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

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 $$$$ 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.
• An explicit example comes from geometry: $\mathrm{GL}_2(\Bbb F_2)$ acts on $\Bbb F_2^2$, a four element set, and has two orbits. Similarly, yet less interestingly, the multiplicative group of units $\Bbb F_4^\times\simeq \Bbb Z/3\Bbb Z$ acts on the four element set $\Bbb F_4$ by homothecies and has two orbits. Aug 7 '15 at 11:19
• The situation is possible with any gorup $G$ having a normal subgroup $N$ of index $2$ or $3$ or with $G/N\cong S_3$. Aug 7 '15 at 11:32
• Your first part is not correct. The group need not be a subgroup of $S_4$ in order to act on a set with $4$ elements. Aug 7 '15 at 11:32
• Indeed-more properly, it must be isomorphic to a quotient of a subgroup of $S_4$ (the action need not be faithful). Aug 7 '15 at 11:36
• @DavidWheeler You mean that a quotient of the group must be isomorphic to a subgroup of $S_4$. Aug 7 '15 at 11:40

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 $$12|34$$.