#### Algorithmically complex residually finite groups

Bull. Math. Sci.
Algorithmically complex residually finite groups
Olga Kharlampovich 0 1 2
Alexei Myasnikov 0 1 2
Mark Sapir 0 1 2
B Mark Sapir 0 1 2
Alexei Myasnikov 0 1 2
Communicated by Efim Zelmanov.
0 Department of Mathematics, Vanderbilt University , Nashville, TN 37240 , USA
1 Stevens Institute of Technology , Hoboken, NJ 07030 , USA
2 Department of Mathematics and Statistics, Hunter College, City University of New York , New York, NY 10065 , USA
We construct the first examples of algorithmically complex finitely presented residually finite groups and the first examples of finitely presented residually finite groups with arbitrarily large (recursive) Dehn functions, and arbitrarily large depth functions. The groups are solvable of class 3. 1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310 1.1 The problem and previous approaches for a solution . . . . . . . . . . . . . . . . . . . . . . 310 1.2 The “yes” and “no” parts of the McKinsey algorithm . . . . . . . . . . . . . . . . . . . . . 313 O. Kharlampovich: Partially supported by NSF Grant DMS-0700811, A. Myasnikov: Partially supported by NSF Grants DMS-0700811 and DMS-0914773. M. Sapir: Partially supported by NSF Grant DMS-1500180 and by BSF (USA-Israel) Grant 2010295.
Word problem; Depth function; Dehn function; Minsky machine
Contents
1.3 The time function of the algorithm Ayes: the Dehn function . . . . . . . . . . . . . . . . . . 315
1.4 The time function of the algorithm Ano: the depth function . . . . . . . . . . . . . . . . . . 315
1.5 Methods of proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316
1.6 Structure of the paper . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316
2 Turing machines and Minsky machines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317
2.1 Turing machines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317
2.2 Universally halting Turing machines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319
2.3 Minsky machines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322
3 Simulation of Minsky machines by semigroups . . . . . . . . . . . . . . . . . . . . . . . . . . . 324
3.1 The construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324
3.2 Residually finite finitely presented semigroups . . . . . . . . . . . . . . . . . . . . . . . . . 326
3.3 Residually finite semigroups with large depth function . . . . . . . . . . . . . . . . . . . . . 328
4 Simulation of Minsky machines in solvable groups . . . . . . . . . . . . . . . . . . . . . . . . 333
4.1 The construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333
4.2 A residually finite finitely presented group with large depth function . . . . . . . . . . . . . 346
5 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347
5.1 Universal theories of sets of finite groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347
5.2 Distortion of pro-finitely closed subgroups of finitely presented groups . . . . . . . . . . . . 349
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 350
1 Introduction
1.1 The problem and previous approaches for a solution
It is well known that finitely presented residually finite algebraic systems of finite
signature (say, semigroups or groups) are much simpler algorithmically than arbitrary
finitely presented algebraic systems. For example, the word problem in every such
algebraic system is decidable. In this paper we discuss the question “how complicated
finitely presented residually finite algebraic systems can be in the cases of groups and
semigroups?”.
The most “common” residually finite groups and semigroups, are linear (say, over
a field). These groups and semigroups are algorithmically “tame”: the word problem
there is decidable in polynomial time and even log-space [
34
]. The Dehn function is
a well-known indicator of complexity of the word problems: the smaller the Dehn
function the easier the word problem. The converse implication does not hold,
however. One can construct groups with decidable word problem and very large Dehn
functions [
39,59
]. However, no examples of finitely presented residually finite groups
or even semigroups with superexponential Dehn function are known. Thus one of the
main open problems in this area is how large could the Dehn function of a
residually finite finitely presented group or semigroup be. The question for groups was
known since early 90s. It was open for so long because all known methods to
construct algorithmically hard groups produced either non-residually finite groups or
groups where the question about their residual finiteness is very difficult. Not much
is known even for linear groups (note that Gersten asked [
19,20
] if there exists a
uniform upper bound for Dehn functions of linear groups). Let us briefly discuss the
previous attempts to solve the problem and the reasons why these methods did not
work.
1.1.1 Method 1. Known groups with large Dehn functions
One could hope that some of the known finitely presented groups with very large
Dehn function may turn out to be residually finite, which would shed some light on
how to produce residually finite groups with even larger Dehn functions.
Unfortunately, this is not the case, all these groups are non-residually finite. For example,
the Dehn function of the one relator group G(1,2) = a, b | b−1a−1bab−1ab = a2
introduced by Baumslag in [2] is bigger than any iterated exponent (see Gersten
[
18
]). Platonov [50] proved that it is equivalent to the function exp(log 2n)(1), where
exp(m)(x ) is the function defined by exp(0)(n) = n and exp(k+1)(n) = exp(exp(k)(n))).
However, G(1,2) is not residually finite (and in fact has very few finite
quotients) [2]. Furthermore, the word problem in G(1,2) is decidable in polynomial
time [
41
].
1.1.2 Method 2. Using subgroups with very large distortion
Consider a finitely presented group G and a “badly” distorted finitely generated
subgroup H . Let T = G, t | t −1ht = h (h ∈ H ) be the HNN extension of G where
the free letter t centralizes H . It was noticed by Bridson and Häfliger [11, Theorem
6.20.III] that the Dehn function of T is at least as large as the distortion function of
H in G. The following result puts some limitations on this method of constructing
complicated residually finite groups.
Lemma 1.1 If the group T is residually finite then H is closed in the pro-finite topology of G.
Proof In the notation above, suppose T is residually finite and u ∈ G\H . Then
w = [u, t ] = 1 in T (by the standard properties of HNN extensions). Hence there
exists a homomorphism φ from T onto a finite group Tw such that φ (w) = 1. But
this implies φ (u) ∈/ φ (H ) (since every element of φ (H ) commutes with φ (t )). Hence
there exists a normal subgroup of finite index N < G such that u does not belong to
N H . In other words H must be closed in the pro-finite topology of G.
Consider the following typical examples of residually finite groups G with highly
distorted subgroups H . The first one is Wise’s version [64] of Rips’ construction [53]
which for every finitely presented group Q gives a finitely presented residually finite
small cancellation group G with a short exact sequence
1 → N → G → Q → 1
where N finitely generated. It is easy to see that the distortion function of N in
G is at least as large as the Dehn function in Q, so choosing Q properly one can
get a finitely presented residually finite group G with a highly distorted subgroup
N . Now, the subgroup N is normal in the HNN extension T . So it is closed in
the pro-finite topology of T only if Q = G/N is residually finite. By Lemma
1.1, T can be residually finite only if Q is residually finite. In other words to
construct a complicated finitely presented residually finite group T one has to
have the initial group Q complicated, finitely presented and residually finite as
well.
