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(OR1)  For a background to this problem, see the survey [G.Baumslag, Some open problems. Summer School in Group Theory in Banff, 1996, 1--9.  CRM Proceedings and Lecture Notes. 17. Amer. Math. Soc., Providence, 1999].

(OR2) For some partial results, see the background to the problem (O4).

(OR5)  J.Harlander [Solvable groups with cyclic relation module,  J. Pure Appl. Algebra 90 (1993), 189--198]  showed that the answer is "yes" in the case where $G$ is finitely generated and solvable.

(OR7) (b), (c) A solution of these two problems was communicated to us by A.Olshanskii. In fact, the commutator subgroup [F,F] here can be replaced by ANY non-cyclic subgroup of a free group F; the answer will still be negative. It follows from a result of [A.Olshanskii, SQ-universality of hyperbolic groups, Mat. Sb. 186 (1995), no. 8, 119--132] that for any m, every non-cyclic subgroup H of F contains a subgroup K, which is a free group of rank m, with the following property: for any normal subgroup U of K, the normal closure of U in F intersects K by the group U. To apply this result to our situation, take two elements, x and y, that generate a subgroup K=F_2 of H = [F,F] with the property described above. Let r be a Baumslag-Solitar relator built on these two elements; for example, take r = xyx^{-1}y^{-2}. Let U be the normal closure (in K) of r. Then, from what is said in the previous paragraph, it follows that the normal closure of U in F (call it V) intersects K by U. Therefore, the (one-relator) group F/V contains a subgroup KV/V which is isomorphic to a Baumslag-Solitar group, hence F/V can be neither residually finite nor automatic.

(OR8)  This problem, as well as (OR7)(a), is motivated by the desire  to find a non-hopfian one-relator group which is essentially different from any of the Baumslag-Solitar groups [G.Baumslag, D.Solitar,  Some two-generator one-relator non-Hopfian groups. Bull. Amer. Math. Soc. 68 (1962), 199--201].

(OR10) Note that hyperbolic groups are automatic, and, in particular, an amalgamated product of two free groups with finitely generated subgroups amalgamated is hyperbolic if at least one of the subgroups is malnormal [O.Kharlampovich and A.Myasnikov,  Hyperbolic groups and free constructions. Trans. Amer.  Math. Soc.  350 (1998),  571--613].
  Furthermore, an amalgamated product of two finitely generated abelian groups is automatic [G.Baumslag, S.M.Gersten, M.Shapiro, H.Short,   Automatic groups and amalgams -- a survey.  Algortims and Classification in Combinatorial Group Theory (Berkeley, CA, 1989), 179--194,  Math. Sci. Res. Inst. Publ. 23. Springer, New York, 1992].
 
 (OR14)  We note that recently, O.Kharlampovich and A.Miasnikov [Irreducible affine varieties over groups, J.Algebra 200 (1998), 517-570] proved that every finitely generated group which is discriminated by a free group can be obtained from a free group by applying finitely many free constructions of a very particular type.
 
 

(FP2)  We note that there are  infinitely presented (but finitely generated) groups with this property - see  [J.M.T.Jones,  Direct products and the Hopf property.  J. Austral. Math. Soc. 17 (1974), 174--196].  Moreover, the same author has constructed, for  any $n \ge 2$,  a (infinitely presented) group $G$ isomorphic to its $n$th direct power $G^n$,  but non-isomorphic to  $G^k$ for any $k, 1<k<n$ -- see  [J.M.T.Jones,   On isomorphisms  of direct powers. Word problems, II (Conf. on Decision Problems in Algebra,  Oxford, 1976),  215--245,  Stud. Logic Foundations Math.,  95, North-Holland,  1980].

(FP3) A paper by M.Bestvina, N.Brady with a negative solution of this problem has been published recently  [Invent. Math. 129 (1997),  445--470].  Explicit presentations of their groups have been given by W.Dicks and I.Leary [Presentations for subgroups of Artin groups, Proc. Amer. Math. Soc. 127 (1999), 343--348].

(FP4)  It is easy to see that the answer is affirmative if the group has rank 2; this is due to the fact that in the free group of rank 2, any element is in the normal closure of a primitive element.

