APPLICATIONS OF THE ALEXANDER IDEALS TO THE ISOMORPHISM PROBLEM FOR FAMILIES OF GROUPS

Abstract. In this paper, we use the Alexander ideals of groups to solve the isomorphism problem for the BaumslagSolitar groups and a family of parafree groups introduced by Baumslag and ClearyAbstract. In this paper, we use the Alexander ideals of groups to solve the isomorphism problem for the BaumslagSolitar groups and a family of parafree groups introduced by Baumslag and Cleary

APPLICATIONS OF THE ALEXANDER IDEALS TO THE ISOMORPHISM PROBLEM FOR FAMILIES OF GROUPS DO VIET HUNG AND VU THE KHOI Abstract. In this paper, we use the Alexander ideals of groups to solve the isomorphism problem for the Baumslag-Solitar groups and a family of parafree groups introduced by Baumslag and Cleary. 1. Introduction 1.1. Problems and results. The isomorphism problem is a fundamental problem in group theory in which we have to decide whether two finitely presented groups are isomorphic. In the most general form, the isomorphism problem was proved to be unsolvable. Therefore it makes sense to restrict the problem to a special class of groups. The isomorphism problem for certain families of groups has been considered by many authors. For those works which are particularly related to our, see [7, 9, 12, 14]. The purpose of this paper is to use the Alexander ideal, an algebraic invariant of groups which was originated from topology, to study the isomorphism problem for families of groups. In many cases, by computing the Alexander ideals of the groups in the family, we can deduce that two groups in that family are not isomorphic. Our main results are the solutions of the isomorphism problems for the Baumslag-Solitar groups and a family of parafree groups. The last part of this section is devoted to reviewing backgrounds on Alexander modules and ideals of a finitely presented group. In section two, we present a proof of a result of Moldavanskii [14] on the isomorphism problem of the Baumslag-Solitar 2010 Mathematics Subject Classification. Primary 20Exx, Secondary 57M05. Key words and phrases. Group isomorphism problem, Alexander ideal, parafree groups. This research is funded by Vietnam National Foundation for Science and Technology Developmen(NAFOSTED) under grant number 101.01-2011.46. 1 2 DO VIET HUNG AND VU THE KHOI groups. Section three solves the isomorphism problem for a family of parafree groups Ki,j which was introduced by Baumslag and Cleary in [6]. In particularly, by using the Alexander ideal, we show that all the groups in that family are distinct. Acknowledgments. The authors would like to express their gratitude to the Vietnam Institute for Advanced Study in Mathematics for support during the writing of the paper. 1.2. Alexander ideals. In this subsection, we will present some backgrounds on the Alexander ideals of finitely presented groups. For more details, the readers are referred to [1, 11, 13, 15]. Let G = x1 , . . . , xk |r1 , . . . , rl be a finitely presented group and ab(G) := H1 (G; Z)/(torsion) be its maximal free abelian quotient. Suppose that (X, p) be a pointed CW- complex ˜ → X be the covering corresponding to φ : G → such that π1 (X, p) = G. Let π : X ˜ p˜; Z) the relative homology. Then the ab(G) and p˜ := π −1 (p). We denote by H1 (X, ˜ makes H1 (X, ˜ p˜; Z) an Z[ab(G)]−module, desk transformation action of ab(G) on X which is called the Alexander module of G. We can also describe the Alexander