Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 151 (Profinite Groups and Group Cohomology)

For an inverse system of sets $(X_j,f_{ij})$ indexed by a directed set $J$, the inverse limit is the set of compatible tuplesIts projection to $X_j$ sends $(x_i{)}_i$ to $x_j$.

Give each nonempty finite set $X_j$ the discrete topology. The product space $X={\prod }_j{X}_j$ is compact by the Tychonoff theorem. For each $i\le j$, the compatibility condition $f_{ij}(x_j)=x_i$ defines a closed subset $C_{ij}\subseteq X$.

These sets have the finite intersection property. Indeed, for finitely many conditions choose an index $k$ above every index occurring in them, choose any $x_k\in X_k$, and use the transition maps from $k$ to define all required coordinates; choose the remaining coordinates arbitrarily. Compactness therefore givesThis is the nonemptiness theorem for inverse limits of finite sets.

The product ${\prod }_j{G}_j$ of the finite groups with their discrete topology is a topological group under coordinatewise multiplication and inversion. The compatibility equations defining $G={\underleftarrow{\lim }}_j{G}_j$ are preserved by both operations, so $G$ is a subgroup. Their restrictions to the subspace topology on $G$ are continuous. Hence $G$ with its standard inverse-limit topology is a topological group.

For fixed $h\in G$, the conjugation map is the compositefollowed by multiplication. Inversion, constant maps, diagonal maps, and multiplication are continuous in a topological group, so $c_h(g)=gh{g}^{-1}$ is continuous.

The profinite group $G$ is compact, and the preceding part shows thatis its image. By the continuous image of a compact space theorem, the conjugacy class is compact. Since a profinite group is Hausdorff, every compact subset is closed, so ${\mathrm{Cl}}_G(h)$ is closed.

If $g=xh{x}^{-1}$ in $G$, applying each projection map gives $p_j(g)=p_j(x)p_j(h)p_j(x{)}^{-1}$ in $G_j$.

Conversely, suppose $p_j(g)$ and $p_j(h)$ are conjugate for every $j$. Define the nonempty finite setEvery transition map $G_j\to G_i$ carries $X_j$ into $X_i$, so the $X_j$ form an inverse system. By the nonemptiness theorem for inverse limits of finite sets, there is a compatible tuple $(x_j)\in {\underleftarrow{\lim }}_j{X}_j\subseteq G$. Coordinatewise equality then gives $xh{x}^{-1}=g$. This proves the finite-quotient criterion for conjugacy in a profinite group.

The canonical image of $\Gamma$ is dense in its profinite completion $\hat{\Gamma }$. Therefore every $x\in \hat{\Gamma }$ is a limit of a net $({\delta }_{\lambda })$ in $\Gamma$. Continuity of conjugation givesso every element of ${\mathrm{Cl}}_{\hat{\Gamma }}(\gamma )$ lies in the closure of ${\mathrm{Cl}}_{\Gamma }(\gamma )$. The reverse inclusion follows because the larger conjugacy class contains the smaller one and is closed by part b(iii). Hence

Suppose first that $\Gamma$ is conjugacy separable. If $h\in \Gamma$ is not conjugate to $\gamma$ in $\Gamma$, some homomorphism to a finite group sends them to nonconjugate elements. This homomorphism factors through a finite quotient of $\hat{\Gamma }$, so $h$ cannot be conjugate to $\gamma$ in $\hat{\Gamma }$. Thus

Conversely, suppose this equality holds and $h$ is not conjugate to $\gamma$ in $\Gamma$. Then they are not conjugate in $\hat{\Gamma }$. By the finite-quotient criterion for conjugacy in a profinite group, their images fail to be conjugate in some finite quotient. This is precisely conjugacy separability.

A subset $S\subseteq G$ topologically generates $G$ exactly whenfor every $j$. In the usual presentation by surjective finite quotients this reads $⟨p_j(S)⟩=G_j$. Indeed, a subgroup is dense exactly when its image in every finite discrete quotient is the whole quotient. This is the finite-quotient criterion for topological generation.

Apply part i toThe element $\alpha$ generates the additive cyclic group $\mathbb{Z}/p^n\mathbb{Z}$ exactly when it is coprime to $p$, which is equivalent to its reduction modulo $p$ being nonzero. Hence $\{\alpha \}$ is a topological generating set of the additive group ${\mathbb{Z}}_p$ if and only if $\alpha \not\equiv 0(\mathrm{mod}p)$.

