(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 = xy^3x^{-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].

(OR9)
This problem has been recently settled in the negative by A.V.Borshev
and D.I.Moldavanskii in [On isomorphism of some groups with one defining
relation, Math. Notes 79 (2006), 31-40]. To describe a counterexample, denote by
G(l,m,k) the group with two generators x and y and one defining relation
y^{-1} x^{-k}y x^l y^{-1} x^k y = x^m. Then the groups
G(18, 2, 2) and G(18, 2, 6) are not isomorphic, but each of them is a homomorphic
image of the other.

Other examples can be constructed based on the same ideas.

(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 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 in [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.

However, if the group has rank 3, then the answer
is negative as was shown by S. V. Ivanov
[On balanced presentations of the trivial group, Invent. Math., 165 (2006), 525--549].
The problem remains open for groups of rank bigger than 3.

(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 [M.Feighn, M.Handel, Mapping tori of free group automorphisms are coherent, Ann. of Math.149 (1999), 1061-1077].

(FP14) No, it is not; see [J. Howie, A proof of the Scott-Wiegold conjecture on free products of cyclic groups, J. Pure Appl. Algebra 173 (2002), 167-176].

(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 [Test ranks of finitely generated abelian groups,
Contemp. Math., Amer. Math. Soc. 296 (2002) 199--206] 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.

(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].

(FP20) D.Osin [Small cancellations over relatively hyperbolic groups and embedding theorems, preprint] recently obtained a much more general result: For any n \ge 2, there is an uncountable set of pairwise non-isomorphic finitely generated groups with exactly n conjugacy classes.

(FP23)
See [J.W.Cannon, W.J.Floyd, and W.R.Parry, Introductory notes on Richard
Thompson's groups. L'Enseignement Mathematique (2) 42 (1996), 215-256]
for a survey on various properties of Thompson's group.

(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 subsequently 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.
155 (2002), 131--156],
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 the preprint
[V.Lin, 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 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. ]

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.

More recently, P. Linnell and T. Schick [Finite group extensions and the
Atiyah conjecture, J. Amer. Math. Soc. 20 (2007), 1003-1051] provided a complete solution to Problem (B6) by
showing
that every braid group B_n is residually torsion-free
nilpotent-by-finite, in particular has plenty of non-trivial
torsion-free quotients. More specifically, if P_n is the pure braid group inside
B_n, and \gamma_k(P_n) its $k$th lower central series subgroup, then
B_n/\gamma_k(P_n) is torsion-free for all sufficiently large k.
Therefore, [B_n, B_n] is residually torsion-free nilpotent-by-finite,
too, which provides a solution to Problem (B6)(b) as well.
Their preprint
is available here.

(B10) Yes, this is true as shown by J.Gonzalez-Meneses [The n-th root of a braid is unique up to conjugacy, Alg. Geom. Topology 3 (2003), 1103-1118].

(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. For a survey on
groups of uniformly exponential growth see
[P. de la Harpe, Uniform growth in groups of exponential growth,
preprint]; available
here.

This problem was settled in the negative 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. 23 (2003), 907--918]

- linear groups [A. Eskin, S. Mozes, H. Oh, Uniform
exponential growth for linear groups. Int. Math. Res.
Not. 2002, no. 31, 1675--1683].

However, the answer turned out to be affirmative in general;
an example of a group of exponential but not uniformly
exponential growth was constructed in [J. S. Wilson, On exponential
growth and uniformly exponential growth for groups, Invent. Math. 155
(2004), 287--303].

Later, L.Bartholdi [A Wilson Group of Non-Uniformly Exponential
Growth, C. R. Math. Acad. Sci. Paris 336 (2003), no. 7, 549--554]
gave a similar but much simpler construction.

(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.

(G9)
It is easy to show that the number of different elements of length n in
the free metabelian group of rank r is bounded from below by the number of
self-avoiding walks (SAW) of length n on the lattice Z^r. This follows from a more general observation
on interpreting trivial elements of a group F/[R,R] by closed walks on the Cayley graph
of the group F/R such that every edge is traversed the same number of times in either direction. See [C.
Droms, J. Lewin, and H. Servatius,
The length of elements in free solvable groups, Proc. Amer. Math. Soc. 119 (1993), 27--33]
for more details.

