One-relator group

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In the mathematical subject of group theory, a one-relator group is a group given by a group presentation with a single defining relation. One-relator groups play an important role in geometric group theory by providing many explicit examples of finitely presented groups.

Formal definition

A one-relator group is a group G that admits a group presentation of the form

G=Xr=1 (1)

where X is a set (in general possibly infinite), and where rF(X) is a freely and cyclically reduced word. If Y is the set of all letters xX that appear in r and X=XY then

G=Yr=1F(X).

For that reason X in (1) is usually assumed to be finite where one-relator groups are discussed, in which case (1) can be rewritten more explicitly as

G=x1,,xnr=1, (2)

where X={x1,,xn} for some integer n1.

Freiheitssatz

Let G be a one-relator group given by presentation (1) above. Recall that r is a freely and cyclically reduced word in F(X). Let yX be a letter such that y or y1 appears in r. Let X1X{y}. The subgroup H=X1G is called a Magnus subgroup of G. A famous 1930 theorem of Wilhelm Magnus,[1] known as Freiheitssatz, states that in this situation H is freely generated by X1, that is, H=F(X1). See also[2][3] for other proofs.

Properties of one-relator groups

Here we assume that a one-relator group G is given by presentation (2) with a finite generating set X={x1,,xn} and a nontrivial freely and cyclically reduced defining relation 1rF(X).

  • A one-relator group G is torsion-free if and only if rF(x1,,xn) is not a proper power.
  • Every one-relator group G is virtually torsion-free, that is, admits a torsion-free subgroup of finite index.[4]
  • A one-relator presentation is diagrammatically aspherical.[5]
  • If rF(x1,,xn) is not a proper power then the presentation complex P for presentation (2) is a finite Eilenberg–MacLane complex K(G,1).[6]
  • If rF(x1,,xn) is not a proper power then a one-relator group G has cohomological dimension 2.
  • A one-relator group G is free if and only if rF(x1,,xn) is a primitive element; in this case G is free of rank n − 1.[7]
  • Suppose the element rF(x1,,xn) is of minimal length under the action of Aut(Fn), and suppose that for every i=1,,n either xi or xi1 occurs in r. Then the group G is freely indecomposable.[8]
  • If rF(x1,,xn) is not a proper power then a one-relator group G is locally indicable, that is, every nontrivial finitely generated subgroup of G admits a group homomorphism onto .[9]
  • Every one-relator group G has algorithmically decidable word problem.[10]
  • If G is a one-relator group and HG is a Magnus subgroup then the subgroup membership problem for H in G is decidable.[10]
  • It is unknown if one-relator groups have solvable conjugacy problem.
  • It is unknown if the isomorphism problem is decidable for the class of one-relator groups.
  • A one-relator group G given by presentation (2) has rank n (that is, it cannot be generated by fewer than n elements) unless rF(x1,,xn) is a primitive element.[11]
  • Let G be a one-relator group given by presentation (2). If n3 then the center of G is trivial, Z(G)={1}. If n=2 and G is non-abelian with non-trivial center, then the center of G is infinite cyclic.[12]
  • Let r,sF(X) where X={x1,,xn}. Let N1=rF(X) and N2=sF(X) be the normal closures of r and s in F(X) accordingly. Then N1=N2 if and only if r is conjugate to s or s1 in F(X).[13][14]
  • There exists a finitely generated one-relator group that is not Hopfian and therefore not residually finite, for example the Baumslag–Solitar group B(2,3)=a,bb1a2b=a3.[15]
  • Let G be a one-relator group given by presentation (2). Then G satisfies the following version of the Tits alternative. If G is torsion-free then every subgroup of G either contains a free group of rank 2 or is solvable. If G has nontrivial torsion, then every subgroup of G either contains a free group of rank 2, or is cyclic, or is infinite dihedral.[16]
  • Let G be a one-relator group given by presentation (2). Then the normal subgroup N=rF(X)F(X) admits a free basis of the form {ui1ruiiI} for some family of elements {uiF(X)iI}.[17]

