Permutation group of a cube in $R^3$

abstract-algebragroup-theorysymmetric-groups

I'm reading Alexander Paulin's Introduction to Abstract Algebra, and could not figure out its page 32 on "The Cube in $R^3$" of Ch 3.7 "Symmetry of Sets with Extra Structure".

Its content I quoted as below.
Notation: Stab means Stabilizer Subgroup, $D_3$ is the Dihedral Group of a triangle.

What I don't understand is the line highlighted:" It can be shown that $τσ = στ$ for all $σ ∈ Rot_\square$."

Let $X ⊂ R^3$ be a solid cube centered at the origin. Again, elements of $Sym(X)$ must fix the
origin, hence, if we wished, we could identify $Sym(X)$ with a subgroup of the group of $3×3$ orthogonal matrices.

Again $Sym(X)$ acts faithfully and transitively on the vertices. If $a ∈ X$ is a vertex, then
$Stab(a)$ can naturally be identified with $D_3$ (figure obmitted) which has size $6$. Hence, by
the orbit-stabilizer theorem, $|Sym(X)| = 48$. The same logic applies to $Rot_\square$, the rotational
symmetries, although the stabilizer of a now has size $3$. This tells us that $|Rot_\square| = 24$.

If $τ ∈ Sym(X)$ is the symmetry sending $x$ to $−x$ (this is not a rotation), then again
$$Sym(X) = Rot_\square \bigsqcup τ Rot_\square$$

It can be shown that $τσ = στ$ for all $σ ∈ Rot_\square$.

… (one can see) $Rot_\square \cong Sym_4$.

Let's say we have a cube, the top 4 vertices, seeing from above, anti-clockwisely, are A,B,C,D; and the lower 4 are E,F,G,H, with E under A, etc.

Choose $\sigma$ as the rotation around the axis goes through vertice $A$ and $G$:
$$\sigma = \pmatrix{A & B & C & D & E & F & G & H\\
A & D & H & E & B & C & G & F} $$

Take AB as the $x$ direction, we have

$$\tau = \pmatrix{A & B & C & D & E & F & G & H\\
B & A & D & C & F & E & H & G} $$

Then we have

$$\tau \sigma = \pmatrix{A & B & C & D & E & F & G & H\\
B & C & G & F & A & D & H & E} $$

and

$$\sigma \tau = \pmatrix{A & B & C & D & E & F & G & H\\
D & A & E & H & C & B & F & G} $$

It looks $\sigma \tau $ is not $\tau \sigma$ at all!

Which part did I missed?

Best Answer

Per the previous line in the text,

If $\tau\in\text{Sym}(X)$ is the symmetry sending $\bf{x}$ to $−\bf{x}$ (this is not a rotation)

$\tau$ refers to a specific permutation, the antipodal map. You're quite right that other permutations do not in general commute with rotations.

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