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?
Best Answer
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 $\#R=\#SO(3,\mathbb{R})=24$.