TITLE: Symmetry of the tetrahedron as a subgroup of the cube
QUESTION [0 upvotes]: Let $\epsilon=(\epsilon_1,\epsilon_2,\epsilon_3)$ with $\epsilon_i\in\{\pm 1\}$ and consider for each $\epsilon$:
$$
D_\epsilon=\begin{pmatrix}
    \epsilon_1 & 0 & 0\\
    0 & \epsilon_2 & 0\\
    0 & 0 & \epsilon_3
\end{pmatrix}
$$
Let $O_h\subseteq GL_3(\mathbb{R})$ be
$$
O_h =\{P_\sigma D_\epsilon\in\mathbb{R}^{(3,3)}\mid \sigma\in S_3, \epsilon\text{ as above}\}
$$
where $P_\sigma$ is the associated permutation matrix with $(P_\sigma)_{ij}=\delta_{i\sigma(j)}$.
This group is a the subgroup of the general linear group for the cube for the symmetries of the cube with the vertices $(\pm1,\pm1,\pm1)$.
Consider the embedded regular tetrahedron with the vertices $(1,1,1),(-1,1,-1),(-1,-1,1),(1,-1,-1)$. What is the subgroup $T_d$ of $O_h$ mapping the tetrahedron to itself. Also what would be the subgroup $T$ of rotational symmetries of this tetrahedron.
For determining $T_d$, notice that the vertices of the tetrahedron are exactly those having an even number of negative components, thus my guess would be to take
$$T_d=\{A=P_\sigma D_\epsilon\in O_h\mid\sigma\in S_3, \epsilon_1\epsilon_2\epsilon_3=1\}$$

REPLY [0 votes]: Your set $\{P_\sigma D_\epsilon : \sigma \in S_3, \prod \epsilon_i = 1\}$ is indeed the symmetry group of the embedded tetrahedron: it has the requisite number of elements (there are six $P_\sigma$, and four such $D_\epsilon$, for a total of $6 \cdot 4 = 24$ symmetries), and permutes vertices of the cube with an even number of $-1$ entries (that is, vertices of the tetrahedron).
For the group of rotational symmetries, recall that the determinant of a matrix detects whether a transformation preserves orientation: for $A \in O_h$, if $\det A = 1$ then $A$ is a rotation, and $A$ is a reflection when $\det A = -1$. All of the $D_\epsilon$ have determinant $\prod \epsilon_i$, thus for $P_\sigma D_\epsilon \in T_d$, we have $\det P_\sigma D_\epsilon = \det P_\sigma$. You probably already know that the determinant of $P_\sigma$ is exactly the sign of the permutation $\sigma$.

Tangentially, you may be interested in a previous answer of mine for the geometric interpretation of writing symmetries of the cube as products of $S_3$-parameterized things, and diagonal matrices with entries $\pm 1$ (the former related to the vertex figure of a cube, the latter in bijection with vertices of the cube).
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