Describe a group $G$ that acts on a set $X$ of 4 elements such that the action of $G$ has 2 orbits.

Question

The motivation of this self made problem is to get a better understanding of Group actions.

Say $G$ is a group that acts on a set $X$ of 4 elements such that the action of $G$ has 2 orbits.

What I want to do is work back and figure out what is the structure of $G$ from the above statement. Also, setup explicit actions ( if possible ).

My results

From definition, $G$ is isomorphic to a subgroup of $S_4$. As there are 2 orbits under the action of $G$ and orbits partition $X$, we can infer that the orbits are of the form i) $\{a_1,a_2\},\{a_3,a_4\}$ or ii) $\{a_1,a_2,a_3\},\{a_4\}$. For both cases let $a_1$ and $a_4$ be representatives of the respective orbits.

If the orbits are of the form (ii), then by the Orbit-Stabilizer theorem we have that $|\text{Stab}_{G}(a_4)| = |G|$ and $|\text{Stab}_{G}(a_1)| = |\text{Stab}_G(a_4)|/3$. By Cauchy's theorem, we have that $|Stab_G(a_4)|$ has an element of order 3. Hence, it has a cyclic subgroup $<g>$ of order 3.

If the orbits are of the form (i), then by the the Orbit-Stabilizer theorem we have that $|\text{Stab}_{G}(a_4)| = |\text{Stab}_{G}(a_1)|$.

Now I am stuck. I am at a loss, what extra information can I deduce that will allow me to achieve my goal ?

By explicitly taking subgroups of $S_4$, and applying them on $X$, I have that $S_3$, $<(123)>$ and $<(12)(34)>$ satisfy the above condition ( there may be more but I have not computed all of them ).

Note

1. I would like to avoid applying Sylow's theorems or Burnside's theorem at this point of time. The class equation is okay.

Answer 1

As you already stated there are orbit partitions into 2 2-element sets and into a singleton and a 3-element set. Furthermore a group action can be considered a group homomorphism of on group into symmetric group of the base set $X$. This does not necessarily have to be injective nor surjective.

Let's consider first the partition $123|4$. Then we can ignore $4$ as this is a fixed point. So we need a group that has a subgroup with a homomorphsim into the cycle $(123)$. Candidates are $\mathbb{Z}_6$ or $\mathbb Z$. Both have a group homomorphism into $\mathbb Z_3$ which we denote by the mapping $a\mapsto a \mod 3$. Now $\mathbb Z_3$ acts on $\{1,2,3\}$ in the usual way: $ψ(a,x) = x + a$. While $ψ(a,4) = 4$ is constant. Thus we have the group actions: $$ ψ_{\mathbb Z}: \mathbb Z\times \{1,2,3,4\}\to \{1,2,3,4\}:\begin{cases} (a,x)\mapsto x&x = 4,\\ (a,x)\mapsto x + (a \mod 3),&\text{else} \end{cases}\\ ψ_{\mathbb Z_6}: \mathbb Z\times \{1,2,3,4\}\to \{1,2,3,4\}:\begin{cases} (a,x)\mapsto x&x = 4,\\ (a,x)\mapsto x + (a \mod 3),&\text{else} \end{cases} $$

The other case can be constructed from two group actions $Ψ_1:G_1\times \{1,2\}\to\{1,2\}$ and $ψ_2:G_2\times \{3,4\}\to \{3,4\}$. $φ:\hat G_1\to G_1$ and $ψ:\hat G_2\to G_2$ two surjective homomorphisms. Then you can do any magic with $\hat G_1$ and $\hat G_2$ in order to form a new group $G$ that preserves the group structure somehow: e.g. group extensions (direct products, subdirect products, semidirect products, …), embeddings or whatever you want. As long as the resulting homomorphisms into $G_1$ and $G_2$ are surjective, you end up with a group action of $G$ with the orbit partition $12|34$.