Second Cohomology and Abelian Kernel Extensions

If $f$ and $f^{\mathrm{\prime }}$ satisfy the cocycle and normalization equations of [F1], then $f+f^{\mathrm{\prime }}$ does too, because each equation is linear in the values of the function. The zero function also satisfies those equations, and so does $-f$. Hence $Z^2(G,M)$ is an abelian group under pointwise addition.

If $u$ and $v$ are normalized one-cochains, then $\delta (u+v)=\delta u+\delta v$ by the formula in [F1], and $\delta 0=0$. So $B^2(G,M)$ is a subgroup of the abelian group from step 1.1.

Therefore $Z^2(G,M)$ is an abelian group and $B^2(G,M)\le Z^2(G,M)$.

Let $f\in C^2(G,M)$ satisfy $d^{2}f=0$, and put $a:=f(1,1)$. Substituting $g=1$ into the cocycle equation gives $f(1,hk)=f(1,h)$ for all $h,k$, so $f(1,g)=a$ for every $g$. Substituting $k=1$ gives $g\cdot f(h,1)=f(gh,1)$ for all $g,h$, and then taking $h=1$ yields $f(g,1)=g\cdot a$.

By [F1], a normalized two-cocycle is exactly a normalized function satisfying the displayed cocycle equation, so every normalized two-cocycle lies in $\ker d^2$.

For an arbitrary one-cochain $u:G\to M$, one has $(d^{1}u)(1,g)=u(1),(d^{1}u)(g,1)=g\cdot u(1)\mathrm{.}$ Hence $d^{1}u$ is normalized if and only if $u(1)=0$, that is, if and only if $u$ is a normalized one-cochain. So the normalized elements of $\mathrm{im}{d}^1$ are exactly $B^2(G,M)$ from [F1].

Let $u:G\to M$ be the constant one-cochain $u(g)=a$. Then $(d^{1}u)(1,g)=a,(d^{1}u)(g,1)=g\cdot a\mathrm{.}$ So the cohomologous two-cochain $f_0:=f-d^{1}u$ satisfies $f_0(1,g)=f_0(g,1)=0$ for every $g$. A direct cancellation shows $d^2(d^{1}u)=0$, hence $d^2{f}_0=0$. Therefore every class in $\ker d^2\mathrm{/}\mathrm{im}{d}^1$ has a normalized representative.

Steps 2.1, 1.2, and 1.3 show that every class in $\ker d^2\mathrm{/}\mathrm{im}{d}^1$ has a normalized representative, and two normalized cocycles represent the same class there exactly when they differ by a normalized two-coboundary. Thus the quotient $\ker d^2\mathrm{/}\mathrm{im}{d}^1$ is naturally the same as $Z^2(G,M)\mathrm{/}{B}^2(G,M)$.

Combining step 3.1 with [F2] gives $H^2(G,M)=Z^2(G,M)\mathrm{/}{B}^2(G,M)=\ker d^2\mathrm{/}\mathrm{im}{d}^1\mathrm{.}$

Because $s(1)=1$, [F2] gives $i(f_s(1,g))=s(1)s(g)s(g{)}^{-1}=1$ and likewise $i(f_s(g,1))=1$. The kernel map $i$ is injective, so $f_s(1,g)=f_s(g,1)=0$.

Compute $(s(g)s(h))s(k)$ and $s(g)(s(h)s(k))$ using [F2]. The left-associated expansion is $i(f_s(g,h))i(f_s(gh,k))s(ghk),$ while the right-associated expansion is $s(g)i(f_s(h,k))s(g{)}^{-1}i(f_s(g,hk))s(ghk)\mathrm{.}$ Translating the conjugation term by the given $G$-action yields $g\cdot f_s(h,k)-f_s(gh,k)+f_s(g,hk)-f_s(g,h)=0.$

Steps 1.1 and 1.2 are exactly the conditions of [F1], so $f_s\in Z^2(G,M)$.

Since $\pi (s^{\mathrm{\prime }}(g)s(g{)}^{-1})=1$, each quotient $s^{\mathrm{\prime }}(g)s(g{)}^{-1}$ lies in the kernel. So there is a unique $u(g)\in M$ with $s^{\mathrm{\prime }}(g)=i(u(g))s(g)$. The normalization $s^{\mathrm{\prime }}(1)=s(1)=1$ gives $u(1)=0$.

Substitute $s^{\mathrm{\prime }}(g)=i(u(g))s(g)$ into the factor-set formula of [F2]. After moving kernel terms past lifts by the prescribed action, the result is $f_{s^{\mathrm{\prime }}}(g,h)=u(g)+g\cdot u(h)-u(gh)+f_s(g,h)=f_s(g,h)+(\delta u)(g,h)\mathrm{.}$

Thus the two factor sets differ by a coboundary; [L1] shows that both are indeed cocycles.

