According to Wikipedia and ncat lab general group extensions
$$N\rightarrow G\rightarrow Q$$
are classified by a group homomorphism
$$\rho: Q\rightarrow \operatorname{Out}(N)$$
subject to a constraint that a certain 3-cocycle in $Z^3(Q,Z(N))$ built from $\rho$ is trivial, and a 2-cocycle
$$\omega\in Z^2(Q,Z(N)).$$
Here, the action of $Q$ on the center $Z(N)$,
$$\hat\rho: Q\rightarrow \operatorname{Aut}(Z(N)),$$
is the restriction of $\rho$ to $Z(N)$.
Question: Given $\rho$ and $\omega$, how do I write down the multiplication of $G$ on the set $N\times Q$?
I know that if $N$ is abelian, we have $Z(N)=N$ and $\hat\rho=\rho$, and the formula is given by
$$(n,q)(n',q')=(n+\rho(q)n'+\omega(q,q'), qq'),$$
writing $N$ additively. How does this generalize to the case where $N$ is non-abelian. I'm assuming the $+\omega(q,q')$ part remains unchanged and is simply restricted to the center. But how do I lift $\rho$ from $\operatorname{Out}(N)$ to $\operatorname{Aut}(N)$ for $n'$ not in the center? Does the triviality of the 3-cocycle somehow ensure that this lift does not matter?
Best Answer
My understanding is that this works as follows. Choose a map $\psi\colon Q\to \mathrm{Aut}(N)$ that "lifts" $\rho$. Then $\psi$ won't be a homomorphism so there is a function $f\colon Q\times Q\to N$ where $\psi(q)\psi(q')=\alpha(f(q,q'))\psi(qq')$ where $\alpha\colon N\to \mathrm{Aut}(N)$ has $\alpha(n)$ conjugation by $n$. Then you would like to define the product on $N\times Q$ by $(n,q)(n',q') = (n\psi(q)(n')f(q,q'),qq')$. The condition you need for this to be associative is that $f(q,q')f(qq',q'')=\psi(g)(f(q',q''))f(q,q'q'')$. According to Brown, the failure of $f$ to satisfy this formula is given by the 3-cocycle $Q\times Q\times Q\to Z(N)$ which measures the "difference" between the two sides (which one can check lies in $Z(N)$) and which coincides with the one you are referring to coming from $\rho$.
Now how does $\omega$ fit in? Because the $3$-cocycle coming from $\psi$ and $f$ as above is trivial, it is the coboundary of a $2$-cocycle $Q\times Q\to Z(N)$ and you can basically multiply by this $2$-cocycle (or maybe its inverse), to modify your $f$ to get one that meets the associativity criterion. That is the vanishing of the 3-cocycle means that some $f$ exists satisfying the associativity condition. Then you can further multiply this $f$ satisfying the asociativity condition by $\omega$ on the correct side to get a new mapping $Q\times Q\to N$ that satisfies the associativity condition and that gives your multiplication. So in fancy language, the set of functions $f$ meeting the associativity is an $H^2(Q,Z(N))$-torsor and you need to fix first some extension giving rise to $\rho$ before you can make explicit the bijection with $H^2(Q,Z(N))$. The vanishing of the $3$-cocycle means exactly that you can find one such choice via the procedure I described above. The choice depends on your lift of $\rho$ to $\psi$ but then there is a map $Q\to Z(N)$ that will measure the difference of your choices and your $f$ will change by a coboundary, i.e., by an element of $B^2(Q,Z(N))$.
I refer you to pages 104-106 of Ken Brown's cohomology book for details.