TITLE: units in distinct division algebras over number fields---are they definitely not isomorphic as abstract groups?
QUESTION [18 upvotes]: This is really an irrelevant question in the sense that the answer isn't remotely "logically crucial for the Langlands programme" or whatever---it's just something that occurred to me when writing some lectures on the subject.
Let $F$ be a number field and let $D$ and $D'$ be division algebras over $F$ with centre $F$. Let me put myself in the following simplified situation: assume $D$ and $D'$ both have dimension $p^2$ over $F$ (so are forms of $M_p(F)$) with $p$ an odd prime, and let me also assume that $D$ and $D'$ are both ramified at precisely the same finite places of $F$ (and at no infinite places of $F$) but that $D$ and $D'$ are not isomorphic.
The question: Are $D^\times$ and $(D')^\times$ sometimes non-isomorphic as groups?
Let me put this slightly strange question into context. Experts will already know what is coming: $D^\times$ and $(D')^\times$ are group-theoretically similar in certain ways; this is what I'm about to explain. This group-theoretic similarity would be trivially explained if the groups were isomorphic! But I think the whole point should be that the groups aren't isomorphic, although I don't know an instance where I can prove this.
OK so here are the details. Given $D$ and $D'$ as above, there is a link between automorphic forms on $D^\times$ and $(D')^\times$. When proving this link using the trace formula, one has to check that two complicated formulae for traces, one associated to $D$ and one to $D'$, coincide. The formulae are of the form "sum over conjugacy classes of certain orbital integrals". The strategy I understand to check these sums coincide is first to use general theory of central simple algebras to explicitly list the conjugacy classes in the groups $D^\times$ and $(D')^\times$, and then to write down an explicit natural bijection between them, and then to check that things add up in the sense that each term on the $D$ side is equal to the corresponding term on the $D'$ side, and then to deduce that certain traces are equal, and then you follow your nose to the answer (modulo a technicality that has to be dealt with using a Plancherel measure argument but which isn't relevant here).
Crucial in this strategy is the identification of the conjugacy classes in $D^\times$ with the conjugacy classes in $(D')^\times$. The way this works in this situation is that a conjugacy class in $D^\times$ is either an element of $F^\times$ (the central classes) or contains some $e\not\in F$; in this case $E:=F(e)$ is a field extension of $F$ of degree $p$ which splits $D$, and the fields that split $D$ of this form are precisely the degree $p$ extensions of $F$ which have one prime above $v$ for all $v\in S$ our bad set. The miracle that has occurred here is that this argument (when fleshed out) shows that the conjugacy classes in $D^\times$ are parametrised by a set that depends only on $F$, $p$ and $S$, and in particular this set is the same for $D$ and $D'$, which have the same degree and which ramify at the same primes.
I will present this argument in detail in class today, and my instinct is to stress that a miracle has occurred, because $D^\times$ and $(D')^\times$ are non-isomorphic groups whose conjugacy classes are completely naturally in bijection with one another. Of course if $D$ and $D'$ are isomorphic then then $D^\times$ and $(D')^\times$ will be isomorphic. Moreover, if $D'=D^{opp}$ then again $D^\times$ and $(D')^\times$ will be isomorphic as groups. In both cases it is hardly surprising that I have managed to canonically biject the conjugacy classes of $D^\times$ and $(D')^\times$! But in general I am still convinced that a miracle has occurred. However to really convince the audience of this I want to assert confidently that the groups $D^\times$ and $(D')^\times$ really are not always isomorphic. Is this definitely true??
{\bf EDIT}: Here's an even stronger question, which is even less likely to be true, and given that I'm hoping/guessing that such things aren't true, it seems worth formulating (the stronger it is, the easier it should be to falsify).
