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.