It turns out however that SAW are quite difficult to count; for general background on
SAW we recommend the monograph by N. Madras and G. Slade [The Self-Avoiding Walk (Probability and its
Applications)] that can be found
here.
Some progress has been made in counting SAW on Z^2, which led to a reasonable
lower bound for the growth rate of the free metabelian group of rank 2. More specifically, in
[A. R. Conway and A. J. Gutmann, Square Lattice Self-Avoiding Walks and Polygons,
Annals of Combinatorics 5 (2001), 319--345], the authors established a lower bound of 2.63815853034.
Note that the growth rate of the free metabelian group of rank 2 should be strictly less than 3
(which is the growth rate of the free group of rank 2)
since a free metabelian group is not free.

(G10) This problem is related to a conjecture of Benjamini and Schramm on percolation on Cayley graphs; see [R. Muchnik and I. Pak, Percolation on Grigorchuk groups, Comm. Algebra 29 (2001), 661–671] for more details.

(G11) For background, we refer to [R.I.Grigorchuk, P. de la Harpe, On problems related to growth, entropy and spectrum in group theory, J. of Dynamical and Control Systems 3 (1997), 51--89] and [P. de la Harpe, Topics in Geometric Group Theory. University of Chicago Press, 2000]. The growth series for the automatic and solvable Baumslag-Solitar groups are rational and can be found in [M.Edjvet, D.Johnson, The growth of certain amalgamated free products and HNN extensions. J. Austral. Math. Soc. Ser. A 52 (1993), 285--298] and [D.Collins, M.Edjvet, C.P.Gill, Growth series for the group $\langle x,y | x^{-1}yx=y^l \rangle$, Arch. Math. (Basel) 62 (1994), 1--11], respectively.

(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 the
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.

(A5)
All algorithmic *decision problems* naturally split into
"yes" and "no" parts. The "yes" part is usually straightforward in
the sense that there is usually a recursive procedure that
enumerates all objects for which the answer to the given decision problem
is "yes". This is the case, in particular, with the word as well as the conjugacy problem for a
given finitely presented group and with the problem of isomorphism to a given finitely presented
group. The situation becomes more complex
if one considers the problem of isomorphism to a group from a given class of
groups. In this case, one may encounter a situation where neither
finitely presented groups in the given class nor finitely presented groups outside of the the
given class are recursively enumerable. The desire to have a class like that is motivated by
cryptography. The class of metabelian groups seems to be the simplest candidate, although the same
question can be asked
about any class of groups that contains infinitely presented finitely generated groups.

(A6) This problem is discussed in [M. Chiodo, Finding non-trivial elements and splittings in groups, preprint], where a special case is settled; namely, it is shown that there is no general procedure to pick a non-trivial generator from a finite presentation of a non-trivial group.

(C1)
The word problem in Aut(F_n) has a straightforward solution (by just acting
on the generators of F_n), but this algorithm is exponential with
respect to the length of the input. (The input here is an automorphism
given as a word in the generators of Aut(F_n). )

S. Schleimer [Polynomial-time word problems, Comment. Math. Helvetici, 83 (2008), 741-765] has settled this
problem in the affirmative.

(C2)
We note that the affirmative answer to Problem (F25a)
would imply the affirmative answer to this problem, but the converse is not
necessarily true.

(C3)
Since the first solution of the conjugacy problem in braid groups was given by
F. A. Garside [The braid group and other groups, Quart. J. Math. Oxford Ser. (2) 20 (1969),
235-254], several other people have come up with different algorithms, e.g. [J. Birman, K. H. Ko and
S. J. Lee, A new approach to the word and conjugacy problems in the braid groups, Adv. Math. 139
(1998), 322--353], [N. Franco, J. González-Meneses, Conjugacy problem for braid groups and
Garside groups, J. Algebra 266 (2003), 112--132], [V. Gebhardt, A New Approach to the
Conjugacy Problem in Garside Groups, preprint],
but none of the algorithms suggested so far has polynomial time
complexity with respect to the lengths of the input words.

