Theorem that $|\ker\varphi|=4$.
Now, a homomorphism is determined by its action on the generators. It is clear that $\varphi_1(1,0)=\gamma^3$, $\varphi_1(0,1)=1$ and $\varphi_2(1,0)=1$, $\varphi_2(0,1)=\gamma^3$ give two possible homomorphisms.
However, why does this describe all possible homomorphisms? For example, why doesn't $\varphi(1,0)=\gamma^3$ , $\varphi(0,1)=\gamma^3$ describe another?