Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 151 (Group Cohomology)

A group extension of $G$ by the $\mathbb{Z}G$-module $M$ is an exact sequencewhose conjugation action on $i(M)$ agrees with the prescribed action of $G$ on $M$. It is a split group extension when $\pi$ has a group-homomorphic section $s:G\to E$. Two such extensions are equivalent group extensions when an isomorphism of their middle groups is the identity on $M$ and induces the identity on $G$. Transporting a section through that isomorphism proves that every extension equivalent to a split extension is split.

Choose a set-theoretic section $s$ with $s(1)=1$. Its failure to preserve multiplication is the normalized two-cocycleAssociativity gives the two-cocycle identity, and replacing $s$ changes $c$ by a group coboundary. The resulting class $[c]\in H^2(G,M)$ is therefore intrinsic to the extension, as expressed by second group cohomology classifies group extensions.

Now write $G=⟨x,y⟩\cong {\mathbb{Z}}^2$ and let $I$ be the augmentation ideal of $\mathbb{Z}G$. The Koszul resolution for a rank-two free abelian group gives, after applying ${\mathrm{Hom}}_{\mathbb{Z}G}(-,M)$, the last coboundaryIts image is $IM$. For $M=\mathbb{Z}G/I^2$ this yields the second cohomology of a rank-two free abelian group with truncated group-ring coefficients calculationThe canonical map $M\to \mathbb{Z}$ induces the identity on these final quotients, so $H^2(G,M)\to H^2(G,\mathbb{Z})$ is surjective; indeed it is an isomorphism.

Let $F_2=⟨x,y⟩$ and let ${\gamma }_i(F_2)$ be its lower central series. The class-two quotient $F_2/{\gamma }_3(F_2)$ is the Integer Heisenberg group. In the class-three free nilpotent group $N=F_2/{\gamma }_4(F_2)$, the module ${\gamma }_2(F_2)/{\gamma }_4(F_2)$ is cyclic over $\mathbb{Z}G$ on $[x,y]$ and is isomorphic to $\mathbb{Z}G/I^2$. Quotienting it by $I/I^2$ gives the central kernel $⟨[x,y]⟩\cong \mathbb{Z}$ of the Heisenberg group. The kernel ofis ${\gamma }_3(F_2)/{\gamma }_4(F_2)$, freely generated by $[[x,y],x]$ and $[[x,y],y]$, and is central. Thus it is ${\mathbb{Z}}^2$. This is the central nonsplit extension of the integer Heisenberg group by ${\mathbb{Z}}^2$. If it split, centrality would give $N\cong H(\mathbb{Z})\times {\mathbb{Z}}^2$, whose abelianization has rank four; but $N$ has abelianization ${\mathbb{Z}}^2$. Hence the extension is nonsplit.

For the free presentation $G=F/R$, conjugation in $F$ gives the relation modulewhere $f\in F$ is any lift of $g\in G$. A different lift differs by an element of $R$, whose inner conjugation acts trivially on the abelianization, so this is a well-defined $\mathbb{Z}G$-module action.

Choose free generators $x_1,\dots ,x_n$ of $F$. The presentation relation sequence becomeswhere $d_1(e_i)={\bar{x}}_i-1$. In Fox calculus, the first map isThe Fox identity $r-1={\sum }_i(\partial r/\partial x_i)(x_i-1)$ gives $d_{1}j=0$, and the standard lifting argument in the free group proves exactness.

Split the sequence at the augmentation ideal $I_G$. Applying ${\mathrm{Hom}}_{\mathbb{Z}G}(-,M)$ toand using projectivity of $(\mathbb{Z}G{)}^n$ givesThe other short exact sequence $0\to I_G\to \mathbb{Z}G\to \mathbb{Z}\to 0$ identifies the last group with $H^2(G,M)$. A one-cocycle on $F$ is a derivation and is determined freely by its values on $x_1,\dots ,x_n$, so $H^1(F,M)$ is $M^n$ modulo principal derivations. Principal derivations vanish on $R$, and restriction sends a derivation $d$ to the $G$-map $r[R,R]\mapsto d(r)$. The preceding cokernel sequence therefore descends to the Mac Lane exact sequence for a free presentation

The left map need not be injective. Take $F=⟨t⟩\cong \mathbb{Z}$, $R=⟨t^m⟩$, $G\cong C_m$, and the trivial module $M=\mathbb{Z}/m\mathbb{Z}$, where $m>1$. Then $H^1(F,M)\cong M$, while restriction sends the derivation determined by $d(t)=a$ toThus the left map is zero although its domain is nonzero, and $R$ is nontrivial.

