$H$ being perfect means $H$ equals its derived group, i.e., $H_1(H,{\mathbb Z}) = 0$. In particular, non-abelian simple groups are perfect. By , central extensions of perfect groups are simpler to study. In fact a perfect group $H$ possesses a universal central extension by $H_2(H,{\mathbb Z})$.

Consider a commuting diagram of sequences as in with $H'=H$. Suppose $\pi = \pi'$, which forces $\phi \in {\mathrm{Aut}}_Z(H)$ and $\iota'(A) = \ker \pi =  \iota(A)$. Fixing $\pi$ and $\iota$, there are $|{\mathrm{Aut}}(Z)|$ choices for $\iota'$, which gives $|{\mathrm{Aut}}(Z)|$ elements of $M(G,A)$. Each different $\iota'$ must induce a different central automorphism $\phi \in {\mathrm{Aut}}_Z(H)$. Thus at most $|{\mathrm{Aut}}_Z(H)|$ of these $|{\mathrm{Aut}}(Z)|$ bottom sequences can lie in the same equivalence class, which proves the first statement.

Adney and Yen [@adney-yen] showed $|{\mathrm{Aut}}_Z(H)| = |\mathrm{Hom}(H,Z)|$ when $H$ has no abelian direct factor. Consequently, ${\mathrm{Aut}}_Z(H) = 1$ when $H$ is perfect.

Reduction to minimal lifts {#sec:reduction}
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Let $G$ be a finite group and $\rho_1, \rho_2$ be two inequivalent irreducible representations of $G$ into $\operatorname{GL}_n({\mathbb C})$. Let $N_i = \ker \rho_i$ and $G_i = \rho_i(G)$ for $i = 1, 2$. We want to reduce the problem of finding lower bounds for $\delta(\rho_1,
\rho_2)$ to the case where 