Semidirect products of $V_4 \ltimes_\alpha C_3$ and isomorphies My task is to find all semidirect products $V_4 \ltimes_\alpha C_3$ and to find those, who are isomorphic.
First of all, I've got to find the automorphism group of $C_3$. I know, that it is isomorphic to $(\mathbb Z / 3 \mathbb Z)^*$, so it has to have $3-1=2$ elements.
The $C_3$ has got three elements $a,a^2,e$, where $e=a^3$. 
For an automorphism of $C_3$ I know that it has to map $e \mapsto e$, so I only have to look how the other two elements can be mapped to eachother. 
The first possibility would be $\alpha_1=Id_{C_3}$ with $e \mapsto e$ $\;$, $a \mapsto a$ and $a^2 \mapsto a^2$.
The second one would be $\alpha_2$ with $e \mapsto e$ $\;$, $a \mapsto a^2$ and $a^2 \mapsto a$.
One thing I know is, if $\alpha$ is the identity then the semidirect product is the normal direct product, so my first one would deliver $V_4 \ltimes_{\alpha_1} C_3$ = $V_4 \times C_3$. Is this one isomorphic to a known group like $C_n$ or $S_n$?
And for $\alpha_2$ I don't know how I have to go on.
 A: Each semidirect products $ ( \mathbb{Z}_2\times \mathbb{Z}_2 )\ltimes_{\alpha} \mathbb{Z}_3$ is characterised by a homomorphism $\alpha : \mathbb{Z}_2\times \mathbb{Z}_2 \to \operatorname{Aut}(\mathbb{Z}_3)$. As you noted there are two automorphisms of $\mathbb{Z}_3$:
\begin{align*}
\operatorname{id}: \mathbb{Z}_3 &\to \mathbb{Z}_3 & f : \mathbb{Z}_3 &\to \mathbb{Z}_3 \\
 1& \mapsto 1 & 1 &\mapsto 2 
\end{align*}
Next we determine the homomorphisms $\alpha :\mathbb{Z}_2 \times \mathbb{Z}_2 \to \operatorname{Aut}(\mathbb{Z}_3)$. Each $\alpha$ is determined by its effecton on $(1,0),(0,1) \in \mathbb{Z}_2 \times \mathbb{Z}_2$. Since $\operatorname{Aut}(\mathbb{Z}_3)$ has two elements, there are $2\cdot 2 = 4$ such homorphisms. First is the trivial map
\begin{align*}
\alpha_1: \mathbb{Z}_2 \times \mathbb{Z}_2 &\to \operatorname{Aut}(\mathbb{Z}_3) \\
x &\mapsto \operatorname{id},
\end{align*}
in which case $\left( \mathbb{Z}_2 \times \mathbb{Z}_2 \right) 
\ltimes_{\alpha_1} \mathbb{Z}_3 \cong \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_3$. There are also the three homomorphisms:
\begin{align*}
\alpha_2: \mathbb{Z}_2 \times \mathbb{Z}_2 &\to \operatorname{Aut}(\mathbb{Z}_3) 
& \alpha_3 : \mathbb{Z}_2 \times \mathbb{Z}_2 &\to \operatorname{Aut}(\mathbb{Z}_3) 
& \alpha_4 : \mathbb{Z}_2 \times \mathbb{Z}_2 &\to \operatorname{Aut}(\mathbb{Z}_3) \\
(0,0) & \mapsto \operatorname{id} & (0,0) & \mapsto \operatorname{id} & (0,0) & \mapsto \operatorname{id} \\
(1,0) & \mapsto \operatorname{id} & (1,0) & \mapsto f & (1,0) & \mapsto f \\
(0,1) & \mapsto f & (0,1) & \mapsto \operatorname{id} & (0,1) & \mapsto f \\
(1,1) & \mapsto f & (1,1) & \mapsto f & (1,1) & \mapsto \operatorname{id} \\
\end{align*}
So there are  four possible semidirect products $ ( \mathbb{Z}_2\times \mathbb{Z}_2 )\ltimes_{\alpha} \mathbb{Z}_3$. But are they all distinct groups?
The answer is no. The homomorphisms $\alpha_2,\alpha_3$ and $\alpha_4$ all induce isomorphic groups. To see this, observe that $\alpha_3 = \alpha_2 \circ \phi$ and $\alpha_4 = \alpha_2 \circ \psi$, where $\phi$ and $\psi$ are the automorphisms:
\begin{align*}
\phi: \mathbb{Z}_2 \times \mathbb{Z}_2 &\to \mathbb{Z}_2 \times \mathbb{Z}_2
& \psi: \mathbb{Z}_2 \times \mathbb{Z}_2 &\to \mathbb{Z}_2 \times \mathbb{Z}_2 \\
(0,0) & \mapsto (0,0) & (0,0) & \mapsto (0,0) \\
(1,0) & \mapsto (0,1) & (1,0) & \mapsto (1,1) \\
(0,1) & \mapsto (1,0) & (0,1) & \mapsto (0,1) \\
(1,1) & \mapsto (1,1) & (1,1) & \mapsto (1,0) \\
\end{align*}
It follows that $\left( \mathbb{Z}_2 \times \mathbb{Z}_2 \right) \ltimes_{\alpha_2} \mathbb{Z}_3 \cong \left( \mathbb{Z}_2 \times \mathbb{Z}_2 \right) \ltimes_{\alpha_3} \mathbb{Z}_3 $ via the map:
\begin{align*}
\Theta:\left( \mathbb{Z}_2 \times \mathbb{Z}_2 \right) \ltimes_{\alpha_2} \mathbb{Z}_3 &\to \left( \mathbb{Z}_2 \times \mathbb{Z}_2 \right) \ltimes_{\alpha_3} \mathbb{Z}_3  \\
((a,b),c) &\mapsto \left( \phi^{-1}(a,b),c \right)
\end{align*}
A similar calculation yields, $\left( \mathbb{Z}_2 \times \mathbb{Z}_2 \right) \ltimes_{\alpha_2} \mathbb{Z}_3 \cong \left( \mathbb{Z}_2 \times \mathbb{Z}_2 \right) \ltimes_{\alpha_4} \mathbb{Z}_3 $.
Thus there are two distict semidirect products $ ( \mathbb{Z}_2\times \mathbb{Z}_2 )\ltimes_{\alpha} \mathbb{Z}_3$.
