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Schur lemma over $\mathbb{C}$ states that an intertwining operator $T:V\to W$ between finite-dimensional irreducible complex representations is either zero or an isomorphism. In particular, every intertwining endomorphism of $V$ is a scalar multiple of the identity.

Indeed, $\ker T\subseteq V$ and $\mathrm{im}T\subseteq W$ are invariant subspaces. If $T\ne 0$, irreducibility gives $\ker T=0$ and $\mathrm{im}T=W$, so $T$ is an isomorphism. For an endomorphism $T:V\to V$, the fundamental theorem of algebra gives an eigenvalue $\lambda \in \mathbb{C}$. Then $T-\lambda I$ is an intertwiner with nonzero kernel, so the first part forces $T-\lambda I=0$. Hence $T=\lambda I$, completing the Proof of Schur lemma.

A representation is completely reducible, or semisimple, when it is a direct sum of irreducible representation. Equivalently, every invariant subspace has an invariant complement.

Maschke's theorem says that every finite-dimensional representation of a finite group $G$ over a field of characteristic zero is completely reducible. More generally, it is enough that the field characteristic does not divide $|G|$.

Schur lemma states the following.

Indeed, $\ker T\subseteq V$ and $\mathrm{im}T\subseteq W$ are invariant subspaces. If $T\ne 0$, irreducibility forcesso $T$ is an isomorphism. For an endomorphism of a finite-dimensional complex representation, choose an eigenvalue $\lambda$ of $T$. The intertwiner $T-\lambda I$ has nonzero kernel, so the first part forcesThis proves the Proof of Schur lemma.

Now let $\rho :G\to \mathrm{GL}(V)$ be faithful and irreducible over $\mathbb{C}$. For every $z\in Z(G)$, the operator $\rho (z)$ commutes with every $\rho (g)$. Schur's lemma therefore givesfor some ${\lambda }_z\in {\mathbb{C}}^{\times }$. Faithfulness makes $z\mapsto {\lambda }_z$ an embeddingSince every finite subgroup of the multiplicative complex numbers is cyclic,This is the cyclic-center obstruction to a faithful irreducible representation.