Corollary 1

Let us retain the assumptions of Theorem 3 and suppose moreover that the elements of A are pairwise permutable. Then E is of dimension 1 over $\mathbf{C}$.

By assumption (ii) of Theorem 3, the space E is not zero. It therefore contains a vector subspace F of dimension 1. This subspace is closed because E is assumed separated. By Theorem 3, every element of A is a homothety, and therefore stabilizes F. It then follows from assumption (i) of Theorem 3 that F is equal to E.