When considering improper rotations (roto-reflections), we can derive that if $n$ is odd, then $C_n$ and $\sigma_h$ (reflection plane normal to $C_n$ axis) must exist. Similarly, we can also derive that if $n$ is even, then $C_{n/2}$ and $\sigma_h$ (reflection plane) must exist.

Further, when considering crystallographic point groups, we can easily derive that there are no rotation operations other than 1-, 2-, 3-, 4- and 6-fold axes. Now, considering the above-mentioned fact that for improper rotations and for $n = even$, $C_{n/2}$ must exist, we can say that $n/2 = 4$ and $6$ must be allowed, and therefore, improper rotations of the kind $S_8$ and $S_{12}$ must be allowed as well. However, that is not the case. Is there any algebraic/geometrical proof for this? I tried looking everywhere including textbooks and various online pages, but no luck. Any help would be much appreciated. Thank you.


1 Answer 1


First of all your statement about the rotoreflection symmetries is not quite right: A $n$-fold rotoreflection symmetry is equivalent to \begin{align} C_{n/2} \oplus \sigma_h \ \ \ &\text{for} \ \ n \ \ odd \\ C_{n/2} \oplus i \ \ \ &\text{for} \ \ n \ \ even \ \ AND \ \ n/2 \ \ odd \end{align} otherwise, i.e. the case where $n$ is even but $n/2$ is even, there are no proper subgroups that you can decompose the point group into. For example, $\tilde{2} \equiv 1 \oplus i$ and $\tilde{10} \equiv 5\oplus i$, $\tilde{3} \equiv 3/m$. BUT $\tilde{4}$ is just $\tilde{4}$.

Second of all, addressing your question regarding admissible $\tilde{n}$ axes, the axis being proper or improper is not relevant to how translational symmetry puts restrictions on $\tilde{n}$. Translational symmetry only puts restrictions on $n$, i.e. $\alpha$. Just look at the following sketch: put a rotoreflection axis on a lattice site. Apply its operations once clockwise, once counterclockwise. It doesn't seem much different than if there was only proper rotation, does it?

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.