Past exam of the Lie algebras and representation theory of symmetric groups course of the University of Oxford / 2007 / B / 5 / c / Solution

For a cycle type $\mu =(m,\nu )$, the Murnaghan–Nakayama rule isThe sum is over partitions $\rho$ whose diagrams remain after removing the specified border strip; $\nu$ is the remaining cycle type, and the recursion ends with the empty character equal to one. At the identity, ${\chi }^{\rho }(1^{|\rho |})=f^{\rho }$. Incompatible total sizes contribute zero.

For shape $(2,2)$ and a transposition, a horizontal strip in the bottom row leaves $(2)$ with positive sign; a vertical strip in the right column leaves $(1,1)$ with negative sign. Both remaining dimensions are one. Thus