1 or 2 according to whether $m$ is odd or even, so there must be a unique nonsplit extension ${\tilde}G \in {\mathrm{Cent}}(G,C_2)$. Then the argument in the proof tells us any cyclic central extension of $G$ is a central product of either $G$ or ${\tilde}G$ with a cyclic group.

However, in general, knowing $H_1(G,{\mathbb Z})$ and $H_2(G, {\mathbb Z})$ is not enough to determine the size of ${\mathrm{Cent}}(G, C_m)$. When $|{\mathrm{Cent}}(G, C_m)| < |H^2(G, C_m)|$, we will sometimes need a way to verify that the central extensions of $G$ by $C_m$ we exhibit exhaust all of ${\mathrm{Cent}}(G, C_m)$. For this, we will use a lower bound on the size of the fibers of $\Phi$, i.e., a lower bound on the number of classes in ${\tilde}M(G, A)$ a given central extension $H \in {\mathrm{Cent}}(G,A)$ appears in.

The central automorphisms of a group $H$ with center $Z$, denoted ${\mathrm{Aut}}_Z(H)$, are the automorphisms $\sigma$ of $H$ which commute with the projection $H \to H/Z$, i.e., satisfy $\sigma(h)h^{-1} \in Z$ for all $h \in H$.

\[prop:32\] Let $A$ be abelian and $H \in {\mathrm{Cent}}(G,A)$ such that $A = Z:= Z(H)$. Then $|\Phi^{-1}(H)| \ge \frac{|{\mathrm{Aut}}(Z)|}{|{\mathrm{Aut}}_Z(H)|}$. Moreover, if $H$ is perfect, then $|\Phi^{-1}(H)| \ge |{\mathrm{Aut}}(Z)|$.

Recall 