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Put the cube's vertices at $(\pm 1,\pm 1,\pm 1)$. Squared distances between distinct vertices are $4,8,12$ according as one, two or three coordinates differ. Four mutually equidistant vertices cannot all differ in one coordinate, since the cube has no triangles, or in three coordinates, since each vertex has only one antipode. They must differ in two coordinates. The two regular tetrahedra are therefore precisely the sets $xyz=1$ and $xyz=-1$.

The full symmetry group $G$ consists of signed coordinate permutations, so $|G|=2^3\cdot 3!=48$. These maps permute the two tetrahedra. The stabilizer $H$ of the first has order $24$ and acts faithfully on its four vertices: a linear map fixing the tetrahedron fixes a spanning set. Thus $H\cong S_4$. Central inversion $-I$ swaps the tetrahedra and commutes with every element of $H$, giving the direct-product decomposition of the cube symmetry groupHere $H$ preserves one tetrahedron; it is not the subgroup consisting only of rotations.

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: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: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). ConsequentlyThe direct-product decomposition of the cube symmetry group supplies the second isomorphism, while the first comes from the specified face action.