Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 4 / 2

A topological group is a group with a topology for which $(x,y)\mapsto xy$ and $x\mapsto x^{-1}$ are continuous; we use the Hausdorff convention. Translations and inversion are then homeomorphisms.

For the final generator-extraction assertion, use the profinite Frattini subgroupFor a finitely generated pro-p group, the Burnside basis theorem says that $G/\Phi (G)$ is a finite-dimensional vector space over ${\mathbb{F}}_p$ and a subset topologically generates $G$ precisely when its image spans that vector space. In particular its dimension is the minimal number $d(G)$ of topological generators. A given $d$-element generating set shows $d(G)\le d$. The image of any other generating set spans the same finite-dimensional vector space, so it contains a basis of $d(G)$ elements. Choose their preimages in that generating set and apply the Burnside basis theorem again. Every generating set contains a generating subset of size $d(G)\le d$. This argument also applies to an infinite generating set.

If $H$ is an open subgroup, each left coset $gH$ is open because translation is a homeomorphism. Its complement is the union of the other cosets, hence is open. Every open subgroup is closed. No compactness is needed.

Let $x,y\in \overline{H}$ and let $W$ be an open neighbourhood of $x{y}^{-1}$. Continuity of $(u,v)\mapsto u{v}^{-1}$ supplies neighbourhoods $U$ of $x$ and $V$ of $y$ with $U{V}^{-1}\subseteq W$. Choose $h\in H\cap U$ and $k\in H\cap V$. Then $h{k}^{-1}\in H\cap W$, so $x{y}^{-1}\in \overline{H}$. The subgroup criterion using $x{y}^{-1}$ proves the closure of a subgroup is a subgroup. The neighbourhood argument avoids assuming the topology is determined by sequences.

For an open subgroup $H$ of a profinite group $G$, the open left cosets form a disjoint cover of a compact space. A finite subcover must include every coset, so $[G:H]$ is finite. Choose representatives $t_1,\dots ,t_m$ for these cosets. The normal core isEach conjugate is open, and the intersection is finite, so $N$ is an open normal subgroup contained in $H$. $[G:H]<\infty$ and $H$ contains an open normal subgroup of $G$.

Choose a finite set $S$ of topological generators of $G$ and let $\Gamma =⟨S⟩$ denote the abstract subgroup it generates. Then $\Gamma$ is dense in $G$. For an open subgroup $H$, the abstract subgroup $K=\Gamma \cap H$ is dense in $H$, since every nonempty open subset of $H$ is open in $G$. Every coset of $H$ meets $\Gamma$, so $[\Gamma :K]=[G:H]=m<\infty$.

Here is the needed Schreier lemma argument explicitly. Choose representatives $t$ in $\Gamma$ for the right cosets $K\backslash \Gamma$, including $1$ for $K$, and write $\overline{\gamma }$ for the representative of $K\gamma$. The finite setis contained in $K$. For $k=s_1\cdots s_r\in K$, set $t_j=\overline{s_1\cdots s_j}$, with $t_0=t_r=1$. Thenand every factor belongs to the displayed finite set. Thus $K$ is finitely generated as an abstract group. Its finite generators topologically generate its closure $H$. Every open subgroup of a finitely generated profinite group is finitely generated.