Because $s(1)=1$, the factor-set formula gives $f_s(1,g)=f_s(g,1)=0$ for every $g\in G$.

Comparing $(s(g)s(h))s(k)$ with $s(g)(s(h)s(k))$ and translating the conjugation term through the prescribed $G$-action yields $g\cdot f_s(h,k)-f_s(gh,k)+f_s(g,hk)-f_s(g,h)=0.$ So $f_s$ is a normalized two-cocycle.

Steps 1.1 and 2.1 show that $f_s\in Z^2(G,M)$, so [F1] defines a class $[f_s]\in H^2(G,M)$.

If $s^{\mathrm{\prime }}$ is another normalized section, then [L1] gives $f_{s^{\mathrm{\prime }}}=f_s+\delta u$ for some one-cochain $u$. Thus $f_s$ and $f_{s^{\mathrm{\prime }}}$ define the same coset in the quotient [F1].

Therefore the extension determines a well-defined class in $H^2(G,M)$ independent of the chosen normalized section.

Assume $f$ is a normalized two-cocycle. Using the normalization equations from [F1], one checks directly from [F2] that $(0,1)$ is a two-sided identity. The inverse of $(m,g)$ is $(-g^{-1}\cdot m-g^{-1}\cdot f(g,g^{-1}),g^{-1})\mathrm{.}$

Still under the cocycle hypothesis, compute both products $((m,g)(n,h))(r,k)$ and $(m,g)((n,h)(r,k))$ from [F2]. Their second coordinates are both $ghk$, and equality of the first coordinates is exactly the cocycle equation from [F1]. So the law is associative.

Conversely, suppose the twisted law makes $M\times G$ a group with identity $(0,1)$. Comparing $(m,g)(0,1)$ and $(0,1)(m,g)$ with $(m,g)$ forces $f(g,1)=f(1,g)=0$. Comparing $((0,g)(0,h))(0,k)$ and $(0,g)((0,h)(0,k))$ then yields the cocycle equation. Hence $f$ is a normalized two-cocycle.

Step 1.2 proves the forward implication, and step 1.3 proves the reverse implication.

Suppose $f^{\mathrm{\prime }}=f+\delta u$. Define $\mathrm{\Phi }:M{\times }_{f}G\to M{\times }_{f^{\mathrm{\prime }}}G$ by $\mathrm{\Phi }(m,g)=(m-u(g),g)$. Using the product formulas from [F1], the identity $f^{\mathrm{\prime }}=f+\delta u$, and [L1], one checks directly that $\mathrm{\Phi }((m,g)(n,h))=\mathrm{\Phi }(m,g)\mathrm{\Phi }(n,h)\mathrm{.}$ The inverse is $(m,g)\mapsto (m+u(g),g)$, so $\mathrm{\Phi }$ is a group isomorphism. It fixes the kernel and quotient maps, so [F2] makes the two extensions equivalent.

Conversely, suppose $\mathrm{\Phi }:M{\times }_{f}G\to M{\times }_{f^{\mathrm{\prime }}}G$ is an extension equivalence. Because [F2] fixes quotient and kernel, $\mathrm{\Phi }$ has the form $\mathrm{\Phi }(m,g)=(m+u(g),g)$ for a unique normalized one-cochain $u$. Comparing the image of $(0,g)(0,h)$ under $\mathrm{\Phi }$ with the product of the images gives $f^{\mathrm{\prime }}(g,h)=f(g,h)-g\cdot u(h)+u(gh)-u(g)=f(g,h)+(\delta (-u))(g,h)\mathrm{.}$ So $f^{\mathrm{\prime }}-f$ is a coboundary.

Step 1.1 proves the forward implication and step 1.2 proves the reverse implication.

For each extension under consideration, apply [A1] to the nonempty fibres of its quotient map over $g\mathrm{\ne }1$, and set $s(1)=1$. This gives a normalized section. Map an extension class to the cohomology class of the factor set of any normalized section. This is well defined by [L1].

Every class $[f]\in H^2(G,M)$ is hit: choose a normalized cocycle representative $f$, build the twisted product $M{\times }_{f}G$, and use its standard section $g\mapsto (0,g)$. The factor set of that section is exactly $f$, so step 1.1 sends the resulting extension to $[f]$.

If an extension $1\to M\stackrel{i}{\to }E\stackrel{\pi }{\to }G\to 1$ has normalized section $s$ with factor set $f_s$, then $\mathrm{\Psi }:M{\times }_{f_s}G\to E,\mathrm{\Psi }(m,g)=i(m)s(g),$ is an extension equivalence. Its inverse sends $x\in E$ to $(i^{-1}(xs(\pi (x){)}^{-1}),\pi (x)),$ and the factor-set identity shows that $\mathrm{\Psi }$ respects multiplication.

