(n, z) \in G_1 = H_1 \times_{Z(H_1)} Z_1
\text{ for some } z \in Z_1 \}$. Then $N' \lhd H_1$. If $N' = 1$, then $N \subset Z_1 = Z(G_1)$, contradicting $N$ being $Z(G_1)$-free. Hence $N' \ne 1$ and must contain a nontrivial $a \in Z(H_1)$. But then $(a,z) \in N \cap Z(G_1)$ for some $z \in Z_1$, which also contradicts $N$ being $Z(G_1)$-free.

This will often allow us to reduce to the case where $G = H \times_{Z(H)} A$ for some cyclic group $A \supset Z(H)$, where we can use the following.

\[lem:35\] Let $H$ be a finite group, $A \supset Z(H)$ an abelian group and $G = H \times_{Z(H)} A$. Then $\delta^\natural_n(G) \ge \min 
\{ \frac 12 \alpha_n(H), \delta^\natural_n(H) \}$.

We may assume $m = |A| > 1$. Let $\rho_1, \rho_2: G \to \operatorname{GL}_n({\mathbb C})$ be distinct primitive representations of $G$. They pull back to $H \times A$, so for $i=1,2$ we can view $\rho_i =  \tau_i \otimes \chi_i$ where $\tau_i: H \to \operatorname{GL}_n({\mathbb C})$ is primitive and $\chi_i: A \to {\mathbb C}^\times$. By a similar argument to the proof of Lemma \[lem:34\], we have that $\delta(\rho_1, \rho_2) \ge \frac{m-1}m \alpha_n(H)$ if $\chi_1 \ne \chi_2$. If $\chi_1 = 