H om ological D ecision P roblem s for F initely G enerated. G ro u p s w ith S o lva b le W o rd P ro b lem. W.A. Bogley Oregon State University

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1 H om ological D ecision P roblem s for F initely G enerated G ro u p s w ith S o lva b le W o rd P ro b lem W.A. Bogley Oregon State University J. Harlander Johann Wolfgang Goethe-Universität 24 May, 2000; revised 17 April, 2001 Abstract We show that for finitely generated groups P with solvable word problem, there is no algorithm to determine whether H 1 (P ) is trivial, nor whether H 2 (P ) is trivial. 1 Introduction This first integral homology group H 1(P ) of a group P is easy to compute from a finite presentation of P. This is because H 1(P ) is isomorphic to the abelianization P ab of P and so one simply interprets the presentation in the category of finitely generated abelian groups. The aim of this paper is to examine algorithmic aspects of integral homology calculations for finitely generated subgroups of finitely presented groups, that is, for recursively presented groups. We find that even if one restricts attention to the finitely generated subgroups P of a single finitely presented group with solvable word problem, then there need not be any algorithm to determine whether H 1(P ) is trivial, nor whether the second homology H 2(P ) is trivial. We conclude 0 AMS Subject classification: Primary 20F10, Secondary 20J06. Keywords: homology of groups, word problem. 1

2 that solvability of the word problem provides no algorithmic leverage in the problem of integral homology calculations in low dimensions. C. M. Gordon [5] proved in 1980 that for finitely presented groups P, there is no algorithm to decide whether the second homology group H 2(P ) is trivial. As a notable consequence, he observed that there is no algorithm to compute the deficiency of finitely presented groups. To prove his result on second homology, Gordon adapted a strategy due to S. I. Adian [1] and M. O. Rabin [8]. Starting with a finitely presented group G with unsolvable word problem, Gordon showed to how recursively enumerate a set of finite presentations for groups P w, parametrized by words in the generators of G, such that H 2(P w) = 0 if and only if w =1inG. The construction is such that P w has unsolvable word problem when w 1inG. C. F. M iller, III [7] later showed that having trivial second homology is a homological Markov property of finitely presented groups and as such is algorithmically unrecognizable. For finitely generated groups P with solvable word problem, G. Baumslag, M iller, and H. Short [2] showed that there is no algorithm to determine whether H 2(P ) is finitely generated. More precisely, they applied a celebrated construction of E. Rips [9] to the direct product of two nonabelian free groups to produce a word hyperbolic group E for which there is no algorithm to determine whether or not any given finitely generated subgroup of E is all of E, or is finitely presented, or has finitely generated second homology. We will prove the following theorem, which combines and includes elements of all of these results. Main Theorem. There exists a finitely presented acyclic group Γ with solvable word problem and finite cohomological dimension such that there is no algorithm to determine whether or not any given finitely generated subgroup P of Γ has trivial first homology H 1(P ), or has trivial second homology H 2(P ). Since Γ has solvable word problem, there is a single algorithm which, given any finite set S of words in the generators of Γ, will solve the word problem for the finitely generated subgroup P = S generated by S. More generally, we say that a class of groups has uniformly solvable word problems if there is a single algorithm that solves the word problem for all groups in the class. The survey article [7] by M iller describes a uniform solution to the word problems for the class of all finitely presented residually finite groups [7, Theorem 5.3] 2

3 and also to the word problems for the class of all recursively presented simple groups [7, Theorem 6.2]. We believe that it should be possible to prove that having trivial second homology is not algorithmically decidable for finitely presented groups with (uniformly) solvable word problems. Indeed, this should be possible even in the presence of the finiteness condition F 3. A group G is of type F 3 if there is a K (G,1)-complex with finite three-skeleton. Of course, if one has an explicit description of a finite three-skeleton for a K (G,1)-complex X, then it is a simple matter to compute H 2(G ) = H 2(X ). But if one is given a finite two-skeleton for a K (G,1)-complex, there is no general algorithm to determine the three-skeleton of a K (G,1). Conjecture. There is a recursively enumerable set of finitely presented groups P i, i I,oftypeF3with uniformly solvable word problems such that there is no single algorithm to determine whether, given any i I, the second homology group H 2(P i) is trivial. Acknowledgement We would like to thank the anonymous referee for suggestions on how to make this paper more attractive to non-specialists. The authors also acknowledge the financial support of Collaborative Research Grant INT from Internation Programs of the National Science Foundation and grant DAAD 315/ab from the Deutscher Akademischer Austauschdienst. 2 The Groups M, G w, and P w The group Γ in the Main Theorem is the direct square of a group M,which we construct in this section. Then, given a countably generated group presentation G for a group G and a word w in the generators of G, we construct a quotient group G w of M and a pull-back P w Γ=M M obtained in a natural way from the projection M G w. We then examine the homology groups H 1(P w) andh 2(P w). Our constructions are descendants of the work of Adian [1] and Rabin [8] and owe much to the papers [3, 5, 7], including M iller s M ain Technical Lemma [7, Lemma 3.6]. To facilitate cancellation arguments in free products, we offer two preliminary lemmas. 3