If two extensions define the same cohomology class, then after choosing normalized sections their factor sets are cohomologous. By [L2], the corresponding twisted products are equivalent extensions. Composing those equivalences with the ones from step 2.2 shows that the original extensions are equivalent. Thus the map of step 1.1 is injective.

Steps 2.1 and 3.1 show that step 1.1 is a bijection.

The semidirect product $M⋊G$ is represented by the zero cocycle, so its class in $H^2(G,M)$ is $0$. Therefore any extension equivalent to $M⋊G$ has class $0$ by [L1] under [A1].

Conversely, if the class of $E$ is $0$, then [L1] under [A1] says that $E$ is equivalent to the extension attached to the zero cocycle, namely the semidirect product $M⋊G$. By [L2], that extension splits.

A split extension is equivalent to a semidirect product by [L2], so steps 1.1 and 1.2 prove all claimed equivalences.

An equivalence of extensions induces an isomorphism of the corresponding pullbacks over $G$, because the pullback is defined by the universal condition that the two quotient maps agree.

Pushing out along the fixed homomorphism $M\oplus M\to M$ respects those pullback isomorphisms. Hence equivalent input extensions produce equivalent pushout extensions.

Therefore the Baer sum depends only on extension classes.

Under [A1], choose cocycle representatives $f_1$ and $f_2$ for the two classes via [L1]. The corresponding twisted-product extensions have a pullback whose kernel is $M\oplus M$, and pushing out along addition sends the pair $(f_1,f_2)$ to the cocycle $f_1+f_2$.

Therefore the extension class of the Baer sum corresponds to the cohomology class $[f_1+f_2]=[f_1]+[f_2]$ in $H^2(G,M)$. By [L2], this description does not depend on the chosen cocycle representatives.

So the classification bijection is an additive identification.

With trivial action, the condition defining an extension inducing the prescribed action says that every lift of every $g\in G$ centralizes the kernel $A$. That is exactly the statement that the kernel is central in $E$.

Under [A1], [L1] applies with $M=A$ and identifies $H^2(G,A)$ with the equivalence classes of central extensions. The split class singled out by [L2] is the direct-product class because the action is trivial.

Hence central extensions are classified by $H^2(G,A)$.

The first three terms are exactly the degree-one inflation-restriction sequence, so they are exact by [L1], with maps described concretely by [F1].

Under [A2] and [L2], a class in $H^2(Q,A^N)$ is an extension of $Q$ by $A^N$. The map to $H^2(G,A)$ first pulls that extension back along $G\to Q$, producing an extension of $G$ by $A^N$, and then pushes out along the inclusion $A^N\hookrightarrow A$. That is the extension-theoretic meaning of the last inflation map in the displayed sequence.

Exactness of [A1] at $H^1(N,A{)}^Q$ says $\ker (\mathrm{Tra})=\mathrm{im}(\mathrm{Res})\mathrm{.}$ Thus $\mathrm{Tra}[u]=0$ exactly when $[u]$ is the restriction of a degree-one class on $G$, which is precisely the asserted extension criterion for the cohomology class.

Steps 1.2 and 1.3 give the two claimed interpretations, while step 1.1 identifies the preceding degree-one maps.

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.

The function $f$ is not normalized, since $f(1,1)=1\mathrm{\ne }0$. It already fails the normalization part of [F1].

Even ignoring normalization, the cocycle equation at $(t,t,t)$ would read $1-1+1-1=0$, but the normalization failure from step 1.1 already prevents $f$ from being a factor set.

Therefore the statement is false.

Let $s(1)=(0,1)$ and $s(t)=(0,t)$. Its factor set is $0$. Let $s^{\mathrm{\prime }}(1)=(0,1)$ and $s^{\mathrm{\prime }}(t)=(1,t)$. Then $s^{\mathrm{\prime }}(t{)}^2=(2,1)$, so the new factor set satisfies $f_{s^{\mathrm{\prime }}}(t,t)=2$.

Step 1.1 shows $f_{s^{\mathrm{\prime }}}\mathrm{\ne }{f}_s$ as functions. This is consistent with [L1], which says only that the two factor sets differ by a coboundary and therefore define the same cohomology class.

Hence the statement is false.

For a nonabelian group $N$, the coefficient $N$ is not an abelian group, hence is not an abelian $G$-module of the type required by [L1]. Consequently the ordinary group $H^2(G,N)$ asserted in the false claim is not defined by the cited construction.

The classification in [L2] concerns extensions with a fixed abelian kernel and fixed action, so it cannot supply a classification for arbitrary nonabelian $N$. The proposed classification by the ordinary group $H^2(G,N)$ is therefore ill-typed and false as stated.

This refutes the false claim without invoking any unproved nonabelian obstruction statement.