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Let $G$ be a finitely presented group.

Are there any algorithm to calculate whitehead group $G$, $Wh(G)$ in terms of presentation of $G$?

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2 Answers 2

There is no algorithm. Indeed, the property that $Wh(G)$ is trivial is Markov that is there exists a finitely presented group (${\mathbb Z}$) with trivial $Wh$ and there exists a finitely presented group not embeddable into any group with trivial Wh (any finite non-trivial group).

Update Recall that Markov proved undecidability of every Markov property for semigroups (see http://iopscience.iop.org/0036-0279/19/3/M05/pdf/0036-0279_19_3_M05.pdf), and later Adyan and Rabin proved it for groups (see Lyndon and Schupp, Combinatorial group theory, Theorem 4.1, Chapter 4).

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Can you explain why the Markov property implies there is no algorithm? Maybe by a reference. –  Wilberd van der Kallen Jun 23 '11 at 8:31
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See reference [6] in eom.springer.de/a/a011860.htm –  Mark Sapir Jun 23 '11 at 8:39
    
Mark, if I understand your argument correctly, it asserts that if a group $G$ has zero Whitehead group, then $G$ is torsion-free. This is not true. For example, $\mathbb Z_2$ has trivial Whitehead group, and if memory serves me, so does $S_n$. –  Igor Belegradek Jun 23 '11 at 12:32
    
I checked, and indeed, $S_n$ does have trivial Whitehead group for any $n$. This can be found e.g in Oliver's book "Whitehead groups of finite groups", Theorem 14.1, also mentioned in example 4 on page 14, which can be viewed in google books. –  Igor Belegradek Jun 23 '11 at 13:19
    
Igor: You are right, the proof is harder than I thought (I thought that zero divisors somewhow produce non-trivial elements in Wh). The reference you give below uses both the Adyan-Rabin theorem and its proof. –  Mark Sapir Jun 23 '11 at 15:43

There is no algorithm to decide from a finite presentation of a group $G$ whether $\mathrm{Wh}(G)$ is zero. See Corollary 5.7 of this paper which studies various computability issues in higher-dimensional topology.

I think the original question that asks whether one can compute $\mathrm{Wh}(G)$ from a presentation of $G$ does not quite make sense because $\mathrm{Wh}(G)$ need not be finitely generated, e.g. Example on page 2 of this paper gives a virtually $\mathbb Z^3$-group whose Whitehead group is infinitely generated.

For finite $G$, the original question does make sense because the group $\mathrm{Wh}(G)$ is finitely generated (thanks to a theorem of Bass; see again Oliver's book mentioned in comments), and abelian (Whitehead groups are abelian by definition), and isomorphism problem is solvable for finitely generated ableian groups. However, the paper mentioned in the first paragraph above shows that even for finite groups there is no algorithm. On the other hand, a lot is known about Whitehead groups of finite groups, see again Oliver's book.

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@Igor: The question makes sense even if the $Wh$ is infinite. One can ask for an algorithm to list the elements of the group one by one. For example, the word problem is decidable in that sense in every finitely presented group because there exists an algorithm listing all words that are equal to 1. In that sense it is not clear if there exists an algorithm that lists all elements of $Wh$. At least I do not have any intuition what the answer should be. It is not even clear if there exists an algorithm that finds a non-trivial element of $Wh$ if there is one. –  Mark Sapir Jun 25 '11 at 8:57
    
@Mark, I do not have a firm grip what an algorithm is, but how could an uncountable abelian group, such as $\mathbb R$, be the output of an algorithm? What if the abelian group $\mathrm{Wh}(G)$ is uncountable? –  Igor Belegradek Jun 25 '11 at 11:44
    
@Igor: $Wh$ of a countable group cannot be uncountable. Right? The algorithm can in principle give elements of $Wh$ one by one as matrices over the group ring $\mathbb{Z}G$ (with integer coefficients). The problem is that it is not clear when a matrix is 1 in $Wh$. We actually can decide when a matrix is 1 (just represent it as a product of elementary matrices in a ring of matrices of some bigger size), but we cannot (I think) decide that the matrix is <b> not</b> 1. This is similar to deciding if a word is (is not) equal to 1 in a f.p. group. –  Mark Sapir Jun 25 '11 at 15:07
    
Since we can prove that an element is 1 if it is equal to 1, we can also prove that a f.p. group is trivial if it is trivial (there is a finite witness to this event). On the other hand I do not see how to prove that $Wh$ is trivial if it is known to be trivial or how to construct a nontrivial element in $Wh$ if it is known that $Wh$ is non-trivial. –  Mark Sapir Jun 25 '11 at 15:11
    
@Mark, your are right, now that I think about it $\mathrm{Wh}$ is always countable. Thanks for clarifying this and other things. –  Igor Belegradek Jun 25 '11 at 16:59

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