An element of the p-adic integers is a unit in a ring exactly when its reduction modulo $p$ is nonzero. More explicitly, if $\alpha \not\equiv 0(\mathrm{mod}p)$, its inverses modulo $p^n$ are unique and compatible, so they define $\beta \in {\mathbb{Z}}_p$ with $\alpha \beta =1$. The converse follows by reducing $\alpha \beta =1$ modulo $p$. Part ii therefore proves that $\alpha$ topologically generates the additive group if and only if $\alpha$ is a p-adic unit.

By the Fundamental theorem of finitely generated abelian groups, writewith $F$ finite. Taking profinite completions givesIf $A\cong \mathbb{Z}$, this is plainly $\hat{\mathbb{Z}}$.

Conversely, suppose $\hat{A}\cong \hat{\mathbb{Z}}$. Reduction modulo a prime $p$ giveswhereas $\hat{\mathbb{Z}}/p\hat{\mathbb{Z}}\cong \mathbb{Z}/p\mathbb{Z}$. Choosing $p\nmid |F|$ first shows $r=1$. If $F\ne 0$, choosing a prime divisor $p$ of $|F|$ makes $F/pF\ne 0$, a contradiction. Thus $F=0$ and $A\cong \mathbb{Z}$.

Suppose $\hat{\Gamma }\cong {\mathbb{Z}}_p$. This completion is an abelian group, so every finite quotient of $\Gamma$ is abelian. Consequently the quotient map to the abelianization $A={\Gamma }^{\mathrm{ab}}$ inducesThe group $A$ is a finitely generated abelian group. If its free rank is zero, $\hat{A}=A$ is finite. If its free rank is positive, $A$ and hence $\hat{A}$ have a nontrivial quotient $\mathbb{Z}/q\mathbb{Z}$ for every sufficiently chosen prime $q$. But ${\mathbb{Z}}_p$ has no nontrivial finite quotient of order coprime to $p$. Both cases are impossible, so $\hat{\Gamma }\ncong {\mathbb{Z}}_p$.

Every element of $A_{\neg p}$ can be written $m/n$ with $p\nmid n$. Since $n$ is a p-adic unit, defineThis is the restriction of the standard embedding $\mathbb{Q}\hookrightarrow {\mathbb{Q}}_p$, so it is an injective group homomorphism. Equivalently, if $m{n}^{-1}=0$ in ${\mathbb{Z}}_p$, then $m=0$ because the characteristic is zero.

If $\pi$ omits a prime $p$, then $A_{\pi }\subseteq A_{\neg p}\subseteq {\mathbb{Z}}_p$ by part i. The reductionsseparate its nonzero elements, so their restrictions separate the elements of $A_{\pi }$. Thus $A_{\pi }$ is residually finite. If $\pi$ contains every prime, then $A_{\pi }=\mathbb{Q}$, which has no nontrivial finite quotient because it is a divisible group.

Let $q\ne p$ and let $f:A_{\neg p}\to \mathbb{Z}/q\mathbb{Z}$ be a homomorphism. Since $q$ is inverted in $A_{\neg p}$, every $x$ has the form $x=qy$. ThereforeHence the only such homomorphism is trivial.

Every finite quotient of the abelian group $A_{\neg p}$ is abelian. Part iii excludes elements of prime order $q\ne p$ by the Cauchy theorem for groups, so any finite quotient is a finite abelian $p$-group. If its exponent divides $p^n$, the quotient map kills $p^n{A}_{\neg p}$ and therefore factors throughEvery quotient of a cyclic group is cyclic, so the finite quotient is isomorphic to $\mathbb{Z}/p^k\mathbb{Z}$ for some $k\le n$. Conversely, reduction modulo $p^k$ gives a surjection $A_{\neg p}\to \mathbb{Z}/p^k\mathbb{Z}$. Thus these are exactly the nontrivial finite quotients, together with the trivial case $k=0$.

The quotient maps define a continuous homomorphismIts kernel is ${\bigcap }_{U\in \mathcal{U}}U=\{1\}$, because $\mathcal{U}$ is a neighborhood basis and $G$ is Hausdorff. Thus $\Phi$ is injective. The standard compactness argument for inverse limits makes it surjective: a compatible family of cosets has the finite intersection property, and the corresponding closed cosets in compact $G$ have nonempty intersection. Finally, a continuous bijection from compact $G$ to the Hausdorff inverse limit is a homeomorphism. Henceas topological groups.

