Matrix representations of particular generators of the full octahedral group

Question

I want to find matrix representations of the generators $a, b, c$ of the full octahedral group in the presentation $$\{a,b,c \mid a^2=1,b^3=1,(ab)^4=1,ac=ca,bc=cb\}.$$ Is there a recipe to write the matrix representations for the generators? And if you know geometrically what are the generators that obey the above the presentation, in terms of reflections/rotations/inversions and how does one come up with them w/o brute force, that will be also helpful.

Answer 1

First, there is a typo in the group presentation (and in the cited reference):A As it is, $c$ generates an infinite, central subgroup. We can fix this immediately: Since $c$ commutes with every element, the group is $\langle a, b \mid a^2 = 1, b^3 = 1, (a b)^4 = 1 \rangle \times \langle c \rangle$, but the first factor is $S_4$ (this entry is correct in the reference), so since the resulting group should have $\#O_h = 48$, $c$ must have order $2$, giving the (correct) presentation \begin{multline} \langle a,b,c \mid a^2=1,b^3=1, c^2 = 1, (ab)^4=1,ac=ca,bc=cb\rangle \\ \cong \langle a,b \mid a^2=1,b^3=1, (ab)^4=1 \rangle \times \langle c \mid c^2 = 1 \rangle \cong S_4 \times \Bbb Z_2. \end{multline}

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Before we get produce a matrix representation, let's describe how we can realize the (full) octahedral group $O_h$ as $S_4 \times \Bbb Z_2$ geometrically: First, the center of $S_4$ is trivial, so the center of $O_h$ is $\{e, c\}$, and we can immediately identify the element of $O_h$ that commutes with every element, namely, the reflection through the center of the octahedron. On the other hand, the oriented octahedral group evidently acts transitively on the four pairs of opposite faces of the octahedron (or dually, on the four long diagonals of the dual cube), and so contains a copy of $S_4$ that does not contain $\Bbb Z_2$.

We can now get to the business of constructing an explicit matrix representation $\phi$: Choose a coordinate system with the center of the octahedron at the origin and such that the coordinate axes run through its vertices. In this system, the reflection $c$ through the origin has representation $$ \phi(c) = \pmatrix{-1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1}, $$ so we need only identify matrix representations of the generators $a, b$ of $S_4 < O_h$. All elements of order $3$ in $S_4 < O_h$ are conjugate, so we can identify any such element with the generator. These are the rotations that fix a pair of opposite faces; for example, the rotation that cyclically permutes the positive $x$, $y$, and $z$ half-axes in that order fixes the face in the first octant, and it has representation $$ \phi(b) = \pmatrix{0 & 0 & 1\\ 1 & 0 & 0 \\ 0 & 1 & 0} . $$ Finally, we need to identify $\phi(a)$. There are two conjugacy classes in $S_4$ whose elements have order $2$, namely the transpositions and the double transpositions. The product of a permutation of order $3$ and a double transposition has order $3$ again, so $a$ must be a transposition. Moreover, $a$ cannot transpose two of the face pairs permuted by $b$, as then $ab$ would fix a face pair and so (by dint of acting on the set of three remaining face pairs) would have order dividing $6$. Thus, $b$ must exchange the face pair including the face in the first octant with some other pair, and a little meditation shows that all three choices are equivalent up to conjugacy. If we take the second face pair to the one containing the face in the octant with $x, y > 0; z < 0$, drawing a diagram shows that the corresponding symmetry is a reflection through the $z$-axis. (By conjugacy, all transpositions are reflections through axes; double transpositions correspond to reflections through lines through midpoints of opposite edges.) Thus, $$ \phi(a) = \pmatrix{-1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1}. $$

Of course, if we just want some matrix representation (and not necessarily a $3$-dimensional representation whose elements are the symmetries of a concrete octahedron), we can produce one immediately using the fact that $O_h \cong S_4 \times \Bbb Z_2$. The group of $n \times n$ permutation matrices comprise a canonical faithful representation of $S_n$. Given any faithful representations $\psi_1, \psi_2$ of groups $G, H$, $\psi_1 \oplus \psi_2$ is a faithful representation of $G \times H$, and applying this construction to the permutation representation of $S_4$ and the sign representation of $\Bbb Z_2$ yields a representation $\psi : O_h \hookrightarrow GL(5, \Bbb R)$ whose first four columns encode the permutations of the face pairs and whose last column encodes the orientation. It is characterized by $$\psi(a) = \pmatrix{0 & 1 & & & \\ 1 & 0 & & & \\ & & 1 & & \\ & & & 1 & \\ & & & & 1}, \quad \psi(b) = \pmatrix{1 & & & & \\ & 0 & 0 & 1 & \\ & 1 & 0 & 0 & \\ & 0 & 1 & 0 & \\ & & & & 1}, \quad \psi(c) = \pmatrix{1 & & & & \\ & 1 & & & \\ & & 1 & & \\ & & & 1 & \\ & & & & -1} .$$

Answer 2

Consider a unit octahedron around the origin. The octahedral symmetry group is the group of symmetries of the octahedron, so why not just identify the group elements with the corresponding linear transformation matrices?

Any bijective linear transformation of the unit octahedron that sends corners to corners must send the three standard unit vectors to three orthogonal axial unit vectors (standard vectors or their negatives). Such matrices are exactly the signed permutations.

A generating set of this group of linear transformations is for example

$$A = \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{pmatrix},~~~ B = \begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{pmatrix},~~~ C = \begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{pmatrix}. $$