# A dodecahedron out of five tetrahedra AKA Partitioning an orbit in X/K into orbits under cosets of H in K

Consider the orbit space $$X/K$$ with $$X$$ a symmetric space and $$K$$ a group. Let $$x$$ represent an orbit $$Kx$$ in $$X/K$$. Now let's introduce a subgroup $$H \subset K$$, split up $$K$$ into cosets $$aH$$ (with $$a \in K/H$$), and look at the parts of the orbit corresponding to each coset, i.e., $$aHx$$. It seems that the coset representative $$a$$ either rotates $$Hx$$ to a distinct subset of $$X$$ or leaves it be. In other words, are there any weird situations where only some points of the coset orbits $$aHx$$ and $$bHx$$ are the same but others are different?

I know of one famous example, the compound of five tetrahedra, in which five tetrahedra (corresponding to the five coset orbits $$aTx$$) together make up a dodecahedron $$Ix$$ (an orbit in $$S^2/I$$). So I was wondering what happens in the general/generic cases for $$S^2$$ and $$H \subset K \subset SO(3)$$.

Let $$Kx$$ be some orbit in the orbit space $$S^2/K$$. Consider the set of points $$Hx$$, i.e., the orbit of $$x$$ under $$H$$. The stabilizer of the set $$Hx$$, $$\text{stab}(Hx)=H$$ generically (we'll do general case later). Applying the orbit-stabilizer theorem, we can then determine the number of elements in the orbit of $$Hx$$ under $$K$$:
$$|\text{orb}(Hx)|=|K|/|\text{stab}(Hx)|=|K|/|H|.$$
Thus, each element of the orbit of $$Hx$$ under $$K$$ is in one-to-one correspondence to a coset $$aH$$ of $$H$$ in $$K$$.
Let's come back to the non-generic cases. Generally, $$\text{stab}(Hx)$$ must be some subgroup $$P$$ such that $$H\subseteq P \subseteq K$$. So if, e.g., $$H$$ is the largest subgroup of $$K$$, then $$P$$ is either $$H$$ (generic case above) or $$K$$ (case at a few special points).
For the tetrahedral compound case, $$H=T$$ and $$K=I$$, and we'll have $$|I|/|T|=5$$ elements in the orbit of $$Tx$$ under $$I$$ for almost all $$x$$. An exception is when $$x=x^\star$$ is a vertex of the icosahedron that is symmetric under $$I$$. That icosahedron has 12 vertices, which can be obtained by applying $$T$$ to $$x^\star$$. Then, $$Tx^\star=Ix^\star$$ and we have $$|I|/|I|=1$$ element only. Since there is no subgroup inbetween $$T$$ and $$I$$, the above two cases are the only ones that can happen.