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 $G$?



Which group extensions are normal?
As a set, $G' = Q' \times \mathbb{Z}$, with the group operation given by $(p,x)(q,y) = (pq, x + y + \zeta'(p,q))$, and $G'' = Q' \times n\mathbb{Z}$ (again, as a set) with group operation $(p,na)(q,nb) = (pq,n(a + b + \zeta''(p,q)) = (pq,na + nb + \zeta'(p,q))$. 


Now, let $p \in Q'$, consider the element $(p,na) \in G''$, and conjugate it by an arbitrary element of $G$, say $(q,x)$ where $q \in Q$. We have


$(q,x)(p,na)(q,x)^{-1} = (qp,x + na + \zeta(q,p))(q^{-1}, - x - \zeta(q,q^{-1}))$


The first coordinate is then $qpq^{-1}$ which is in $Q'$ since $Q'$ is normal in $Q$ and $p \in Q'$. The second coordinate is


$x + na + \zeta(q,p) - x - \zeta(q,q^{-1}) + \zeta(qp,q^{-1})$


So the condition you need is exactly that for any $p \in Q'$ and any $q \in Q$, $\zeta(q,p) - \zeta(q,q^{-1}) + \zeta(qp,q^{-1})$ is divisible by $n$. I'm pretty sure this condition is necessary and sufficient (even if it's not exactly a condition on $\zeta''$, as requested).