4 Lemma 2.1 Let u be an element of infinite order in a free product G H and suppose that there is a factor Φ {G,H} such that the first and last letters of u, when written in normal form, both lie in the factor Φ. Then for each nonzero integer k, the first and last letters of u k also lie in Φ. Proof: It suffices to prove the result for k > 0. We proceed by induction on the free product length n of u, the result being clear if n = 1. Suppose for convenience that the first and last letters of u are nontrivial elements of H. When n > 1, we can write u = hvh where 1 h, h H and the first and last letters of v lie in G. If h h 1inH then no cancellation occurs between bracketed terms in the factorization u k =[h][v ][h h][v ]...[h h][v ][h ] of u k. So the first and last letters of u k are h and h in this case. If h h = 1 in H, then u k = hv k h 1 and the first and last letters of v k belong to G by induction. So in this case the first and last letters of u k are h and h 1 respectively. QED We use the notation g h = h 1 gh for conjugation and [g,h]=g 1 h 1 gh for commutators. Lemma 2.2 Let G and H be groups. Let A be a subset of G and let f : A G H be a function such that for each a A, (i) f (a) has infinite order in G H and (ii) when written as a normal form in the free product G H, the first and last letters of f (a) are nontrivial elements of H. Then the elements f (a) a,a A, form a basis for a free subgroup of G H. Proof: Consider the free group F (A ) with basis A and the homomorphism φ : F (A ) G H given by φ(a) = f (a) a. We show that φ is injective. View F (A ) as the free product of infinite cyclic groups a generated by the elements a A. Let w be a nontrivial element of F (A ) of free product length n 1; we show that φ(w ) 1. When n = 1 this follows from the condition (i) since φ(a k )=(f(a) a ) k has infinite order in G H for all a A. 4

5 When n > 1, we can write w = a k a kn n where k i 0 for all i and ai ai+1 for 1 i n 1. Applying φ, weobtain φ(w ) =...(f(ai) a i ) ki (f (a i+1) a i+1 ) ki+1... =...[ai 1a 1 i ][f (ai) k i ][a ia 1 i+1 ][f (a i+1) k i+1 ][a i+1a 1 i+2 ]... Lemma 2.1 implies that the first and last letters of each f (ai) k i lie in H. Since aia 1 i+1 1 in G for all i, there is no cancellation between successive bracketed terms in the last factorization of φ(w ). The normal form theorem applied to the free product G H implies that φ(w ) 1inG H. QED The group M Let M be the group given by the finite presentation M with generators a, b, c, α, β and with defining relators as follows. (0) (1) (2) (3) (4) [b, c] = β [b, c] a = (α β ) α (a [b,c] ) a2 = (α β ) α2 (b[b, c]) a3 = β α3 (c[b, c]) a4 = β α4 Proposition 2.3 The two-complex K (M) modeled on the presentation M is aspherical. The group M decomposes as the free product with amalgamation M = F (a, b, c) C F (α,β) where C is a free group of rank five. The group M is acyclic, has cohomological dimension two, and has solvable word problem. Proof: The defining relations show that each of the generators a, b, c, α, β is conjugate in M to the commutator [b, c]. Thus M is perfect: H 1(M ) = 0. Using Lemma 2.2, one checks that the words on the left sides of the defining relations determine a basis for a free subgroup of rank five in F (a, b, c). Similarly, the right sides determine a basis for a free subgroup of rank five in F (α,β). As a free product of finitely generated free groups with a finitely generated free group amalgamated, the group M has solvable word problem since free groups have solvable generalized word problem. (See, e.g. [6].) The remaining claims follow by standard arguments. QED 5

6 The groups G w Now suppose that we are given a group presentation G = x1, x2,... : r with a countable generating set. Let G be the group presented by G. Given any word w in the generators of G, letg w be the largest quotient of the free product G M in which the following relations hold. (5) (6) [w,[b, c]] a5 = β α5 (xi[b, c]) a5+i = β α5+i,i =1, 2,... Proposition 2.4 For any given word w, the natural map M G w is surjective. In addition, we have the following. 1. If w =1in G,thenG w is trivial. 2. If w 1 in G, then G w decomposes as the free product with amalgamation G w = (G F (a, b, c)) D F (α,β) where D is a free group that has a basis consisting of a set of six elements together with a set in one-to-one correspondence with the set x of generators in G. Proof: The relations (6) show that the map M G w is surjective. When w =1inG, the relation (5) shows that β =1inG w and it is straightforward to use the remaining relations to show that G w = 1 in this case. If w 1in G, then the words on the left sides of the relations (0)-(6) determine a basis for a free subgroup of G F (a, b, c) by Lemma 2.2. Similar remarks apply to the right sides when viewed in F (α,β). QED The groups P w Given G and w, we now consider the group P w M M = Γ given by P w = {(x, y) M M : x = y in G w}. Theorem 2.5 Given the countably generated presentation G = x1, x2,... : r and a word w in the generators of G, we have the following: 6

