Let $G\leq S_4$ and consider the natural action of $G$ on the set $\{1,2,3,4\}$. For each of the following choices of $G$, write down the orbits of the action and find the stabilizer of each point.

  1. $G=\big <(123)\big >$

  2. $G=\{1,(12)(34),(13)(24),(14)(23)\}$

  3. $G=A_4$

For $x\in \{1,2,3,4\}$ $$G_x:=\mathrm{Stab}_G(x)=\{g \in G | gx=x\} $$ is the stabilizer of x under the action of $G$ and

$$Gx=G.x:=\{ gx | g\in G \} $$ is the orbit of x under the action of $G$

Take $G=\{1,(123),(132)\}=\langle(123)\rangle$

Now to find, for example, the orbit of $1\in \{1,2,3,4\}$ under the action of G, I know I should compute $$G.1=\{ g.1 | g\in G \} $$ which is $$G.1=\{ 1.1,1.(123),1.(132) \} $$ But how do we compute these $1.1,1.(123),1.(132)$?

Same difficulty shows up in computing stabilizer of x's as well. Can somebody show me how to do these calculations?

Edit: I computed 1. and 2.

  • $\begingroup$ For example, $(123)$ acting on $1$ will give you $2$ because the cycle sends $1$ to $2$, while $(132)$ acting on $1$ will give you $3$ and the same element acting on $2$ will give you $1$. Does that make sense? $\endgroup$ – stressed out Dec 5 '17 at 13:33
  • $\begingroup$ So is the answer $G.1=\{1,2,3\}$ ? @stressed-out $\endgroup$ – Leyla Alkan Dec 5 '17 at 13:35
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    $\begingroup$ $G \cdot 1 = \{1,2,3\}$. Please note that $G\cdot1$ is not the same as $G$ itself. $1$ here does not represent the identity element, but merely a symbol. $\endgroup$ – stressed out Dec 5 '17 at 13:36

As we discussed in the comments, the first two parts are manageable by hand.

To determine the orbit of the case when $G=A_4$, note that the permutation:

$$(i, k)(i,j) \in A_4$$

and it maps $i$ to $j$. It doesn't matter where $k$ is sent to, but it matters that $k \neq i,j$. This will tell you that $A_4$ is transitive, i.e. $GX=X$.

For example, for $1$, you have the following relations:

$$(1,3)(1,2).1=2$$ $$(1,4)(1,3).1=3$$ $$(1,2)(1,4).1=4$$

And of course, $1$ is sent to $1$ by the identity element which is in $A_4$. You can apply the same logic to $2$, $3$ and $4$.

To determine the stabilizer of a subgroup, note that if two cycles leave something alone, then their product will also not affect it.

For example:

$$(2,3).1=1$$ $$(2,4)(4,3).1=1$$

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    $\begingroup$ Definitely makes sense! $\endgroup$ – Leyla Alkan Dec 5 '17 at 14:14

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