= Solution
Place the <cube> at the origin with face normals $\pm e_1,\pm e_2,\pm e_3$, and label opposite faces by $(1,2),(3,4),(5,6)$. Every cube symmetry is a <signed permutation matrix>: there are $2^3\,3!=48$ choices. Its <group action> on faces is faithful because fixing all faces fixes all their normal directions. Central inversion $-I$ exchanges every opposite pair and therefore acts as $h$. Since $-I$ commutes with every linear cube symmetry, the face-action image lies in $C_{S_6}(h)$. The preceding count gives $|C_{S_6}(h)|=48$, so the faithful face action is onto this <centraliser>:
$$
\boxed{G\cong C_{S_6}(h).}
$$
For the remaining isomorphism, let $R$ be the <rotational symmetry group of a cube>. Half the signed permutation <matrices> have <determinant> $1$, so $|R|=24$. Central inversion has <determinant> $-1$, and every orientation-reversing symmetry is $(-I)r$ for a unique $r\in R$. Since $-I$ is central and $R\cap\{I,-I\}=\{I\}$, multiplication gives a <direct product of groups>:
$$
G\cong C_2\times R.
$$
The <group action> of $R$ on the four body diagonals is faithful. To prove this, choose direction <vectors> $(1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1)$. They span $\mathbb R^3$, and their only linear relation is that their sum is zero. A rotation fixing all four diagonal lines sends each listed <vector> to itself or its negative. Applying the relation forces all four signs to agree. The all-negative choice is $-I$, which is not in $R$, so the rotation is the identity. Thus $R$ embeds in $S_4$; both have order $24$, giving $R\cong S_4$. Finally $C_{S_6}(g)\cong C_2\times S_4$ from part (i). Consequently
$$
\boxed{G\cong C_2\times S_4\cong C_{S_6}(g).}
$$
The <direct-product decomposition of the cube symmetry group> supplies the second isomorphism, while the first comes from the specified face action.
Back to article page
