Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 160 / 1 / c

The equality $h\{t\}=h^{\ast }\{t^{\ast }\}$ says that $h(i)$ and $h^{\ast }(i)$ lie in corresponding rows for every entry $i$. For $i$ in the leftmost column, both $h(i)$ and $h^{\ast }(i)$ remain in that column. Corresponding rows therefore force $h(i)=h^{\ast }(i)$.

Remove the leftmost entry of every row and repeat the argument on the shortened tableaux. Induction across the columns gives equality on every entry, so $h=h^{\ast }$.

The nonzero coefficients and part a(ii) give unique elements $h\in C(t)$ and $h^{\ast }\in C(t^{\ast })$ such thatPart i gives $h=h^{\ast }=:k$, so $k\in C(t)\cap C(t^{\ast })$. Both coefficients are then $\mathrm{sgn}(k)$ and hence are equal.

By part ii, the common tabloids in the supports of $e(t)$ and $e(t^{\ast })$ are precisely $k\{t\}$ with $k\in C(t)\cap C(t^{\ast })$, and each contributes $\mathrm{sgn}(k{)}^2=1$. ConsequentlyFor the $a_j$ rows of length $j$, an element of the intersection applies the same arbitrary permutation of those rows independently in each of their $j$ columns. Hence that row length contributes $(a_j!{)}^j$, and