Past exam of the Lie algebras and representation theory of symmetric groups course of the University of Oxford / 2008 / B / 6

If the cycle partition contains a part $r$, the Murnaghan–Nakayama rule isA border strip is a connected skew Young diagram with no $2\times 2$ block; its height is its number of occupied rows minus one. The remaining boxes must be a partition diagram. The recursion starts with character one for the empty diagram at the empty cycle type, and an empty sum is zero. Repeated removal for all cycle lengths computes the character value.

Write the hook cycle partition as $\mu =(k,1^{n-k})$, $1\le k\le n$. At the identity, the representation dimension is $n-1$: a standard tableau of shape $(2,1^{n-2})$ is determined by the entry in its top-right box, any one of $2,\dots ,n$.

For $2\le k\le n-2$, the only removable border strip of length $k$ is the vertical bottom strip. Its height is $k-1$, and the remaining hook has size $n-k$ and dimension $n-k-1$. A strip of length $n-1$ would leave the top-left box but disconnect the removed top-right box from the removed column, so none exists and the value is zero. For $k=n$, the whole hook is a border strip of height $n-2$ and contributes $(-1{)}^{n-2}$. These cases combine asIn particular the formula includes zero at $k=n-1$ and the full-cycle sign at $k=n$.

The two partition diagrams are transposes, so their characteristic-zero Specht modules differ by tensoring with the sign representation. A transposition has sign minus one. ConsequentlyThe middle character is $n-3$, since the standard representation is the point-permutation representation minus its trivial line, and a transposition fixes $n-2$ points.