 Solution

Schur's lemma states that an intertwiner $T:V\to W$ between irreducible representations is either zero or an isomorphism. In particular, inequivalent irreducible representations have no nonzero intertwiners. For a finite-dimensional irreducible representation over the complex numbers, every commuting endomorphism is scalar:Indeed, the kernel and image of an intertwiner are invariant subspaces, proving the first statement. For the second, choose an eigenvalue $\lambda$ of $T$. The intertwiner $T-\lambda I$ has nonzero kernel, so it must vanish. The scalar conclusion uses an algebraically closed coefficient field: over the real numbers, planar rotation through $\pi /2$ gives an irreducible real representation of ${\mathbb{Z}}_4$ and commutes with itself without being a real scalar matrix.