Let $G$ be the group of symmetries of the cube, and consider the action of $G$ on the set of edges of the cube. Determine the stabilizer of an edge and its orbit. Hence compute the order of $G$.

The symmetric group $S_n$ acts on the set $X=\{1,2,3,\ldots,n\}$, and hence acts on $X\times X$ by $g(gx,gy)$. Determine the orbits of $S_n$ on $X\times X$.

note: When I approach the first part, it should be obvious that the orbit contains 12 elements since and edge can go to any other edge through rotation and combinaton of rotation, but I don't know how to show this in an organised way. e.g if I decided to use the axes though the centre of opposite faces of the cube as my rotational axes and call them a, b, c, how can I describe the type of rotation that send the origial edge to others using a b c so that each case is considered and no repetition? And the same question for stabilizer. It seems to me that many rotations can eventually send the edge back to its original place, but I don't know how to categorise them and recognize those ones that are essentially the same in property.

p.s It may be a bit troublesome but I would be very thankful if you can answer with a sketch of the cube so I can understand better.

  • $\begingroup$ I assume what you meant by X*X was $X\times X$? If so, do you mean that $G$ acts on $X\times X$ by $$g(x,y)=(gx,gy)\quad ?$$ $\endgroup$ – Zev Chonoles Dec 30 '12 at 2:58
  • $\begingroup$ @ZevChonoles yes that's what I mean $\endgroup$ – Neptune Dec 30 '12 at 15:12

It's enough to show that any edge can be rotated into one of the neighbouring edges; by composing such operations, you can move any edge to any other edge. To rotate an edge into a neighbouring edge, rotate through $2\pi/3$ about an axis through the vertex they share.

For an edge to be rotated into itself, the axis has to pass through the centre of the edge and the angle has to be $\pi$; this is the only kind of rotation that leaves a line segment invariant, other than a rotation about an axis along the line segment, which isn't an option in this case.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.