This is a problem from an old qualifying exam I am reviewing:
Calculate: $\langle h_3p_6, s_{(5,4)}\rangle$ where $\langle \cdot,\cdot \rangle$ is the Hall scalar product defined by $\langle s_\lambda, s_\mu\rangle = \delta_{\lambda,\mu}$.
Here $s_\lambda,$ etc.. are the symmetric functions in the notation of, e.g., Sagan's Symmetric Group. Can someone provide some insight? The way I've done this for (simpler) problems is to write everything in the basis $s_\lambda$; but I am getting stuck on doing this for $h_3$ specifically. Thanks!
===== Update =====
Here is an (attempted) solution. I got an unexpected non-integral answer so I am very curious what, if anything, went wrong. Per the hint I started by writing
$$h_3 = \frac{1}{3!}\text{det}\begin{pmatrix}p_1 & -1 & 0\\ p_2&p_1&-1\\p_3&p_2&p_1\end{pmatrix} = \frac{1}{6}\left(p_1^3 + 2p_2p_1 + p_3\right)$$
and hence
$$h_3 p_6 = \frac{1}{6}\left(p_{(6,1,1,1)}+2p_{(6,2,1)}+p_{(6,3)}\right)$$
Then, separately, we write
$$s_{(5,4)} = \sum_{\mu\vdash 9}\frac{1}{z_\mu}\chi^{(5,4)}_\mu p_\mu$$
So that, since $\langle p_\nu,p_\lambda\rangle = z_\lambda\delta_\lambda$ where $\delta_\lambda$ is the Kronecher delta, we have
$$ \langle h_3 p_6, s_{(5,4)}\rangle = \frac{1}{6}\left(\chi^{(5,4)}_{(6,1,1,1)}+2\chi^{(5,4)}_{(6,2,1)}+\chi^{(5,4)}_{(6,3)}\right)$$
which is promising. It remains to use Murnaghan-Nakayama rule to calculate the character values, which I did iteratively with rim-hook removal. To this end, some calculations:
$$\begin{split} \chi^{(5,4)}_{(6,1,1,1)} &= \chi^{(3)}_{(1,1,1)} = \chi^{(2)}_{(1,1)} = \cdots= -\chi^{(0)}_{(0)} = -1\\ \chi^{(5,4)}_{(6,2,1)} &= \chi^{(3)}_{(2,1)} = \cdots = -1\\ \chi^{(5,4)}_{(6,3)} &= \chi^{(3)}_{(3)} = -\chi^{(0)}_{(0)} = -1 \end{split} $$
So that, in the end, $\langle h_3 p_6, s_{(5,4)}\rangle = -2/3$. This is an answer (Lol!) but I would have expected an integer. Maybe I am naïve.