Z}/n_i {\mathbb Z}, A) = \bigoplus A/n_i A$, so $\mathrm{Ext}(H_1(G,{\mathbb Z}), C_m) = \mathrm{Ext}(H_1(G,{\mathbb Z}), C_{m'})$. An analogous statement is true for $\mathrm{Hom}(H_2(G,{\mathbb Z}), -)$ so $|H^2(G, C_m)| = |H^2(G, C_{m'})|$.

Assume $m \ne m'$. Consider a sequence as in with $A = C_{m'}$. This gives a sequence $$0 \to C_m \to H \times_{C_{m'}} C_m \to G \to 1$$ in $M(G,C_m)$ by extending $\iota : C_{m'} \to H$ to be the identity on $C_m$. Note if one has an equivalence $\phi$ of two sequences in $M(G, C_m)$ constructed in this way, then commutativity implies $\phi(H) = H$ so restricting the isomorphism $\phi$ on the middle groups to $H$ yields and equivalence of the corresponding sequences in $M(G, C_{m'})$. Hence all elements of ${\tilde}M(G, C_m)$ arise from “central products” of sequences in $M(G, C_{m'})$, and thus no elements of ${\mathrm{Cent}}(G, C_m)$ can be minimal lifts.

When $H_1(G,{\mathbb Z}) \times H_2(G, {\mathbb Z}) \simeq 1$, then $H^2(G,A) = 0$ for any abelian group $A$, which means all central extensions are split, i.e., ${\mathrm{Cent}}(G, A) = \{ G \times A \}$ for any $A$. When $H_1(G,{\mathbb Z}) \times H_2(G, {\mathbb Z}) \simeq {\mathbb Z}/2{\mathbb Z}$, then tells us that $|H^2(G, C_m)|$ has size 