Transformations commuting in 3D (crystallography)

Question

I'm studying symmetry operations and trying to show that the symmetry operations of a crystal* are closed, so they form a group. A friend of mine said that these operations commute but I can't justify how - I have tried showing that the eigenspaces of the transformations are preserved but have not been able to.

*the operations are: $C_n = $ rotation about principle axis of $2pi/n$ and $S_n =$ rotation about principle axis of $2\pi/n$ followed by a reflection in the plane perpendicular to the axis. My goal is to show $C_n \circ S_m = S_m \circ C_n$ (intuitively I think it is true but I haven't actually proven it). Any help would be much appreciated.

More on symmetry operations here

Answer 1

By the Bieberbach Theorem, every space group contains a commutative subgroup of finite index, consisting of translations. There are $219$ different space groups (see the wikipedia link in the first comment). In general, space groups are not commutative; also the finite point groups are not commutative in general.

Answer 2

Looking at eigenspaces isn’t really going to work because the only real eigenspace of a rotation is the rotation axis. (Even if you allow complex vectors and matrices, you’ll still have to combine two of the eigenspaces to match the reflection.) If you instead look at invariant subspaces of the transformations, though, you’ll find what you’re looking for.

The invariant spaces of a rotation in 3-D are the rotation axis, which is also the eigenspace of $1$, and its orthogonal complement—the plane in which the rotation takes place. These obviously correspond to the eigenspaces of the given reflection, so I expect you’ll be able to take it from here.

You could also proceed via a direct calculation. You can choose an orthonormal basis in which the principal axis is the $z$-axis, so that the reflection is represented by the matrix $\operatorname{diag}(1,1,-1)$. It’s a simple calculation to show that $[C_n,S_m]$ vanishes in this basis.