Schur lemma

An intertwiner between irreducible complex representations is either zero or an isomorphism; an endomorphism of one irreducible is scalar.

The kernel and image of an intertwiner are invariant subspaces. Irreducibility therefore makes a nonzero intertwiner injective and surjective. For an endomorphism $T$ of a finite-dimensional irreducible complex representation, choose an eigenvalue $\lambda$. The noninvertible intertwiner $T-\lambda I$ must be zero, so $T=\lambda I$.

Let a cyclic-group generator act on a real plane by a rotation through an angle other than zero or $\pi$. The representation is irreducible, but its commuting endomorphisms areThe non-scalar complex structure $J$ shows that the scalar endomorphism conclusion of Schur lemma fails over $\mathbb{R}$.

If a finite group has a faithful irreducible complex representation, every central element acts as a scalar by Schur lemma. Faithfulness embeds the centre into ${\mathbb{C}}^{\times }$, and every finite subgroup of ${\mathbb{C}}^{\times }$ is cyclic. Hence the group centre must be cyclic.