Groups of order 56 with a normal Sylow 7-subgroup

Question

I am trying to classify the groups of order 56 containing a normal Sylow 7-subgroup. It is easy enough to see that any such group $G$ must be a semidirect product $Z_7 \rtimes_\varphi P_2$ where $P_2 \in \mathrm{Syl_2}(G)$ and $\varphi:P_2 \rightarrow \mathrm{Aut}(Z_7)$. I am having trouble analyzing the case when $P_2 \cong Z_4 \times Z_2=\langle a\rangle \times \langle b \rangle$.

I am trying to determine the possible homomorphisms from $\varphi:Z_4 \times Z_2 \rightarrow \mathrm{Aut(Z_7)}$. If we let $Z_7=\langle h \rangle$, $\mathrm{Aut}(Z_7) \cong \langle \alpha\rangle$ is cyclic of order 6 where $\alpha$ is the map that sends $h \mapsto h^5$. By Lagrange, the image of any nontrivial homomorphism $\varphi$ has order $2$, and thus must be $\{1,\alpha^3\}$. It seems to me that there are three possible nontrivial homomorphisms determined by the following relations:

\begin{align} \varphi_1 = \left.\begin{cases} a \mapsto \alpha^3\\ b \mapsto 1 \end{cases}\right\rbrace \qquad \varphi_2 = \left.\begin{cases} a \mapsto 1\\ b \mapsto \alpha^3 \end{cases}\right\rbrace \qquad \varphi_3= \left.\begin{cases} a \mapsto \alpha^3\\ b \mapsto \alpha^3 \end{cases}\right\rbrace. \end{align}

According to my textbook (D&F) there should be $2$ isomorphism types for nonabelian groups in this case, but there seem to be three nontrivial homomorphisms that yield groups with seemingly different relations. Why are there only two isomorphism types, and not three?

Answer 1

There is an automorphism $\psi$ of $Z_4 \times Z_2$ such that $\psi(a)= ab$ and $\psi(b) = b$. Since $\varphi_2 \circ \psi = \varphi_3$, we have that $\varphi_2$ and $\varphi_3$ give isomorphic groups.