Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2019 / Paper 2 / 10

The symmetric group $S_n$ consists of all bijections of $\{1,\dots ,n\}$, with composition as its operation; its group order is $n!$.

An $m$-dimensional complex group representation of $G$ is a group homomorphism $\rho :G\to GL(V)$, where $V$ is an $m$-dimensional complex vector space. After a choice of basis, each $\rho (g)$ is an invertible $m\times m$ complex matrix, with $\rho (gh)=\rho (g)\rho (h)$.

An irreducible representation has $V\ne 0$ and no invariant vector subspace other than $\{0\}$ and $V$. Two group representations $\rho$ on $V$ and $\sigma$ on $W$ are isomorphic representations if there exists an invertible linear map $S:V\to W$ such that $S\rho (g)=\sigma (g)S$ for every $g$. They are inequivalent if no such map exists; in equal-dimensional matrix form this says no fixed change of basis conjugates every $\rho (g)$ to $\sigma (g)$.

In a symmetric group, two permutations are in the same conjugacy class exactly when they have the same cycle type. For $S_4$ the five possibilities and their sizes areThere are $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ transpositions, three pairings into two unordered pairs, $\left(\genfrac{}{}{0ex}{}{4}{3}\right)(3-1)!=8$ three-cycles, and $(4-1)!=6$ four-cycles. Together with the identity these counts sum to $24=|S_4|$.

The two one-dimensional representations are the trivial representation, ${\rho }_{+}(g)=1$, and the sign representation, ${\rho }_{-}(g)=\mathrm{sgn}(g)$. The first respects multiplication immediately. For the second, the sign of a permutation is the determinant of its permutation matrix, and multiplicativity of the determinant proves ${\rho }_{-}(gh)={\rho }_{-}(g){\rho }_{-}(h)$. They are inequivalent since a transposition has images $1$ and $-1$, and conjugation by a nonzero scalar cannot change a one-dimensional matrix.

Their group homomorphism kernels areBoth are subgroups, indeed normal subgroups, because they are kernels. In the second case $A_4$ is the alternating group of even permutations, with twelve elements.

The characters are their scalar values, since the trace of a one-dimensional matrix is its only entry. In the class order from part (b),The character inner product is the group average of ${\chi }_{+}\overline{{\chi }_{-}}$, soEach has squared norm $(1+6+3+8+6)/24=1$, in agreement with character orthogonality for distinct irreducible representations.

For a finite group over $\mathbb{C}$, the number of inequivalent irreducible representations is its number of conjugacy classes, and the sum of squares of irreducible representation dimensions equals the group order. There are therefore five irreducibles for $S_4$, with squared dimensions summing to $24$.

First establish that part (c) found every one-dimensional representation. In such a representation, conjugate elements have the same scalar image. All transpositions are conjugate and square to the identity, so they all have one common image $s$ with $s^2=1$. Transpositions generate $S_4$, hence $s=1$ gives exactly the trivial representation, while $s=-1$ gives exactly the sign representation. There are precisely two degree-one irreducibles.

The other three dimensions $d_1,d_2,d_3$ are integers at least two and satisfyNo one of them can be at least four, since $4^2+2^2+2^2=24>22$. They are therefore all two or three. Starting from three twos gives sum of squares $12$; each replacement of a two by a three adds $5$. To reach $22$ requires exactly two replacements. Thus the complete list isThe count of conjugacy classes guarantees that these are all the inequivalent complex irreducible representations, not just dimensions of a proposed incomplete list.