Permutations whose images are a basis of the endomorphism algebra of a module
If \(G\) is a finite group and \(S\) is a \(\mathbb{C}G\)-module then the linear maps \(\rho_g: s \mapsto gs\), as \(g\) varies, span a subspace of \(\operatorname{End}_\mathbb{C}(S)\). We can ask for a “nice” subset of \(G\) such that the corresponding linear maps are a basis of that subspace. In the case that \(S\) is simple, the subspace will be all of \(\operatorname{End}_\mathbb{C}(S)\), by the Jacobson density theorem.
For example, take \(G = S_n\) and \(S\) the natural permutation module. The images of permutations \(e, (1, r), (1, r, s)\) are a basis of the span of the span of the maps \(\rho_g\) as I showed here. But there’s a much nicer basis arising from a (very) special case of some work of Steve Doty: the images of the set of all permutations having an increasing subsequence of length at least \(n-1\). (Here we are thinking of a permutation as the sequence of numbers on the bottom row of its two row form.)
When I learned this it set off my Robinson-Schensted radar. The length of the longest increasing subsequence of a permutation is equal to the number of columns in the first row of its \(P\)-symbol under the RS correspondence. Furthermore, the quotient of the natural rep by its trivial submodule is the simple Specht module \(V^{(n-1, 1)}\). The dimension of the Specht module \(V^\lambda\) is the number of standard tableaux of shape \(\lambda\), so the dimension of its \(\mathbb{C}\)-endomorphism algebra is the square of this. That’s the same as the number of permutations whose \(P\)-symbol has shape \(\lambda\). Could the images of those permutations in \(\operatorname{End}_\mathbb{C}(V^\lambda)\) be a basis?
Sadly, the answer is no. It’s not even true for \(n=4\): the pointwise stabilizer of \(V^{(2, 2)}\) is \(V_4\) and two elements of \(V_4\) have Robinson-Schensted correspondent of shape \((2, 2)\) so their images in the endomorphism algebra are linearly dependent.