Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 160 / 2

Write $a=\lambda -\mathrm{id}$. The two modified entries satisfyThus $\mu -\mathrm{id}$ is obtained from $\lambda -\mathrm{id}$ by the adjacent transposition $\tau =(i,i+1)$. In the alternating definition of $\psi$, reindexing $\pi$ by $\tau \pi$ preserves every induced permutation character and reverses every sign. Therefore the straightening of a symmetric-group character indexed by a composition gives

The restriction form of the restriction branching rule for a symmetric group iswhere ${\lambda }^{-}$ contains the distinct partitions obtained by deleting one Removable node of a Young diagram. In particular, the restriction is multiplicity-free.

Restrict the alternating expressionThe supplied restriction formula for a Young permutation character, with $k=1$, says that each term restricts by subtracting one from each possible component. After collecting the alternating sums, this givesIf row $i$ has no removable node, part i straightens ${\psi }^{\lambda -{ϵ}_i}$ against the adjacent term with the opposite sign, or makes it zero when two shifted entries coincide. The surviving terms are exactly ${\psi }^{\mu }$ for $\mu \in {\lambda }^{-}$. Since $\lambda$ and each surviving $\mu$ are partitions, ${\psi }^{\lambda }={\chi }^{\lambda }$ and ${\psi }^{\mu }={\chi }^{\mu }$. We obtainComplex representations of a finite group are semisimple, so equality of characters proves the asserted module decomposition.

Repeated use of the induction branching rule giveswhileOn the five conjugacy classes $1,(12),(12)(34),(123),(1234)$, their characters are respectivelyThe tensor-product character is their pointwise product $(48,-4,0,0,0)$. Taking inner products with the five irreducible characters of $S_4$ gives multiplicities $1,5,4,7,3$. Therefore the induced-tensor decomposition for the symmetric group on four points is

Use the character formula for an induced representation. For $g\in G$,The middle equality uses that a character of a representation is constant on conjugacy classes. This proves the tensor identity for an induced character.

The point-permutation character isBy the tensor identity for an induced character and Frobenius reciprocity,The restriction branching rule for a symmetric group is multiplicity-free with one constituent for each member of ${\lambda }^{-}$, so the right side is $|{\lambda }^{-}|$. Also $⟨{\chi }^{\lambda }{\chi }^{\lambda },{\chi }^{(n)}⟩=1$ because every symmetric-group character is real and irreducible. Subtracting the trivial constituent proves the standard-character multiplicity in a Specht self-product formula

Suppose ${\chi }^{\alpha }{\chi }^{\beta }$ is irreducible. Since symmetric-group characters are real,Both self-products contain the trivial character once. They can therefore have no other common irreducible constituent. By part i, the standard character occurs in the two self-products with multiplicities $|{\alpha }^{-}|-1$ and $|{\beta }^{-}|-1$. Hence one of these numbers is zero; say $|{\alpha }^{-}|=1$.

A partition has exactly one removable node precisely when all its nonzero rows have equal length, so $\alpha =(a^b)$ is rectangular. Since $ab=n$ and $n$ is prime, either $a=1$ or $b=1$. Thus $\alpha =(1^n)$ or $(n)$. The same argument applies with $\alpha$ and $\beta$ interchanged, proving the prime-degree irreducible Kronecker product criterion for a symmetric group.