Nonabelian extension obstruction and torsor under choice

Inner automorphisms fix $Z$ pointwise. Consequently any representative of $\alpha (p)$ acts on $Z$ in the same way; because $\alpha$ is a homomorphism, these restrictions give a genuine $Q$-module structure on $Z$. By Choice select automorphisms $t_p\in \mathrm{Aut}(N)$ representing $\alpha (p)$, with $t_1=\mathrm{i}\mathrm{d}$. Choose $f(p,q)\in N$, normalized by $f(1,q)=f(p,1)=1$, such that $t_p{t}_q=\mathrm{Inn}(f(p,q))t_{pq}\mathrm{.}$ This is possible because the two automorphisms have the same outer class. For triples, the two ways of composing $t_p{t}_q{t}_r$ show that $z(p,q,r):=t_p(f(q,r))f(p,qr)(f(p,q)f(pq,r){)}^{-1}\mathrm{.}$ belongs to $Z$. It is normalized whenever an argument is $1$.

Here is the cocycle calculation without suppressing the noncentral factors. Set $T=t_p(t_q(f(r,s))),t_p(f(q,rs)),f(p,qrs)$. Applying the triple relation first to $(q,r,s)$, then to $(p,qr,s)$, then to $(p,q,r)$ gives $T=(p\cdot z(q,r,s))z(p,qr,s)z(p,q,r),f(p,q)f(pq,r)f(pqr,s)\mathrm{.}$ Applying it instead to $(p,q,rs)$ and $(pq,r,s)$, using $t_p{t}_q=\mathrm{Inn}(f(p,q))t_{pq}$, gives $T=z(p,q,rs)z(pq,r,s),f(p,q)f(pq,r)f(pqr,s)\mathrm{.}$ Cancel the identical ordered product of the three noncentral $f$ factors on the right. The remaining central factors commute, yielding $p\cdot z(q,r,s);z(pq,r,s{)}^{-1};z(p,qr,s);z(p,q,rs{)}^{-1};z(p,q,r)=1.$ This is exactly $dz=1$ for the inhomogeneous degree-three differential, so $z$ defines a class in $H^3(Q,Z)$.

If $f^{\mathrm{\prime }}(p,q)=c(p,q)f(p,q)$ for a normalized map $c:Q^2\to Z$ with the same representatives $t$, direct substitution gives $z^{\mathrm{\prime }}=zdc$. If representatives change to $t_p^{\mathrm{\prime }}=\mathrm{Inn}(b_p)t_p$ with $b_1=1$, use $f^{\mathrm{\prime }}(p,q)=b_p\cdot t_p(b_q)\cdot f(p,q)\cdot b_{pq}^{-1}\mathrm{.}$ Substitution cancels the $b$ factors and gives the same $z$; any other factor choice for $t^{\mathrm{\prime }}$ differs by a central $c$ and hence changes $z$ by a coboundary. Thus $o(\alpha ):=[z]$ is independent of every choice.

If an extension inducing $\alpha$ exists, Choice supplies a normalized section $s:Q\to E$. Conjugation by $s(p)$ yields representatives $t_p$, and $s(p)s(q)=f(p,q)s(pq)$ gives factors $f(p,q)\in N$. Associativity in $E$ says $t_p(f(q,r))f(p,qr)=f(p,q)f(pq,r)$, so $z=1$ and $o(\alpha )=0$.

Conversely, if $o(\alpha )=0$, a normalized central two-cochain $c$ can be chosen with $zdc=1$; replace $f$ by $cf$. Here $c$ is normalized because $H^3$ was defined from normalized cochains. On the set $N\times Q$ define $(n,p)(m,q)=(n\cdot t_p(m)\cdot f(p,q),pq)\mathrm{.}$ The identity is $(1,1)$. The conjugation identity of step 1.1 and the associativity equation imposed at the start of this step make this multiplication associative. Left and right translations have inverses because each $t_p$ is an automorphism and $Q$ is a group; explicitly, the unique right inverse of $(n,p)$ has second coordinate $p^{-1}$ and first coordinate $t_p^{-1}(n^{-1}f(p,p^{-1}{)}^{-1})$, and associativity makes it a two-sided inverse. Thus $N\times Q$ is a group, and projection to $Q$ is an extension whose conjugation action has outer class $\alpha$.

Fix one associative normalized factor system $(t,f_0)$ from step 4.2. Any other extension realizing $\alpha$ has a normalized section by Choice. Its representatives differ from $t_p$ by inner automorphisms; using Choice again, adjust the section by elements of $N$ to make its representatives exactly $t_p$. Its factors then have the form $f=c{f}_0$ for a unique normalized map $c:Q^2\to Z$, because both factor systems implement $t_p{t}_q$. The associativity equation for $f$ reduces precisely to $dc=1$. Conversely every normalized $Z$-valued two-cocycle $c$ gives an associative factor system $c{f}_0$ and hence an extension by step 4.2.

Two such extensions with the same $t$ are equivalent while fixing $N$ and $Q$ exactly when their factors differ by a central coboundary. Indeed, an equivalence sends the standard section $(1,p)$ to $(b_p,p)$; compatibility with conjugation by that section forces $b_p\in Z$, and compatibility with products gives $f^{\mathrm{\prime }}(p,q)=f(p,q)b_{pq}[b_p(p\cdot b_q){]}^{-1}$, namely $f^{\mathrm{\prime }}\mathrm{/}f=(db{)}^{-1}$. Conversely the map $(n,p)\mapsto (n{b}_p,p)$ is an equivalence whenever this relation holds. Therefore the difference between two extension classes is a unique element of $H^2(Q,Z)$, and multiplication of factors by representatives gives a free transitive $H^2(Q,Z)$-action. The initial choice of $f_0$ identifies the torsor with the group, but no preferred origin is implied.