7 1. If w =1in G,thenP w = M M is acyclic. 2. If w 1in G,thenH 1(P w) 0. If, in addition, the presentation G is finitely related, then the groups P w are finitely generated, recursively presented, and have uniformly solvable word problems. Proof: If w = 1 in G,thenG w = 1 by Proposition 2.4 and so P w = M M is acyclic by Proposition 2.3. Let N w be the kernel of the natural projection M G w. We have a commutative diagram with exact rows. 1 N w j P w p2 M 1 = p1 1 N w M G w 1 Here, the maps pi : P w M, i = 1, 2, are restrictions of the product projections M M M and j (x) = (x, 1) for all x N w. Note that the right hand square is a pullback square and that the top row is split exact. Since H 1(M )=H2(M) = 0, the five-term homology sequence associated to the top row shows that H 1(P w) N w = [P w, Nw] where we idenity N w with its image under j in P w. p2 : P w M determines a surjection Now the projection N w [P w, Nw] N w [M,Nw] and the five-term homology sequence for the bottom row shows that N w [M,Nw] = H 2(G w). Suppose now that w 1 in G. We will show that H 2(G w) 0 and the above remarks then suffice to show that H 1(P w) 0. The group G w is a 7

8 amalgamated free product as described in Proposition 2.4. A portion of the associated long exact homology sequence has this form. H 2(G w) H 1(D ) i H 1(G ) H 1(F (a, b, c)) H 1(F (α,β)) The amalgamating relations (0) and (5) determine a nontrivial element in the kernel of the map labeled i. This implies that H 2(G w) 0. Now suppose that the presentation G = x1, x2,... : r1,...,rm for G is finitely related. By a Tietze equivalence, we can expand the presentation M for M, adjoining the generators x1, x2,... of G and the relators (6) for G w to the presentation M. A presentation for G w is then obtained by adjoining the relations r1,...,rm for G and the relator (5) for G w. With this we see that N w = ker(m G w) is normally generated in M by a finite subset of M. This implies that P w = N w M is finitely generated. Since M has solvable word problem, so does its direct square Γ = M M. For each word w in the generators of G, one can effectively determine a finite set of generators and a recursive set of relators for P w in terms of generators for Γ. In particular, a solution to the word problem for Γ can be used to uniformly solve the word problems for the groups P w. QED 3 Proof of the Main Theorem First homology For finitely presented groups, it is a simple matter to determine whether the first homology is trivial. The portion of the Main Theorem dealing with first homology shows that this is no longer the case if we move on to finitely generated recursively presented groups, even among those with solvable word problems. The proof proceeds by applying the constructions of the previous section beginning with a finitely presented group G with unsolvable word problem. An explicit finite presentation for such a group appears in [10]. As seen in Theorem 2.5, the construction yields a family of finitely generated and recursively presented groups P w with the property that H 1(P w) = 0 if and only if w =1inG. By enumerating words w in the generating set for G,we see that the groups P w are defined by a recursively enumerable set of recursive presentations. However, since the word problem for G is unsolvable, there is no algorithmic procedure for deciding, for a given w, whether or not H 1(P w) 8

9 is trivial. We summarize this portion of the Main Theorem from a slightly more general perspective. Corollary 3.1 There is a recursively enumerable set {P w} of finitely generated recursively presented groups with uniformly solvable word problems such that there is no algorithm to determine, for a given w, whether or not H 1(P w) is trivial. Second homology The main homological tool that we needed for the result on first homology was the five-term exact homology sequence associated to a group extension. For the result on second homology, we will need to use a spectral sequence argument. This argument is embedded in the following general result on pairs of aspherical two-complexes and associated groups, which may be of independent interest. Theorem 3.2 Suppose that an aspherical two-complex Y is obtained from a finite connected aspherical two-complex X by attaching a nonempty collection of two-cells. Let N be the kernel of the inclusion-induced homomorphism π1(x ) π1(y ) and let P be the split extension P = N π1(x ). The following are equivalent. 1. Y is contractible. 2. P is finitely presented. 3. H 2(P ) is finitely generated as an abelian group. Proof: Let G = π1(x )andq = π1(y )sothatg/n = Q and P = N G. If Y is contractible, then N = G so P = G G = G G is finitely presented. On the other hand, all finitely presented groups have finitely generated second homology, so it remains to show that if Y is not contractible, then H 2(P )is not finitely generated. The Lyndon-Hochschild-Serre spectral sequence for the split extension 1 N P G 1 is given by E 2 pq = H p(g,hq(n )) H p+q(p ). Asphericity of X implies that H 3(G ) = 0, and this in turn implies that E 11 = E 2 11 = H 1(G,H1(N )). 9