module in a purely algebraic way as (see [13]) M = m(G)/m(ker φ) · m(G). Where M(H), for a subgroup H ⊂ G, is the augmentation ideal of Z[G] generated by (h − 1) : h ∈ H . Suppose that we fix an isomorphism χ : ab(G) → Zn , then the group ring ±1 ±1 Λ := Z[ab(G)] can be identified with Z[t±1 1 , t2 , . . . , tn ]. It is well-known that the ˜ p˜; Z) is a finitely generated Λ-module, so we choose Alexander module M = H1 (X, a presentation: A → Λl → Λk → M. (1) Let A be the presentation matrix of the Alexander module as above. The ith Alexander ideal of G is the ideal generated by all the (k − i) × (k − i) minors of the presentation matrix A. APPLICATIONS OF THE ALEXANDER IDEALS TO THE ISOMORPHISM PROBLEM 3 The Alexander ideals do not depend on the choices of the CW-complex X and the presentation (1). However we have a freedom in choosing the isomorphism χ above, so the Alexander ideals are invariants of the group G up to a monomial automorphism of Λ. that is an automorphism such that ϕ(ti ) = t1ai1 t2ai2 . . . tanin , i = 1, . . . , n, where (aij ) is a matrix belonging to GL(n, Z). There is an effective algorithm to compute the Alexander modules and ideals by using Fox’s free differential calculus [10] which we will describe briefly below. Suppose that Fk = x1 , . . . , xk | is the free group on k generators. Let ǫ : ZFk → Z be the augmentation homomorphism defined by ǫ( ni gi ) = ni . The j th partial Fox derivative is a linear operator ∂j : ZFk → ZFk which is uniquely determined by the following rules: ∂j (1) = 0; ∂j (xi ) = δij ; ∂j (uv) = ∂j (u)ǫ(v) + u∂j (v). As consequences of the above rules we get: i) ∂i (xni ) = 1 + xi + x2i + · · · + xin−1 for all n ≥ 1. −1 −2 −n ii) ∂i (x−n for all n ≥ 1. i ) = −xi − xi − · · · − xi Let G = x1 , . . . , xk |r1 , . . . , rl be a group as above. From the quotient map φ Fk → G → ab(G) we get a map Φ : ZFk → Λ. The results of Fox’s free differential calculus say that the Jacobian matrix J := (∂j ri )Φ : Λl → Λk is a presentation matrix for the Alexander module of G. Thus, we have an effective method to find the Alexander ideals. 2. Isomorphism problem for the Baumslag-Solitar Groups The Baumslag-Solitar group B(m, n), where m, n are non-zero integers, can be defined by: B(m, n) = a, b a−1 bm a = bn = a, b a−1 bm ab−n = 1 . 4 DO VIET HUNG AND VU THE KHOI The Baumslag-Solitar groups are first given by Baumslag and Solitar [8] as an example of one-relator non-Hopfian groups. The Baumslag-Solitar groups attract attention of many authors since they serve as a rich source of examples and counterexamples for many questions in group theory. The isomorphism problem for Baumslag-Solitar groups was solved by Moldavanskii [14] in 1991. As a first illustration of our method, we will show that the solution of the isomorphism problem for the Baumslag-Solitar groups can be deduced quickly from the computation of Alexander ideals. Proposition 2.1. The first Alexander ideal of the group B(m, n) is ±1 i) I = ((1 + t2 + ... + t2n−1 )(1 − t1 ), 1 − tn2 ) ⊆ Z[t±1 1 , t2 ] in the case m = n > 0; ±1 ±1 ii) I = ((1 + t2 + ... + t2−n−1 )(1 − t1 ), 1 − t−n 2 ) ⊆ Z[t1 , t2 ] in the case m = n < 0; iii) I = (m − nt1 ) ⊆ Z[t1 ] in the case m = n. Proof. To shorten the computations, we note without proof the following fact: B(m, n) ∼ = B(−n, −m). = B(−m, −n) ∼ = B(n, m) ∼ i) If m = n > 0, we can deduce that ab(B(n, n)) ∼ = Zt1 ⊕ Zt2 a → t1 b → t2 . ±1 So the group ring Λ = Z[ab(B(n, n)] ∼ = Z[t±1 1 , t2 ]. Applying Fox’s free differential calculus to the relation r = a−1 bn ab−n , we have: ∂r = −a−1 + a−1 bn , ∂a ∂r = a−1 (1 + b + b2 + ... + bn−1 ) − a−1 bn a(b−1 + ... + b−n ). ∂b Therefore Φ( ∂r ∂r n ) = t−1 ) = (1 + t2 + ... + t2n−1 )(t−1 1 (t2 − 1), Φ( 1 − 1). ∂a ∂b We see that the first Alexander ideal of B(n, n) is I = ((1 + t2 + ... + t2n−1 )(1 − t1 ), 1 − tn2 ). APPLICATIONS OF THE ALEXANDER IDEALS TO THE ISOMORPHISM PROBLEM 5 ii) The case m = n < 0, by using the fact that B(n, n) ∼ = B(−n, −n) and the result of i), we find that the Alexander ideal is I = ((1 + t2 + ... + t2−n−1 )(1 − t1 ), 1 − t−n 2 ). iii) We will divide this case into several subcases: a) m = n and m, n > 0. We see that ab(B(m, n)) ∼ = Zt1 a → t1 b → 0 We get Λ = Z[ab(B(m, n))] ∼ = Z[t±1 1 ]. Applying Fox’s free differential calculus to r = a−1 bm ab−n , we have: ∂r = −a−1 + a−1 bm ∂a ∂r = a−1 (1 + b + b2 + ... + bm−1 ) − a−1 bm a(b−1 + ... + b−n ). ∂b From this computation we get Φ( ∂r ∂r ) = 0, Φ( ) = mt−1 1 − n. ∂a ∂b So the first Alexander ideal is I = (m − nt1 ). b) m = n and m, n < 0. Since B(m, n) ∼ = B(−m, −n), it follows from the subcase a) that the first Alexander ideal is I = (−m + nt) = (m − nt). c) m > 0 and n < 0. This case is almost identical the subcase a) except a small change : ∂r = a−1 (1 + b + b2 + ... + bm−1 ) + a−1 bm a(1 + b + ... + b−n−1 ). ∂b However, straightforward computations show that the first Alexander ideal is still the same I = (m − nt1 ). d) m < 0 and n > 0. We use the fact that B(m, n) ∼ = B(−m, −n) to reduce the computation to the subcase c). So I = (−m + nt1 ) = (m − nt1 ). 6 DO VIET HUNG AND VU THE KHOI Theorem 2.2. (Moldavanskii [14]) The group B(m, n) and B(p, q) are isomorphic if and only if for a suitable ǫ = ±1 either m = pǫ and n = qǫ or m = qǫ and n = pǫ. Proof. We only prove the necessity since the sufficiency is obvious as we already notice above. Now suppose that B(m, n) ∼ = ab(B(p, q)) = B(p, q). As ab(B(m, n)) ∼ we see that m = n implies p = q. So we only need to prove Theorem in the following cases i) The case m = n and p = q. We denote by V (I) = {(t1 , t2 ) ∈ (C∗ )2 | f (t1 , t2 ) = 0 ∀ f ∈ I} the zero locus of the first Alexander ideal I. By the results of Proposition 2.1 part i) and ii), for the group B(n, n), V (I) consists of (|n| − 1) lines {t2 = e 2πik |n| }, k = 1, . . . |n| − 1, and an isolated point {t1 = 1, t2 = 1}. In particular, V (I) has |n| connected components. Therefore B(n, n) ∼ = B(q, q) implies |n| = |q|. So the Theorem is proved in this case. ii) The case m = n and p = q. Using Proposition 2.1, we get the equality of ideals (m − nt1 ) = (p − qt1 ) in Z[t1 ] up to a monomial automorphism. In this case, the monomial automorphism is just changing t ↔ t−1 . So this implies that either m − nt1 = ±(p − qt1 ) or m − nt1 = ±(pt − q1 ). We get the required conclusion. 3. Isomorphism problem for a family of parafree groups Parafree group is a class of groups which shares many similar properties with free groups. However they may not necessarily be free. To give the precise definition, we will need some notations. For a group G, let G = γ1 (G) ≥ γ2 (G) ≥ · · · ≥ γn (G) ≥ · · · be the lower central series which is defined inductively by γn+1(G) := [G, γn (G)]. A group is called residually nilpotent if ∩∞ i=1 γi (G) = 1. A group G is called parafree if it is residually nilpotent and has the same nilpotent quotient as a given free