Finite group $ G=\{x_1,x_2,…x_n\}$ . Consider $ G$ ‘s multiplication table to be an $ n\times n$ matrix $ A$ . Set $ x_i=1$ , $ x_j=0$ ($ j≠i$ ), $ 1≤i≤n$ , then we get $ n$ permutation matrices $ S_i$ ($ 1≤i≤n$ ) s.t. $ A=∑x_iS_i$ . DetA is called groupdet, which is a polynomial of n variables. DetA is reducible with at least a factor ∑xi. So we’re able to find a common invariant subspace of $ S_i$ (considering them to be linear transformation on $ V$ ). If we can find another common invariant subspace $ V_1$ ($ V=V_1⊕W$ ), there must exist a linear transformation $ T$ s.t. $ T(V_1)=V_1$ , $ T(W)=V_1$ . Let $ T’=∑S_iTS_i^{-1}$ . Then $ V=T'(V_1) \oplus \ker T’$ and $ \ker T’$ is a common invariant subspace. Thus there exists matrix $ H$ s.t. $ HAH^{-1}=\operatorname{diag}(A_1,A_2,….A_m)$ and it’s unable to find smaller blocks in $ A_k$ .

I find it difficult to prove detAk are irreducible without group representation. Let Ak=∑xiHSiH^(-1)=(aij) If aij are linearly independent over C, add some other characters bj(1≤j≤n-k^2) s.t. ring C[x1,x2,… xn]≌C[a11, a12,… akk, b1,…], then aij are algebraically independent over C. If detAk=fg and f contains a11, a1i and ai1 mustn’t be contained in g. Similarly, all aij are not contained in g, so g is constant. Thus detAk is irreducible.

My question: how to prove aij are linearly independent? I think we just need to prove that aij≠0 for all i and j. I know Frobenius established the theories with concepts like group character, but I still hope that the proof does not use group character and group representation theory or use only a little of them. Thanks.