Suppose $G$ is topologically generated by $d$ elements. An open subgroup $H$ of index $n$ gives a continuous transitive coset actionA continuous homomorphism $G\to S_n$ is determined by the images of the $d$ topological generators, so there are at most $|S_n{|}^d$ such homomorphisms. Each has only finitely many point stabilizers. Therefore $G$ has only finitely many open subgroups of index $n$.

By part ii, $G_n$ is the intersection of finitely many open subgroups, so it is open. Conjugation permutes the subgroups of each index, hence $G_n$ is also a normal subgroup. The groups $G_n$ form a descending family. If $U$ is any open normal subgroup and $[G:U]=m$, then $G_m\subseteq U$. Thus $(G_n)$ is cofinal among the open normal neighborhoods of the identity. Part i now gives

A topological group has the topological Hopf property when every continuous surjective endomorphism is a topological automorphism.

Let $\varphi :G\to G$ be a continuous surjective endomorphism. For each $n$, inverse image under $\varphi$ permutes the finite set of open subgroups of index at most $n$: surjectivity preserves the index, and injectivity of the inverse-image operation follows from surjectivity. HenceIf $x\in \ker \varphi$, then $x\in G_n$ for every $n$. Part 3(a)(iii) implies ${\bigcap }_n{G}_n=\{1\}$, so $x=1$. Thus $\varphi$ is bijective. A continuous bijection from compact $G$ to Hausdorff $G$ is a homeomorphism, proving the Hopf property of a topologically finitely generated profinite group.

If $\gcd (m,|H|)=1$, choose $r$ with $mr\equiv 1(\mathrm{mod}|H|)$ by the Bezout identity. The Lagrange theorem gives $x^{|H|}=1$ for every $x\in H$, so $(x^m{)}^r=x$ and $(x^r{)}^m=x$. Thus $x\mapsto x^m$ is bijective, even though it need not be a homomorphism.

Conversely, if a prime $p$ divides both $m$ and $|H|$, the Cauchy theorem for groups gives $x\ne 1$ with $x^p=1$. Then $x^m=1$, so the power map sends both $x$ and the identity to the identity and is not injective. This proves the power-map criterion for a finite group.

Necessity follows by applying $p_j$ to an equality $x^m=g$. Conversely, suppose every finite quotient contains an $m$th root of $p_j(g)$ and defineThese are nonempty finite sets, and the transition maps preserve them. The nonemptiness theorem for inverse limits of finite sets supplies a compatible tuple $x\in {\underleftarrow{\lim }}_j{X}_j=G$, for which $x^m=g$. This is the finite-quotient criterion for roots in a profinite group.

If $m$ is coprime to every $|G_j|$, the power-map criterion for a finite group says that every finite-quotient power map is bijective. Part ii gives surjectivity of $f_m:G\to G$. If $x^m=y^m$, then $p_j(x)=p_j(y)$ for every $j$ by injectivity in $G_j$, and hence $x=y$. Thus the continuous power map is bijective.

The inhomogeneous group cochain group iswith $C^0(G,M)=M$. Its group coboundary $d:C^r(G,M)\to C^{r+1}(G,M)$ isOne checks that $d^2=0$, and group cohomology is $H^r(G,M)=\ker d/\mathrm{im}d$.

A crossed homomorphism is a map $\varphi :G\to M$ satisfyingFor a one-cochain, the formula in part i givesso the crossed homomorphisms are exactly the one-cocycles. A zero-cochain $m\in M$ has coboundary $g\mapsto gm-m$, the principal crossed homomorphism associated with $m$. Therefore

For $g,h\in G$, substitute the definition of $\psi$ and in the first sum replace $\gamma$ by $\delta g$:The two-cocycle identity at $(\delta ,g,h)$ says that the expression in parentheses is $\delta \varphi (g,h)$. Every summand is therefore $\varphi (g,h)$, and

Let $\varphi \in Z^2(G,\mathbb{Q})$. Part iii gives $d\psi =|G|\varphi$. Since division by $|G|$ is possible in the rational numbers,is a two-coboundary. HenceThis is a degree-two instance of the vanishing of finite-group cohomology when the group order is invertible on the coefficient module.