(FP7) It is -- see  [O.Kharlampovich, A.Myasnikov, Irreducible affine varieties over groups. I, II.  J.Algebra  200 (1998),  472--516, 517--570].

(FP8) It is;  see the preprint [M.Feighn, M.Handel, Mapping tori of free group automorphisms are coherent]  at  http://andromeda.rutgers.edu/~feighn/

(FP14) No, it is not; see a recent preprint [J. Howie, A proof of the Scott-Wiegold conjecture on free products of cyclic groups] at http://www.ma.hw.ac.uk/~jim/

(FP15) A single element g \in G is called a test element (see [V. Shpilrain, Recognizing automorphisms of the free groups, Arch. Math. 62 (1994), 385--392]) if, whenever \phi(g)=g for some endomorphism \phi of the group G, this \phi is an automorphism of G. Thus, if G has a test element, the test rank of G is 1. For example, any free group of finite rank has test rank 1. On the other hand, there are groups (for example, free abelian groups of finite rank) whose test rank equals their rank. (Obviously, it cannot be bigger than that.)
  E. I. Timoshenko [Test elements and test rank of a free metabelian group, Siberian Math. J. 41 (2000), 1200--1204] proved that a free metabelian group of rank > 2 has test rank 2.
  C.F.Rocca Jr. and E.Turner have shown that for any pair of integers (k, n) with 1 \le k \le n, there are finitely generated abelian groups of rank n and test rank k. Their preprint is available here.

(FP18) R.Hirshon himself [Some properties of endomorphisms in residually finite groups, J. Austral. Math. Soc. Ser. A 24 (1977),  117--120 ] proved that in the case where f(G) has finite index in G. However, the answer is negative in general [D.Wise, A continually descending endomorphism of a finitely generated residually finite group,  Bull. London Math. Soc. 31 (1999),  45--49].

(FP19) J.Makowsky [On some conjectures connected with complete sentences, Fund. Math. 81 (1974), 193-202] pointed out that the affirmative answer to this problem would give an example of a complete finitely axiomatizable theory T which is categorical in uncountable cardinals but not $\omega$--categorical. Subsequently, examples of such theories were given by Peretyat'kin (1980) and others (see [Hodges, Model theory, p. 619]).
  We also note that S.V.Ivanov has constructed examples, for big numbers p, of finitely generated (but infinitely presented) infinite groups of period p with precisely p conjugacy classes. These examples are included as Theorem 41.2 in [A.Yu.Ol'shanskii, Geometry of defining relations in groups. Mathematics and its Applications (Soviet Series), 70. Kluwer Academic Publishers Group, Dordrecht, 1991].
 

(B1), (B2) There are two canonical representations of braid groups by matrices over Laurent polynomial rings - the Burau and Gassner representations (the latter is actually a representation of the pure braid group which is a subgroup of finite index in the whole braid group). Both of these representations are faithful for n = 2,3 (a general reference here is [J.S.Birman, Braids, links and mapping class groups, Ann. Math. Studies 82, Princeton Univ. Press, 1974]). A proof of the Gassner representation being faithful for every n (which implies braid groups being linear) was recently claimed in [S.Bachmuth, Braid groups are linear groups, Adv. Math. 121 (1996), 50--61]. However, there is a controversy around this paper since several people believe they have found essential gaps in the proof (see J. S. Birman's review article 98h:20061  in Math. Reviews).  This makes us consider Problem (B2) open.
  Problem (B1) has been recently settled in the affirmative by S.Bigelow [Braid groups are linear, J. Amer. Math. Soc. 14 (2001), 471--486] and D. Krammer [Braid groups are linear, Ann. of Math., to appear], who proved that the Krammer representation of the braid group B_n is faithful for every n, and therefore all braid groups are linear. For n=4, see also [D. Krammer, The braid group B_4 is linear, Invent. Math. 142 (2000), 451--486].

(B3) The Burau representation was shown to be non-faithful for n \ge 10 in [J.Moody, The faithfulness question for the Burau representation, Proc. Amer. Math. Soc. 119 (1993), 671--679], and then for n \ge 6 in [D.Long and M.Paton, The Burau representation is not faithful for n \ge 6, Topology 32 (1993), 439-447].  More recently, S.Bigelow [The Burau representation is not faithful for n=5, Geom. Topol. 3 (1999), 397--404 (electronic)] has shown that the answer is negative for  n = 5 as well.
  On the other hand, it is known that the Burau representation is faithful for n = 3 [W.Magnus, A.Peluso,  On a theorem of V. I. Arnold. Comm. Pure Appl. Math.  22 (1969),  683-692].

