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This is, of course, a well-known fact, but here's one proof. Let $H$ be a finite group, $Z(H)$ be its center, and $\lambda$ be a linear character of $Z(H).$ Then the number of irreducible characters of $H$ which lie over $\lambda$ is $\sum_{\chi \in {\rm Irr(H)}} \frac{\langle {\rm Res}^{H}_{Z(H)}(\chi), \lambda \rangle}{\chi(1)}.$ Now for each $z \in Z(H),$ the quantity $\sum_{\chi \in {\rm Irr}(G)} \frac{\chi (z)}{\chi(1)}$ was shown by Burnside to be non-negative, and non-zero if and only if $z$ is a commutator in $H$. It follows easily that for any irreducible character $\lambda$ of $Z(H),$ we have$ \sum_{z \in Z(H)} \sum_{\chi \in {\rm Irr}(H)} \frac{\chi(z)}{\chi(1)} \lambda(z^{-1}) \leq \sum_{z \in Z(H)}\sum_{\chi \in {\rm Irr}(H)} \frac{\chi(z)}{\chi(1)} = |Z(H)| k(H/Z(H)),$ where $k(X)$ denotes the number of conjugacy classes of $X.$ The last equality follows because the irreducible characters of $H$ with $Z(H)$ in their kernels are precisely those which contain the trivial character on restriction to $Z(H).$ Hence $\sum_{\chi \in {\rm Irr(H)}} \frac{\langle {\rm Res}^{H}_{Z(H)}(\chi), \lambda \rangle}{\chi(1)} \leq k(H/Z(H)),$ as claimed. Furthermore, if (and only if) there is a non-identity element $z \in Z(H) \backslash {\rm ker}(\lambda)$ which is a commutator in $H$, then the inequality becomes strict.