= Solution
A representation is completely reducible, or <semisimple representation>[semisimple], when it is a <direct sum> of <irreducible representations>[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.
* If $V$ and $W$ are irreducible $G$-representations, every <intertwining operator> $T:V\to W$ is either zero or an isomorphism.
* If $V$ is finite-dimensional over an algebraically closed field, every $G$-endomorphism of $V$ is a scalar multiple of the identity.
Indeed, $\ker T\subseteq V$ and $\operatorname{im}T\subseteq W$ are invariant subspaces. If $T\ne0$, irreducibility forces
$$
\ker T=0,
\qquad
\operatorname{im}T=W,
$$
so $T$ 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 forces
$$
T-\lambda I=0.
$$
This proves the <Proof of Schur lemma>.
Now let $\rho:G\to\operatorname{GL}(V)$ 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 gives
$$
\rho(z)=\lambda_zI
$$
for some $\lambda_z\in\mathbb C^\times$. Faithfulness makes $z\mapsto\lambda_z$ an embedding
$$
Z(G)\hookrightarrow\mathbb C^\times.
$$
Since every <finite subgroup of the multiplicative complex numbers> is cyclic,
$$
\boxed{Z(G)\text{ is cyclic}}.
$$
This is the <cyclic-center obstruction to a faithful irreducible representation>.
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