(B5), (B6), (B7) For a background and discussion on these problems, we refer to a recent preprint by V.Lin on "Braids, permutations, polynomials.I" which can be either found on the Max Planck Institut f\"{u}r Mathematik electronic preprint server, or requested from the author at vlin@techunix.technion.ac.il
  Here we only note that automorphisms of braid groups were described in [J.Dyer,  E.Grossman, The automorphism groups of the braid groups, Amer. J. Math. 103 (1981), no. 6, 1151--1169. ]
  Recently, S.Humphries [Torsion-free quotients of braid groups. Internat. J. Algebra Comput. 11 (2001), 363--373] has constructed a representation of the group B_n which is shown to provide torsion-free non-abelian factor groups of B_n  as well as of the commutator subgroup  [B_n, B_n]  for n < 7.  It is likely that the same representation should work for other values of n as well.
 
(B11) An affirmative answer to this problem  would also imply a solution of the problem (B6).

(B12) The groups B_3 and B_4 do have non-elementary hyperbolic factor groups (note that B_4 maps onto B_3, but, say, B_5 does not map onto B_4).
  We also note that the pure braid group P_n can be mapped onto the free group F_2 for any n > 2.
 

(G1) The point here is that there are examples of groups of intermediate growth (between polynomial and exponential), but all these groups are infinitely presented -- see [R.I.Grigorchuk, On the Milnor problem of group growth. (Russian) Dokl. Akad. Nauk SSSR 271 (1983), 30--33]; [R.I.Grigorchuk, Construction of p-groups of intermediate growth that have a continuum of factor-groups. (Russian) Algebra i Logika 23 (1984), 383--394, 478]; [R.I.Grigorchuk, Degrees of growth of p-groups and torsion-free groups. (Russian) Mat.Sb. 126(168) (1985), 194--214, 286]; [R.I.Grigorchuk, A.Maki, On a group of intermediate growth that acts on a line by homeomorphisms. (Russian) Mat. Zametki 53 (1993), 46--63; translation in Math. Notes 53 (1993), 146--157].

(G3) A group G has uniformly exponential growth if there is   e>0   such that the growth rate of G with respect to any generating set is   > 1+e.
  This problem was settled affirmatively for the following classes of groups:
- hyperbolic groups [M. Koubi, Croissance uniforme dans les groupes hyperboliques, Ann. Inst. Fourier 48 (1998), 1441--1453]
- one-relator groups [R. I. Grigorchuk, P. de la Harpe, One-relator groups of exponential growth have uniformly exponential growth, Math. Notes 69 (2001), 628--630]
- solvable groups [D.V. Osin, The entropy of solvable groups, Ergodic Theory and Dynam. Sys., to appear]

(G8) R. Sharp [Local limit theorems for free groups, Math. Ann. 321 (2001), 889--904] showed that the asymptotics of f(n) is   C (2r-1)^n / n^{r/2}   for some constant C.

(E2)   It is clear from considering abelianization that, if $G=F_n/R$ is a counterexample, then $G$ must be perfect, i.e., $G=[G,G]$.
Also, it suffices to consider the case where $G$ is an infinite simple group.
  There are several related problems about (systems of) equations over groups. We only give one of them here; it appears as Problem 2a on Lyndon's list  [R.Lyndon, Problems in combinatorial group theory. Combinatorial group theory and topology (Alta, Utah, 1984), 3--33, Ann.
of Math. Stud., 111, Princeton Univ. Press, 1987]:

   If, in the above notation, the sum of exponents on $x_{n+1}$ in $s$  is not 0, does the equation $s=1$ always have a solution over $G$?

   For recent results on the latter problem, we refer to [A.A.Klyachko,  A funny property of sphere and equations over groups. Comm. Algebra 21 (1993),  2555--2575] and  [A.Clifford,   R.Z.Goldstein,  Tesselations of $S\sp 2$ and equations over torsion-free groups. Proc. Edinburgh Math. Soc. (2)  38 (1995),  485--493].
  For other related  problems, we refer to [J.Howie, How to generalize one-relator group theory. Combinatorial group theory and topology (Alta, Utah, 1984), 53--78,  Ann. of Math. Stud.,  111, Princeton Univ. Press, 1987].

