Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 3 / 19I (Representation Theory)

For a character $\psi$ of a subgroup $H\le G$, the induced character isIt is the character of the induced representation $\mathbb{C}[G]{\otimes }_{\mathbb{C}[H]}{V}_{\psi }$, and has degree $[G:H]\psi (1)$. The character inner product is $⟨\alpha ,\beta {⟩}_G=|G{|}^{-1}{\sum }_g\alpha (g)\overline{\beta (g)}$. Substituting the displayed formula and putting $g=xh{x}^{-1}$ yieldsThus Frobenius reciprocity isNow assume index two, so $H$ is normal. Choose $g\notin H$ and write ${\psi }^g(h)=\psi (g^{-1}hg)$. Directly from the induction formula,If ${\psi }^g\ne \psi$, this norm is one, so induction is irreducible of degree $2d$: $\psi$ is monogamous. If ${\psi }^g=\psi$, the norm is two. Since a norm is the sum of squared nonnegative integer multiplicities, induction is the sum of two distinct irreducibles, each once. Each is related to $\psi$, so its degree is at least $d$ by restriction and reciprocity; their degrees sum to $2d$, forcing both to have degree $d$. Thus $\psi$ is bigamous.

Every irreducible $\chi$ of $G$ restricts to a nonzero sum of irreducibles. Choose a constituent $\psi$. If $\psi$ is bigamous, $\chi$ is one of its two degree-$d$ partners; since its restriction already contains a degree-$d$ constituent, $\mathrm{Res}\chi =\psi$. If $\psi$ is monogamous, $\chi =\mathrm{Ind}\psi$ and $\mathrm{Res}\chi =\psi +{\psi }^g$, giving precisely two distinct monogamous partners of equal degree.

The complex irreducible character degrees of $A_5$ are $1,3,3,4,5$, with squared sum $60$. Conjugation by the outside coset preserves degrees, so it fixes the unique degree-$1$, degree-$4$ and degree-$5$ characters and either fixes or swaps the two degree-$3$ characters. The two possibilities for $G$ areBoth lists have squared sum $120=|G|$. They occur for $A_5\times C_2$ and $S_5$ respectively; in $S_5$, odd conjugation swaps the two classes of $5$-cycles and hence the two degree-$3$ characters.