Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 3 / 8D (Groups)

The orbit-stabilizer theorem states that for a finite group $G$ acting on a set $X$ and $x\in X$,Define $G/G_x\to Gx$ by $g{G}_x\mapsto gx$. This is well-defined because $g{G}_x=h{G}_x$ exactly when $h^{-1}g\in G_x$, equivalently $gx=hx$. The same equivalence proves injectivity, and the definition of the orbit proves surjectivity. Thus $|Gx|=[G:G_x]$, giving the formula.

The group acts transitively on the eight vertices. At a chosen vertex, an isometry fixing it may permute its three incident edges arbitrarily, and each of the six permutations is realized by an isometry. Thus the vertex stabilizer is isomorphic to $S_3$ and has order six. Orbit-stabilizer givesThis begins the three stabilizer descriptions in the symmetry group of a cube.

There are twelve edges and the action on them is transitive, so an edge stabilizer has orderFor an edge parallel to the $z$-axis with midpoint $(1,1,0)$, its stabilizer is generated by interchanging $x$ and $y$ and by replacing $z$ by $-z$. These are commuting involutions, so

There are four main diagonals and the action is transitive, so the stabilizer of one has orderFor the diagonal spanned by $(1,1,1)$, the stabilizing signed permutation matrices are $P$ and $-P$ with $P$ a permutation matrix. HenceThe action is not faithful: both $I$ and the central inversion $-I$ fix every main diagonal setwise. In fact its kernel is exactly $\{I,-I\}$.

The action on the four diagonals identifieswhere the $C_2$ factor is generated by central inversion. An element of order three must have trivial $C_2$ component and a three-cycle in $S_4$. Since all three-cycles in $S_4$ are conjugate,Elements of order two are not all conjugate. For example, $-I$ is central and therefore has a singleton conjugacy class, while a reflection is a different element of order two. Thus