(E3)  The answer was recently shown to be negative [T.Coulbois , A.Khelif,  Equations in free groups are not finitely approximable, Proc. Amer. Math. Soc. 127 (1999),  963-965].

(E5) R.Bryant [The verbal topology of a group. J.Algebra {\bf 48} (1977), 340--346] and V.Guba [Equivalence of infinite systems of equations in free groups and semigroups to finite subsystems. Math. Notes USSR  40 (1986),   688--690] proved that free groups are equationally noetherian. See also  [J.Stallings, Finiteness properties of matrix representations. Ann. of Math. (2) 124 (1986),   337--346].
  For a general discussion on this and related problems, we refer to [G.Baumslag, A.Myasnikov, V.Remeslennikov, Algebraic geometry over groups I: Algebraic sets and ideal theory, J. Algebra 219 (1999), 16--79].
 
 

(A4) We just note here that importance of this problem is enhanced by a well-known fact that the knot group of a knot K is (infinite) cyclic if and only if K is isotopic to the unknot.
See also [ W.Whitten, Knot complements and groups, Topology 26 (1987), 41--44] for a generalization of this fact implying, in particular, that the knot group of a prime knot determines the knot type up to the mirror reflection.

 
(MA2) M.Evans [Primitive elements in the free metabelian group of rank 3, J.Algebra 220 (1999), 475--491] has recently found such matrices.  It was previously known due to [S.Bachmuth, H.Mochizuki,   $E\sb{2}\not={\rm SL}\sb{2}$ for most Laurent polynomial rings. Amer. J. Math. 104 (1982),  1181--1189]  that such matrices do exist.
 

(H1)  I.Kapovich and  D.Wise [The equivalence of some residual properties of word-hyperbolic groups, J. Algebra 223 (2000), 562--583] proved the equivalence of (a) and (b).

(H3) We note that Sela [Ann. of Math. (2) 141 (1995), 217--283] has solved the isomorphism problem for torsion-free hyperbolic groups that do not split (as an amalgamated product or an HNN extension) over the trivial or the infinite cyclic group.

(H5) Papasoglu [An algorithm detecting hyperbolicity.  Geometric and computational perspectives on infinite groups (Minneapolis, MN and New Brunswick, NJ, 1994), 193--200,  DIMACS Ser. Discrete Math. Theoret. Comput. Sci., 25,
Amer. Math. Soc., Providence, RI, 1996] gave  a partial algorithm to recognize hyperbolic groups. Given a finite presentation $\langle S,R\rangle$, the algorithm terminates if the group $G=\langle S,R\rangle$ is hyperbolic and gives an estimate of the hyperbolicity constant $\delta$.

(H6)-(H10)  For a background on problems (H6) through (H10) we refer to  [S. M.Gersten,  Problems on automatic groups. Algorithms and classification in combinatorial group theory (Berkeley, CA, 1989), 225--232, Math. Sci. Res. Inst. Publ., 23, Springer, New York, 1992.]

(H12)  For a background on  equationally noetherian groups, we refer to  [G.Baumslag,  A.Myasnikov,  V.Roman'kov,  Two theorems about equationally Noetherian groups,  J. Algebra 194 (1997), 654--664.]

(H14) We note that malnormality is decidable in free groups -- see  [G.Baumslag,  A.Myasnikov, V.Remeslennikov, Malnormality is decidable in free groups, Internat. J. Algebra Comput. 9 (1999), 687--692].
  Recently, M.Bridson and D.Wise [Malnormality is undecidable in hyperbolic groups] have settled the general case in the negative. Their preprint is available here.
 
