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.