The second example is the standard Mikhailova construction. In this case highly
distorted subgroups of the direct product of two free groups F2 × F2 can be obtained
as equalizers of two homomorphisms φ1 : E1 → M and φ2 : E2 → M where E1, E2
are finitely generated subgroups of F2 and M is finitely presented (see Sect. 5.2
below). But by Lemma 5.3 below the equalizer is closed in the pro-finite topology
only if M is residually finite. Thus as in the previous example, in order to construct an
algorithmically complex residually finite finitely presented group using Mikhailova’s
construction and the HNN extensions as above, one needs to have already a finitely
presented residually finite algorithmically complex group M .
The third example is Cohen’s [12] construction of highly distorted subgroups
employing the modular machines. One can also prove that in that construction the
subgroup will be pro-finitely closed only if the modular machine is very easy.
One can also try to use the hydra groups [
10,14
] to construct HNN extensions
as above with Dehn functions bigger than any prescribed Ackermann function. The
question of whether these groups are residually finite was open when the first version
of this paper was written, and is now answered in negative in [49].
1.1.3 Method 3. Boone–Novikov constructions
One of the standard ways to produce algorithmically complicated groups is by
simulating Turing machines using free constructions (HNN extensions and amalgamated
products) which goes back to the seminal papers by Boone and Novikov (see, for
example, [54]). There are currently many versions of that construction (for a recent
survey see [57]). But in fact, it can be shown that for each known version of the proof
of Boone–Novikov theorem using free constructions, even for easy Turing machines
the corresponding group is non-residually finite. This is, for instance, the main idea
of the example in [
31
]. Here is an even easier example. Let G = G(M ) be a group
constructed by any of these constructions. Then for every input word u of the Turing
machine M there exists a word w = w(u) obtained by inserting some copies of u
in w(∅) (here ∅ denotes the empty word), so that u is accepted by M if and only if
w(u) = 1 in G. Now consider M that accepts a word an if and only if n = 0 (that
machine is actually one of the basic building blocks in [59]). Then w(an) = 1 in G if
and only if n = 0. Suppose that there exists a homomorphism φ onto a finite group
H that separates w(∅) = w(a0) from 1. Then φ (a) has finite order, say, s, in H .
Therefore φ (w(∅)) = φ (w(a0)) = φ (w(as )) = φ (1) = 1, a contradiction. Hence
G(M ) is not residually finite.
1.1.4 Method 4. Residually finite groups obtained by free constructions
In general, the question about residual finiteness of free constructions is very difficult.
Currently there are only two large classes of groups where the question was settled:
these are ascending HNN extensions of free groups [6] and certain groups acting
“nicely” on CAT(0)-cubical complexes including small cancelation groups (see the
recent work of Wise, for example, [
25
] and references therein). All these groups have
easy word problem and uniformly bounded Dehn functions. The reason for the lack
of more examples is that groups obtained by free constructions from “nice” groups
contain a lot of extra elements and it is not at all clear how to separate these elements
from 1 by homomorphisms onto finite groups. In the two cases when it could be done,
it was possible to reformulate the problem in the language of algebraic geometry and
geometric topology, respectively.
1.1.5 Method 5. Other groups with complicated word problem
There are several other constructions of groups which simulate various algorithmic
problems including undecidable ones, but each of these also always either produce
non-residually finite groups or groups with simple word problem. For example, the
group in [
37
] is based on the R. Thompson group V which is infinite and simple (hence
not residually finite).
1.1.6 The main results of the paper.
In this paper, we construct finitely presented residually finite semigroups and groups
with arbitrarily complex word problem, and also easy word problem but arbitrarily
large (of course recursive) Dehn function, and arbitrarily large (recursive) depth
function. We also give applications of these results to the questions about decidability of
the universal theory of finite solvable groups and to distortion of pro-finitely closed
subgroups of residually finite groups.
1.1.7 What next?
We expect the approach used in this paper to be useful in solving other problems that
are still open. For example, the residually finite version of the Higman embedding
theorem [
24
] would be very desirable. It is known [
15,21,38
] that a finitely generated
recursively presented residually finite group may have undecidable word problem, and
hence cannot be embedded into a finitely presented residually finite group. But it is
not known whether undecidability of the word problem is the only obstacle for such
an embedding. Thus it would be very interesting to find out whether every finitely
generated residually finite group with decidable word problem embeds into a finitely
presented residually finite group. Note that usually a version of Boone–Novikov
theorem precedes a version of Higman theorem, hence we can consider this paper as a
step toward the residually finite version of Higman’s theorem.
1.2 The “yes” and “no” parts of the McKinsey algorithm
One of the initial motivations for studying residually finite groups, semigroups and
other algebraic structures was McKinsey’s algorithm solving the word problem in
finitely presented residually finite algebraic structures.
Let G = X | R be a residually finite finitely presented algebraic structure of finite
signature (say, groups, semigroups, rings, etc.) Let us recall McKinsey’s algorithm
solving the word problem in G (see [
36,40
]). Let F (X ) be the free algebraic structure
of signature freely generated by X . To solve the word problem in G one runs in
parallel two separate algorithms Ayes and Ano, such that starting with a given pair of
elements w, w ∈ F (X ) Ayes stops if and only if w = W in G and Ano stops if and
only if w = w in G.
The algorithm Ayes enumerates one by one all consequences of the defining
relations R and waits until w = w appears in the list.
The algorithm Ano enumerates all homomorphisms φ1, φ2, . . . , of G into finite
algebraic structures of signature and waits until φi (w) = φi (w ).
Note that although Ayes and Ano enumerate complimentary sets of elements of
F (X ) × F (X ), these algorithms are quite different and their complexity functions can
be very different too.
To estimate the complexity of algorithms and algorithmic problems, recall the basic
definitions of the complexity theory. Let S be a set of objects (words, numbers, etc.)
equipped with a size function S → N so that there are only finitely many objects in S
of any given size. For every algorithm A computing some partial function φ : S → N,
by TA we denote the time function of A which is defined as the maximal running time
of A on inputs of size at most n from the domain of φ (thus we simply ignore inputs
not from the domain of φ). The time complexity of an algorithmic problem, i.e., the
membership problem in a subset X ⊆ S is the minimum of TA for all algorithms A
solving the problem, i.e., computing the indicator function of X . It is not that easy to
define what “minimum of a set of functions” means. But we can define a pseudo-order
on the set of functions as follows. For two functions f, g we write f g ( f is smaller
than g) is there exists a constant C such that for any n, f (n) ≤ Cg(C n) + C n + C.