 

(N1) V.Bludov has communicated the following example of a non-trivial element  g   of a free nilpotent group of rank 2  and nilpotency class  4k \ge 8,  which is fixed by every automorphism:
g = [a,[a,b],[a,b,b], [a,b],...,[a,b]],  where there are  (2k-3)  occurrences of  [a,b]  after  [a,b,b].  (Here  a  and  b  are generators of the free nilpotent group).
  Recently, A. Papistas [A note on fixed points of certain relatively free nilpotent groups, Comm. Algebra 29 (2001), 4693--4699] and, independently, E.Formanek [Fixed Points and Centers of Automorphism Groups of Free Nilpotent Groups, Comm. Algebra, to appear] have solved this problem completely by classifying all pairs   (r,c)   for which F(r,c), the free nilpotent group of rank   r   and class   c, has nontrivial elements fixed by all automorphisms. Formanek's preprint is available here.

(N2) Let $P = \langle X \mid R \rangle$ be a finite presentation of a group $G$, $F(X)$ a free group on $X$, and $ncl(R)$ the normal closure of $R$ in $F(X)$. The ``area" of $w \in ncl(R)$ is defined by $$A(w) = min\{\ m \ \mid \ w = \prod_{i = 1}^{m}c_i^{-1}r_i^{e_i}c_i , \ c_i \in F(X), r_i \in R, e_i = \pm 1\}.$$ Now, the {\em isoperimetric function} of the presentation $P$ is given by $$ \Phi_P(n) = max\{\ A(w) \ \mid \ w \in ncl(R), \ |w| \leq n \}, $$ where $|w|$ is the length of $w$ in $F(X).$ Let $N$ be the set of all non-negative integers. For functions $f,h : N \rightarrow N$ we define a relation $f \preceq h$ iff there exists a constant $K$ such that $f(n) \leq K \cdot h(Kn) + Kn$ for every $n \in N.$ We write $f \simeq h$ iff $f \preceq h$ and $h \preceq f$. It is not hard to show that if $P$ and $Q$ are two finite presentations of a group $G$, then $\Phi_P \simeq \Phi_Q$. Any function equivalent to $\Phi_P$ is called the {\em Dehn function } of $G$. From now on, we shall denote the Dehn function of a group $G$ by $\Phi_G$. S.Gersten [{\it Isodiametric and isoperimetric inequalities in group extensions}. Preprint, University of Utah, 1991] proved that for any finitely generated nilpotent group $G$, $\Phi_G$ is bounded by a polynomial of degree $2^h$, where $h$ is the Hirsch length of $G$. G.Conner [{\it Central extensions of word hyperbolic groups satisfy a quadratic isoperimetric inequality}, Arch. Math. {\bf 65} (1995), 465--479] improved the bound on the degree to $2^c$, where $c$ is the nilpotency class of $G$. Recently, C.Hidber [{\it Isoperimetric functions of finitely generated nilpotent groups and their amalgams}, Ph.D. thesis] proved that $\Phi_G \preceq n^{2c}$. It is known that if $G$ is a free nilpotent group of class $c$, then $\Phi_G \simeq n^{c+1}$, in particular, $\Phi_G$ is equivalent to a polynomial. Whether or not this is true in general, is still an open problem.

 (N4) Hamkins [Every group has a terminating transfinite automorphism tower, Proc. Amer. Math. Soc. 126 (1998), 3223--3226] established the property in the title.
 

 (M1) There is an algorithm to determine whether or not a given finitely generated metabelian group is free metabelian -- see [J.R.J.Groves, C.F.Miller, III, Recognizing free metabelian groups, Illinois J. Math. 30 (1986), 246--254] and the paper by Noskov cited in the background to the problem (F14).
We also note that "most" algorithmic problems about finitely presented metabelian groups have solutions by now - see  [G.Baumslag, F.Cannonito,  D.Robinson,  The algorithmic theory of finitely generated metabelian groups, Trans. Amer. Math. Soc. 344 (1994),  629--648]  and references thereto.

(M2) The automorphism group of a free metabelian group of finite rank is known to be finitely generated unless the rank equals 3 -- see   [S.Bachmuth, H.Mochizuki,  ${\rm Aut}(F)\to{\rm Aut}(F/F")$ is surjective for free group $F$ of rank $\geq 4$,  Trans. Amer. Math. Soc. 292 (1985),  81--101]   and  [S.Bachmuth, H.Mochizuki,  The nonfinite generation of ${\rm Aut}(G)$, $G$ free metabelian of rank $3$, Trans. Amer. Math. Soc. 270 (1982),  693--700.]