Being in the class NP is a weaker property, i.e., it might be the case
that the conjugacy problem in braid groups has no polynomial time solution,
but it is in the class NP. The latter would follow, in particular, from: given two
conjugate elements of B_n represented by braid words of lengths \le m,
there is a conjugator whose length is bounded by a polynomial function of m.

We note that recently, M. Calvez and B. Wiest [A fast solution to the conjugacy problem in the 4-strand
braid group, preprint] showed that for the braid group B_3, the
conjugacy problem is actually in the class P. In fact, it has a cubic-time solution with respect to the lengths
of the input words.

(C4)A description of Dehornoy's algorithm
can be found in [P. Dehornoy, A fast method for comparing braids, Adv. Math.
125 (1997), 200-235]. We note that there is a quadratic time
algorithm for solving the word problem in braid groups, which is
based on the fact that braid groups are automatic. Dehornoy's
algorithm seems to outperform, in real-life implementations, this as well as other known
algorithms for solving the word problem in braid groups.
However, no subexponential upper bound for the time complexity of
Dehornoy's algorithm has been established theoretically.

(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.

(MA3) For SL_4(Z), Problem (a) is unsolvable since a direct product of two free groups embeds into SL_4(Z). For SL_2(Z), the problem is solvable since SL_2(Z) is virtually free.

(MA6) It is
well known that for any a, |a| \ge 2, the matrices (1, a ; 0, 1) and
(1, 0 ; a, 1) generate a free group. It is also known that for a=1 (and therefore also for a=1/n for any nonzero integer n) the group is not free.

(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).

(H2)
The following way of constructing a
hyperbolic group which is not isomorphic to any group of matrices
over C has been suggested by M. Kapovich. We note however that
under a more general definition, a group is called linear if
it is isomorphic to a group of matrices over *some* field, not
necessarily over C. The corresponding more general form of the
Problem (H2) remains open. We also note that recently,
M. Sapir and C. Drutu [Non-linear residually finite groups,
preprint]
constructed examples of non-linear (in the more general sense)
residually finite
one-relator groups. Their groups however are not hyperbolic.

Start with a uniform lattice \Gamma in the quaternionic
hyperbolic space HH^n of dimension n \ge 2.
Since HH^n is negatively curved, its boundary is a sphere of
dimension \ge 7, the group \Gamma is nonelementary word-hyperbolic, and,
therefore, it has a proper infinite hyperbolic quotient group
G=\Gamma/(r), where r is an element of \Gamma. Then we claim that G
is a nonlinear group. Moreover, if h: G \to GL_m(R) is any homomorphism, then
the image of h is finite.

Proof. **(1)** It is known that lattices \Gamma satisfy both
"archimedian" and "nonarchimendian" superrigidity, which is due to K.Corlette
[Archimedean superrigidity and hyperbolic geometry, Ann. of Math. 135 (1992),
165--182] (archimedian case) and M. Gromov and R. Schoen
[Harmonic maps into singular spaces and p-adic superrigidity for lattices
in groups of rank one, Publ. Math. IHES 76 (1992), 165--246]
(nonarchimedian case). The "archimedian" superrigidity means that
for each representation h: \Gamma \to GL_m(R)
which has the image that is not relatively compact, there exists a finite
index subgroup \Gamma' \subset \Gamma such that the restriction h|\Gamma'
is induced by a representation of the entire automorphism group of HH^n.
In particular, h has to be faithful. The "nonarchimendian" superrigidity
means the same if one considers representations h: \Gamma \to GL_m(F),
where F is a local field with discrete valuation (e.g. p-adic numbers);
since GL_m(F) is totally disconnected, it follows that
in this case, the image of h would have to be a bounded subgroup of
GL_m(F).

**(2)** Combining the "archimedian" and "nonarchimendian"
superrigidity implies
that every linear representation h of \Gamma either has finite image or is
faithful. (The proof is the standard adelic argument.)

