Why aren't there any $S_8$ or $S_{12}$ point groups in crystallography? 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.
 A: 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?

