# Why are there 48 symmetries of a cube?

I'm trying to prove that there are a total of 24 rotation and 24 reflection symmetries of a cube. I can show the first part, but I don't have a good proof for why there are also 24 reflections.

The argument I have so far is that we can pick a reflection $$A$$, and then $$A$$ acting on any element of the rotation group $$SO(3)$$ gives us a distinct reflection. But I don't know why this generates the full set of reflections.

In other words, why would another reflection $$A'$$ acting on the rotation group give us the same set of reflections? I guess I'm trying to show that for $$B\in SO(3)$$, there exists $$B'\in SO(3)$$ such that $$AB=A'B' \Rightarrow (A')^{-1}AB=B'.$$ Is it enough to say that this holds since $$(A')^{-1}A$$ is the composition of two reflections and thus a rotation, and composed with $$B$$, this means $$(A')^{-1}AB$$ is also a rotation?

• It depends on your definition of a reflection whether you want to call all the symmetries $AB$, with $B$ ranging over the set of rotational symmetries "reflections"? Among them you find transformations that flip one axis, but rotate the plane orthogonal to that axis by 90 degrees. Anyway, counting the neighbors of vertices of the cube shows that there cannot be more than 48 symmetries total. So you are basically done by exhibiting those 24 rotations and a single non-rotation. Commented Oct 31, 2018 at 16:34
• (cont'd) Those rotations form a subgroup of symmetries. By Lagrange, this implies that the total number of symmetries is a multiple of $24$. It is strictly larger than $24$, and cannot be higher than $48$, so... Commented Oct 31, 2018 at 16:36
• Why can't there be more than $48$? Are you saying it's because there are $8$ ways to assign vertex 1, and then $3!=6$ ways to order the adjacent vertices? Commented Oct 31, 2018 at 17:20
• That's exactly the reason, Glassjawed! You do need to convince yourself f the fact if you know where a fixed corner goes, and what happens to its neighbor, the symmetry is fully determined. Hint: the rest of the corners are 3D-diagonally opposite to the four corners we have already dealt with. Commented Oct 31, 2018 at 17:22
• Ah perfect -- so the only reason Lagrange was needed was to show that any of those 48 rearrangements can be achieved via suitable rotations and reflections. Very cool. Commented Oct 31, 2018 at 18:06

Let $$R$$ be the set of all reflections. Fix $$r_0\in R$$. For each $$r\in R$$, $$r\circ{r_0}^{-1}$$ is an isometry of the cube. But, since $$r_0$$ and $$r$$ both have determinant $$-1$$, $$r\circ{r_0}^{-1}$$ has determinant $$1$$. In other words, $$r\circ{r_0}^{-1}$$ belongs to $$SO(3,\mathbb{R})$$. So, $$r=g\circ r_0$$, for some $$g\in SO(3,\mathbb{R})$$. And, if $$r'\in R$$ and $$g'\in SO(3,\mathbb{R})$$ are such that $$r'=g'\circ r_0$$, then $$r=r'\iff g=g'$$. On the other hand, if $$g\in SO(3,\mathbb{R})$$, then $$g\circ r_0\in R$$. This proves that $$\#R=\#SO(3,\mathbb{R})=24$$.

Hint: The composition of two reflections is a rotation.

• So my sketch is the way to go here? Commented Oct 31, 2018 at 16:21
• Your wording is a bit off; you want to show that there exists $B'\in\operatorname{SO}(3)$ such that $AB=A'B'$. It follows that this must be $B'=(A')^{-1}AB$, which is indeed also a rotation because the composition of two reflections is a rotation. Commented Oct 31, 2018 at 16:26

Here is a proof that there are at least 48 symmetries. (Goal: make it obvious and easy to understand.)

Go to a craps table and borrow one of the dice.

There are eight corners on your die, and so you can position the die in any of eight ways based on the corner you are looking at.

For each of those eight corners, there are three rotations. For example, if you are looking at the corner where you can see sides with one dot, two dots, and three dots, you can rotate the die so that any of these three sides is on top.

So, that gets you 24 symmetries.

Now, the twist others have alluded to about a non-physical move. If you got the die in Las Vegas, the one/two/three corner will go counter-clockwise. If you got the die in Macau, the one/two/three corner will go clockwise. This gives us an additional factor of two in our count of symmetries. "Change" the die from left-handed to right-handed or vice versa.

This additional factor of 2 gives us 48 symmetries.

That 48 is an upper bound: pick any of the 8 corners. The three adjacent corners can be permuted in 3-factorial ways. So, there can be no more than 48 permutations.