(The functions are equivalent if f g and g f.) We do not need to address the
question whether every set of functions has a minimum with respect to this order. We
can certainly say whether the time complexity is polynomial (exponential, etc.).
Let now G be a finitely presented residually finite group. Although it is universally
assumed that Ayes and Ano are very slow in general, there were no examples of groups
G for which these algorithms were actually very slow.
Moreover it was not known if for some universal recursive function f (n) the time
complexity of the word problem in any finitely presented residually finite group (or
semigroup) does not exceed f (n).
In the case of finitely presented linear groups it is well known that the word problem
can be solved in deterministic polynomial time [
34,61
]. This applies to most finitely
presented groups (where “most” means “with overwhelming probability” in one of
several probabilistic models): recent results of Agol [1] and Ollivier and Wise [45]
together with the older result of Olshanskii [46] imply that most finitely presented
groups are linear (even over Z).
One of our main results is the following theorem (an immediate corollary of
Theorem 4.21 below):
Theorem For every recursive set of natural numbers X there exists a finitely presented residually finite solvable group G of class 3 such that the word problem in G is as hard as the membership problem in X .
Remark 1.2 Note that if we replace “finitely presented” assumption by “recursively
presented”, then residually finite groups are known to be very complicated. As we have
mentioned before, recursively presented finitely generated residually finite groups
may have undecidable word problem [
15,21,38
]. See also the survey [
16
] where it
is shown how to construct complicated residually finite groups using the method of
Golod–Shafarevich.
Note also that although our groups are not linear, they are (Abelian of prime
exponent)-by-linear since they are solvable of class 3 with the second derived
subgroup Abelian of prime exponent (one can assume that the exponent is equal to any
prime number, say, 2).
1.3 The time function of the algorithm Ayes: the Dehn function
In [
39
] Madlener and Otto constructed finitely presented groups with arbitrarily large
Dehn functions. For residually finite groups, the situation is different. Nilpotent groups
are examples of residually finite groups with arbitrarily high polynomial Dehn
function [4]. The Baumslag–Solitar groups x , y | x y = x k , k ≥ 2, are examples of
residually finite (even linear) groups with exponential Dehn function. No examples of
residually finite groups with bigger Dehn functions were known. This gap is filled by
the following theorem (see Theorem 4.21).
For every recursive function g(n) there exists a finitely presented residually finite solvable group G of class 3 such that the Dehn function of G is bigger than g(n).
In addition we can assume that the time complexity of the word problem is
polynomial.
1.4 The time function of the algorithm Ano: the depth function
The time function of the algorithm Ano can be estimated in terms of the depth function
introduced by Bou-Rabee [9]. Recall that if G = X is a finitely generated residually
finite group or semigroup, the depth function ρG (n) is the smallest function such that
every two words w =G w of length at most n are separated by a homomorphism to a
finite group (semigroup) H with |H | ≤ ρG (n). That function does not depend on the
choice of finite generating set X (up to the natural equivalence).
It is easy to see that for every finitely generated linear group or semigroup G, ρG
is at most polynomial. Since finitely generated metabelian groups are subgroups of
direct products of linear groups [62] the depth function of every finitely generated
metabelian group is at most polynomial. By the recent result of Agol [1] based on the
earlier results of Wise [63], every small cancelation group is a subgroup of a Right
Angled Artin group, hence linear and has polynomial depth function. In fact one can
have much smaller bounds for many linear groups. For example, for the free group F2,
n3 by [9]. A lower bound for the depth function for a free group is equivalent
ρF2 (n)
2
to n 3 by a result of Kassabov and Matucci [
26
]. There are some finitely presented
groups for which the depth function is unknown and very interesting. For example the
ascending HNN extensions of free groups are known to be residually finite and even
virtually residually nilpotent (proved by Borisov and the third author [6,7]) but the
only upper bound one can deduce from the proof is exponential. Although many of
these groups have small cancelation presentations and so are covered by the results
from [1], there are some groups of this kind for which the depth function is not known.
For finitely generated infinitely presented groups (even amenable ones) the situation
is much clearer. Using the method of Kassabov and Nikolov [
27
] and the result of
Nikolov and Segal [44] one can construct a finitely generated residually finite group
with arbitrary large recursive depth function.
In this paper, we show that a similar result holds for finitely presented solvable of
class 3 groups (see Theorem 4.22).
For every recursive function f one can construct a residually finite finitely presented solvable of class 3 group with depth function greater than f .
In addition, again, one can assume that the time complexity of the word problem
in G is polynomial.
1.5 Methods of proof
As we have shown above (see Sect. 1.1.3) all versions of the Boone–Novikov
construction ([
12,37,54,59
]) do not produce complicated residually finite groups. Instead,
we simulate Minsky machines. We first simulate Minsky machines in semigroups and
then “embed” these semigroups into solvable groups of class 3. The first simulation
of Minsky machines in semigroups were studied by Gurevich [
23
] (see also [55,58]).
Simulations of Minsky machines in groups, including solvable groups could be found
in [60] and [
28
] (see also [
32
]). In this paper we closely follow the construction by
the first author from her unpublished thesis [
29
].
Of course that construction also often leads to non-residually finite groups. But it
turned out that the difficulty can be overcome by modifying the Minsky machine first.
In this paper, we use the fact that every Turing machine recognizing a recursive set is
equivalent to a sym-universally halting Minsky machine, i.e., Minsky machine whose
symmetrization halts on every non-accepted configuration (see Theorem 2.7).
1.6 Structure of the paper
The paper is organized as follows. Section 2 contains preliminary results about Turing
and Minsky machines that are needed further.
In Sect. 3, we simulate sym-universally halting Minsky machines in residually
finite finitely presented semigroups and prove the analogs of the above theorems for
semigroups. The advantage of using Minsky machines instead of the general
Turing machines is that the resulting semigroups are much “smaller”, and all non-zero
elements of these semigroups are basically subwords of words corresponding to
configurations of the Minsky machines. In the construction, we use the ideas from the
papers of the third author [55,58] who proved that these semigroups are minimal in the
following sense: the varieties generated by these semigroups are minimal varieties of
semigroups containing finitely presented semigroups with undecidable word problem
(this leads to a complete description of varieties with decidable word problem where
periodic groups are locally finite [55]).
In Sect. 4 we use semigroups from Sect. 3 to construct the groups with large Dehn
and depth functions. The idea of such a construction came from the paper [
28
] of the
first author solving a problem formulated by Adyan [
33
] by constructing a finitely
presented group with undecidable word problem and satisfying a non-trivial identity.
