Classifing groups of order 56: problems with the semidirect product While I was doing an exercise about the classification of groups of order 56, I had some problems concerning the semidirect product.
Let $G$ a group of order 56 and let us suppose that the 7-Sylow is normal (let's call it $H$). Then we want to construct the non abelian group whose 2-Sylow $S$ is $S \cong \mathbb Z_8$.
First of all, we have to determine the homomorphism $\phi \colon \mathbb Z_8 \to \text{Aut}(\mathbb Z_7)$. 
It is known that $\text{Aut}(\mathbb Z_7) \cong \mathbb Z_6$ and the isomorphism is given by
$$
\begin{split}
& \mathbb Z_6 \to \text{Aut}(\mathbb Z_7) \\
& a \mapsto \psi_a \colon \mathbb Z_7 \ni n \mapsto an \in \mathbb Z_7
\end{split}
$$
So we can start by finding the homomorphism $\mathbb Z_8 \to \mathbb Z_6$. There are exacly $(6,8)=2$ such homomorphism. Who are they? Simply the one who sends everything to $0$ and the multiplication by $3$ (which is the only element in $\mathbb Z_6$ whose order - 2 - divides 8). In multiplicative terms, they are the homomorphism which sends everything to $1$ and the homomorphism which sends $n \mapsto 6^n=(-1)^n$.
So, by composition, we have the two homomorphism 
$$
\begin{split}
\phi_1 \colon & \mathbb Z_8 \to \text{Aut}(\mathbb Z_7)\\
& n \mapsto \text{id}
\end{split}
$$
and
$$
\begin{split}
\phi_2 \colon & \mathbb Z_8 \to \text{Aut}(\mathbb Z_7)\\
& n \mapsto \psi_n \colon \mathbb Z_7 \ni x \mapsto 6^nx = (-1)^nx \in \mathbb Z_7 
\end{split}
$$
Am I right? 
Now, if we take $\phi_1$ we simply get the direct product. What if we take $\phi_2$?
For sake of simplicity, let's assume additive notation (this is stupid, I know but it has helped me somehow to understand). If I'm not wrong, we obtain that $H \rtimes_{\phi_2} \mathbb Z_8$ is the set $H \times \mathbb Z_8$ with the operation given by 
$$
(a,b) + (c,d) = (a+(-1)^bc,b+d)
$$
Now if I do $(0,k)+(h,0)-(0,k) = ((-1)^k h, 0) = \phi_k(h)$ which is exactly what I want. 
Now I must pass to the much more confortable multiplicative notation: so let's $C_7=\langle s \rangle$ and $C_8=\langle r \rangle$ be the cyclic groups of order 7 and 8. Then we define the automorphisms 
$$
\begin{split}
\phi_n \colon & C_7 \to C_7 \\
& x \mapsto x^{(-1)^n}
\end{split}
$$
and the homomorphism 
$$
\begin{split}
\psi \colon & C_8 \to \text{Aut}(C_7) \\
& n \mapsto \phi_n
\end{split}
$$
In other words, we can simply say that $\psi$ is the homomorphism which sends the generator $r$ to the inversion $x^{-1}$. Am I right so far? 
Well, now $C_7 \rtimes_{\psi} C_8$ is the set $C_7 \times C_8$ with the operation given by 
$$
(a,b)(c,d) = (ac^{(-1)^b},bd)
$$
I do again the calculation $(1,k)(h,1)(1,k)^{-1}=(h^{(-1)^k},1) = \phi_k(h)$ which is exactly what I want (also according to ineff's answer). 
Are there any mistakes? 
May I ask one more question? Who is this mysterious group I've built up? Is it isomorphic to some other (simpler) group? How can I do to write down a presentation? 
I thank you in advance for your kind help.
 A: I suppose your confusion is due to the double representation of the semidirect product: internal vs external.
The semidirect product's theorem states that if you have a group $G$ having two subgroups $H,K < G$ such that $H$ is normal in $G$, $H \cap K = \{1_G\}$ and $G=HK$ then there's an isomorphism $G \cong H \rtimes_\psi K$, for a certain $\psi \colon K \to \text{Aut}(H)$. 
By the theorem  we can represent every element of $G$ as a pair $(h,k) \in H \times K$ (which is the support of the group $H \rtimes K$). 
Consider the two subgroups $\bar H = \{(h,1_K) | h \in H\} \leq H \rtimes K$ and $\bar K = \{(1_H,k)|k \in K\} \leq H \rtimes K$, these subgroups correspond, via the isomorphism, to the subgroups $H$ and $K$ of $G$.  
In $H \rtimes K$ we have that for every $h \in H$ and $k \in K$
$$(1_H,k) * (h,1_K) *(1_H,k)^{-1} = (\psi_k(h),1_K)$$
if we identify every $h \in H$ with its corresponding element $(h,1_K)$ and every $k \in K$ with $(1_H,k)$ then this equality become (internally in $G$)
$$k*h*k^{-1}=\psi_k(h)$$
The $\psi$ which determine the operation in the semidirect product is exactly the homomorphism sending every $k \in K$ in the (restriction to $H$ of the) automorphism $\psi_k \colon H \to H$ which send $h \in H$ in $khk^{-1}$ (this is clearly well defined because $H$ is normal in $G$.
Hope this help.
