Classification of Finite Rotation Groups - Problem understanding the proof

Question

I'm working through a proof of the classification of finite rotation groups in the Euclidean Space (i.e. finite subgroups of $SO_3$) and am not understanding a particular step. My proof is from M.A. Armstrong's Groups and Symmetry in case anybody has it available, but the particular step I'm struggling with is also executed the same way in this paper.

In case (d) (the icosahedral case - bottom of page 14 in the PDF linked here and bottom of p.108 / top of p.110 in Armstrong), it is claimed that we can find $u,v$ in the orbit of $z$ such that the following holds:

$ 0 < \lVert z -u \rVert < \lVert z -v \rVert < 2. $

Now it is clear that the only point with distance $2$ from $z$ is $-z$, but what I don't understand is why all the points in $G(z) \setminus \{z,-z\}$ can't have the same distance from $z$. Can anybody clarify?

Answer 1

I didn't read everything up to that point, and it may be that I missed something simpler. Because I concentrated on that case only, I do agree with you that the point you are concerned about is not fully justified there.

But, aided by the surrounding text, it is not too difficult to come up with an argument bridging this gap:

• In this case the orbit $G(z)$ has 12 points and the stabilizer $G_z$ of $z$ is cyclic of order five, generated by $g$. The orbits of $G_z$ must have orders one or five. Therefore the 12 poinst must be split into two orbits of size five and two of size one. The latter orbits must be $z$ and $-z$, because they are the only fixed points of $g$.
• The points in the 5-point orbits $G_z(u)$ and $G_z(v)$ are all equidistant from both $z$ and $-z$ because they are gotten from each other by rotating the sphere about the axis from $z$ to $-z$.
• Because $-z$ is in the orbit $G(z)$ there exists an element $\pi\in G$ such that $-z=\pi(z)$. Because the points of $G_z(u)$ are equidistant from $z$, the points in $\pi(G_z(u))$ must be equidistant from $-z=\pi(z)$. By the previous bullet we must have $\pi(G_z(u))=G_z(u)$ or $G_z(v)$.
• If all the ten points in $G_z(u)\cup G_z(v)$ were at the same distance from $z$, then, by the previous bullet, all those ten points must lie on the equator corresponding to the axis of $g$. IOW $||z-u||=\sqrt2$ for all $u\neq\pm z$.
• But then some pairs of points within those ten would be closer to each other than to $z$ (the length of the equator divided by ten is less than the length of the quarter circle along a meridian from the pole to the equator). This is a contradiction, because the distance to the closest neighbor is the same for all the points in the orbit.