TITLE: Sylow theorem and tetrahedron
QUESTION [5 upvotes]: According to Sylow's theorem, the group of isomorphism (rotations and mirror symmetries) of tetrahedron has a subgroup of order 8. How does one find it？ Moreover, is there a method to find a Sylow p-subgroup, or a subgroup of any order, of any regular polyhedron?

REPLY [4 votes]: Consider the set $E$ of pairs of opposite edges. Its cardinal is three, and the group $G$ of automorphisms of the tetrahedron acts transitively on it. Consider the stabilizer $H$ of an element of $E$. It is of order $24/3 = 8$.
A geometric description could be given as follows: consider two opposite edges $e_1$ and $e_2$. There is a unique symmetry $s_1$ fixing $e_2$ and reversing $e_1$; there is a unique symmetry $s_2$ fixing $e_1$ and reversing $e_2$; there are symmetries exchanging $e_1$ and $e_2$. The group generated by these symmetries is the stabilizer of $\{e_1,e_2\}$. Continuing on this, one could show that it is dihedral.
Finally, the other $2$-Sylow can be found by considering the other two pairs of opposite edges.
EDIT: A general suggestion, for problems of this kind, would be the following. Assume $G$ is given as the group of automorphisms of some geometrical object $O$, and that you want to find a subgroup of $G$ of cardinal $m$. Then look for a set of $m$ things on $O$ that are preserved under the action of $G$ and such that the induced action is transitive; now just consider the stabilizer of one of these things.