We use the simplification of that construction from the unpublished dissertation of
the first author [
29
]. It turned out that if we start with residually finite semigroups
from Sect. 3, we often get residually finite finitely presented groups whose Dehn
and depth functions resemble the corresponding functions of the semigroups we start
with. Finally, Sect. 5 contains two applications of our main results and methods. In
Sect. 5.1, we strengthen the well-known result of Slobodskoi about undecidability
of the universal theory of finite groups by showing that the universal theory of any
set of finite groups that contains all finite solvable groups of class 3 is undecidable
(see Theorem 5.1). This gives the first proper variety of groups where the set of all
finite groups has undecidable universal theory. In Sect. 5.2, we construct pro-finitely
closed subgroups of the direct product of two free groups with arbitrary large time
complexity of the membership problem, distortion function and ther relative depth
function (defined there).
2 Turing machines and Minsky machines
2.1 Turing machines
In this paper, we shall consider several types of machines. A machine M in general
has an alphabet and a set of words in that alphabet called configurations. It also has a
finite set of commands. Each command is a partial injective transformation of the set
of configurations. A machine is called deterministic if the domains of its commands
are disjoint. A machine usually has a distinguished configuration sacc called the accept
configuration and a set I = I (M ) of input configurations which recursively enumerate
words in some finite alphabet A.
A machine halts if none of the commands is applicable to the configuration. We
always assume that a machine halts on every accept configuration but it may halt on
non-accept configurations too.
A computation of M is a finite or infinite sequence of configurations and
commands from P:
w1 −θ→1 w2 −θ→2 . . . −θ→l wl+1, . . .
such that θi (wi ) = wi+1 for every i = 1, . . . , l, . . . .
If the computation is finite and wl+1 is the last configuration, then l is called
the length of the computation, and we say that wl+1 is obtained from w1 by applying
. A configuration is called accepted by M if there exists a computation connecting
that configuration with sacc (that computation is called accepting). The time function
TM (n) of M is the minimal function such that every accepted configuration of length
≤ n has an accepting computation of length ≤ TM (n).
Let us give a definition of a Turing machine (see [59]). A Turing machine M with
k tapes consists of hardware (the tape alphabet A = ik=1 Ai , and the state alphabet
Q = iK=1 Qi )1 and program P (a list of commands, defined below). A configuration
of a Turing machine M is a word
α1u1q1v1ω1 α2u2q2v2ω2 · · · αK u K qK vK ωK
(we included spaces to make the word more readable) where ui , vi are words in Ai ,
qi ∈ Qi and αi , ωi are special symbols (not from A ∪ Q). This machine has k tapes.
For every configuration c the content of tape number i is the subword αi ui qi vi ωi .
A command simultaneously replaces subwords ai qi bi by words a q b
i i i where
qi , qi ∈ Qi , ai , ai , are either letters from Ai ∪ {αi } or empty, bi , bi are either
letters from Ai ∪ {ωi } or empty. A command cannot insert or erase αi or ωi , so if, say,
ai = αi , then ai = αi .
Note that with every command θ one can consider the inverse command θ −1 which
undoes what θ does.
For the Turing machine we choose stop states qi0 in each Qi , then a configuration
w is accepted if there exists a computation starting with w and ending with a
configuration where all state symbols are qi0 and all tapes are empty (which is the accept
configuration for the Turing machine). Thus for a k-tape Turing machine sacc is equal
0 0
to α1q1 ω1 · · · αk qk ωk .
Also we choose start states qi1 in each Qi . Then an input configuration
corresponding to a word u over A1 is a configuration inp(u) of the form
α1uq11ω1 α2q21ω2 · · · αK qK1 ωK .
Thus the input set I consists of all these words. We say that a word u over A1 is
accepted by M if the configuration inp(u) is accepted. The set of all words accepted
by M is called the language accepted by M .
For every machine M , the machine Sym(M ) is made from M by adding the inverses
of all commands of M . Two configurations w, w are called equivalent, written w ≡M
w , if there exists a computation of Sym(M ) connecting these configurations. Clearly,
≡M is an equivalence relation. For example, every two accepted input configurations
are equivalent, because both are equivalent to sacc.
The following general lemma is easy but useful.
Lemma 2.1 Suppose that M is deterministic. Then
(a) Any reduced computation of Sym(M ) is a concatenation of two (possibly empty)
parts 1 2−1, both 1 and 2 uses only commands of M .
(b) Two configurations w, w of M are equivalent if and only if there exist two
computations of M connecting w, w with the same configuration w of M .
1
denotes disjoint union.
Proof Part (a) of the lemma follows from the fact that in any reduced computation of
Sym(M ) inverses of commands of M cannot be followed by commands of M (since
M is deterministic).
For Part (b), if w is equivalent to w , then by Part (a) there is a computation of the
form 1 2−1 connecting w and w . Then applying 1 to w, and 2 to w produces
the same configuration w .
Definition 2.2 (a) We say that an algorithmic problem A is as hard as an algorithmic
problem B if for any decision algorithm for A which solves the problem in time
TA there exists an algorithm for B that solves it in time TA .
(b) We say that a language Y polynomially reduces to a language X if there exists a
deterministic Turing machine C checking membership in Y which uses an oracle
checking membership in X and runs in polynomial time (in terms of the length of
the word being checked.
(c) We say that languages X and Y are polynomially equivalent if there are polynomial
reductions of X to Y and vice versa.
(d) We say that a machine M polynomially reduces (resp. is polynomially equivalent)
to a machine M if the configuration equivalence problem of M polynomially
reduces (resp. is polynomiall equivalent) to the configuration equivalence problem
for M .
2.2 Universally halting Turing machines
A deterministic machine M is called universally halting if it does not have infinitely
long computations (see [8]). We say that a computation of Sym(M ) is reduced if
no command is followed by its inverse. We call a deterministic machine M
symuniversally halting if Sym(M ) does not have infinitely long reduced computations
that start at a non-accepted configuration. It is proved in [13] that for every recursive
set X of natural numbers, that is accepted by a deterministic Turing machine M there
exists a universally halting deterministic Turing machine M with one tape accepting
X . From the construction, it is clear that M polynomially reducible to M .
One can also convert a sym-universally halting Turing machine into a 1-tape
symuniversally halting Turing machine:
Lemma 2.3 Let M be a deterministic sym-universally halting Turing machine
recognizing a language X . Then there exists a one-tape deterministic sym-universally
halting Turing machine M recognizing X and polynomially equivalent to M .
