LHS for a split group extension

The displayed operators satisfy $T^2=S^2=TS=ST=0$. In characteristic two this gives $(1+T{)}^2=(1+S{)}^2=1$ and commuting actions, so $M$ is a well-defined $G$-module. The map $s\mapsto s$ is a group section, and its conjugation action on $N$ is trivial by commutativity, as in F3. It need not act trivially on the coefficient module.

For either cyclic factor use the rank-one free integral group-ring resolution with alternating differentials $t-1,1+t,t-1,1+t,\dots$ (replace $t$ by $s$ for $Q$). It is exact: in $\mathbb{Z}[C_2]$, the kernels are respectively $\mathbb{Z}(1+t)$ and $\mathbb{Z}(1-t)$, equal to the preceding images, and the augmentation kernel is $\mathbb{Z}(1-t)$. Hom into a characteristic-two module replaces every differential by $T$, or by $S$. Thus its positive cohomology is $\ker T\mathrm{/}\mathrm{im}T$, or $\ker S\mathrm{/}\mathrm{im}S$. F4 licenses this computation; each rank-one free module is projective by a single generator lift.

Here $W=M^N=⟨b,c,d⟩$, and $H^1(N,M)=⟨[b],[d]⟩$. The quotient action is trivial on these two classes, since $Sb=c$ is a $T$-boundary and $Sd=0$. To see the action agrees with F1's resolution convention, let $Q$ act trivially on the cyclic $N$-resolution and by its given action on $M$; the Hom-to-injective-total comparison in F4 commutes with these actions and its augmentation. On $W$, $Sb=c$ and $Sc=Sd=0$, so $H^1(Q,W)=H^2(Q,W)=⟨[d]⟩$. Consequently the three page entries in F2 have dimensions $1,2,1$.

Tensor the two cyclic resolutions over $\mathbb{Z}$ and take the signed total. This is a free $\mathbb{Z}[G]$-resolution of $\mathbb{Z}$: each bidegree is rank one over that ring. For exactness, each augmented factor, as an abelian complex, splits into its degree-zero copy of $\mathbb{Z}$ and contractible two-term complexes. Indeed its successive boundary groups have the single displayed generator in step 1.2, and each surjection to that generator has the explicit lift $1$ or $-1$; the augmentation also has lift $1$. These splittings decompose the differentials into identity maps on adjacent summands. Tensoring such a contractible summand with the other complex stays contractible: the homotopy $h\otimes 1$ has cross terms cancelling under the tensor sign. There are finitely many summands in each degree. Thus the total has homology $\mathbb{Z}$ in degree zero and zero above it, proving the resolution claim.

Hom of this total into $M$ has degree zero $M$ and degree one $M\oplus M$, with coboundary $w\mapsto (Tw,Sw)$. Its degree-one cycles $(u,v)$ satisfy $Tu=0$, $Sv=0$ and $Su=Tv$; the three equations come from bidegrees $(2,0),(0,2),(1,1)$ and signs disappear over $k$. Here $u\in ⟨b,c,d⟩$ and $v\in ⟨c,d⟩$, so the third equation forces the $b$-coefficient of $u$ to vanish. Cycles are therefore $⟨c,d⟩\oplus ⟨c,d⟩$, of dimension four. Boundaries are generated by $(c,d)$ and $(0,c)$, of dimension two. F4 gives $H^1(G,M)\cong k^2$.

Restriction to $N$ sends $[(u,v)]$ to $[u]\in \ker T\mathrm{/}\mathrm{im}T$. Indeed inclusion of the $N$-resolution at degree zero of the other factor lifts the identity augmentation, so its Hom map is this projection; the canonical comparison of F4 identifies it with F2's restriction. Its image is exactly $⟨[d]⟩$: all allowable $u$ lie in $⟨c,d⟩$, and $(d,0)$ is a cycle. Thus F2 forces the kernel of transgression to be $⟨[d]⟩$ in $⟨[b],[d]⟩$. Its target is the one-dimensional $⟨[d]⟩$ from step 2.1, so $d_2([b])=[d]\mathrm{\ne }0$. The five-term portion is $0\to k\to k^2\to k^2\to k\to H^2(G,M)$, with middle restriction of rank one, transgression of rank one, and the last inflation zero. This proves noncollapse despite the group section.

Restriction to the section subgroup projects a cycle to $[v]\in H^1(Q,M)$. Here $\ker S=\mathrm{im}S=⟨c,d⟩$, so this target is zero. It cannot retract the nonzero injection $H^1(Q,M^N)\hookrightarrow H^1(G,M)$: the coefficient modules in those two quotient-group cohomologies differ. With trivial coefficients $k$ instead, $T=S=0$, the same resolution gives $H^1(G,k)=k^2$ and $H^1(N,k)=H^1(Q,k)=k$. Inflation is $z\mapsto (0,z)$, restriction is $(u,v)\mapsto u$, and restriction to the section is $(u,v)\mapsto v$. To verify the inflation formula, project the tensor resolution onto the $Q$ factor by augmentation of the $N$ factor; its Hom map is the displayed inclusion and lifts the quotient fixed-point map in F2. This is a valid split degree-one sequence and has zero transgression by exactness.

F1 gives finite strong convergence for both coefficient modules. The calculations establish only the stated low-degree portion; other page differentials and the full degree-two target in the first example are not claimed computed. A group section imposes no bidegree vanishing on those uncomputed arrows. All displayed resolutions, bases and linear equations are explicit and require no AC; resolution independence retains the supplied-data/DC convention.