I was wondering if in the adeles ring over $\mathbb{Q}$, viewed as an additive group, any closed subgroup H is such that $((1/2)H)/H$ is finite or compact, where $1/2H$ is the group of adeles $x$ such that $2x \in H$. My first guess is to answer yes but in fact I noticed that classifying the closed subgroups of the adeles is not so obvious (at least for me)... Any idea is welcome, thanks in advance!
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So the answer seems to be yes, such a quotient is always finite. We can follow a reasonning similar to Closed subgroups of $\hat{\mathbb{Z}}\cong\prod_{p}\mathbb{Z}_{p}$ as subsets?
If $H$ denotes a subset of the adèle ring $\mathbb{A}$, $h = (h_p)$ an element of $H$ with $h_p \in \mathbb{Q}_p$ for any place (finite prime or $\infty$ for the real component), I define $H^\perp$ to be the (closed) subgroup of $\mathbb{A}$ defined as $\{h' = (h'_p) \in \mathbb{A} , h'_p.h_p \in \mathbb{Z}_p ~~\forall p \leq \infty, ~~\forall h=(h_p) \in H \}$. When $H$ is a closed subgroup of $\mathbb{A}$ one has $(H^\perp)^\perp = H$ (from Pontryagin duality).
If for a finite prime $p$ we write $h_p = p^{n(p,h)}z(p,h)$ with $z(p,h) \in \mathbb{Z}_p^\times$, the condition $h'_p.h_p \in \mathbb{Z}_p$ means that $n(h',p) \geq -n(h,p)$ for all $p< \infty$ and $h \in H$, so that $H'$ is made of any element $(h'_p)\in \mathbb{A}$ such that $n(h',p) \geq N'(p) = \max(-n(h,p),~~h\in H)$ where $N'(p)$ can possibly be infinite, meaning that the $p$ component of $h'$ vanishes. In case $h_p$ is only $0$, that is to say $n(p,h)=\infty$ then $n(h',p) \geq -\infty$ means that it can take any value and in that case $h'_p$ can be any element of $\mathbb{Q}_p$.
For the archimedean case, I denote by ${}_{a}\mathbb{Z}$ a closed subgroup of $\mathbb{R}$, where $a$ is an extended real number, that is ${}_{a}\mathbb{Z} = a\mathbb{Z}$ when $a$ is finite and different from $0$, ${}_{0}\mathbb{Z} = 0$ and ${}_{\infty}\mathbb{Z} = \mathbb{R}$ (not to be confused with $\mathbb{Z}_{\infty} = \mathbb{Z}$). With this notation we have that the condition $h'_\infty.h_\infty \in \mathbb{Z}$, where $h_\infty$ describes a closed subgroup ${}_{a}\mathbb{Z}$ of $\mathbb{R}$, gives that $h'_\infty$ describes the subgroup ${}_{a^{-1}}\mathbb{Z}$.
Therefore $H' = {}_{a^{-1}}\mathbb{Z}\times \prod_{p< \infty} p^{N'(p)}\mathbb{Z}_p$ ($p^{-\infty}\mathbb{Z}_p = \mathbb{Q}_p$ by convention, only a finite number of $N'(p)$ can be equal to $-\infty$) and since $(H^\perp)^\perp = H$ for a closed subgroup, following the same reasonning we get that $H$ is also of this form, that is $H = {}_{r}\mathbb{Z}\times\prod_{p< \infty} p^{N(p)}\mathbb{Z}_p$.
But now, when considering $1/2H$,
we have that $1/2{}_{r}\mathbb{Z}= {}_{r/2}\mathbb{Z}$ with the natural convention that $\infty/2 = \infty$,
$1/2p^{N(p)}\mathbb{Z}_p = p^{N(p)}\mathbb{Z}_p$ when $p>2$ since $1/2$ is in that case a unit padic integer, and
$1/2 (2^{N(2)}\mathbb{Z}_2) = 2^{N(2)-1}\mathbb{Z}_2$, and $(1/2{}_{r}\mathbb{Z})/{}_{r}\mathbb{Z} = \mathbb{Z}/2\mathbb{Z}$
for $r$ finite different from $0$, trivial otherwise, $(1/2p^{N(p)}\mathbb{Z}_p)/p^{N(p)}\mathbb{Z}_p$ is always trivial
for $p>2$ and $1/2 (2^{N(2)}\mathbb{Z}_2)/2^{N(2)}\mathbb{Z}_2 = \mathbb{Z}/2\mathbb{Z}$ if $-\infty < N(2)<\infty$, trivial otherwise.
Finally we get that $1/2H/H$ is a isomorphic to a subgroup of $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ and is
therefore finite.
Eric Chopin
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