Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 151 / 2

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 $k=0$.