One-relator groups with torsion

Suppose a one-relator group G given by presentation (2) where r=sm where m2 and where 1sF(X) is not a proper power (and thus s is also freely and cyclically reduced). Then the following hold:

  • The element s has order m in G, and every element of finite order in G is conjugate to a power of s.[18]
  • Every finite subgroup of G is conjugate to a subgroup of s in G. Moreover, the subgroup of G generated by all torsion elements is a free product of a family of conjugates of s in G.[4]
  • G admits a torsion-free normal subgroup of finite index.[4]
  • Newman's "spelling theorem"[19][20] Let 1wF(X) be a freely reduced word such that w=1 in G. Then w contains a subword v such that v is also a subword of r or r1 of length |v|=1+(m1)|s|. Since m2 that means that |v|>|r|/2 and presentation (2) of G is a Dehn presentation.
  • G has virtual cohomological dimension 2.[21]
  • G is a word-hyperbolic group.[22]
  • G has decidable conjugacy problem.[19]
  • G is coherent, that is every finitely generated subgroup of G is finitely presentable.[23]
  • The isomorphism problem is decidable for finitely generated one-relator groups with torsion, by virtue of their hyperbolicity.[24]
  • G is residually finite.[25]
  • G is virtually free-by-cyclic, i.e. G has a subgroup H of finite-index such that there is a free normal subgroup FH with cyclic quotient F/H.[26]

Magnus–Moldavansky method

Starting with the work of Magnus in the 1930s, most general results about one-relator groups are proved by induction on the length |r| of the defining relator r. The presentation below follows Section 6 of Chapter II of Lyndon and Schupp[27] and Section 4.4 of Magnus, Karrass and Solitar[28] for Magnus' original approach and Section 5 of Chapter IV of Lyndon and Schupp[29] for the Moldavansky's HNN-extension version of that approach.[30] Let G be a one-relator group given by presentation (1) with a finite generating set X. Assume also that every generator from X actually occurs in r. One can usually assume that #X2 (since otherwise G is cyclic and whatever statement is being proved about G is usually obvious). The main case to consider when some generator, say t, from X occurs in r with exponent sum 0 on t. Say X={t,a,b,,z} in this case. For every generator xX{t} one denotes xi=tixti where i. Then r can be rewritten as a word r0 in these new generators X={(ai)i,(bi)i,,(zi)i} with |r0|<|r|. For example, if r=t2btat3b2a2t1at1 then r0=b2a1b22a22a1. Let X0 be the alphabet consisting of the portion of X given by all xi with m(x)iM(x) where m(x),M(x) are the minimum and the maximum subscripts with which xi±1 occurs in r0. Magnus observed that the subgroup L=X0G is itself a one-relator group with the one-relator presentation L=X0r0=1. Note that since |r0|<|r|, one can usually apply the inductive hypothesis to L when proving a particular statement about G. Moreover, if Xi=tiX0ti for i then Li=Xi=Xi|ri=1 is also a one-relator group, where ri is obtained from r0 by shifting all subscripts by i. Then the normal closure N=X0G of X0 in G is

N=iLi.

Magnus' original approach exploited the fact that N is actually an iterated amalgamated product of the groups Li, amalgamated along suitably chosen Magnus free subgroups. His proof of Freiheitssatz and of the solution of the word problem for one-relator groups was based on this approach. Later Moldavansky simplified the framework and noted that in this case G itself is an HNN-extension of L with associated subgroups being Magnus free subgroups of L. If for every generator from X0 its minimum and maximum subscripts in r0 are equal then G=Lt and the inductive step is usually easy to handle in this case. Suppose then that some generator from X0 occurs in r0 with at least two distinct subscripts. We put Y to be the set of all generators from X0 with non-maximal subscripts and we put Y+ to be the set of all generators from X0 with non-maximal subscripts. (Hence every generator from Y and from Y occurs in r0 with a non-unique subscript.) Then H=Y and H+=Y+ are free Magnus subgroups of L and t1Ht=H+. Moldavansky observed that in this situation

