Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was doing a combinatorics problem which states this definition of symmetry: for a subset $S$ of $\mathbb{R}^3$ a symmetry is "rigid motion" $f:\mathbb{R^3}\rightarrow \mathbb{R^3}$ such that any $x\in S$ has its image $f(x)$ also in S. (To provide some fodder for the problem, the author dumbs it down for me that a symmetry is some operation which leaves $S$ in the same place although the individual points may be rearranged)

From this working definition it is asked to find the number of rotational symmetries of the cube. So far I have seen two different arguments and I still dont know because drawing on paper is getting confusing. Also my solution was incorrect and I dont know where I undercounted.

Ignore my method.Drawing 4 axis perpendicular with the faces, each with 4 rotational symmetries $(0^\circ,90^\circ,180^\circ,270^\circ)$ and I get $4\times 4 =16$ symmetries. Plus the four body diagonals$=20$. This is wrong and the answer is $24$.

The book gives the short solution: "move any vertex to any other and rotate the edges leading to it in three ways". I don't get it. There are 8 vertices, and 3 ways of what?

I google and get another argument from wikipedia: "The group of orientation-preserving symmetries is S4, or the group of permutations of four objects, since there is exactly one such symmetry for each permutation of the four pairs of opposite sides of the octahedron." Indeed $4!=24$ but I cannot convince myself that aall permutations of the pair of opposite edges is all the symmetries there can be (I clearly undercounted when I got 16 and still have no insight why, so this does not help me as I dont know for sure there cant be more)

My imagination and visualization skills are very weak. Is there a good online software that allow me to visualize and play around with a labelled cube?

share|improve this question
How did you get 4 axes perpendicular to the faces? There are 6 faces, right...? –  Alon Amit Jun 18 '11 at 6:30
Yes.. so there must be 3 axes. That makes it 12+4 body diagonals =16 –  kuch nahi Jun 18 '11 at 6:33
There is a paper P. J. Morandi. Computing the symmetry groups of the platonic solids with the help of Maple sierra.nmsu.edu/morandi notes/PlatonicSolids.pdf It contains some code, but it is more about groups, not about visualization. –  Martin Sleziak Jun 18 '11 at 6:34
Another approach is to show that each cube symmetry corresponds to a permutation of body diagonals. See e.g. Stillwell's Mathematics and History: books.google.com/… –  Martin Sleziak Jun 18 '11 at 6:41
@martin thanks that explains the $4!$ neatly –  kuch nahi Jun 18 '11 at 6:56

5 Answers 5

up vote 9 down vote accepted

The point of the comment in the book is this: given a cube, pick one corner and call it $A$. You can certainly rotate the cube so that $A$ either stays put or moves to any chosen other corner, of which there are 7. So you have 8 choices for where you want $A$ to be.

Once you've done this, the other corners are obviously not free to move as they wish. Let $B$ be some corner adjacent to $A$, meaning it shares an edge of the cube with $A$. If A moves to $A'$, $B$ must move to some corner adjacent to $A'$ - otherwise the edge $AB$ gets distorted. How many corners are adjacent to $A'$? That's where the 3 comes from.

Finally you'll need to convince yourself that once $A$ and $B$ have been placed, all other corners are "forced" - their final location is already fixed. See if you can do that just using the relationships of adjacency, sharing a face etc.

share|improve this answer
Re: Two vertices completely determining the orientation of a cube. Two vertices would determine an edge AB. This edge cannot be rotated in any way without moving out of $S$ in the definition.. would it constitute as a valid argument? –  kuch nahi Jun 18 '11 at 6:41
Well, I guess it depends on how formal you wish to get. Are you convinced by that argument? I think it may initially help you to work out in more detail why every other vertex is forced. Start with the other neighbors of $A$ perhaps? –  Alon Amit Jun 18 '11 at 7:11

A cube has 12 edges, so it has 24 oriented edges (each edge can be oriented in exactly two ways) It is pretty obvious, if you have a cube to play with, that the group $G$ of rotations acts simply transitively on these orientied edges. Therefore $G$ has 24 elements.

Exactly the same argument counts the number of rotational symmetries of each regular polyhedron, of course.

share|improve this answer

"Drawing 4 axis perpendicular with the faces, each with 4 rotational symmetries (0∘,90∘,180∘,270∘) and I get 4×4=16 symmetries"

The cube has 6 faces, so 6 axis, and gives $6\cdot 4 = 24$ as needed. (Sorry, this seemed like something only worth a comment, but I don't see where to comment... It could be I don't have enough rep to do so yet? Either that or I'm blind :) )

Edit: If you want to visualize objects, and view different symmetries, it may be helpful to give each vertex a name (so for the cube, a,b,c,d,e,f,g,h could be your vertices). This is going to be easier than looking at a single object and imagining rotating it.

share|improve this answer
6 faces would incorporate 3 axes, right? –  kuch nahi Jun 18 '11 at 6:34
Sure, you could look at it as if the cube was centered at the origin of a cartesian plane (i.e., x,y,z axis). Then the ways to "look" at the cube would be positive x, positive y, positive z, negative x, negative y and negative z. –  Alex Jun 18 '11 at 6:38
You do need 50 points to comment, as indicated in the faq. If you are actually answering the question, though, it is usually better to answer. I don't understand your answer. For example, what 8 symmetries are you associating with the positive and negative x axes in your comment? –  Jonas Meyer Jun 18 '11 at 6:44
Fair enough. I'm saying there are 6 faces, so if we can rotate each face 90, 180 and 270 degrees, there are 4 ways to be looking straight at any individual face (it could be easier to write this as looking down an oriented axis), and 6 faces, so 24 in total. So the four symmetries from the positive x axes are the four 90 degree rotations that leave all the vertices with positive x value as vertices with positive x value. Then rotating the face "on" the negative x axis to the positive x axis, the same four 90 degree rotations are another 4 symmetries. –  Alex Jun 18 '11 at 7:16

Counting the possible orientations of the cube we know that there are 24 rotational symmetries, by considering faces, edges or corners and the their respective number of orientations, giving 6 * 4 = 12 * 2 = 8 * 3 = 24.

It seems you also want to explicitly know the rotations, instead of just counting the rotational symmetries. The rotations are:

  • 1 identity rotation that does nothing.

  • 9 rotations around axes through the middle of one of 3 pairs of opposing faces, with rotations of 90°, 180° and 270°.

  • 6 rotations around axes through the middle of one of 6 pairs of opposing edges, with a rotation of 180°.

  • 8 rotations around axes through one of 4 pairs of opposing corners, with rotations of 120° and 240°.

share|improve this answer

The threefold rotations about a corner/long diagonal can be quite hard to visualise. The count which is most intuitive to me is that, with the cube on the table in front of me I can place any one of the six faces on the table, then choose any one of four faces to be facing me.

However, it is useful to work on the harder to visualise perspectives in a simple example such as this where the answer is clear, because then intuition is sharpened for more complex situations.

Another way of looking at this, which is more complex, is to analyse what happens to the two tetrahedra which are formed by picking out alternate vertices of the cube, or to the set of long diagonals of the cube.

share|improve this answer
"The threefold rotations about a corner/long diagonal can be quite hard to visualise." - that is, unless you're familiar with being able to embed a tetrahedron within a cube. –  J. M. Oct 17 '11 at 0:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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