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.
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.
The three coordinate axes are all on the same footing here; there's no reason to treat one of them preferentially. In three dimensions, a reflection isn't a rotation; I'm not sure what you mean by a "reflection along a line", but in case you mean a reflection in the plane normal to the line, that's not a rotation.