What is the group of rotations of a volleyball(pyritohedron)? Practice test for Abstract algebra final, very stuck on this particular question.
 A: 
Here is the pyritohedron; let's call it $P$. 
It should be clear that its symmetry group is a subgroup of the symmetry group of the cube, which I assume you know is $\operatorname{Sym}(4) \times C_2$, the product of the symmetric group of degree four and a cyclic group of order $2$. As a reminder, this is because rotations of the cube permute the $4$ diagonals, and the remaining symmetries (all orientation-reversing) are a rotation composed with the antipode map $x \mapsto -x$.
Fast and Loose 
The corners come in two varieties (orbits, if we're being fancy), a "clockwise" and "anticlockwise" version. This suggests that the symmetry group of $P$ is "half" of the symmetry group of the cube, and is more than likely isomorphic to $\operatorname{Alt}(4) \times C_2$ (it doesn't feel like $S_4$, at least!). I'm sure there's some nice observation we could make that confirms this, but I don't know what it is, offhand.
Slow and Methodical
Let $G$ be the group of rotational symmetries of P. We claim that $G \cong \operatorname{Alt}(4)$, the alternating group of degree $4$. More specifically, a rotation is a symmetry of $P$ if and only if its corresponding permutation on the diagonals of the cube is even.
Let's examine nontrivial rotations of the (unmarked) cube by cases:


*

*The rotation is through opposite vertices is also a symmetry of $P$. The corresponding permutation of diagonals is a $3$-cycle, hence even.

*The rotation is through opposite edges. This is clearly not a symmetry of $P$. The corresponding permutation of the diagonals is a single $2$-cycle, hence odd.

*The rotation is through opposite faces. There are two further cases to consider.


*

*The rotation has order $4$. These are not symmetries of $P$, and the corresponding permutation of the diagonals are $4$-cycles, hence odd.

*The rotation has order $2$. These are symmetries of $P$, and the corresponding permutations have the format $(i\ j)(k\ \ell)$, and are even.
Thus, the group of rotational symmetries is $\operatorname{Alt}(4)$. Now, the antipode map $x \mapsto -x$ is still a symmetry of $P$, which implies that the symmetry group of $P$ is $\operatorname{Alt}(4) \times C_2$ for the same reason the cube's is $\operatorname{Sym}(4) \times C_2$.