10 In order to show that H 2(P ) is not finitely generated as an abelian group, it suffices to show that H 1(G,H1(N )) is not finitely generated as an abelian group. Let p : Ỹ Y be the universal covering projection and let X = p 1 (X ). Since H 2(Ỹ )=H 1(Ỹ ) = 0, the homology sequence for the pair (Ỹ,X)shows that H 1(N ) = H 1(X ) = H 2(Ỹ,X ). Since Y is obtained from X by attaching a nonempty set of two-cells, we find that H 1(N ) is a nontrivial free ZQ - module with a ZQ -basis in one-to-one correspondence with the two-cells of Y X. Additivity implies that as an abelian group, H 1(G,H1(N )) is a nontrivial direct sum of copies of H 1(G,ZQ ). Now by Shapiro s Lemma, H 1(G,ZQ ) = H 1(G,Z ZN ZG ) = H 1(N ). Since H 1(N )isafreezq-module, we learn that as an abelian group E 11 = E 11 2 = H 1(G,H1(N )) is a nontrivial direct sum of copies of ZQ. Since Y is an aspherical two-complex with nontrivial fundamental group Q, the integral group ring ZQ is not finitely generated as an abelian group, and the result is proved. QED Theorem 3.3 There is a recursively enumerable set of finitely generated recursively presented groups P w with uniformly solvable word problems such that there is no algorithm to determine, for a given w, whether or not H 2(P w) is trivial. Proof: Once again we employ the construction of the previous section, beginning with a finite presentation G = x1,...,xn : r1,...,rm for a group G with unsolvable word problem. This time we use the fact, due to D. J. Collins and M iller [4], that there is a finite presentation with unsolvable word problem and aspherical cellular model. We obtain a recursively enumerable family of recursively presented groups P w, parametrized by words w in the generators of G. By Theorem 2.5, H 2(P w) =0ifw = 1 in G. We now assume that w 1 in G. It suffices to show that H 2(P w) 0 in this case. By Proposition 2.3, the presentation M for M has aspherical cellular model. By a Tietze equivalence that adjoins the generators x1,...,xn of G and the relators (6), we obtain another finite presentation for M with aspherical cellular model. We denote this two-complex by X. Given a word w in x1,...,xn, we then adjoin the relators r1,...,rm and (5) to obtain a finite presentation for G w. We denote the cellular model of this presentation by Yw. Using the asphericity of the cellular model of G, the free product 10

11 description of G w appearing in Proposition 2.4 shows that Yw is aspherical. Applying Theorem 3.2 to the pair (Yw, X), the fact that π1(yw) 1 implies that H 2(P w) = H 2(N w M ) is not finitely generated as an abelian group. QED References [1] S. I. Adian, On algorithmic problems in effectively complete classes of groups, Dokl. Akad. SSSR 123 (1958), [2] G. Baumslag, C. F. M iller, III, and H. Short, Unsolvable problems about small cancellation and word hyperbolic groups, Bull. London Math. Soc. 26 (1994), [3] G. Baumslag and J. E. Roseblade, Subgroups of direct products of free groups, J. London Math. Soc. (2) 30, [4] D. J. Collins and C. F. M iller, III, The word problem in groups of cohomological dimension 2, Groups St. Andrews in Bath, I, London Math. Soc. Lecture Note Series, 260 (CUP, 1999), [5] C. M. Gordon, Some embedding theorems and undecidability questons for groups, in: Combinatorial and Geometric Group Theory, A. J. Duncan, N. D. Gilbert, and James Howie, eds., London Math. Soc. Lecture Note Series 204 (CUP, 1995), [6] W. Magnus, A. Karass, and D. Solitar, Combinatorial Group Theory, (Dover, 1976). [7] C. F. M iller, III, Decision problems for groups: survey and reflections, in: Algorithms and Classification in Combinatorial Group Theory, G. Baumslag and C. F. M iller, III, eds., M SRI Publ. 23 (Springer-Verlag, 1992), [8] M. O. Rabin, Computable algebra, general theory and theory of computable fields, Trans. Amer. Math. Soc. 95 (1960), [9] E. Rips, Subgroups of small cancellation groups, Bull. London Math. Soc. 14 (1982),

12 [10] J. J. Rotman, An Introduction to the Theory of Groups, 4thed.,Graduate Texts in Mathematics 148 (Springer-Verlag, 1995). 12