Moreover there exists a map φ from the set C (M ) of configurations of M to the set C (M )
of configurations of M such that
(1) For every word u in the alphabet of X , φ (inpM (u)) = inpM (u)
(2) For every c ∈ C (M ), |φ (c)| = O(|c|)
(3) φ (c) ≡M φ (c ) if and only if c ≡M c
(4) If a command θ of M takes c to c , then there exists a computation of M of length
at most O(|c|) that takes φ (c) to φ (c ).
Proof The proof is basically by inspection of the proof from [48, Theorem 2.1]. Recall
the way to convert a k-tape Turing machine M into a 1-tape Turing machine M . The
tape alphabet of M consists of all letters occurring in the configurations of M . A
configuration of M is a word αc1qc2ω where c = c1c2 is a configuration of M or
differs from a configuration of M by at most two letters: the left and right neighbor of
some state letter. The map φ takes each configuration c to which a command of M is
applicable to the configuration αcqθ ω where qθ is a state letter of M corresponding
to the command θ that is applicable to c (note that since M is deterministic, θ is
determined by c). If no command of M is applicable to c, we set φ (c) = αcqω where
q is a distinuished state letter of M .
A command θ of M that substitutes ai qi bi by ai qi bi , i = 1, . . . , k, is simulated
as follows: the letter qθ moves from right to left, and every time it meets qi , it checks
if it is a part of the subword ai qi bi , and if so, replaces it by a q b . After all these
i i i
substitutions the letter qθ returns to the right end of the configuration (next to the letter
α) and becomes ready to simulate the next command of M or becomes the distinguished
state letter q (for more details see [48]). It is easy to check (using Lemma 2.1) that the
new machine is polynomially equivalent to the old one and properties (1)–(4) hold. It
is also easy to check that if the original machine is sym-universally halting, the new
one is also sym-universally halting.
Theorem 2.4 For every recursive language X there exists a deterministic symuniversally halting Turing machine M with one tape recognizing X . Moreover if M is any deterministic Turing machine recognizing X then we can additionally assume that M polynomially reduces to M .
Proof Let M be a deterministic universally halting Turing machine with k tapes
recognizing X . Consider the new Turing machine M constructed as follows. M has
one more tape than M , called the history tape. The alphabet A of this tape is in
one-to-one correspondence with the set of commands P of M : A = {[θ ], θ ∈ P}.
An input configuration of M does not have letters from A and its subword written
on the first k tapes is an input configuration of M . With every command θ of M we
associate a command θ of M . It does what θ would do on the first k tapes of M
and inserts [θ ] on the history tape of M next to the right of qk+1. After the first k
tapes of M form the accept configuration sacc(M ), the machine erases letters from
the tape alphabet A on the history tape and halts, producing the accept configuration
sacc(M ) of M (thus sacc(M ) = s0αk+1qk0+1ωk+1).
Let P be the program of M . We shall modify M further to obtain a new (k +
1)tape Turing machine M . It has the same tape alphabets as M and all state letters of
M are also state letters of M . The input and accept configurations are also the same.
The program P of M contains a copy P˜ of P (the set of the main commands) and
some new commands. After each main command θ˜ of P˜ which does the same as the
corresponding command θ of M , but changes the state letters to state letters which
are not in M , M executes the history written on the history tape backward, without
erasing the history tape. It just scans the history tape from left to right, reading the
symbols written there one by one and executing on the first k tapes the inverses of the
commands written on the history tape. If at the end of the scanning the history tape,
the word written on the first k tapes is an input configuration of M , M executes on
the first k tapes the history written on the history tape in the natural order, scanning the
history tape from right to left. After that M changes the state letters to what θ would
do, and is ready to execute the next main command. We do not give precise definition
of the program of M because it is obvious on the one hand and long on the other hand.
The machine M is deterministic and universally halting. Moreover for every input
configuration c of M , M accepts c if and only if c is accepted by M , hence if and
only if c corresponds to a word from X .
Note that since M is deterministic, and no commands of it are applicable to the
accept state, M accepts the same language as M .
Let c be a configuration of M that is not accepted by M . By Lemma 2.1(a) every
reduced computation of Sym(M ) starting at c is a concatenation of a computation of
M followed by a computation of the machine M −1 obtained from M by replacing
every command with its inverse. Since M is universally halting, there are only finitely
many computations of M starting with c.
Claim 1 There are finitely many computations of M −1 starting with any non-accepted
configuration c. Equivalently, there are only finitely many computations of M ending
with c.
Indeed, since c is not accepted, none of the tape letters on the history tape is
erased during any computation ending in c. Therefore by the definition of M every
computation ending at c and having length ≥|c| must arrive at a configuration
c = c1c2 where c1 is an input configuration of M and c is the content of the history
tape of c with the state letter moved next to ωk+1. And, moreover, this should happen
at most |c| steps before arriving to c. The suffix c2 of c is completely determined
by c. The sequence of commands from used to get from c to c is in one-to-one
correspondence with the sequence of tape letters of c2 to the right of the state letter.
Therefore c is completely determined by c. If the length of the computation is at
least 2|c|, then a configuration of the form c1c2 must occur in it before c , where c2
differs from c2 only by the state letter (which, as in c2 is next to ωk+1. This is impossible
because between every two arrivals at such configurations, every computation of M
must execute one of the main commands, and increase the number of tape letters on
the history tape. Thus we proved
Claim 2 Every computation ending at c is of length less than 2|c|.
Claim 2 implies Claim 1 and the fact that M is sym-universally halting.
To prove that M polynomially reduces to M , let c, c be two configurations of M .
In order to check whether c and c are equivalent first check whether c is accepted. For
this we need to run the program of M for at most 2|c| steps and see whether we first
get a configuration c of the form c1c2 where c1 is an input configuration of M and
c2 is of the form αk+1vqk+1ωk+1 (qk+1 ∈ Qk+1) and then a configuration of the form
c1αqk+1vωk+1. If so, then check the equivalence c1 ≡M sacc(M ) using the oracle
that checks equivalence of configurations of M . The answer is “yes” if and only if c
is accepted. That process takes linear (in |c|) number of steps and one oracle query.
Similarly, we check if c is accepted. If Both c and c are accepted, then c ≡M c .
If one of them is accepted and another one is not, then these configurations are not
equivalent.
