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This is a just-for-fun question inspired by this one. Let $P$ be a property of finitely presentable groups. Suppose that

  1. The truth of $P(G)$ only depends on the isomorphism class of $G$.

  2. Given a finite presentation of $G$, the truth of $P(G)$ is computable.

Let $\hat{G}$ denote the profinite completion of $G$. Is it possible to have groups $G$ and $H$, and such a property $P$, so that $\hat{G} = \hat{H}$ but $P(G) \neq P(H)$?

For example, is there a computable property which separates Higman's group from the trivial group?

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I'm glad to see that you've picked up the thread, David. This question is related to understanding the equivalence relation \equiv at conclusion of my question, to which you link. mathoverflow.net/questions/16532. A related issue: It seems that the relation of having the same profinite completion is likely not decidable. Do we have proof of this? –  Joel David Hamkins Feb 27 '10 at 0:53
    
Do you know of interesting properties satisfying 1. and 2.? The ones I can think of involve calculating the abelianization of G (or other nilpotent quotients). These certainly won't work. On an unrelated-but-feels-a-bit-related note, Bridson (Paper I <a href="people.maths.ox.ac.uk/~bridson/papers/profinite/…) constructs examples of injective homomorphism $i:H \hookrightarrow G$ so $i$ induces an isomorphism on profinite completions, but you can't decide if $G$ and $H$ are isomorphic. You can take H and G to both be residually finite, or alternatively you can take $H=\{1\}$. –  Daniel Groves Feb 27 '10 at 0:55
    
Oh, I just read the linked question, and so I guess you don't know of interesting properties satisfying 1. and 2. ... –  Daniel Groves Feb 27 '10 at 0:58
    
Not to say that the answers given in that question aren't interesting, just that I know about them. (I realised that my second comment was quite rude, for which I apoligize.) –  Daniel Groves Feb 27 '10 at 1:03
    
Joel, Regarding whether or not having isomorphic profinite completions is decidable... In the second paper on the webpage that Daniel linked to above, Bridson exhibits pairs of a group and a subgroup such that no algorithm can determine whether or not they have isomorphic profinite completions. Unfortunately, it may be impossible to compute a presentation for the subgroup. I discuss this sort of thing in my answer to this question: mathoverflow.net/questions/15957/… –  HJRW Feb 27 '10 at 2:19

3 Answers 3

OK, I think I have an example of two groups with the same profinitization and a computable property which distinguishes them. The point is that very fine detail about the commutator subgroups can't be seen in the profinitization.

Let $q$ be prime and let $K$ be the $q$-th cyclotomic field. Choose $q$ such that the class group of $K$ is not trivial. Let $I$ be a trivial ideal of $\mathcal{O}_K$ and $J$ a nontrivial ideal. Our groups $G$ and $H$ will be $(\mathbb{Z}/q) \ltimes I$ and $(\mathbb{Z}/q) \ltimes J$.

For any group $B$, let $B' = [B,B]$ and $B'' = [B', B']$. Note that $B/B'$ acts on $B'/B''$ by conjugation. Our computable criterion is the following:

$B/B' \cong \mathbb{Z}/q \times \mathbb{Z}/q =: A$, the action of the group ring $\mathbb{Z}[A]$ on $B'/B''$ factors through a map $\mathbb{Z}[A] \to \mathcal{O}_K$ and, as such, $B'/B''$ is a free $\mathcal{O}_K$ module.

We leave it as an exercise that $G$ satisfies this condition and $H$ does not.

I believe this condition should be computable. We can go from a finite presentation of $B$ to one of $B'$. (UPDATE I have revised this argument.) Abelianizations are computable, so we can check whether $B/B'$ has the right format. If it does, then $B'$ has finite index in $B$. I think we can use this to get a finite presentation of $B'$: Let $\Delta$ be a two-dimensional $CW$-complex with one vertex, an edge for each generator of $B$ and a two cell for each relation. Let $\Delta'$ be the cover of $B$ corresponding to $B'$. Since $B$ has finite index in $B'$, $\Delta'$ will have finitely many cells, and we get a finite presentation of $B'$.

We can the compute the abelianization of $B'$ and, I think, the action of the abelianization of $B$ on that of $B'$ should be computable. Note that there are only $q^2$ maps from $\mathbb{Z}[A]$ to $\mathcal{O}_K$, so we can just check them each in turn. The class of a finite generated module for a Dedekind domain should be computable by standard number theory methods, although I admit I couldn't describe them.

The fact that these two groups have the same profinitization is relatively well known. Let $\hat{I}$ and $\hat{J}$ denote the profinite completions of $I$ and $J$. The profinite completions of $G$ and $H$ are $\mathbb{Z}/n \ltimes \hat{I}$ and $\mathbb{Z}/n \ltimes \hat{J}$.

