# Show: group $G$ has one orbit on $X$, stabilizer of $z$ is $3$…

If $X$ denotes the set of corners of a cube and let $G$ denote the group of permutations of $X$ which correspond to rotations of the cube.

(i) $G$ has just one orbit

(ii) if $z$ is any corner then $|G_z| =3$

(iii) $|G|=24$

In this link, a guy have given an hint how to solve the problem and i did solve the exersice as the hint he gave before i saw it. But, if i want to show that the orbit of G is equal to 1, should i first write down every possible permutations and then check the equivalence class $$Gx=\{y \in X | y=g(x) , g \in G\}$$ If I do this i will see that $G1=G2=...=G8=\{1,2,3,4,5,6,7,8\}$ which means that we have one orbit. But is it any better way to prove this ? Can I use the techniques for counting sets of pairs? $$r_g(S):=\{(g,y)\in S\}, g\in G$$ $$c_y(S):=\{(g,y)\in S\}, y\in X$$ $r_g(S)$ is the total pairs $(g,y) \in S$ there $g$ is a permutation. So this means that there is just on $y$ such that $g(x)=y$ in each row. This means that in every row we have exactly one pair$(g,y)$ which is in $S$. Therefore $G$ has only one orbit on $X$. And by the way, are these permutations automorphism of the cube?

How do i prove that the stabilizer of $z$ is equal to 3 without drawing a picture?

To prove that $|G|=24$ i can use the orbit-stabilizer theorem or else i can just write down all the permutations which correspond to rotations of the cube and count them.

This was my first thought how to solve the exercise. Byt i could not prove that (ii) is right without drawing the cube.

• Hint: rotations of a cube correspond to permutations of its' 4 diagonals. – Marcin Łoś May 25 '14 at 23:34
• is it a hint to prove (ii) ?:) @MarcinŁoś – Angelica May 25 '14 at 23:35
• Yes, I suppose it requires at least some visualization, but it should be possible to identify some subset generating $S_4$ with certain "simple" rotations, and proceed purely algebraically from there. – Marcin Łoś May 25 '14 at 23:44
• ok, thanks :) But did i do everything else ((i) and (iii)) right? @MarcinŁoś – Angelica May 26 '14 at 8:30

## 1 Answer

For $i)$ just note that showing there is one orbit is just showing that the action is transitive, i.e. given any two corners of the cube, there is a rotation that sends one to the other. This is in some sense geometrically "obvious". You could make this more formal by explicitly describing some of the rotations of the cube, but I don't think that this is really necessary. After all, if you actually write out the rotations for this, you may as well write out the whole group, which is clearly not the point of the exercise.

For $ii)$, suppose some rotation fixes $z$. What does this rotation do to the edges on which $z$ lies?

For $iii)$ you can just apply orbit-stabiliser as you mention.

• Yes, so dumb of me not to think of the transitive. I did think about that "...there is a rotation that sends one to the other" but i had forget about the transitive. I will try to do (ii) again it later when I come home from school in one hour or something like that. I can post my solution here and you can (if you have time) check my solution/argument. Thanks for your answer :) – Angelica May 26 '14 at 11:06
• @Angelica No problem, happy to help. – Tom Oldfield May 26 '14 at 11:12