If \(S=\{1\le i_1<i_2<\cdots<i_k\le n\}\) is a subset of \(\{1,2,\ldots,n\}\) and \(\sigma=(\sigma_1,\sigma_2,\ldots,\sigma_n)\) is a permutation, let \(\sigma(S)\) denote the subset \(\sigma(S)=\{\sigma_{i_1},\ldots,\sigma_{i_k}\}\). In this manner we can define an action of \(S_n\) on the \(k\)-subsets of \(\{1,2,\ldots,n\}\) and induce a representation \(A^{(k,n)}\) such that if \(S_1,\ldots,S_{\binom{n}{k}}\) is a list of the \(k\)-element subsets of \(\{1,\ldots,n\}\), then \[\begin{align} \sigma\langle S_1,\ldots,S_{\binom{n}{k}}\rangle &=\langle\sigma(S_1),\ldots,\sigma(S_{\binom{n}{k}})\rangle\\ &=\langle S_1,\ldots,S_{\binom{n}{k}}\rangle A^{(k,n)}(\sigma). \end{align}\] Let \(\chi^{(k,n)}\) be the character of \(A^{(k,n)}\).
Find the Frobenius image of \(\chi^{(2,4)}\).
Use your result in (a) to compute the decomposition of \(\chi^{(2,4)}\) into a sum of irreducible characters of \(S_4\).
Solution
1. The permutation representation \(A^{(2,4)}\) is isomorphic to\[ \operatorname{Ind}_{S_2\times S_2}^{S_4}\mathbf 1. \]
The group \(S_4\) acts transitively on the six \(2\)-element subsets of \(\{1,2,3,4\}\). The stabilizer of the subset \(\{1,2\}\) consists of the permutations preserving \(\{1,2\}\) and \(\{3,4\}\) setwise, hence is the Young subgroup \[ S_{\{1,2\}}\times S_{\{3,4\}}\cong S_2\times S_2. \] A transitive permutation representation on \(G/H\) is the induced representation \(\operatorname{Ind}_H^G\mathbf 1\), which gives the claim.
2. The Frobenius image of \(\chi^{(2,4)}\) is\[ \boxed{\operatorname{ch}\chi^{(2,4)}=h_2h_2=s_{(2)}s_{(2)}.} \]
Under the Frobenius characteristic map, induction from a Young subgroup corresponds to multiplication of symmetric functions. The trivial representation of \(S_2\) corresponds to \[ s_{(2)}=h_2. \] Therefore \[ \operatorname{ch}\left(\operatorname{Ind}_{S_2\times S_2}^{S_4}(\mathbf1\boxtimes\mathbf1)\right) =s_{(2)}s_{(2)}=h_2h_2. \] By 1, this is the Frobenius image of \(\chi^{(2,4)}\).
3. By the Pieri rule,\[ h_2s_{(2)}=s_{(4)}+s_{(3,1)}+s_{(2,2)}. \]Hence\[ \boxed{\chi^{(2,4)}=\chi^{(4)}+\chi^{(3,1)}+\chi^{(2,2)}.} \]
Multiplication by \(h_2\) adds a horizontal \(2\)-strip to the Young diagram \((2)\). The possible resulting partitions of \(4\) are exactly \[ (4),\qquad(3,1),\qquad(2,2), \] each with coefficient \(1\). Applying the inverse Frobenius characteristic map yields the stated irreducible-character decomposition.