We can identify $\hat{I}$ and $\hat{J}$ with submodules of $\mathbb{A}^0_K$, the integral adeles of $K$. Since $I$ and $J$ are locally principal, these are principal ideals in the ring $\mathbb{A}^0_K$. They are thus equivalent as $\mathbb{A}^0_K$ modules, and thus as $\mathcal{O}_K$ modules.

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The commutator subgroup of a finitely presented group need not be a finitely presented group. For example, the commutator subgroup of the free group on 2 generators is not even finitely generated. –  Bjorn Poonen Feb 27 '10 at 15:17
    
David, you're correct that a finite-index subgroup of a fp group is fp, so if the abelianization of B is finite and B is fp then B' is also fp. So why is your G fp? –  HJRW Feb 27 '10 at 23:50
    
To add a little bit: the first construction of a pair of non-isomorphic, fp, residually finite groups with isomorphic profinite completions was given by Bridson and Grunewald in 2004. Their construction is still essentially the only known one. Your groups seem to be obviously residually finite, so if they're fp then they this is very interesting. –  HJRW Feb 28 '10 at 0:05
    
@Henry: A semidirect product of finitely presented groups is finitely presented. –  Bjorn Poonen Feb 28 '10 at 1:51
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@Henry: I think it was known <=1964 that there exist 2 residually finite fp groups with isomorphic profinite completions. If one starts with a smooth projective variety over a number field k, extends the base via two embeddings k --> C, and takes the fundamental group, then the two resulting groups are fp groups with isomorphic profinite completions (étale fundamental group of V_kbar). In 1964 Serre gave an example in which these groups were not isomorphic. And if I remember correctly, they were residually finite. In fact, I think they were exactly the groups David has constructed. –  Bjorn Poonen Feb 28 '10 at 2:27

In this paper, Owen Cotton-Barratt and I construct two finitely presentable groups with isomorphic profinite completions, but such that one is conjugacy separable (implying solvable conjugacy problem) and the other has unsolvable conjugacy problem.

(The construction is very much in the spirit of the paper of Bridson that Daniel Groves mentioned in the comments.)

EDIT:

Sorry, I only just noticed requirement 2. Since almost no properties are computable from a finite presentation, and yet the class of properties computable from a finite presentation is mysterious (eg does it include having a proper finite-index subgroup?), I don't see how you'll get any interesting answers with condition 2.

FURTHER EDIT:

As Bjorn explained to me in the comments to David's answer, it's not nearly as hard as I had thought to build two fp groups with the same profinite completion. Indeed, there are virtually abelian examples. As one can solve the isomorphism problem for virtually abelian groups, it follows that there examples of computable properties that are not determined by the profinite completion, as David's answer shows.

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Is it decidable from a presentation whether or not a group is large, i.e. admits a homomorphism onto the nonabelian free group on two letters? This seems totally unlikely, and surely either Henry or Daniel would know, but I like the following theorem anyway, so I'll advertise. Lackenby showed (`Detecting large groups', GR/0702571) that largeness is a property of the profinite completion of discrete finitely presented groups.

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Huh, interesting! It's unknown whether this is decidable. It's equivalent to asking whether a group has a proper finite-index subgroup, as G has a proper finite-index subgroup if and only if GGG is large. –  HJRW Feb 27 '10 at 2:40
    
But how does this answer the question? For a negative answer, what is needed is a decidable property that doesn't respect the profinite completion. (And for a positive answer to the question, you need to grapple with the collection of all decidable properties.) –  Joel David Hamkins Feb 27 '10 at 14:36
    
Joel, as far as I can tell, there are no genuinely interesting properties that are known to be computable. Indeed, I think 'largeness' is one of the few real candidates. So it's certainly relevant, even if it doesn't answer the question in the strictest sense. –  HJRW Feb 27 '10 at 16:42
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Actually, what Stover asks is decidable, whether a group has a homomorphism to a free group, by a result of Razborov. The usual definition of large is that there is a finite-index subgroup which maps onto a (non-cyclic) free group, which is actually the condition that Marc shows is determined by the profinite completion. If there is a non-residually finite hyperbolic group, then I suspect the answer to the question is no. ams.org/mathscinet-getitem?mr=755958 –  Ian Agol Feb 27 '10 at 23:26
    
Right, good point, Ian. In my comment above, I presumed Matt had given the usual definition. Actually, you don't need the full power of Razborov, you just need Makanin. I mentioned this in an answer to this question: mathoverflow.net/questions/16532/… –  HJRW Feb 28 '10 at 0:07

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