group, that is, there exists a free group F such that APPLICATIONS OF THE ALEXANDER IDEALS TO THE ISOMORPHISM PROBLEM 7 G/γk (G) ∼ = F/γk (F ) for all k. Parafree group is studied for the first time by Baumslag [2]. Since then, several explicit families of parafree groups have been introduced, see [3, 4, 5]. As the parafree groups enjoys many common properties with free groups, the isomorphism problem for parafree groups is very hard. There have been some partial results in the isomorphism problem for groups in the families of parafree groups mention above. In [9] the authors study the isomorphism problem for the family of parafree groups Gi,j := a, b, t| a−1 = [bi , a][bj , t] introduced by Baumslag in [4], here [x, y] := x−1 y −1xy. In particular, they show that Gi,1 ∼ = G1,1 for i > 1 and Gi,1 ∼ = Gj,1 for distinct primes i, j. So far, no other theoretical attack has been carried out. Computational approach to distinguish parafree groups by enumerating homomorphism to a fixed finite group has been done by some authors, see [7, 12]. However, this method can only distinguish finitely many groups in the family. In the following, we use the Alexander ideals to distinguish parafree groups in the family Ki,j which was introduced by Baumslag and Cleary in [6]: Ki,j := a, s, t| ai [s, a] = tj , where i > 0, j > 1 are relatively prime. We compute the second Alexander ideals of the groups in this family in the following proposition. Proposition 3.1. The second Alexander ideal of the group Ki,j above are given by (i−1)j I = (1 − tj2 , (1 + tj2 + . . . + t2 (i−1)j −1 t1 ) − t2 (i−1)j + t2 Proof. Note that ab(Ki,j ) ∼ = Zt1 ⊕ Zt2 s → t1 a → tj2 . t → ti2 . (j−1)i , 1 + ti2 + . . . + t2 ±1 We deduce that the group ring Λ = Z[ab(Ki,j )] ∼ = Z[t±1 1 , t2 ]. ). 8 DO VIET HUNG AND VU THE KHOI Applying Fox’s free differential calculus to the relation r = ai s−1 a−1 sat−j , we have: ∂r = −ai s−1 + ai s−1 a−1 , ∂s ∂r = (1 + a + . . . + ai−1 ) − ai s−1 a−1 + ai s−1 a−1 s, ∂a ∂r = −ai s−1 a−1 sa(t−1 + ... + t−j ). ∂t So, we find that Φ( ∂r (i−1)j −1 −1 ) = −tij t1 , 2 t1 + t2 ∂s ∂r (i−1)j (i−1)j −1 (i−1)j ) = (1 + tj2 + . . . + t2 ) − t2 t1 + t2 , ∂a ∂r (j−1)i ). Φ( ) = −(1 + ti2 + . . . + t2 ∂b From that, the proposition follows. Φ( Now we can solve the isomorphism problem for the family Ki,j . Theorem 3.2. For i > 0, j > 1 relatively prime, all the group Ki,j in the family are distinct. Proof. We first find the zero set V (I) of the second Alexander ideal of Ki,j . It follows from Proposition 3.1 that V (I) is the solution set of the system:    = 0 (1) 1 − tj2    (i−1)j (i−1)j −1 (i−1)j (1 + tj2 + . . . + t2 ) − t2 t1 + t2 = 0 (2)     1 + ti + . . . + t(j−1)i = 0 (3) 2 2 Using the fact that i, j are relatively prime, we get:    tj = 1   2 2πik (1) + (3) ⇐⇒ tij = 1 ⇐⇒ t2 = e j , k = 1, · · · , j − 1. 2    i  t2 = 1 Combining with equation (2), we find that V (I) consists of j − 1 points {(t1 , t2 ) = ( 2πik 1 , e j )}k=1,··· ,j−1 . i+1 APPLICATIONS OF THE ALEXANDER IDEALS TO THE ISOMORPHISM PROBLEM 9 Suppose that Ki,j ∼ = Km,n . By counting the cardinality of V (I), we deduce that j = n. Now, suppose that ϕ : C∗ × C∗ → C∗ × C∗ t1 → ta1 tb2 t2 → tc1 td2 . be the monomial automorphism that maps the zero set of the second Alexander ideal of Ki,j to that of Km,n . That is, (ϕ(t1 ), ϕ(t2 )) ∈ {( 2πik 1 , e n )}k=1,··· ,n−1 m+1 ∀(t1 , t2 ) ∈ {( 2πik 1 , e j )}k=1,··· ,j−1. i+1 1 c From that, we get |ϕ(t2 )| = ( i+1 ) = 1 