Finally suppose that both c and c are not accepted. By Lemma 2.1, c ≡M c if
and only if there exist two computations 1 starting with c1 and 2 starting with c2
such that the end configurations of these computations are the same. We can assume
that either 1 or 2 has length >2(|c| + |c |) (that can be checked in time linear
in |c| + |c |). Without loss of generality assume that the length of 1 is bigger than
2|c|. Then after at most 2|c| steps of 1 we arrive at a configuration of the form c1c2
where c1 is a configuration of c such that there exists a computation of length
<|c2| of M starting with an input configuration d of M and ending at c1, and the
sequence of commands used in this computation is in one-to-one correspondence
with the tape letters in c2. Therefore c is equivalent to a configuration of the form
dαk+1qk+1ωk+1 where qk+1 ∈ Qk+1 which is an input configuration of M (and we
need linear time in terms of |c| to find this configuration). Hence we can assume
that c = dαk+1qk+1ωk+1. If the longest computation of M starting at c has length
<2|c |, then the longest configurations we can reach by one of these computations
is at most 2|c |, and it would take at most O((|c | + |c|)2) steps of M to reach any
of these configurations starting at c. Thus it would take polynomial time to check
whether c ≡M c in this case. Thus we can assume that there exists a computation
of M that starts at c and has length >2|c |. Then, as before we can assume that
c = d αk+1qk+1ωk+1, where d is an input configuration of M . Now if order to check
if c ≡M c it is enough to check whether d ≡M d which is one oracle query.
It remain to apply Lemma 2.3 and convert M to a 1-tape Turing machine.
2.3 Minsky machines
The hardware of a k-glass Minsky machine MMk , k ≥ 2, consists of k glasses
containing coins. We assume that these glasses are of infinite height. The machine can
add a coin to a glass, and remove a coin from a glass (provided the glass is not empty).
The commands of a Minsky machine are numbered started at 0. A configuration
of a k-glass Minsky machine is a k + 1-tuple (i ; 1, . . . , K ) where i is the number
of the command that is to be executed, j is the number of coins in the glass # j . We
can write a number in the unary notation: the number n is written as 1 . . . 1 (n ones).
Clearly then we can view a configuration as a word in the alphabet consisting of digits
1 and symbols (,), ; and comma “,”. The accept configuration is s0 = (0; 0, . . . , 0)
(the command number is 0, all glasses are empty) and input configurations have the
form (1; m, 0, . . . , 0).
Let us describe commands of Minsky machines more precisely. Each basic
command has one of the following forms:
• Put a coin in each of the glasses ##n1, . . . , nl and go to command # j . We shall
encode this command as
i ; → Add(n1, . . . , nl ); j
where i is the number of the command;
• Provided the glasses ##n1, . . . , nl are not empty, take a coin from each of these
glasses and go to instruction # j . This command is encoded as
• Stop. This command is encoded as i ; → 0;
Remark 2.5 This defines deterministic Minsky machines. We will also need
nondeterministic Minsky machines. Those will have two or more commands with the
same number.
Remark 2.6 We can also use (composite) commands that are not literally commands
described above but can be easily split into a few basic commands, such as “Put coins
in glasses ##i1, . . . , il provided glasses ##n1, . . . , nm are empty”.
Theorem 2.7 Let X be a recursively enumerable set of natural numbers. Then the
following holds:
(a) there exists a 2-glass deterministic Minsky machine MM2 which “enumerates”
X in the following sense: for every m ∈ N, if m ∈ X , then MM2 takes
configuration (1; 2m , 0) to the accept configuration (0; 0, 0), if x ∈/ X , then starting with
(1; 2m , 0) it works infinitely long time.
(b) If X is recursive, then there exists a 2-glass deterministic Minsky machine MM2
that recognizes X and is sym-universally halting.
(c) In (a) and (b) we can also assume for every computation of MM2 starting with a
configuration c and each of the glasses, that glass is emptied after at most O(|c|)
steps (here |c| denotes the size of configuration c, i.e., the total number of coins
in all glasses of the configuration).
(d) If M is a deterministic Turing machine recognizing X , then we can assume that
MM2 polynomially reduces to M .
Proof The proof of Part (a) can be found in [
35
]. To prove (b) let us first recall the
way to convert a 1-tape Turing machine M into a 2-glass Minsky machine MM2 [
35
],
and then prove that if M is sym-universally halting, then so is MM2.
Suppose that the tape alphabet of M has m letters. For simpliciy consider the case
when m = 2. The general case is absolutely similar. So suppose that the set of tape
letters is {1, 2}. Let q1, . . . , qs be the set Q of state letters. With every configuration
c = αuqi vω of M we associate the following configuratiion of a 3-glass Minsky
machine MM3: φ (s) = (i ; nu , nv, 0) where nu is the word u viewed as a natural
number written in base 3, and v is the word v read from right to left. Now every
command aqi b → a q j b of M is interpreted by MM3 as follows. If a, a are not
empty, the machine MM3 needs to check whether nu ≡ a (mod 3), nv ≡ b (mod 3)
and if so, then replace the last digit of nu by a and the last digit of nv by b . If, say, a
is empty, then the machine should just multiply nu by 3 and add a . For example, the
command θ of the form 1qi 2 → q j 1 is interpreted by a sequence P(θ ) of commands
of MM3 as follows. The commands of P(θ ) will be numbered i.1 through i.l for some
l. In order to check that nu ≡ 1 (mod 3), the machine should remove one coin from
the first glass, then repeatedly keep removing 3 coins from the first glass while adding
1 coin to the third glass. If at the end the first glass is empty, then, indeed, nu ≡ 1
(mod 3). If this is the case (otherwise the command is not applicable and the machine
halts), we remove the coins from the third glass, and check if nv ≡ 2 (mod 3). If so,
we multiply the number of coins in the second glass by 3 (keep adding 3 coins to the
second glass while removing a coin from the third glass until the third glass is empty),
and then add two coins to the second glass.
It is easy to see that if c is a configuration of M , then after applying P(θ ) to φ (c),
we get φ (θ (c)). The length of the computation of P(θ ) connecting φ (c) with φ (θ (φ))
is O(|c|).
Let MM3 be the resulting 3-glass Minsky machine. In order to convert it into a
2-glass Minsky machine, we associate with every configuration (c = i, m, n, p) of
MM3 the following configuration ψ (c) of a 2-glass Minsky machine: (i ; 2m 3n5 p, 0).
Now removing (adding) a coin from (to) glass number j of MM3 where j = 1, 2 or
3 is simulated by dividing (multiplying) the number of coins in the first glass by 2,3
or 5 respectively.
Notice that the following property of MM3 holds:
(*) If is a computation of MM3 and c1, c2, . . . , cm are configurations occurring
in that computation, and m > |c1| for some universal constant , then one of the
configurations c2, . . . , cm must be of the form φ (c ) where c is a configuration of M .
Moreover if φ (c ) and φ (c ) are two consecutive configurations of that form in ,
then there exists a command θ of M such that θ (c ) = c .
Therefore if there exists an infinite computation of MM3 or MM3−1 starting with
some configuration c of MM3, then there exists an infinite computation of M (resp.
