# 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.

• Is $V_4$ the Klein $4$-group? Direct and semidirect products are generally considered "known groups" and often have no simpler description, the point here is just to identify how many there are and when they're isomorphic. – Jim Oct 19 '17 at 11:18
• Also, the map you really need to look at is $V_4 \to \mathrm{Aut}(C_3)$. You get a direct product when this map is trivial. So now you know that $\mathrm{Aut}(C_3) \simeq C_2$. You need to start by finding all the homomorphisms $V_4 \to C_2$. – Jim Oct 19 '17 at 11:21
• Yes, it is the Klein-4-group. Is their a theorem that gives me the number of homomorphisms betwenn two groups? So I know how many I have to find. – Myrkuls JayKay Oct 19 '17 at 11:31
• I know that for a homomorphism $\phi : V_4 \to Aut(C_3)$, the neutral element of $V_4$ has to be mapped to $Id_{C_3}$. – Myrkuls JayKay Oct 19 '17 at 11:38
• A hint regarding $\alpha_2$: What can you say in that case about (a) the set of Sylow-$3$ subgroups of $G$, (b) the action of $G$ on that set and (c) the kernel of that action? – jpvee Oct 19 '17 at 12:51

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 $( \mathbb{Z}_2\times \mathbb{Z}_2 )\ltimes_{\alpha} \mathbb{Z}_3$.

• I really thank you for this solution! In your definition of $\alpha_4$ are mistakes in the last two mappings. You have there another two times the element $(0,0)$. These should be $(0,1)$ and $(1,1)$. – Myrkuls JayKay Oct 19 '17 at 19:00