# Rotational symmetries of a simple rectangular solid with application

Determine the group of rotational symmetries of the following solid. The faces of the solid can be painted in three different colours. How many different boxes can be produced?:

The length, breadth, width of the solid are all different from one another. Let $G$ be the group of rotational symmetries acting on the set of faces of the solid. Then I found $|G|$ = $4$. Reasoning: Take any one face (say $1$ for example). Then under any symmetry it is fixed or is mapped to face $6 \Rightarrow \operatorname{|Orb_G(1)|} = 2$ Similarly, $\operatorname{|Stab_G(1)|} = 2$. By Orbit Stabilizer we have $4$ symmetries.

It has been verified this is correct, but I am having doubts about a couple of things. We were told that the four symmetries corresponded to the identity, $g$ (which I think is rotation by $\pi$), $h$ (reflection along line shown) and $gh$. First question: Why $gh$? Does this not correspond to the identity? Second question: How about a reflection axis through the faces $3$ and $5$?

Why are $e,g,h,gh$ necessarily the only symmetries? This is what I have for the individual mappings under each symmetry (written for g and h):

$g: 1 \mapsto 6,\,2\mapsto 2,\,3 \mapsto 5,\,4 \mapsto 4,\,5 \mapsto 3,\,6 \mapsto 1$

$h: 1 \mapsto 1,\,2 \mapsto 4,\,3 \mapsto 5,\,4 \mapsto 2,\,5 \mapsto 3,\,6 \mapsto 6$

Many thanks.

• @Arthur: $(\mathbb Z_2)^3$ has $8$ elements, not $6$. But anyway there are only $4$ symmetry elements here: Face $1$ can be left invariant, it can be mapped onto itself rotated by $\pi$, and it can, as you write, be mapped to face $6$ in two different ways; that exhausts all $4$ symmetries; there's no non-trivial rotation that leaves a face invariant. – joriki Feb 24 '13 at 11:01

The coordinate axes through the centre of the solid are the only symmetry axes. You can turn through $\pi$ about any of them. In coordinates, that corresponds to negating the other two coordinates. You don't get any new operations from combining any of these three, since e.g. negating the two coordinates other than $x$ and then negating the two coordinates other than $y$ results in negating the two coordinates other than $z$, and negating the same pair of coordinates twice results in the identity.
• Align the solid such that it's centre is at the origin of the standard Cartesian coordinate system. Then, one rotation is about the z axis, another about the y axis and finally another about x axis (all rotations through $\pi$). So these correspond to the 3 symmetries? (Plus identity which gives the 4 in total) If I understand your post correctly, you are saying that indeed $gh$ corresponds to the identity? – CAF Feb 24 '13 at 11:26
• @CAF: Yes to everything except the last question. I don't know which part of my post you interpreted in that sense. As I said, I don't know what you mean by $h$ being a reflection along a line, so I can't say anything about your $h$ or its products. I did say, however, what combining two of the non-trivial symmetries yields, namely, the third non-trivial symmetry. Thus, if, as it seems, $1\ne g\ne h\ne1$, then $gh\ne1$. – joriki Feb 24 '13 at 11:33
• Sorry, I think I meant that h is a rotation about the y axis of pi. Perhaps this element $gh$ discussed in class was actually a rotation about the x axis. – CAF Feb 24 '13 at 11:45