= Solution
The <Murnaghan–Nakayama rule> computes an <irreducible character> on a <permutation> with a specified cycle of <length> $r$ by
$$
\boxed{\chi_\lambda((r)\rho)=\sum_{\substack{\nu\vdash n-r\\\lambda/\nu\text{ a rim hook of size }r}}(-1)^{\ell(\lambda/\nu)}\chi_\nu(\rho).}
$$
Here the leg <length> is one less than the number of occupied rows, and iteration ends at the empty <partition of an integer>[partition] with <character> value one. Conjugating a <partition of an integer>[partition] twists its <character> by the sign <character>:
$$
\boxed{\chi_{\lambda'}(\sigma)=\operatorname{sgn}(\sigma)\chi_\lambda(\sigma).}
$$
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 of a Young diagram>[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 therefore
$$
\boxed{\begin{array}{c|r|r|r}
\lambda&\prod h(u)&\chi_\lambda(1)&\chi_\lambda((12))\\\hline
(5)&120&1&1\\
(4,1)&30&4&2\\
(3,2)&24&5&1\\
(3,1,1)&20&6&0\\
(2,2,1)&24&5&-1\\
(2,1,1,1)&30&4&-2\\
(1^5)&120&1&-1
\end{array}}
$$
As 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>.
Back to article page