The Artin–Wedderburn theorem says that every central simple algebra $A$ over $k$ has the form $M_r(D)$ for a finite-dimensional central division algebra $D$, uniquely up to the evident data. If $A$ and $B$ are central simple, extend scalars to an algebraic closure $\bar{k}$. Both become full matrix algebras, hencefor suitable $a,b$. Any nonzero proper ideal of $A{\otimes }_{k}B$ would extend to one in this simple matrix algebra, and faithful flatness prevents it from vanishing or becoming the whole algebra. The same scalar-extension argument shows that the center is $k$. This proves the tensor product of central simple algebras theorem.

The Brauer group $\mathrm{Br}(k)$ consists of Morita equivalence classes of central simple $k$-algebras. Its product is $[A][B]=[A{\otimes }_{k}B]$, its identity is $[k]$, and $[A{]}^{-1}=[A^{\mathrm{op}}]$ because $A{\otimes }_k{A}^{\mathrm{op}}\cong {\mathrm{End}}_k(A)$ is a full matrix algebra.

Let $L/k$ be a Finite Galois extension with Galois group $G_L$, and let $\varphi :G_L\times G_L\to L^{\times }$ be a normalized two-cocycle. The crossed-product algebra of a Galois extension has underlying left $L$-vector spaceand multiplicationThe cocycle identity is exactly associativity. After scalar extension to $L$, the algebra acts by the twisted regular representation and becomes $M_{|G_L|}(L)$; Galois descent shows that it is central simple over $k$. If $\varphi$ is multiplied by the coboundary of a one-cochain $b_{\sigma }$, rescaling $u_{\sigma }$ by $b_{\sigma }$ gives an isomorphic algebra. Hence the cohomological construction of a Brauer class gives a well-defined map

It remains to show that every Brauer class is torsion. For a finite group $G$, restriction and corestriction on normalized bar cochains satisfyon cohomology: the first equality follows by summing the translated cochain over coset representatives, and each $g^{\ast }$ is the identity because an inner automorphism is cochain-homotopic to the identity. Restriction to the trivial subgroup is zero in positive degree, so this proves that finite-group cohomology is annihilated by the group order. In multiplicative notation, every $\alpha \in H^2(G_L,L^{\times })$ therefore satisfies ${\alpha }^{|G_L|}=1$.

By the permitted assumption, $[A]$ is the image of such an $\alpha$ for some $L$. Consequently $[A{]}^{|G_L|}=1$ in $\mathrm{Br}(k)$. By the definition of Brauer equivalence, this says that for some $n$,as required.

For $1\to H\to G\to Q\to 1$, filter a bar resolution of $G$ by the number of quotient variables, or equivalently use the double complex obtained from projective resolutions over $\mathbb{Z}H$ and $\mathbb{Z}Q$. Taking cohomology first in the $H$-direction and then in the $Q$-direction produces the Lyndon–Hochschild–Serre spectral sequenceThe quotient acts on $H^q(H,\mathbb{Z})$ through conjugation. Differentials have bidegree $(r,1-r)$; after determining the $E_2$-page, one follows these differentials and then resolves the filtration extensions on each total degree.

For the dihedral group of order ten, write $G=D_{10}=C_5⋊C_2$, with $H=C_5$ and $Q=C_2$. The integral cohomology of a finite cyclic group isIf $u$ generates $H^2(C_5,\mathbb{Z})$, inversion acts by $u\mapsto -u$, and hence by $(-1{)}^r$ on $u^r\in H^{2r}(C_5,\mathbb{Z})$. Since multiplication by $2$ is invertible on $\mathbb{Z}/5$, every positive-degree cohomology group of $C_2$ with coefficients in this module vanishes. Its invariants are $\mathbb{Z}/5$ when $q$ is divisible by four and zero when $q\equiv 2(\mathrm{mod}4)$.

Thus the only nonzero terms aretogether with the bottom rowDegree considerations leave no possible nonzero differential, so the spectral sequence collapses. In total degrees divisible by four, the $\mathbb{Z}/2$ and $\mathbb{Z}/5$ filtration factors combine uniquely as $\mathbb{Z}/10$ because their orders are coprime. Therefore the integral cohomology of the dihedral group of order ten is