Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 151 / 3

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.