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 $G$?

  • $\begingroup$ Typo: $Q$ -> $Q'$ in "$\zeta'' \in H^2(Q;\mathbb{Z})$." Also, you're assuming that 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
  • $\begingroup$ @JoshuaGrochow: I fixed the typo. I also rephrased things to resolve the ambiguity that led to your confusion: I'm not fixing $n$ (impossibly since the cohomology class might not be divisible by $n$); rather, I'm choosing a cohomology class which is a multiple for some $n$. This clearly exists (e.g. for $n=1$). $\endgroup$ – Edward Cooper Dec 3 '15 at 4:26

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).

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  • $\begingroup$ As you saw from your many edits, it's a subtle question! This is probably as good an answer as I'm going to get, so I'll accept it. Thanks! $\endgroup$ – Edward Cooper Dec 4 '15 at 3:15
  • $\begingroup$ @EdwardCooper: Indeed :). I think there may be some fiddling one can do by trying to relate the final condition to the 2-cocycle condition, which might lead to a "cleaner" condition. For example, by playing a bit, you can get that it's equivalent to have $\zeta(q,p) + \zeta(qp^{-1} q^{-1}, qp)$ divisible by $n$ for all $q \in Q, p \in Q'$. This might seem a little "more plausible", since in both of the terms here, at least one of the arguments is in $Q'$... (using the cocycle condition with $a_0=qp^{-1}q^{-1}$, $a_1=qp$, $a_2=q^{-1}$) $\endgroup$ – Joshua Grochow Dec 4 '15 at 4:34
  • $\begingroup$ Ah, in fact, here's a slightly cleaner sufficient condition: $\zeta(q_1,q_2)$ should be divisible by $n$ whenever at least one of $q_1,q_2$ is in $Q'$. $\endgroup$ – Joshua Grochow Dec 4 '15 at 4:38
  • $\begingroup$ Also note that I haven't taken into account 2-coboundaries here, so the above should be taken to read that there is some representative of the cohomology class $[\zeta]$ such that $\zeta(q,p) +\zeta(qp^{-1}q^{-1}, qp)$ is divisible by $n$ for all $q \in Q, p \in Q'$. Adding a coboundary to the preceding expression changes it by $b(p)+ b(qp^{-1}q^{-1})$ for some 1-cochain (i.e., function) $b\colon Q \to \mathbb{Z}$. Interestingly, the latter expression only depends on the restriction of $b$ to $Q'$. $\endgroup$ – Joshua Grochow Dec 6 '15 at 4:21

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