what did he mean ? 8! elements in the symmetry group of the cube?

Question

This question concerns a cube $C$ and its symmetry group $G$.

(i) Let $X$ represent any object, and denote by $\operatorname{Sym}(X)$ the group of symmetries of $X$. Show that the subset $S \subset \operatorname{Sym}(X)$ consisting of all symmetries of $X$ which fix a particular point $x_0 ∈ X$, that is $S = \{\sigma \in \operatorname{Sym}(X) \mid \sigma(x_0) = x_0\}$, is a subgroup of $\operatorname{Sym}(X)$.

(ii) Now consider the case $X = C$. Choose a corner $c_0$ of the cube $C$ and consider the subset H of $G$ which fixes the point at $c_0$. By considering the symmetries of the cube, identify the order of this subgroup.

The question goes on to ask what do the cosets of $H$ in $G$ represent and then by using Lagrange calculate the order of $G$.

My question is this. I was under the impression that the symmetry group of the cube had 48 elements and was isomorphic to $S_4\times \Bbb Z_2$ after a meeting with my lecturer today he told me this assumption was overly simplistic and that in fact there were 8! elements in G as in the case of $S_8$ and that it was not isomorphic to $S_4 \times \Bbb Z_2$ because this didn't have enough elements. I can't picture this and have only found contradictory answers on the internet . Did I misunderstand him ? I really need someone to explain to me what he meant !

Answer 1

If there were $8!$ elements, then the symmetry group of the cube would be isomorphic to $S_8$. However, if we label the vertices $\{ a, b, c, d, e, f, g, h \}$ and assume that $a$ and $e$ are at opposing corners, then clearly the permutation $(\, a\;\, e\, )$ that swaps $a$ and $e$ but keeps the other vertices fixed is invalid. So there are at most $8! - 1$ elements in the symmetry group $\Rightarrow$ it is not isomorphic to $S_8.$

Answer 2

The group $H$ has order $3$ for direct symmetries and $6$ if reflections are allowed. The easiest way to see this is that it must permute the vertices adjacent to $x_0$.

As for the cosets of $H$ there is a very simple geometric description. Whether or not reflections are allowed $H$ has index $8$ in the whole group, so you expect $8$ cosets. Suppose $Ha=Hb$ then $Hab^{-1}=H$ and $ab^{-1}\in H$ - that is very basic for any coset. But think about what it means geometrically in terms of the special point $x_0$.