M −1) starting with some configuration c . Moreover if c is not accepted by MM3,
then c is not accepted by M . Thus if M is sym-universally halting, then MM3 is
sym-universally halting. The proof for MM2 is similar. This gives Part (b).
Part (c) of the theorem immediately follows from the construction.
Part (d) is proved as follows. Suppose that c, c are two configurations of MM3
(for MM2 the proof is similar). By (*) in at most O(|w|) steps of MM3 either c turns
into a configuration of the form φ (d) for some configuration d of M or MM3 halts.
In the latter case, we check whether c is equivalent to c in O(|c|) steps. So we can
assume that both c and c are equivalent to configurations φ (d) and φ (d ) for some
configurations d, d of M , and the lengths of d, d are O(|c| + |c |). Again by (*), c
is equivalent to c if and only if d and d are equivalent configurations of M . Thus we
need to use the oracle once.
3 Simulation of Minsky machines by semigroups
3.1 The construction
Here we will show how to simulate a Minsky machine by a semigroup. The
construction is based on the following general idea which applies also in the case of solvable
groups considered later.
(1)
First, with every configuration ψ we associate a word w(ψ ). Then with every
command κ of the Minsky machine M we associate a finite set of defining relations
Rκ . The semigroup S(M ) is defined by the relations from the union R of all Rκ (which
is finite since we have only a finite number of commands) and usually some auxiliary
relations Q which are in a sense “independent” of R but make the semigroup “smaller”.
We need Q, for example, to make sure S(M ) satisfies a particular identity.
We say that the semigroup S(M ) simulates M if the following holds for arbitrary
configurations ψ1, ψ2 of M :
ψ1 ≡M ψ2 if and only if w(ψ1) = w(ψ2) in S(M ).
Usually, in order to prove the property (1) one has to prove the following two
properties.
Property 3.1 If a configuration c can be obtained from a configuration c by a command
κ of M then the word w(c ) can be obtained from the word w(c) by applying defining
relations of S(M ) from the set Rκ .
Property 3.2 If a word w(c ) can be obtained from a word w(c) by applying the
defining relations of S(M ) then c ≡M c .
It is easy to see that Properties 3.1 and 3.2 imply property (1).
There is an easy way to interpret Minsky machines in a semigroup S(M ). Let MMk
be a Minsky machine with k glasses and commands ##1, 2, . . . , N , 0 (here the
command number 0 is the stop command, it is the command with domain {0; 0, . . . , 0)}).
Then S(MMk ) is generated by the elements q0, . . . , qN and {ai , Ai | i = 1, . . . , k}.
The set of defining relations of S(MMk ) consists of all relations
ai a j = a j ai , ai A j = A j ai , Ai A j = A j Ai , i = j,
(2)
which we shall call commutativity relations, all relations of the form x y = 0 where x y
is a two-letter word which is not a subword of a word of the form qi a11 . . . akk A1 . . . Ak
modulo the commutativity relations (2) (for example qi q j = Ai ai = ai q j = Ai q j =
0), which we shall call 0-relations, and relations associated with commands of M
according to the following table,
Command of MMk
i, n1 > 0, . . . , nm > 0 → Sub(n1, . . . , nm ); j qi an1 . . . anm = q j
These will be called the Minsky relations.
(3)
The words in S(MMk ) corresponding to configurations of M are the following:
w(i ; 1, . . . , k ) = qi a11 . . . akk A1 . . . Ak .
The proof that Properties 3.1 and 3.2 hold in S(MMk ) follows easily from Lemma
2.1, see [
32,55
].
3.2 Residually finite finitely presented semigroups
The following obvious lemma shows that the auxiliary 0-relations make the semigroup
S(MMk ) really small: it just does not have too many elements which are not related
to configurations of MMk .
Lemma 3.3 Every word W in the generators of S(MMk ) that is not equal to 0 in
S(MMk ) is, modulo the commutativity relations, a subword of some word of the form
w(i ; 1, . . . , k ).
Lemma 3.4 Suppose that a word W is not 0 in S(MMk ). By Lemma 3.3 W is a
subword of a word of the form w(i ; 1, . . . , k ) (up to the commutativity relations).
Suppose that W does not contain either qi or one of the A j . Then there are at most
O(|W |) different (up to the commutativity relations) words that are equal to W in
S(MMk ). All these words are subwords of words of the form w(i ; 1, . . . , k ) such
that the configurations (i ; 1, . . . , k ) and (i , 1, . . . , k ) of M are equivalent.
Proof If W does not contain qi , then the only relations that apply to W are the
commutativity relations, so the only words that are equal to W in S(M ) are the words
obtained from W by the use of commutativity relations.
Suppose that W contains qi but does not contain one of the A j .
Without loss of generality, we can assume that W contains every letter from
w(i ; 1, . . . , k ) except some of the A j ’s.
Every application of a Minsky relation to W corresponds to a command of the
Minsky machine, applied to the configuration c = (i ; 1, . . . , k ). Let c = c1 →
c2 → . . . be any computation of Sym(MMk ) starting with c. Then the sequence of
commands of MMk applied in that computation has the form 1 2−1 where 1, 2
are computations of MMk (by Lemma 2.1). Each computation i corresponds to
a sequence Y of applications of Minsky relations and commutativity relations (by
Property 3.1). If that sequence of relations can be applied to W , then this computation
never checks whether glass # j is empty. By Property (c) of Theorem 2.7, the lengths
of 1 and 2 must be at most O(|W |). This implies the statement of the lemma.
Recall that s0 is the accept configuration of MMk , i.e., s0 = (0; 0, . . . , 0).
Lemma 3.5 Suppose that the Minsky machine MMk is sym-universally halting. Then
(a) Every element z of S(MMk ) which is not equal to 0 or w(s0) has finitely many
divisors, i.e., elements y such that z = pyq for some p, q ∈ S(MMk ) ∪ {1}.
(b) For every configuration c of MMk the word w(c) is equal to w(s0) in S(MMk ) if
and only if ψ is accepted by MMk .
Proof (a) If z is represented by a word w that contains one of the qi and all letters
A j , then it must be equal to one a word of the form w(i ; 1, . . . , k )
modulo commutativity relations (by Lemma 3.3). In that case applying relations of
S(MMk ) to w amounts to applying commands of Sym(MMk ) to the configuration
c = (i ; 1, . . . , k ) (by Properties 3.1 and 3.2). Thus every divisor of z is
represented by a subword of one of the words w(c ) such that c ≡MMk c . Also note
that c cannot be an accepted configuration of MMk because otherwise z would be
equal to w(s0) in S(MMk ). Since MMk is sym-universally halting, the number of
configurations that are equivalent to c is finite. Hence the number of divisors of z
is finite too.
