hand, $\operatorname{tr}\rho_2(gk) = \operatorname{tr}\rho_2(g)$ for all $k \in K$. So provided $\operatorname{tr}\rho_1(g) \ne 0$, $\operatorname{tr}\rho_1$ and $\operatorname{tr}\rho_2$ can agree on at most $\frac{1}m |K|$ values on the coset $gK$. Then note that the fraction of elements $g \in G$ for which $\operatorname{tr}\rho_1(g) \ne 0$ is the same as the fraction of elements in $h \in H_1$ for which $\operatorname{tr}h \ne 0$.

We say a subgroup $H_0$ of a group $H$ is $Z(H)$-free if $H_0 \ne 1$ and $H_0 \cap Z(H) = 1$. The above lemma implies that if $G_1$ has no $Z(G_1)$-free normal subgroups, then $\delta(\rho_1, \rho_2) \ge \frac {\alpha_n(H_1)} 2$ or $N_2 = 1$ (as the $K$ in the proof must be nontrivial). This will often allow us to reduce to the case where $N_2 = 1$, and similarly $N_1 = 1$, i.e., $G = G_1 = G_2$, when we can check this property for $G_1$ and $G_2$. The following allows us to simply check it for $H_1$ and $H_2$.

If $H_1$ has no $Z(H_1)$-free normal subgroups, then $G_1$ has no $Z(G_1)$-free normal subgroups.

Suppose $H_1$ has no $Z(H_1)$-free normal subgroups, but that $N$ is a $Z(G_1)$-free normal subgroup of $G_1$. Let $N' = \{ n \in H_1 : 