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

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 $k=0$.