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

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 Develop-men(NAFOSTED) under grant number 101.01-2011.46.

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 writ-ing 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 = hx1, , xk|r1, , rli 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 such that π1(X, p) = G Let π : ˜X → X be the covering corresponding to φ : G → ab(G) and ˜p := π− 1(p) We denote by H1( ˜X, ˜p; Z) the relative homology Then the desk transformation action of ab(G) on ˜X makes H1( ˜X, ˜p; Z) an Z[ab(G)]−module, 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(h − 1) : h ∈ Hi

Suppose that we fix an isomorphism χ : ab(G) → Zn, then the group ring

Λ := Z[ab(G)] can be identified with Z[t± 1

1 , t± 1

2 , , t± 1

n ] It is well-known that the Alexander module M = H1( ˜X, ˜p; Z) is a finitely generated Λ-module, so we choose

a presentation:

→ Λl A→ Λk→ M (1) Let A be the presentation matrix of the Alexander module as above The ithAlexander ideal of G is the ideal generated by all the (k − i) × (k − i) minors of the presentation matrix A

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) = tai1

1 tai2

2 ta in

n , 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 = hx1, , xk|i is the free group on k generators Let ǫ : ZFk→ Z

be the augmentation homomorphism defined by ǫ(P nigi) = P ni The jth 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(xn

i) = 1 + xi+ x2

i + · · · + xn−1i for all n ≥ 1

ii) ∂i(x− n

i ) = −x− 1

i − x− 2

i − · · · − x− n

i for all n ≥ 1

Let G = hx1, , xk|r1, , rli be a group as above From the quotient map

Fk → G→ ab(G) we get a map Φ : ZFφ k → Λ

The results of Fox’s free differential calculus say that the Jacobian matrix

J := (∂jri)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:

The Baumslag-Solitar groups are first given by Baumslag and Solitar [8] as an ex-ample of one-relator non-Hopfian groups The Baumslag-Solitar groups attract at-tention of many authors since they serve as a rich source of examples and counterex-amples for many questions in group theory

The isomorphism problem for Baumslag-Solitar groups was solved by Moldavan-skii [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

i) I = ((1 + t2+ + tn−12 )(1 − t1), 1 − tn

2) ⊆ Z[t± 1

1 , t± 1

2 ] in the case m = n > 0; ii) I = ((1 + t2+ + t− n−1

2 )(1 − t1), 1 − t− n

2 ) ⊆ Z[t± 1

1 , t± 1

2 ] in the case m = n < 0; iii) I = (m − nt1) ⊆ Z[t1] in the case m 6= 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)) a b

∼ 7→

7→

Zt1⊕ Zt2

t1

t2

So the group ring Λ = Z[ab(B(n, n)] ∼= Z[t±11, t± 1

2 ]

Applying Fox’s free differential calculus to the relation r = a− 1bnab− n, we have:

∂r

∂a = −a

− 1+ a− 1bn,

∂r

∂b = a

− 1(1 + b + b2+ + bn−1) − a− 1bna(b− 1+ + b− n)

Therefore

Φ(∂r

∂a) = t

− 1

1 (tn2 − 1), Φ(∂r

∂b) = (1 + t2 + + t

n−1

2 )(t−11− 1)

We see that the first Alexander ideal of B(n, n) is I = ((1 + t2 + + tn−12 )(1 −

t1), 1 − tn

2)

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 + + t− n−1

2 )(1 − t1), 1 − t− n

2 )

iii) We will divide this case into several subcases:

a) m 6= n and m, n > 0 We see that

ab(B(m, n)) a b

∼ 7→

7→

Zt1

t1

0

We get Λ = Z[ab(B(m, n))] ∼= Z[t±11]

Applying Fox’s free differential calculus to r = a− 1bmab− n, we have:

∂r

∂a = −a

− 1+ a− 1bm

∂r

∂b = a

− 1(1 + b + b2+ + bm−1) − a−1bma(b−1 + + b−n)

From this computation we get

Φ(∂r

∂a) = 0, Φ(

∂r

∂b) = mt

− 1

1 − n

So the first Alexander ideal is I = (m − nt1)

b) m 6= 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

∂b = a

− 1(1 + b + b2+ + bm−1) + a− 1bma(1 + b + + b− n−1)

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)

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) ∼= B(p, q) As ab(B(m, n)) ∼= ab(B(p, q))

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 = e2πik|n| }, k =

1, |n| − 1, and an isolated point {t1 = 1, t2 = 1} In particular, V (I) has |n| con-nected components Therefore B(n, n) ∼= B(q, q) implies |n| = |q| So the Theorem

is proved in this case

ii) The case m 6= n and p 6= 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=1γi(G) = 1

A group G is called parafree if it is residually nilpotent and has the same nilpo-tent quotient as a given free group, that is, there exists a free group F such that

G/γk(G) ∼= F/γk(F ) for all k Parafree group is studied for the first time by Baum-slag [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 := ha, b, t| a− 1 = [bi, a][bj, t]i introduced by Baumslag in [4], here [x, y] := x− 1y− 1xy In particular, they show that Gi,1 6∼= G1,1 for i > 1 and

Gi,1 6∼= 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 := i[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 − tj2, (1 + tj2 + + t(i−1)j2 ) − t(i−1)j2 t− 1

1 + t(i−1)j2 , 1 + ti

2+ + t(j−1)i2 ) Proof Note that

ab(Ki,j) s a t

∼ 7→

7→

7→

Zt1⊕ Zt2

t1

tj2

ti

2

Applying Fox’s free differential calculus to the relation r = ais− 1a− 1sat− j, we have:

∂r

∂s = −a

i

s− 1+ ais− 1a− 1,

∂r

∂a = (1 + a + + a

i−1) − ais− 1a− 1+ ais− 1a− 1s,

∂r

∂t = −a

is− 1a− 1sa(t− 1+ + t− j)

So, we find that

Φ(∂r

∂s) = −t

ij

2t− 1

1 + t(i−1)j2 t− 1

1 , Φ(∂r

∂a) = (1 + t

j

2+ + t(i−1)j2 ) − t(i−1)j2 t−11+ t(i−1)j2 , Φ(∂r

∂b) = −(1 + t

i

2 + + t(j−1)i2 )

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:















(1 + tj2+ + t(i−1)j2 ) − t(i−1)j2 t−11+ t(i−1)j2 = 0 (2)

1 + ti

2+ + t(j−1)i2 = 0 (3) Using the fact that i, j are relatively prime, we get:

(1) + (3) ⇐⇒















tj2 = 1

tij2 = 1

ti

2 6= 1

⇐⇒ t2 = e2πikj , k = 1, · · · , j − 1

Combining with equation (2), we find that V (I) consists of j − 1 points

{(t1, t2) = ( 1

i + 1, e

2πik

j )}k=1,··· ,j−1

Suppose that Ki,j ∼= Km,n By counting the cardinality of V (I), we deduce that

j = n Now, suppose that

ϕ : C∗

× C∗

t1

t2

→ 7→

7→

C∗× C∗

ta

1tb 2

tc

1td

2

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)) ∈ {( 1

m + 1, e

2πik

n )}k=1,··· ,n−1 ∀(t1, t2) ∈ {( 1

i + 1, e

2πik

j )}k=1,··· ,j−1 From that, we get |ϕ(t2)| = ( 1

i+1)c = 1 and therefore c = 0 Because the matrix

a b

c d

!

belongs to GL(2, Z), from c = 0, we find that a = ±1 So we must have:

|ϕ(t1)| = ( 1

i+1)± 1 = 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

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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