The Murnaghan–Nakayama character rule

Reduction to standard blocks. Since ${\chi }^{\lambda }$ is constant on conjugacy classes [F5], let $\gamma \in S_n$ be the permutation that carries $D$ order-preservingly onto $B_2=\{n-r,\dots ,n-1\}$ and the complement of $D$ order-preservingly onto $B_1=\{0,\dots ,n-r-1\}$; put ${\rho }^{\mathrm{\prime }}:=\gamma \rho {\gamma }^{-1}$ and ${\sigma }^{\mathrm{\prime }}:=\gamma \sigma {\gamma }^{-1}$. Then ${\rho }^{\mathrm{\prime }}$ is cyclic on $B_2$ and fixes $B_1$, the permutation ${\sigma }^{\mathrm{\prime }}$ lies in $\mathrm{Sym}(B_1)$ and fixes $B_2$ pointwise, the element ${\sigma }^{\mathrm{\prime }}$ is exactly the image of $\sigma$ under the order-preserving identification with $S_{n-r}$ on $B_1$ used in the statement, and ${\chi }^{\lambda }(\sigma \rho )={\chi }^{\lambda }(\gamma \sigma \rho {\gamma }^{-1})={\chi }^{\lambda }({\sigma }^{\mathrm{\prime }}{\rho }^{\mathrm{\prime }})$. It therefore suffices to prove the displayed identity in the reduced case $D=B_2$, $\sigma \in \mathrm{Sym}(B_1)$.

Reduced-case expansion. Assume now $D=B_2$ and $\sigma \in \mathrm{Sym}(B_1)$, so that $\sigma \rho$ is the element $(\sigma ,\rho )$ of the two-block subgroup $K$. By the restriction identity [F1] and the evaluation rule for external tensor products [F2], ${\chi }^{\lambda }(\sigma \rho )={\sum }_{\nu ⊢n-r}{\sum }_{\mu ⊢r}{c}_{\nu \mu }^{\lambda }{\chi }^{\nu }(\sigma ){\chi }^{\mu }(\rho )$.

Substitution and cancellation. In the display of step 1.2 the full-cycle lemma [F3] gives ${\chi }^{\mu }(\rho )=(-1{)}^{r-x}$ when $\mu =(x,1^{r-x})$ and $0$ for every other $\mu$, so ${\chi }^{\lambda }(\sigma \rho )={\sum }_{\nu ⊢n-r}{\chi }^{\nu }(\sigma )[{\sum }_{x=1}^r(-1{)}^{r-x}{c}_{\nu ,(x,1^{r-x})}^{\lambda }]$. By the cancellation lemma [F4] the bracket is $(-1{)}^{r-c(\lambda \mathrm{/}\nu )}=(-1{)}^{\mathrm{ht}(\lambda \mathrm{/}\nu )}$ when $\lambda \mathrm{/}\nu$ is a removable border strip of size $r$ occupying $c(\lambda \mathrm{/}\nu )$ columns, and is $0$ for every other $\nu$; hence in the reduced case ${\chi }^{\lambda }(\sigma \rho )={\sum }_{\lambda \mathrm{/}\nu }(-1{)}^{\mathrm{ht}(\lambda \mathrm{/}\nu )}{\chi }^{\nu }(\sigma )$, the sum over the removable $r$-strips of $\lambda$ and empty (equal to $0$) when there is none.

Undoing the reduction of step 1.1, the identity of step 2.1 holds for arbitrary specified cyclic sets $D$, permutations $\rho$ cyclic on $D$, and compatible $\sigma$, which is the displayed rule of the statement. For the final assertion, prove by induction on $n$ that the rule computes all values ${\chi }^{\lambda }(\sigma )$: for $n=0$ one has $\lambda =\mathrm{\varnothing }$, $\sigma =1$ and ${\chi }^{\mathrm{\varnothing }}(1)=1$ [F6]; for $n\ge 1$ choose a cycle of $\sigma$, including a specified fixed point as a one-cycle when necessary, and let $D$ be its $r$ points and $c$ its cyclic permutation, so that $\sigma ={\sigma }^{\mathrm{\prime }}c$ with ${\sigma }^{\mathrm{\prime }}$ fixing $D$ pointwise and lying in $S_{n-r}$ under the identification of the statement, and apply the rule just proved: ${\chi }^{\lambda }(\sigma )={\sum }_{\lambda \mathrm{/}\nu }(-1{)}^{\mathrm{ht}(\lambda \mathrm{/}\nu )}{\chi }^{\nu }({\sigma }^{\mathrm{\prime }})$, where each right-hand value involves the strictly smaller degree $n-r$, so by induction each ${\chi }^{\nu }({\sigma }^{\mathrm{\prime }})$ is computed by iterating the rule along the remaining cycles of $\sigma$; substituting computes ${\chi }^{\lambda }(\sigma )$.