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Consider the following two conditions for a group $G$:

(1) $G$ does not satisfy a nontrivial law.

(2) $G$ contains a non-abelian free subgroup.

Obviously (2) implies (1) and it is easy to construct torsion groups that do not satisfy any law (e.g., the direct product of all finite groups). Thus (1) does not imply (2) in general. However the following question seems open:

Are (1) and (2) equivalent for profinite groups?

Here is a similar question:

Suppose that a residually finite group does not satisfy a law. Does its profinite completion contain a nonabelian free subgroup?

These questions may be thought of as possible generalizations of the Tits alternative to residually finite (or profinite) groups.

The answer to the second question is positive for finitely generated p-groups. This follows from [Wilson, Zelmanov, Identities for Lie algebras of pro-p groups, JPAA 81(192), 103-109], where the authors prove that if $G$ is a finitely generated residually finite p-group, then either it is finite (and hence satisfies a law), or its profinite completion contains a nonabelian free group. In the general case they only prove that if the profinite completion of a residually finite group does not contain a nonabelian free subgroup, then some Lie algebra associated to the group satisfies a law.

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Could you possibly briefly expand on what it means for a group to satisfy a law, or provide a reference? Thanks! –  Alex B. Jan 23 '11 at 5:23
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Denis, can you explain the result of Zelmanov? The p-adic integers are infinite and commutative; maybe you mean that any pro-p groups which satisfies a law is virtually solvable? –  Andreas Thom Jan 23 '11 at 10:03
    
Alex, to satisfy a law means that there exists a non-trivial word in a free group which is satisfied by all tuples of elements in $G$. Example: $G$ satisfies $xyx^{-1}y^{-1}$ iff $G$ is commutative. –  Andreas Thom Jan 23 '11 at 10:07
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In de Cornulier, Yves; Mann, Avinoam, Some residually finite groups satisfying laws. Geometric group theory, 45–50, Trends Math., Birkhäuser, Basel (2007), it was shown: Corollary 10. There exists a pro-p-group, topologically finitely generated, that satisfies a non-trivial group law, and is not virtually solvable. –  Andreas Thom Jan 23 '11 at 10:29
    
Andreas, thanks a lot for your comments. I've corrected the question. My original statement of Zelmanov's result was wrong. –  Denis Osin Jan 24 '11 at 2:29
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2 Answers 2

up vote 5 down vote accepted

Since Henry asked, here is a reference: Aner Shalev in the first chapter (Lie Methods in the Theory of pro-$p$ Groups) in New Horizons in pro-$p$ Groups posed 4 conjectures in decreasing order of strength:

  1. Let $G$ be a finite $p$-group satisfying some identity $w$ with probability $\epsilon>0$. Then $G$ satisfies some identity depending only on $w$, $p$, and $\epsilon$.

  2. Let $G$ be a finitely generated pro-$p$ group satisfying some identity with positive probability. Then $G$ satisfies some identity.

  3. Let $G$ be a finitely generated pro-$p$ group satisfying some coset-identity. Then $G$ satisfies some identity.

  4. Let $G$ be a $k$-generated pro-$p$ group satisfying some identity on all generating $k$-tuples. Then $G$ satisfies some identity.

There is a discussion there about what is known. I am not sure what developments occurred since the book appeared and what is known about the profinite case in general. I think it may have been mentioned by Shalev or Mann in other surveys.

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Yiftach, thanks a lot, I did not know about this paper. –  Denis Osin Feb 4 '11 at 19:41
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Here is a partial solution to the conjecture. I don't know how strong it is, but I feel like posting it.

Every profinite group $G$ is a compact Hausdorff space, and therefore a Baire space, and so is $G \times G$. This exactly means that the intersection of countably many open dense sets is still dense. Now a word $w(a,b)$ in two generators $a$ and $b$ yields a function $$w:G \times G \to G.$$ The word is satisfied when $w(a,b) = 1$. It is a law if $w(a,b) = 1$ always, i.e., if $w$ sends all of $G \times G$ to 1. In general $w$ is satisfied on some closed set of $G \times G$. Suppose that every word is only satisfied on a nowhere dense set in $G \times G$. Then the Baire category theorem says that there is a comeager set of pairs $(a,b)$ that don't satisfy any word, and there then is a free subgroup of rank 2.

On the other hand, suppose that a word $w$ is satisfied on a closed set $C$ that is somewhere dense. Then because the topology on $G \times G$ is induced by homomorphisms to finite groups, $C$ contains the inverse image of some point under a homomorphism $$(\alpha,\alpha):G \times G \to A \times A,$$ where $A$ is a finite group. Suppose that $C$ contains $(\alpha^{-1}(1),\alpha^{-1}(1))$, and let $n$ be the exponent of $A$. Then the word $w'(a,b) = w(a^n,b^n)$ is a law on $G$.

Where I get stuck is that $C$ might instead contain the inverse image of some different element of $A \times A$. For instance, suppose that $G = C_2 \ltimes C_3^\infty$ is a semidirect product, where the $C_2$ acts (by conjugation) on each $C_3$ by inverting it. Then the word $w(a) = a^2$ is satisfied on an open set that does not contain the identity. Given such a word (in two generators), I tried to make another word that contains an open set around the identity, but I did not succeed.

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Greg, this is a well known (and probably very difficult) problem and your observation has already been noticed. I think it appeared in a survay by Shalev somewhere and maybe in other places. Please notice also that an open set in a profinite group contains a coset of an open subgroup. So you actually have a left coset of some open subgroup such that its elements staisfy $w$. –  Yiftach Barnea Jan 30 '11 at 18:22
    
To check I understand your final example correctly: the identity does of course satisfy $a^2=1$, but your point is that it's not in the interior of the set of elements that satisfy $a^2=1$. –  HJRW Jan 30 '11 at 18:34
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Yiftach, if this is a well known open problem, then it would be good to post that as an answer (with references). –  HJRW Jan 30 '11 at 18:35
    
@Henry Yeah, that's right. –  Greg Kuperberg Jan 30 '11 at 20:08
    
@Yiftach I had the feeling that it might be so, but I was not sure what was meant by "the following question seems open". I gather that it means, "the following question has been open for a long time; has there been major progress?" –  Greg Kuperberg Jan 30 '11 at 20:11
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