Can anyone give me any hints of how to proceed: I have scoured this forum and exhausted all online notes.
I am trying to prove that there is no room for an irreducible permutation representation, $\rho_{\text{perm}}$.
Attempt 1 By dimensionality theorem, $$ |G| = \sum_i \dim{V_i}^2 $$ I then ran into issues as thus, $$ |S_n| = n! = \sum_i \dim{V_i}^2 =?\ V_{\text{ident}}^2 + V_{\text{alt}}^2 + V_{\text{perm}}^2 = 1^2 + 1^2 + n^2 $$ where $=?$ indicates my confusion since it appears as though there is room for the permutation representation in certain cases for large n and so it is not general.
Attempt 2 I followed the examples I have worked out by using the trace or character of the representation, $\chi(\rho)$ such that, $||\chi(\rho)||^2 = 1$ iff irreducible. My further issue is then that whilst I can practically calculate this trace for a specific example I do not know how to compute for $S_n$ generally.
Attempt 3 I then tried looking for invariant subspaces, I know that $\rho_{\text{perm}}$ acts on an n-dimensional subspace of orthogonal vectors but I can't see how I can then obtain a second invariant subspace. I have seen a few posts mentioning a conjugate subspace but I don't understand the theory of how this arises / can't find anything that sticks out online / my notes.
Attempt 4 I know $\rho_{\text{perm}}$ decomposes into other representations and I thought of proving it decomposes into $\rho_{\text{alt}}$ and $\rho_{\text{indent}}$. However, I realised that this doesn't satisfy $S_2$ and furthermore I found notes suggesting this is not the only decomposition of $\rho_{\text{perm}}$ either.