and therefore c = 0. Because the matrix a b belongs to GL(2, Z), from c = 0, we find that a = ±1. So we must have: c d 1 ±1 1 |ϕ(t1 )| = ( i+1 ) = m+1 . It follows that a = 1 and i = m. In conclusion, we find that if Ki,j ∼ = Km,n then i = m and j = n. Remark. For other families of parafree groups which were investigated in [7], the Alexander ideals are found to be trivial. So the method described here does not work. References [1] Alexander, James W. Topological invariants of knots and links. Trans. Amer. Math. Soc 30 (1928), 275-306. [2] Baumslag, Gilbert. Groups with the same lower central sequence as a relatively free group. I. The groups. Trans. Amer. Math. Soc 129 (1967), 308-321. [3] Baumslag, Gilbert. Some groups that are just about free. Bull. Amer. Math. Soc. 73 (1967), 621-622. [4] Baumslag, Gilbert. Groups with the same lower central sequence as a relatively free group. II. Properties. Trans. Amer. Math. Soc 142 (1969), 507-538. [5] Baumslag, Gilbert. Musings on Magnus. The mathematical legacy of Wilhelm Magnus: groups, geometry and special functions (Brooklyn, NY, 1992), 99106, Contemp. Math., 169, Amer. Math. Soc., Providence, RI, 1994. [6] Baumslag, Gilbert, and Sean Cleary. Parafree one-relator groups. J. Group Theory 9 (2006), no. 2, 191-201. 10 DO VIET HUNG AND VU THE KHOI [7] Baumslag, Gilbert, Cleary, Sean and Havas, George. Experimenting with infinite groups, I. Experiment. Math. 13 (2004), no. 4, 495-502. [8] Baumslag, Gilbert, and Donald Solitar. Some two-generator one-relator non-Hopfian groups. Bull. Amer. Math. Soc 68 (1962), 199-201. [9] Fine, Benjamin, Gerhard Rosenberger, and Michael Stille. The isomorphism problem for a class of para-free groups. Proc. Edinburgh Math. Soc. (2) 40 (1997), no. 3, 541549. [10] Fox. R. H. Free differential calculus I. Annals of Math. 57 (1953), 547560. [11] Hironaka, Eriko. Alexander stratifications of character varieties. Ann. Inst. Fourier (Grenoble) 47 (1997), no. 2, 555583. [12] Lewis, Robert H., and Sal Liriano. Isomorphism classes and derived series of certain almostfree groups. Experiment. Math. 3 (1994), no. 3, 255-258. [13] McMullen, Curtis T. The Alexander polynomial of a 3-manifold and the Thurston norm on cohomology. Ann. Sci. cole Norm. Sup. (4) 35 (2002), no. 2, 153171. [14] Moldavanskii, D. I. On the isomorphism of Baumslag-Solitar groups. Ukrainian Math. J. 43 (1991), no. 12, 15691571 [15] Suciu, A., Fundamental groups, Alexander invariants, and cohomology jumping loci. Topology of algebraic varieties and singularities, 179223, Contemp. Math., 538, Amer. Math. Soc., Providence, RI, 2011. Do Viet Hung, Ha Giang College of Education, Ha Giang, Viet Nam E-mail address: viethunghg81@gmail.com Vu The Khoi (Corresponding author), Institute of Mathematics, 18 Hoang Quoc Viet, 10307, Hanoi, Vietnam E-mail address: vtkhoi@math.ac.vn  ... (k − i) minors of the presentation matrix A APPLICATIONS OF THE ALEXANDER IDEALS TO THE ISOMORPHISM PROBLEM The Alexander ideals not depend on the choices of the CW-complex X and the presentation... that the solution of the isomorphism problem for the Baumslag-Solitar groups can be deduced quickly from the computation of Alexander ideals Proposition 2.1 The first Alexander ideal of the group... properties with free groups, the isomorphism problem for parafree groups is very hard There have been some partial results in the isomorphism problem for groups in the families of parafree groups mention