Past exam of the mathematics course of the University of Cambridge / 2003 / ia / Paper 1 / 7B (Algebra and Geometry)

For a group action of $G$ on a set $X$ and $x\in X$, the orbit-stabilizer theorem statesIf $G$ is finite, this becomes $\boxed{|G|=|Gx||G_x|}$. The underlying bijection sends the left coset $g{G}_x$ to $gx$. Equality of two image points is equivalent to $g_2^{-1}{g}_1\in G_x$, so this map is both well-defined and injective, and it is surjective by the definition of the group orbit.

Place the cube's vertices at $(\pm 1,\pm 1,\pm 1)$. Every symmetry fixes its centre, and permutes the three perpendicular pairs of face normals, so it is a signed permutation matrix. Conversely every such matrix preserves the cube. The symmetry group of a cube acts transitively on its eight vertices: independent coordinate sign changes take $(1,1,1)$ to any vertex.

A symmetry fixing $(1,1,1)$ must permute its three incident edges. This gives an injective map from its stabilizer subgroup into $S_3$, since those three edge directions determine the linear transformation. All six permutations are realized by coordinate permutation matrices, which fix $(1,1,1)$. Thus the vertex stabilizer has order six. The orbit-stabilizer theorem now yieldsThe use of all symmetries includes reflections as well as the twenty-four elements of the rotational symmetry group of a cube.

There are nontrivial proper normal subgroups. The determinant defines a surjective group homomorphism $G\to \{1,-1\}$. Its kernel is the rotational symmetry group of a cube, a normal subgroup of order twenty-four. Also $\{I,-I\}$ is a central subgroup of order two: central inversion preserves the cube and commutes with every matrix. Both provide affirmative examples.

For the body diagonal through $(1,1,1)$, write a general cube symmetry as $D_{\epsilon }P$, with $P$ a permutation matrix and $D_{\epsilon }=\mathrm{diag}({\epsilon }_1,{\epsilon }_2,{\epsilon }_3)$, each ${\epsilon }_j=\pm 1$. Since $P(1,1,1{)}^T=(1,1,1{)}^T$, preserving the unoriented diagonal requires $({\epsilon }_1,{\epsilon }_2,{\epsilon }_3)$ to be either $(1,1,1)$ or $(-1,-1,-1)$. ThusThe map $(\pi ,\epsilon )\mapsto \epsilon P_{\pi }$ from $S_3\times C_2$ to $H$ is a group homomorphism, because central inversion commutes with every coordinate permutation. It is surjective by the displayed description and injective because a permutation matrix cannot equal the negative of another permutation matrix. Consequently the body-diagonal stabilizer in the cube symmetry group isThis also agrees with the orbit-stabilizer theorem for the transitive action on the four body diagonals. The line is preserved as a set, so exchanging its opposite endpoints is allowed.