OK so same set-up: $D$ and $D'$ division algebras of dimension $p^2$, $p$ prime, and ramified at the same set of finite primes $S$, but $D$ and $D'$ not isomorphic. Let $G$ be the form of $PGL_p$ associated to $D^\times/Z(D^\times)$ and let $G'$ be the form corresponding to $D'$. If $\mathbf{A}^S$ denotes the adeles of $F$ away from $S$ then choosing isomorphisms $D_v=D'_v$ for all $v\not\in S$, sending $O_v$ to $O'_v$ for two chosen maximal orders $O$ and $O'$ in $D$ and $D'$ when $v$ is finite, gives us an induced isomorphism $G(\mathbf{A}^S)=G'(\mathbf{A}^S)$. Call this group $X$. Now $\Gamma:=D^\times/F^\times$ and $\Gamma':=(D')^\times/F^\times$ are two discrete subgroups of $X$ and using the trace formula one can prove that not only do the conjugacy classes of $D^\times$ and $(D')^\times$ match up, but (if my understanding is correct) the covolumes of $\Gamma$ and $\Gamma'$ are the same. This would be explained by the highly unlikely statement that $\Gamma$ and $\Gamma'$ were actually conjugate within $X$! This is perhaps even stronger than just being abstractly isomorphic, in this setting? One might also ask to give an explicit example where this cannot possibly be the case?
I mention this stronger statement because people might find it easier to use high-powered methods to refute it.

REPLY [6 votes]: I know this is a very old question but I saw it just a few days ago. Super-rigidity can indeed be applied not for "higher rank lattices"  but for the global points $D^*/F^*$ (which form a lattice in the adelic group $ PGL_1(D\otimes_{\mathbb{Q}} {\mathbb A})$ (the latter is certainly a "higher rank group". 
First of all, as was observed in the comments (by Emerton?), the groups $D^*$ and $(D^{opp})^*$ are isomorphic under the map $x\mapsto x^{-1}$. The result is then as follows.
Proposition:  Suppose $\theta: D^* \rightarrow D'^*$ is an isomorphism of groups. Then one of the following two alternatives hold.
[1] There is an isomorphism $\sigma: D \rightarrow  D'$ of central division algebras over $F$, and a homomorphism $\phi: D^*\rightarrow F^* \subset D'^*$ such that the isomorphism $\theta $ is given by 
$$\theta (x)=\sigma (x)\phi (x) \quad \forall  \quad x\in D^*.$$  
[2] There is an isomorphism $\tau : D^{opp} \rightarrow D'$ of division algebras and a homomorphism $\psi :(D^{opp})^*\rightarrow F^*$ such that 
$$\theta (x)= \tau (x^{-1})\psi (x) \quad \forall x\in D^*.$$ Since $x\mapsto x^{-1}$ maps $D^* $ into $(D^{opp})^*$, the rhs makes sense. 
So $D^*$ and $D'^*$ are isomorphic if and only if $D'$ is isomorphic as a division algebra either to $D$ or to $D^{opp}$. 
The proposition follows from superrigidity (see (C) on page 259 of Chapter VIII (title of the chapter: Normal Subgroups and "Abstract" homomorphisms of semi-simple algebraic groups over global fields) of Margulis 's book "discrete subgroups of semi-simple Lie groups", where he states the superrigidity for an arbitrary set $S$ (not necessarily finite) of finite places of $F$ (If you take $S$ to be all finite places, then you get global points).
Let me state the (consequence of) superrigidity that I am using. If $\theta: G(F)\rightarrow G'(F)$ is an isomorphism of groups over $F$, with $G,G'$ absolutely simple and adjoint, then there exists an automorphism $a$ of $F$ and a morphism of algebraic groups $\alpha :^a (G)\rightarrow G'$ such that $\theta =\alpha \circ a$. This is deduced in Margulis' book from the "usual" superrigidity.  
The isomorphism from $D^*$ to $(D')^*$ gives an isomorphism from $PGL_1(D)$ (a lattice in a semi-simple adelic group) to $PGL_1(D')$. Then superrigidity applies for this case. To go from the theorem stated in Margulis' book to the proposition  needs some work; the essential point is that the only algebraic automorphisms of $PGL_p({\mathbb C})=GL_1(D\otimes {\mathbb C})$ are inner conjugations (then we get into the case of [1] of the proposition), or else are inner conjugate to the map $x\mapsto (^tx)^{-1}$, and then we land in case [2].