Reprensentation Theory, Pieri s rule

I can give you two solutions.

First solution: we can see that, by construction, the space spanned by the $\alpha_{ij}$ with the action $\sigma \alpha_{ij} = \alpha_{\sigma(i), \sigma(j)}$ is the induced representation of the trivial representation of $S_2 \times S_{n-2}$. This should be clear by the contents of your course.

In case it's not, notice that \[ \mathsf{Ind}^{S_n}_{S_2 \times S_{n-2}} \coloneqq \bigoplus_{[\sigma] \in S_n / (S_2 \times S_{n-2})} [\sigma] \mathbb{C}, \] where $[\sigma]$ runs over a complete set of representatives of the coset, which are permutations of the form $ij12 \dots n$ in one-line notation. These are exactly in equivariant bijection with the $\alpha_{ij}$, so we can get the character via the Pieri rule, and get $s_2 \cdot s_{n-2} = s_n + s_{n-1,1} + s_{n-2,2}$.


Second solution: let $\chi$ be the character of $H_1(F_2(\mathbb{C}, \mathbb{C}))$. We have that \[ \langle \chi, \chi \rangle = \frac{1}{n!} \sum_{\sigma \in S_n} \chi(\sigma)^2 \] and by Burnside's theorem this is the number of orbits of the group action of $S_n$ on the set of $\{ \alpha_{ij} \}^2$. The action has three orbits (pairs $(\alpha_{ij}, \alpha_{ij})$, pairs $(\alpha_{ij}, \alpha_{ik})$ with $j \neq k$, and pairs $(\alpha_{ij}, \alpha_{hk})$ where all four indices are distinct. So the scalar product is $3$ and the action has three irreducible components. We definitely have a copy of the standard representation (which is isomorphic to $S^{(n)} \oplus S^{(n-1,1)}$), given by the action on the elements $\alpha_i = \sum_{j \neq i} \alpha_{ij}$.

The character of the remaining component can be computed explicitly by difference, it being $\chi - \mathsf{std}$, and it is equal to $\chi_{n-2,2}$, so the thesis follows.


Let me know if you need any further clarification.