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Here's what I can get from

Second try. Victor points out that my original application of super-rigidity for was faulty. This edit is to remove the false claim and leave only the lemma I believe. While I'm at it, I have added some stuff I don't understand.

Let $H$ and $G$ be algebraic groups of units of division algebras over $\mathbb Q$ (because super-rigidity is of limited use over other fields, it buys rather less)bigger fields). It shows If $f\colon H(\mathbb Q)\to G(\mathbb Q)$ is an isomorphism of abstract groups, then super-rigidity implies that for all the local factors are isomorphic $S$, $H(\mathbb Z[S^{-1}])$ and that the isomorphism approximately preserves $G(\mathbb Z[S^{-1}])$ contain finite index subgroups identified by $f$. One can't ask for more, since the $S$-integral points . The isomorphism of local factors gives examples of groups that $H$ and $G$ are not isomorphiconly well-defined up to finite index. These groups are insensitive to inclusion of the ramified places in $S$.

I think integral points should don't know everything so that isomorphism requires how much this buys us.

Perhaps one can use the congruence subgroup property to reconstruct the local points of the group? I believe that the CSP is conjectured but open for these groups are algebraically isomorphic(specifically, that there should be a finite congruence kernel, but perhaps for large $S$). Worse, I'm not sure .

Once we know the local factors are isomorphicwhat it says, we're lead to it particular whether it allows reconstruction of the same question for local fieldscompact factors. The original question doesn't really motivate global fields(If this does work, as opposed to general fieldsthen the compact factors would be the profinite completion of the rational points and one would not need the first paragraph. Local fields Maybe this should be easier! Boyarsky's comments on Victor's answer addresses have been a separate answer.)

If we can reconstruct the compact local case: continuous isomorphism must be algebraic; so factors from the algebras are isomorphic or opposed; ($S$-)integral points, and thus from the rational points, we can extract reconstruct the invariantlocal invariants, up to a sign (the local version of this question, addressed by Victor and Boyarsky). So if we have This would not be enough to show that the two units discrete groups over $\mathbb Q$ that of rational points are isomorphic as algebraic groups, they have but it would be enough to show that some examples of groups of rational points are not isomorphiclocal factors, so their . A division algebra with local invariants are equal, up to local signs. Thus, all $\frac 15$ is not isomorphic \frac15$ and $\frac45$ has non-isomorphic factors to all one with $\frac 25$, though this doesn't distinguish \frac25$ and $\frac35$. But a division algebra with invariants $(\frac 15,\frac 25,\frac 35,\frac 45)$ from is not isomorphic or opposed to one with invariants $(\frac 45,\frac 25,\frac 35,\frac 25)$; , even though locally it is always isomorphic or opposed. Thus, such a method could not distinguish $10$ at $\frac15$ from $5$ at $\frac15$ and $5$ at $\frac45$; or do anything for $p=3$.their unit groups.

Proof of the claim in the first paragraph that isomorphism of rational points must preserve integral points: Super-rigidity says that maps of lattices extend to maps of their ambient locally compact groups. There's probably a version groups It requires that the $S$-rank of rational points, which are lattices the source is least $2$, where the $S$-rank is the sum over places in adelic groups, but I've never heard. That would immediately give $S$ of the local factorsranks. But I do know that there's a version for Under Kevin's assumptions, we already have rank $S$-integral groups. 2$ from the infinite place; in general we would need to use large $S$ to draw conclusions about small $S$. $H(\mathbb Q)$ is not finitely generated, but it is filtered by groups of $S$-integers, which are finitely generated. Thus an isomorphism $f\colon H(\mathbb Q)\cong G(\mathbb Q)$ yields a homomorphism $H(\mathbb Z)\to G(\mathbb Z[S^{-1}])$, for some $S$. By super-rigidity, $H(\mathbb Z)$ contains a finite index subgroup on which this extends to a (continuous) homomorphism $H(\mathbb R)\to G(\mathbb R\times \mathbb Q_p\times\ldots)$, which has to land in $G(\mathbb R)$. And similarly back, so we get an isomorphism $H(\mathbb R)\cong G(\mathbb R)$ which is the given isomorphism restricts to be $f$ on a finite index subgroup of $H(\mathbb Z)$. That is, $f$ preserves the integral points. Similarly , for each $S$, the $S$-integral pointsare, up to finite index, preserved by the given isomorphism and the local factors are isomorphic. (That proves some kind of adelic/$G(\mathbb Q)$ version of the theorem, but I'm not sure what the statement is.A weak statement is that an isomorphism of $\mathbb Q$-points implies an isomorphism of adelic groups.)
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Here's what I can get from super-rigidity for groups over $\mathbb Q$ (over other fields, it buys rather less). It shows that all the local factors are isomorphic and that the isomorphism approximately preserves the $S$-integral points. The isomorphism of local factors gives examples of groups that are not isomorphic. I think integral points should know everything so that isomorphism requires that the groups are algebraically isomorphic, but I'm not sure.

Once we know the local factors are isomorphic, we're lead to the same question for local fields. The original question doesn't really motivate global fields, as opposed to general fields. Local fields should be easier! Boyarsky's comments on Victor's answer addresses the local case: continuous isomorphism must be algebraic; so the algebras are isomorphic or opposed; we can extract the invariant, up to a sign. So if we have two units groups over $\mathbb Q$ that are isomorphic, they have isomorphic local factors, so their local invariants are equal, up to local signs. Thus, all $\frac 15$ is not isomorphic to all $\frac 25$, though this doesn't distinguish $(\frac 15,\frac 25,\frac 35,\frac 45)$ from $(\frac 45,\frac 25,\frac 35,\frac 25)$; or distinguish $10$ at $\frac15$ from $5$ at $\frac15$ and $5$ at $\frac45$; or do anything for $p=3$.

Proof: Super-rigidity says that maps of lattices extend to maps of their ambient locally compact groups. There's probably a version groups of rational points, which are lattices in adelic groups, but I've never heard. That would immediately give the local factors. But I do know that there's a version for $S$-integral groups. $H(\mathbb Q)$ is not finitely generated, but it is filtered by groups of $S$-integers, which are finitely generated. Thus an isomorphism $H(\mathbb Q)\cong G(\mathbb Q)$ yields a homomorphism $H(\mathbb Z)\to G(\mathbb Z[S^{-1}])$, for some $S$. By super-rigidity, $H(\mathbb Z)$ contains a finite index subgroup on which this extends to a (continuous) homomorphism $H(\mathbb R)\to G(\mathbb R\times \mathbb Q_p\times\ldots)$, which has to land in $G(\mathbb R)$. And similarly back, so we get an isomorphism $H(\mathbb R)\cong G(\mathbb R)$ which is the given isomorphism on a finite index subgroup of the integral points. Similarly, for each $S$, the $S$-integral points are, up to finite index, preserved by the given isomorphism and the local factors are isomorphic. (That proves some kind of adelic/$G(\mathbb Q)$ version of the theorem, but I'm not sure what the statement is. A weak statement is that an isomorphism of $\mathbb Q$-points implies an isomorphism of adelic groups.)