For the trivial action, crossed homomorphisms $G\to \mathbb{Q}$ are ordinary group homomorphisms, while principal crossed homomorphisms vanish. The image of a finite group in the torsion-free additive group $\mathbb{Q}$ must be trivial, soDegree-zero cohomology is the invariant submodule; the action is trivial, so

The short exact sequence of $G$-modules induces the long exact sequence in group cohomology

Apply part i to the short exact sequence of trivial $G$-modulesThe relevant segment isBoth outer groups vanish by parts a(iv) and a(v), so the connecting homomorphism is an isomorphism:

The trivial action givesbecause $D_n$ is finite. By part b(ii),In the abelianization the relation $aba=b^{-1}$ becomes $b=b^{-1}$. HenceEach finite cyclic group is naturally isomorphic to its character group in $\mathbb{Q}/\mathbb{Z}$, so

Write every element as $a^k{b}^l$ with $k\in \{0,1\}$. The action of this element on $M=\mathbb{Z}$ is multiplication by $(-1{)}^k$. A direct check of the four possibilities for the two exponents of $a$ showsso $\varphi$ is a crossed homomorphism and hence a one-cocycle.

It cannot be principal: if $\varphi (g)=gm-m$ for some $m\in \mathbb{Z}$, then at $g=a$ one would havewhich is impossible in the integers. Thus $[\varphi ]\ne 0$ and

Choose a normalized set-theoretic section $s:H\to E$ of $\pi$, so $s(1)=1$. Define the extension cocycleAssociativity of $s(h)s(k)s(l)$ gives, in multiplicative notation for $M$,which is exactly the two-cocycle identity. Thus $\varphi$ represents the class of the group extension in $H^2(H,M)$.

The fiber product of groupsis a group under componentwise multiplication. The maps $m\mapsto (m,1)$ and $(e,g)\mapsto g$ give an exact sequenceThe section $g\mapsto (s(f(g)),g)$ has extension cocycleTherefore this pullback extension represents $f^{\ast }([\varphi ])\in H^2(G,M)$.

Assume $f$ is injective. If two elements $(e,g),(e^{\prime },g^{\prime })\in E^{\prime }$ have the same image under the first projection $E^{\prime }\to E$, then $e=e^{\prime }$. Applying $\pi$ gives $f(g)=f(g^{\prime })$, so injectivity of $f$ gives $g=g^{\prime }$. Hence the natural map $E^{\prime }\to E$ is injective.

For every open normal subgroup $U◃G$, the compositehas finite image and therefore factors uniquely through the profinite completion $\hat{\Gamma }$. These factor maps are compatible as $U$ varies. The universal property of an inverse limit consequently produces a continuous homomorphismwith $\hat{f}\iota =f$. It is unique because $\iota (\Gamma )$ is dense in $\hat{\Gamma }$ and two continuous maps into the Hausdorff group $G$ that agree on a dense subset agree everywhere. This is the universal property of profinite completion.

Choose $\hat{\zeta }\in H^2(\hat{\Gamma },M)$ with ${\iota }^{\ast }(\hat{\zeta })=\zeta$, and letbe its extension. By the given fact, $E$ is a profinite group, hence is residually finite. The extension $E_{\zeta }^{\prime }$ over $\Gamma$ is the pullback of $E$ along the injective map $\iota$. Part 5(a)(iii) embeds $E_{\zeta }^{\prime }$ into $E$. Since every subgroup of a residually finite group is residually finite, so is $E_{\zeta }^{\prime }$.

Let $\hat{\zeta }\in H^2(\hat{\Gamma },M)$ lie in the kernel of ${\iota }^{\ast }$, and represent it by a profinite extensionIts pullback to $\Gamma$ is split, so there is a homomorphism $s:\Gamma \to E$ satisfying $\pi s=\iota$. By part b, $s$ extends uniquely to a continuous homomorphism $\hat{s}:\hat{\Gamma }\to E$. The continuous maps $\pi \hat{s}$ and the identity of $\hat{\Gamma }$ agree on the dense image of $\Gamma$, hence agree everywhere. Thus $\hat{s}$ is a section of $\pi$, the extension splits, and $\hat{\zeta }=0$. Therefore