**(3)** Now a linear representation of G would yield a linear
representation h of \Gamma by composition \Gamma \to G\to GL_n(R).
Hence, since h would have to be nonfaithful, it follows that
h would have a finite image by (2) above. Thus, G cannot have
a faithful representation in GL_n(R). It follows that G cannot have
a faithful representation in GL_n(C) either because GL_k(C) embeds
into GL_{2k}(R) by means of a+bi -> (a, b; -b, a).

(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].

M.Bridson and D.Wise
[Malnormality is undecidable in hyperbolic groups, Israel J.of Math.
124 (2001), 313-316] have settled the general case in the
negative.

(H18)
See [J.W.Cannon, W.J.Floyd, and W.R.Parry, Introductory notes on Richard
Thompson's groups. L'Enseignement Mathematique (2) 42 (1996), 215-256]
for a survey on various properties of Thompson's group.

(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).

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 30 (2002), 1033--1038]
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.

(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
[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 $G$. C. Hidber
[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.

More recently, S.Gersten, D.Holt, and T.Riley [Isoperimetric
inequalities for nilpotent groups, GAFA 13 (2003), 795-814]
showed that the Dehn function of any finitely generated nilpotent group of
class c is bounded by a polynomial of degree c+1.

However, S. Wenger [Nilpotent groups without exactly polynomial Dehn function, J. Topology 4 (2011),
141 - 160] has recently settled the problem in the negative. His preprint is available here.

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

(N7) Long time ago, H. Heineken [Engelsche Elemente der Länge drei. (German) Illinois J. Math. 5 (1961), 681--707] proved that every 3-Engel group (i.e., a group with the identity [x,y,y,y] = 1) is locally nilpotent. More recently, G. Havas and M. Vaughan-Lee [4-Engel groups are locally nilpotent, Internat. J. Algebra and Comput. 15 (2005), 649-682] have proved that every 4-Engel group is locally nilpotent.

(N8)(a) V. A. Roman’kov in [Diophantine questions in the class of finitely generated nilpotent groups, J. Group Theory 19 (2016), 497–514] constructed finitely generated nilpotent groups of nilpotency class 2 where equations of this form are undecidable.

(b) In the same paper, V. A. Roman’kov showed that equations of this form are decidable in free nilpotent groups of class 2 of any finite rank. The problem remains open for groups of larger nilpotency class.

(N9) (a) V. A. Roman’kov proved in [Diophantine questions in the class of finitely generated nilpotent groups, J. Group Theory 19 (2016), 497–514] that there is no algorithm that would solve this problem for any finitely generated nilpotent group of nilpotency class 2. However, the problem of existence of a *particular* finitely generated
nilpotent group with unsolvable retract problem remains open.

(b) Yes, it is, see [V. A. Roman'kov, N. G. Khisamiev, A. A. Konyrkhanova, Algebraically and verbally closed subgroups and retracts of finitely generated nilpotent groups, Sib. Math. J. 58 (2017), 536–545].

(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, Aut(F) \to 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 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 answered in the negative by M. Kassabov
[An automorphism of a free metabelian
group without fixed points, Commun. Algebra 32 (2004), 3297--3303].

(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. ]

V.Roman'kov [Test elements for free solvable
groups of rank 2, Algebra and Logic 40 (2001), 106--111] has constructed test elements in the
free solvable group of rank 2 and derived length 3.
It is plausible that the same method can be used for constructing test
elements in 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, Siberian Math. J. 41 (2000), 1200--1204]
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].

Finally, a complete solution was recently given by E.I.Timoshenko in
[Computing test rank for a free solvable
group, Algebra and Logic 45 (2006), 254-260]. He showed that the test rank of a free solvable non-abelian
group of finite rank is 1 less than the rank of that group. In particular, the free solvable group of rank 2
of
any derived length d>2 has test elements.

(GA5) For a general setup that motivated this problem, we refer to the paper [V. Nekrashevych, S. Sidki, Automorphisms of the binary tree: state-closed subgroups and dynamics of 1/2-endomorphisms, In Groups: Topological, Combinatorial and Arithmetic Aspects, LMS Lecture Notes Series 311 (2004), 375-404].