[proofplan] Let $\rho: G \to \operatorname{GL}(V)$ be a faithful irreducible complex representation. For each $z \in Z(G)$, multiplication by $z$ is a $G$-endomorphism of $V$ (it commutes with every $g \in G$ because $z$ does), so by part 2 of Schur's lemma it is a scalar multiple of the identity, say $\mu_z \iota_V$. The assignment $z \mapsto \mu_z$ is a group homomorphism $Z(G) \to \mathbb{C}^\times$, and it is injective because $\rho$ is faithful: distinct central elements act as distinct scalars. Hence $Z(G)$ embeds into $\mathbb{C}^\times$ as a finite subgroup, and every finite subgroup of $\mathbb{C}^\times$ is cyclic. [/proofplan] [step:Construct a $G$-endomorphism of $V$ from each central element] Let $\rho: G \to \operatorname{GL}(V)$ be a faithful irreducible complex representation, where $V$ is a finite-dimensional $\mathbb{C}$-vector space. Fix $z \in Z(G)$. Define \begin{align*} \phi_z : V &\to V \\ v &\mapsto zv, \end{align*} where $zv$ denotes the action $\rho(z)(v)$. This is $\mathbb{C}$-linear because $\rho(z) \in \operatorname{GL}(V)$. We check $\phi_z$ is a $G$-homomorphism. For any $g \in G$ and $v \in V$, using $zg = gz$ in $G$ (since $z \in Z(G)$): \begin{align*} \phi_z(gv) = z(gv) = (zg)v = (gz)v = g(zv) = g\phi_z(v). \end{align*} Hence $\phi_z \in \operatorname{End}_G(V)$. [guided] We want to extract information about $Z(G)$ from the representation $\rho$. The natural object to consider is the action of a central element on $V$: since $z$ commutes with every $g \in G$, the action of $z$ on $V$ should also commute with the action of every $g$. Concretely, define $\phi_z: V \to V$ by $v \mapsto zv$ (here $zv := \rho(z)(v)$). Linearity over $\mathbb{C}$ is immediate: $\rho(z)$ is a linear map. The crucial property is $G$-equivariance: \begin{align*} \phi_z(gv) = \rho(z)\rho(g)v = \rho(zg)v = \rho(gz)v = \rho(g)\rho(z)v = g\phi_z(v), \end{align*} where the third equality uses $zg = gz$ in $G$ (the defining property of $Z(G)$). So $\phi_z \in \operatorname{End}_G(V)$. This is the bridge: each central element of $G$ produces a $G$-endomorphism of the irreducible $V$. Schur's lemma will now classify these endomorphisms tightly. [/guided] [/step] [step:Apply Schur's lemma to obtain a scalar $\mu_z$ for each $z \in Z(G)$] Since $V$ is an irreducible complex $G$-representation and $\mathbb{C}$ is algebraically closed, part 2 of [Schur's Lemma](/theorems/2414) applies to $\phi_z \in \operatorname{End}_G(V)$ and gives a unique scalar $\mu_z \in \mathbb{C}$ with \begin{align*} \phi_z = \mu_z \iota_V. \end{align*} Equivalently, $\rho(z) = \mu_z \iota_V$ as elements of $\operatorname{GL}(V)$, so $zv = \mu_z v$ for every $v \in V$. Since $z \in Z(G)$ acts invertibly (it acts as $\rho(z) \in \operatorname{GL}(V)$), we have $\mu_z \neq 0$, so $\mu_z \in \mathbb{C}^\times$. [guided] Schur part 2 requires (i) $V$ irreducible as a $G$-representation and (ii) the base field algebraically closed. Both hold: $V$ is irreducible by hypothesis, and the base field is $\mathbb{C}$. The conclusion is that every $G$-endomorphism of $V$ is a scalar multiple of $\iota_V$. Applied to $\phi_z$, this gives $\phi_z = \mu_z \iota_V$ for a unique $\mu_z \in \mathbb{C}$. Uniqueness of $\mu_z$ matters: it allows us to define a function $\sigma: Z(G) \to \mathbb{C}^\times$ unambiguously in the next step. The scalar $\mu_z$ is non-zero because $\rho(z)$ is invertible — $\rho$ takes values in $\operatorname{GL}(V)$, never landing on the zero map. So $\mu_z \in \mathbb{C}^\times$ as required for our target group. [/guided] [/step] [step:Show $\sigma: z \mapsto \mu_z$ is a group homomorphism $Z(G) \to \mathbb{C}^\times$] Define \begin{align*} \sigma : Z(G) &\to \mathbb{C}^\times \\ z &\mapsto \mu_z, \end{align*} where $\mu_z$ is the scalar produced in the previous step. We verify $\sigma$ is a group homomorphism. For $z, z' \in Z(G)$ and any $v \in V$, \begin{align*} \mu_{zz'} v = (zz')v = z(z'v) = z(\mu_{z'}v) = \mu_{z'}(zv) = \mu_{z'} \mu_z v = \mu_z \mu_{z'} v, \end{align*} using $\mathbb{C}$-linearity of $\rho(z)$ to pull the scalar $\mu_{z'}$ through, and commutativity of multiplication in $\mathbb{C}$. Choosing any non-zero $v \in V$ (which exists since $V \neq 0$, as the trivial representation is reducible by convention or by inspection — irreducibles are non-zero by definition) and cancelling, we obtain $\mu_{zz'} = \mu_z \mu_{z'}$. Hence $\sigma(zz') = \sigma(z)\sigma(z')$. [guided] We have a function $\sigma: Z(G) \to \mathbb{C}^\times$, $z \mapsto \mu_z$. The natural question is whether it respects the group operations on each side. For $z, z' \in Z(G)$, the composite endomorphism $\phi_{z} \circ \phi_{z'}$ acts as \begin{align*} \phi_z \phi_{z'}(v) = z(z'v) = (zz')v = \phi_{zz'}(v). \end{align*} On the scalar side, $\phi_z = \mu_z \iota_V$ and $\phi_{z'} = \mu_{z'} \iota_V$, so \begin{align*} \phi_z \phi_{z'} = (\mu_z \iota_V)(\mu_{z'} \iota_V) = \mu_z \mu_{z'} \iota_V. \end{align*} Comparing both expressions, $\mu_{zz'} \iota_V = \mu_z \mu_{z'} \iota_V$, and since $\iota_V \neq 0$ this forces $\mu_{zz'} = \mu_z \mu_{z'}$. So $\sigma$ is a group homomorphism. (In particular $\mu_e = 1$ since $\phi_e = \iota_V$.) [/guided] [/step] [step:Show $\sigma$ is injective using faithfulness of $\rho$] Suppose $z \in \ker \sigma$, i.e., $\mu_z = 1$. Then $\phi_z = 1 \cdot \iota_V = \iota_V$, so $\rho(z) = \iota_V$, the identity element of $\operatorname{GL}(V)$. Since $\rho$ is faithful, $\rho$ is an injective group homomorphism $G \to \operatorname{GL}(V)$, so $\rho(z) = \rho(e)$ forces $z = e$. Hence $\ker \sigma = \{e\}$ and $\sigma$ is injective. [guided] Recall $\rho$ is faithful means $\ker \rho = \{e\}$ — distinct elements of $G$ act as distinct linear maps on $V$. We use this to upgrade "$\sigma$ is a homomorphism" to "$\sigma$ is injective". If $z \in \ker \sigma$, then $\mu_z = 1$, so $\rho(z) = \mu_z \iota_V = \iota_V$. Faithfulness gives $z = e$. So $\ker \sigma = \{e\}$, which is the kernel-criterion for injectivity of a group homomorphism. The image $\sigma(Z(G)) \subseteq \mathbb{C}^\times$ is therefore an isomorphic copy of $Z(G)$ inside $\mathbb{C}^\times$. Since $G$ is finite, so is $Z(G)$, hence $\sigma(Z(G))$ is a finite subgroup of $\mathbb{C}^\times$. [/guided] [/step] [step:Conclude $Z(G)$ is cyclic by classifying finite subgroups of $\mathbb{C}^\times$] The injective homomorphism $\sigma: Z(G) \hookrightarrow \mathbb{C}^\times$ identifies $Z(G)$ with the finite subgroup $H := \sigma(Z(G)) \subseteq \mathbb{C}^\times$. We claim every finite subgroup $H$ of $\mathbb{C}^\times$ is cyclic. [claim:Every finite subgroup of $\mathbb{C}^\times$ is cyclic] Let $H \leq \mathbb{C}^\times$ be finite of order $m := |H|$. By Lagrange's theorem, every $h \in H$ satisfies $h^m = 1$. Hence $H$ is contained in the set $\mu_m := \{\zeta \in \mathbb{C} : \zeta^m = 1\}$ of $m$th roots of unity in $\mathbb{C}$. The polynomial $X^m - 1 \in \mathbb{C}[X]$ has at most $m$ roots in $\mathbb{C}$ (a polynomial of degree $m$ over a field has at most $m$ roots), and the elements $1, e^{2\pi i/m}, e^{2 \cdot 2\pi i/m}, \ldots, e^{(m-1) \cdot 2\pi i/m}$ are $m$ distinct roots. Hence $|\mu_m| = m = |H|$ and $H = \mu_m$. The element $\zeta_m := e^{2\pi i/m}$ has order exactly $m$ in $\mathbb{C}^\times$: $\zeta_m^k = 1$ iff $m \mid k$. Hence $\langle \zeta_m \rangle$ is a subgroup of $\mu_m$ of order $m$, and since $|\mu_m| = m$ we conclude $\mu_m = \langle \zeta_m \rangle$ is cyclic of order $m$. Therefore $H$ is cyclic. [/claim] By the claim, $H \cong Z(G)$ is cyclic, hence $Z(G)$ is cyclic. This completes the proof. [guided] We have $Z(G) \cong H$ for a finite subgroup $H$ of $\mathbb{C}^\times$. The proof reduces to a classical fact: finite subgroups of the multiplicative group of an algebraically closed field — and more generally of any field — are cyclic. In our setting $\mathbb{C}$ is algebraically closed. If $|H| = m$, every element of $H$ satisfies $h^m = 1$ by Lagrange's theorem, so $H \subseteq \mu_m$, the group of $m$th roots of unity. The polynomial $X^m - 1$ has at most $m$ roots in $\mathbb{C}$ (degree-$m$ polynomial over a field), and exactly $m$ are exhibited by $\zeta_m^k$ for $k = 0, \ldots, m-1$ where $\zeta_m = e^{2\pi i/m}$. So $|\mu_m| = m$ and $H = \mu_m = \langle \zeta_m \rangle$ is cyclic. Pulling back through $\sigma$, $Z(G)$ is cyclic. **Where was each hypothesis used?** Faithfulness gave injectivity of $\sigma$; irreducibility together with the algebraic closedness of $\mathbb{C}$ made Schur 2 applicable, producing the scalars $\mu_z$ in the first place; finiteness of $G$ ensured $Z(G)$ is finite, putting it inside $\mathbb{C}^\times$ as a finite subgroup. [/guided] [/step]