Past exam of the representation Theory of Symmetric Groups course of the University of Oxford / 2011 / 1

The Murnaghan–Nakayama rule computes an irreducible character on a permutation with a specified cycle of length $r$ byHere the leg length is one less than the number of occupied rows, and iteration ends at the empty partition with character value one. Conjugating a partition twists its character by the sign character:For a transposition, remove a rim domino and evaluate the remaining character at the identity of $S_3$. The three possible remaining shapes $(3),(2,1),(1^3)$ have degrees $1,2,1$ by the hook formula. For $(5)$ the single removable horizontal domino leaves $(3)$, giving one. For $(4,1)$ the only rim domino leaves $(2,1)$, giving two. For $(3,2)$ the removable bottom-row domino leaves $(3)$, giving one. For $(3,1,1)$, a horizontal domino leaves $(1^3)$ and a vertical domino leaves $(3)$; their contributions are $1-1=0$. The conjugate-partition identity gives the remaining three values with reversed sign.

The Hook-length formula is ${\chi }_{\lambda }(1)=5!/{\prod }_{u\in [\lambda ]}h(u)$. Both requested columns are thereforeAs a check, the squared degrees sum to $1+16+25+36+25+16+1=120=|S_5|$. The transposition column above was obtained from rim-hook removal and conjugation, independently of permutation characters.

Use the question's indexing convention: $m_{\lambda ,\mu }$ is the multiplicity of the irreducible Specht module $S_{\mathbb{C}}^{\mu }$ in the permutation module $M_{\mathbb{C}}^{\lambda }$. In notation for semistandard-tableau Kostka numbers this is often written $K_{\mu \lambda }$, with the indices reversed.

Let $S_{\lambda }=S_{{\lambda }_1}\times S_{{\lambda }_2}\times \cdots$ be the Young subgroup. The permutation character isand ${\xi }^{\lambda }(g)$ counts the $\lambda$-tabloids fixed by $g$. Equivalently it counts ordered partitions into labelled rows of the prescribed sizes, each row invariant under $g$. Character inner products and Frobenius reciprocity give the purely character-theoretic computationThe characters of symmetric groups are real-valued, so no conjugation is needed in the displayed subgroup average.

For $\lambda =(2,1^3)$, the Young subgroup has two elements, the identity and one transposition. Thus $m_{(2,1^3),\mu }=({\chi }_{\mu }(1)+{\chi }_{\mu }((12)))/2$, and part (a) givesThe resulting dimension is $1+3\cdot 4+3\cdot 5+3\cdot 6+2\cdot 5+4=60=5!/2$, the dimension of the permutation module.

This is the character average for Kostka numbers.

The Young subgroup for $(1^n)$ is trivial. Hence $M_{\mathbb{C}}^{(1^n)}$ is the regular representation of $S_n$. The subgroup-average formula reduces toThis is the usual multiplicity of an irreducible representation in the regular representation.