G=L,tt1Ht=H+

is an HNN-extension of L. This fact often allows proving something about G using the inductive hypothesis about the one-relator group L via the use of normal form methods and structural algebraic properties for the HNN-extension G. The general case, both in Magnus' original setting and in Moldavansky's simplification of it, requires treating the situation where no generator from X occurs with exponent sum 0 in r. Suppose that distinct letters x,yX occur in r with nonzero exponents α,β accordingly. Consider a homomorphism f:F(X)F(X) given by f(x)=xyβ,f(y)=yα and fixing the other generators from X. Then for r=f(r)F(X) the exponent sum on y is equal to 0. The map f induces a group homomorphism ϕ:GG=Xr=1 that turns out to be an embedding. The one-relator group G' can then be treated using Moldavansky's approach. When G splits as an HNN-extension of a one-relator group L, the defining relator r0 of L still turns out to be shorter than r, allowing for inductive arguments to proceed. Magnus' original approach used a similar version of an embedding trick for dealing with this case.

Two-generator one-relator groups

It turns out that many two-generator one-relator groups split as semidirect products G=Fm. This fact was observed by Ken Brown when analyzing the BNS-invariant of one-relator groups using the Magnus-Moldavansky method. Namely, let G be a one-relator group given by presentation (2) with n=2 and let ϕ:G be an epimorphism. One can then change a free basis of F(X) to a basis t,a such that ϕ(t)=1,ϕ(a)=0 and rewrite the presentation of G in this generators as

G=a,tr=1

where 1r=r(a,t)F(a,t) is a freely and cyclically reduced word. Since ϕ(r)=0,ϕ(t)=1, the exponent sum on t in r is equal to 0. Again putting ai=tiati, we can rewrite r as a word r0 in (ai)i. Let m,M be the minimum and the maximum subscripts of the generators occurring in r0. Brown showed[31] that ker(ϕ) is finitely generated if and only if m<M and both am and aM occur exactly once in r0, and moreover, in that case the group ker(ϕ) is free. Therefore if ϕ:G is an epimorphism with a finitely generated kernel, then G splits as G=Fm where Fm=ker(ϕ) is a finite rank free group. Later Dunfield and Thurston proved[32] that if a one-relator two-generator group G=x1,x2r=1 is chosen "at random" (that is, a cyclically reduced word r of length n in F(x1,x2) is chosen uniformly at random) then the probability pn that a homomorphism from G onto with a finitely generated kernel exists satisfies

0.0006<pn<0.975

for all sufficiently large n. Moreover, their experimental data indicates that the limiting value for pn is close to 0.94.

Examples of one-relator groups

Generalizations and open problems

  • If A and B are two groups, and rAB is an element in their free product, one can consider a one-relator product G=AB/r=A,Br=1.
  • The so-called Kervaire conjecture, also known as Kervaire–Laudenbach conjecture, asks if it is true that if A is a nontrivial group and B=t is infinite cyclic then for every rAB the one-relator product G=A,tr=1 is nontrivial.[33]
  • Klyachko proved the Kervaire conjecture for the case where A is torsion-free.[34]
  • A conjecture attributed to Gersten[22] says that a finitely generated one-relator group is word-hyperbolic if and only if it contains no Baumslag–Solitar subgroups.

See also

Sources

  • Wilhelm Magnus, Abraham Karrass, Donald Solitar, Combinatorial group theory. Presentations of groups in terms of generators and relations, Reprint of the 1976 second edition, Dover Publications, Inc., Mineola, NY, 2004. ISBN 0-486-43830-9. MR2109550
  • Lyndon, Roger C.; Schupp, Paul E. (2001). Combinatorial group theory. Classics in Mathematics. Springer-Verlag, Berlin. ISBN 3-540-41158-5. MR 1812024.

References

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  2. Lyndon, Roger C. (1972). "On the Freiheitssatz". Journal of the London Mathematical Society. Second Series. 5: 95–101. doi:10.1112/jlms/s2-5.1.95. hdl:2027.42/135658. MR 0294465.
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