Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2020 / Paper 1 / 10

The sign homomorphism $\mathrm{sgn}:S_n\to \{+1,-1\}$ records parity of a permutation. The kernel of a group homomorphism in this case is the alternating group $A_n$, consisting of even permutations. For $n\ge 2$ it has index two in the symmetric group and order $n!/2$.

The cycle type is the multiset of lengths of the disjoint cycles, including fixed points if desired. For $A_4$ the complete count isA single transposition and a four-cycle are odd and are excluded. Thus

A rigid rotation of a regular tetrahedron permutes its four vertices, giving an injective group homomorphism from its rotation group to $S_4$: fixing every vertex forces the rotation to be the identity. Its seven nontrivial rotation axes are four lines through a vertex and the opposite face centre, and three lines through midpoints of opposite edges.

Each vertex-face axis admits rotations by $\pm 2\pi /3$, giving eight three-cycles on the vertices. Each opposite-edge axis admits a half-turn, giving three double transpositions. Along with the identity there are twelve rotations. All their induced permutations are even, so the injective image is a twelve-element subgroup of $A_4$, hence the whole group. ThereforeIts action on ${\mathbb{R}}^3$ supplies a three-dimensional representation. To prove irreducibility even after complexification, choose verticesThe group contains the half-turns $D_x=\mathrm{diag}(1,-1,-1)$ and their cyclic variants, as well as the coordinate cycle $(x,y,z)\mapsto (y,z,x)$. If a nonzero invariant subspace over the complex numbers contains a vector with nonzero $x$ component, applying $(I+D_x)/2$ shows it contains the $x$ axis. If another component is nonzero, use the corresponding projection instead. The coordinate cycle then supplies the other two axes. Every nonzero invariant subspace is therefore the full space, provingThis also gives an elementary proof without presupposing the character table.

For a finite group over the complex numbers, the sum of the squared dimensions of its inequivalent irreducible representations equals the group order. There is always a one-dimensional trivial representation, and part (b) supplies an irreducible representation of dimension three. Thus the remaining squared dimensions sum toThey must be two further dimensions equal to one. HenceThe number of inequivalent irreducible representations equals the number of conjugacy classes, so $A_4$ has four classes. There are only three cycle types in part (a); consequently cycle type cannot classify conjugacy within $A_4$.

More explicitly, the identity is a singleton class, and conjugation by three-cycles permutes the three double transpositions transitively. For a three-cycle, the centralizer in $A_4$ consists of its three powers, so its conjugacy class has $12/3=4$ elements. The eight three-cycles therefore split into two classes of four:For example, $(123)$ and $(132)$ have the same cycle type but are not conjugate in $A_4$. Any permutation conjugating the first to its inverse must fix its unique fixed vertex and reverse the orientation of the remaining three labels, so it is odd. They are conjugate in $S_4$, illustrating exactly why restricting the ambient group changes the answer.