- \frac 1{n^2} ) |G|$ elements of $G$. On the remaining $2n$ elements of $Z(G)$, $\operatorname{tr}\rho$ and $\operatorname{tr}\rho'$ must differ on precisely $n$ elements, giving $G$ with $\delta_n(G) = \frac 1{2n^2}$ as desired.

Finally that $\rho$ and $\rho'$ so constructed are induced for $n > 1$. It suffices to show $\tau$ is induced. Since $\bar H$ is solvable, there is a subgroup of prime index $p$, so there exists a subgroup $K$ of $H$ of index $p$ which contains $Z=Z(H)$. Put $\chi = \operatorname{tr}\tau$. Now $\sum_K |\chi(k)|^2
= \sum_Z |\chi(k)|^2 = |H|$. On the other hand $\sum_K |\chi(k)|^2 \le
\sum_{i=1}^r \sum_K |\psi_i(k)|^2 = r |K|$, where $\chi |_K = \psi_1 + \cdots + \psi_r$ is the decomposition of $\chi|_K$ into irreducible characters of $K$. Thus $r \ge p$ and we must have equality, which means $\tau$ is induced from a $\psi_i$. We note that, more generally, Christina Durfee informed us of a proof that $\rho$, $\rho'$ must be induced if $\delta(\rho, \rho') = \frac 1{2n^2}$.

General Methods {#sec:general}
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Central extensions and minimal lifts
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The first step in the proof of Theorem \[thm1\] is the determination of the minimal lifts of irreducible finite subgroups of $\operatorname{PGL}_2({\mathbb C})$ and $\operatorname{PGL}_3({\mathbb 