Consider a central extension $$1 \longrightarrow \mathbb{Z} \longrightarrow G \longrightarrow Q \longrightarrow 1$$ with Euler class $\zeta \in H^2(Q;\mathbb{Z})$. Let $Q'$ be a normal subgroup of $Q$ and let $\zeta' \in H^2(Q';\mathbb{Z})$ be the restriction of $\zeta$. Finally, consider some $\zeta'' \in H^2(Q';\mathbb{Z})$ such that there exists some $n \geq 1$ with $n \cdot \zeta'' = \zeta'$. Corresponding to $\zeta''$, there is a central extension $$1 \longrightarrow \mathbb{Z} \longrightarrow G'' \longrightarrow Q' \longrightarrow 1$$ which fits into a commutative diagram $$\begin{array}{ccccccccc} 1 & \longrightarrow & \mathbb{Z} & \longrightarrow & G'' & \longrightarrow & Q' & \longrightarrow & 1\\ & & \downarrow \times n & & \downarrow & & \downarrow & & \\ 1 & \longrightarrow & \mathbb{Z} & \longrightarrow & G & \longrightarrow & Q & \longrightarrow & 1 \end{array}$$ The group $G''$ is thus a subgroup of $G$.

Question : What assumptions can I place on $\zeta''$ which would ensure that $G''$ is a normal subgroup of $G$?

assumingthat there exists a $\zeta''$ such that $n \zeta'' = \zeta'$, right? Given $\zeta'$, such $\zeta''$ need not always exist... $\endgroup$ – Joshua Grochow Dec 3 '15 at 4:10