(M4) For groups of finite rank, the answer is affirmative -- see [V. A.Artamonov,  Projective metabelian groups and Lie algebras. (Russian) Izv. Akad. Nauk SSSR Ser. Mat. 42 (1978),  226--236, 469. ]

(M5) For a survey on homological properties of metabelian groups, we refer to [Yu.V.Kuz'min, Homology theory of free abelianized extensions, Comm. Algebra 16 (1988), 2447--2533].

(M6) For a background, we refer to [V.Shpilrain,  Fixed points of endomorphisms of a free metabelian group, Math. Proc. Cambridge Philos. Soc. 123 (1998),  77--85.]
The problem was recently answered in the negative by M. Kassabov; his preprint is available here.

(M7) No, there is no such group - see [R.Goebel, A.Paras, Outer automorphism groups of metabelian groups, J. Pure Appl. Algebra 149 (2000), 251--266.]

(S1) The automorphism group of a free solvable group of derived length > 2 and rank > 2 cannot be generated by elementary Nielsen automorphisms -- see [C. K.Gupta,  F. Levin, Tame range of automorphism groups of free polynilpotent groups, Comm. Algebra 19 (1991),  2497--2500]  and [V.Shpilrain, Automorphisms of $F/R'$ groups, Internat. J. Algebra Comput. 1 (1991), 177--184. ]  Moreover, every free solvable group of derived length d > 2 and rank  r> 2  has automorphisms that cannot be lifted to automorphisms of the free solvable group of derived length d+1  and the same rank  r  -- see  [V.Shpilrain,   Non-commutative determinants and automorphisms of groups, Comm. Algebra 25 (1997),  559--574.]
   It is not known however whether or not those automorphism groups are finitely generated.

(S2) We note that the word problem for groups admitting finitely many defining relations in the variety of all solvable groups of a given derived length >2,  is, in general, unsolvable - see [O. G. Kharlampovich,  A finitely presented solvable group with unsolvable word problem. (Russian) Izv. Akad. Nauk SSSR Ser. Mat. 45 (1981), 852--873, 928.]

(S3) E.Timoshenko [Center of a group with one defining relation in the variety of $2$-solvable groups (Russian), Sibirsk. Mat.  Z. 14 (1973), 1351--1355, 1368]  settled this problem in the affirmative for metabelian groups.  C. K.Gupta and  V.Shpilrain [The centre of a one-relator solvable group, Internat. J. Algebra Comput. 3 (1993),  51--55]  settled the problem (also in the affirmative) for solvable groups of arbitrary derived length, under an additional assumption that the relator is not a proper power modulo any term of the derived series.

(S4) The answer is "yes" for free metabelian groups -- see [Kh. S.Allambergenov,  V.A.Romankov,  Products of commutators in groups (Russian), Dokl. Akad. Nauk UzSSR 1984,  14--15] and for free solvable groups of derived length 3 -- see [A. H. Rhemtulla,  Commutators of certain finitely generated soluble groups. Canad. J. Math. 21 (1969), 1160--1164.]

(S9) The most obvious candidate for a test element in a group generated by x and y,  would be  u=[x,y].  This however is not a test element in a free  solvable group of derived length d>2 - see [N.Gupta,  V.Shpilrain,  Nielsen's commutator test for two-generator groups,  Math. Proc. Cambridge Philos. Soc. 114 (1993),  295--301. ]
 Recently, V.Roman'kov has constructed test elements in the  free  solvable group of rank 2 and derived length 3  - see his preprint. It is plausible that the same method can be used for constructing test elements in the  free  solvable group of any bigger rank as well, but technically it is getting more complicated.
 We also mention a related result of  E.I.Timoshenko [Test elements and test rank of a free metabelian group, Sib. Mat. Zh. , to appear]  who proved that a free metabelian group of rank > 2 does NOT have any test elements. It was previously known [V.G.Durnev, The Mal'tsev-Nielsen equation on a free metabelian group of rank 2. Math. Notes USSR 46 (1989), 927--929]  that the free metabelian group of rank  2 does have test elements, for example, u=[x,y].

(GA5) For a general setup that motivated this problem, we refer to a recent preprint [V. Nekrashevych, S. Sidki, Automorphisms of the binary tree: state-closed subgroups and dynamics of 1/2-endomorphisms], which is available here.