If a word w representing z does not contain a q-letter or one of the Ai , then we
can apply Lemma 3.4.
(b) This follows from Properties 3.1 and 3.2.
Remark 3.6 Note that for every element z of any semigroup S the set of all
nondivisors of z in S is an ideal (denoted by N (z)). By definition of a divisor, N (z) does
not contain z. The Rees quotient semigroup S/N (z) consists of all divisors of z and
0 with a natural multiplication. In particular, if z has only finitely many divisors then
S/N (z) is finite. Also note that for every ideal I of S, if z ∈/ I , then z is separated
from every other element of S by the natural homomorphism from S to S/I .
Lemma 3.7 If MMk is sym-universally halting, then S(MMk ) is residually finite.
Proof Suppose that MMk is sym-universally halting. Let z1 = z2 be two different
elements of S(MMk ). We need to show that there exists a homomorphism from S(MMk )
to a finite semigroup separating z1 and z2.
First suppose that either z1 or z2 is not in {0, w(s0)}. Then by Remark 3.6 z1 and z2
are separated by one of the natural homomorphisms from S(MMk ) to S(MMk )/N (z1)
or S(MMk )/N (z1) which are finite semigroups by Lemma 3.5 (a). Thus we can assume
that z1 = 0, z2 = w(s0). Consider the (finite) set U of all subwords of the word
A1 A2 . . . Ak , including the empty word ∅. We identify words in U which are equal
modulo the commutativity relations. For each u ∈ U let us introduce a symbol κu .
Now consider the finite set L = {0, αi , Ai , κu | i = 1, . . . , k, u ∈ U } with the
following operation: κu αi = κu if u does not contain the letter Ai otherwise κu αi = 0,
κu Ai = κu Ai , if u does not contain Ai and κu Ai = 0 otherwise, αi Ai = Ai , αi2 =
αi , αai A j = A j αi , αi α j = α j αi , Ai A j = A j Ai for every i = j between 1 and k,
all other products are equal to 0. Then it is easy to see that L is a finite semigroup and
the map qm → κ, ai → αi , Ai → Ai extends to a homomophism from S(MMk ) to L
separating z1 and z2.
Recall that the Dehn function of a finite semigroup presentation X | R is the
minimal function f (n) such that for any words u, v which are equal in S and such that
|u| + |v| ≤ n, there exists a derivation of length at most f (n) of this equality from
the defining relations. For a finitely presented semigroup, a Dehn function does not
depend on the choice of finite presentation (up to equivalence), and the equivalence
class of that function is called the Dehn function of the semigroup.
Remark 3.8 The time complexity of the word problem is bounded from above in terms
of the Dehn function of a finitely presented semigroup S: given the Dehn function f (n)
of a semigroup presentation P, in order to check whether w = w (mod P) with
|w| + |w | ≤ n, we just need to check all sequences of length ≤ f (n) of applications
of defining relations w → w1 → . . .. moreover the word problem is decidable if and
only if the Dehn function is recursive [
39
].
Theorem 3.9 For every recursive set of natural numbers X and every recursive func
tion g(n) there exists a finitely presented residually finite semigroup S such that the
word problem in S is as hard as the membership problem in X and polynomially
reduces to it; the Dehn function S is bigger than g(n).
Proof By Theorems 2.4 and 2.7 there exists a sym-universally halting 2-glass Minsky
machine that recognizes X and whose configuration equivalence problem
polynomially reduces to the membership problem in X . By Lemma 3.5, the problem of
recognizing equality to w(s0) in S(MM2) is at least as hard as the membership
problem in X . By Lemma 3.7, S(MM2) is residually finite.
Remark 3.10 The proof of Theorem 3.9 could be simplified a little if instead of the
semigroup S(MMk ) we consider the semigroup S˜(MMk ) obtained from S(MMk ) by
adding one relation q0 = 0. That is S˜(MMk ) is the Rees quotient of S(M Mk ) by the
ideal generated by q0. Indeed, if MMk is sym-universally halting, then in S˜(MMk )
every non-zero element has only finitely many divisors, and so it is residually finite
by [
22
]. The word problem in S˜(MMk ) and the word problem in S(MMk ) are
polynomially equivalent. That follows from the fact that no command of MMk apply to
a configuration of the form (0; m, n). The semigroup S˜(MMk ) is used in the next
subsection.
3.3 Residually finite semigroups with large depth function
Recall the definition of the depth function ρ: for every finitely generated residually
finite semigroup S and every number n, ρS(n) is defined as the smallest number such
that for every two different elements z, z in S of word length ≤ n there exists a
homomorphism φ from S onto a finite semigroup B of cardinality at most ρS(n) such
that φ (z) = φ (z ).
The following lemma from [
22
] follows from Remark 3.6.
Lemma 3.11 Suppose that every non-zero element of a semigroup S with 0 has finitely many divisors. Then S is residually finite.
Theorem 3.12 For every recursive set of natural numbers X and every recursive
function g(n) there exists a finitely presented residually finite semigroup S such that
the depth function of S is bigger than g(n). In addition, the word problem in S and
the membership problem in X polynomially reduce to each other.
Proof Let MM2 be a sym-universally halting 2-glass Minsky machine with N + 1
commands numbered 0, . . . , N . We need the following property of MM2:
Thus we can assume that both w1 and w2 start with a q-letters. For every
where αi = αi (w1) ∈ {0, 1} we denote li by li (w), and αi by αi (w).
Then for every word w obtained from w by applying the relations of S˜(MM4)
we have α j (w ) = α j (w), j = 1, 2, 3, 4, u(w) is obtained from u(w1) by applying
relations of S(MM2).
Claim There are only finitely many words that are equal to u(w1) in S˜(MM2).
Indeed, if one of the numbers α1 or α2 is 0, the Claim is true by Theorem 2.7 (c). If
both α1 and α2 are equal to 1, since w1 is not equal to 0 in S˜(MM4), the configuration
(i ; l1, l2) is not accepted by MM2 (here we use the relation q0 = 0), and the Claim is
true because MM2 is sym-universally halting.
Let Y be the set of all words that are equal to w1 in S˜(MM4). Then the Claim implies
that the set of numbers l3(w) − l4(w), w ∈ Y , is finite. Let D(w1) be the maximum
of all these numbers. The number D(w2) is defined similarly. Let D be the maximum
of D(w1), D(w2).
Let us add the relations a3D = a32D, a4D = a42D to S˜(MM4). Let S¯ be the resulting
semigroup, and ψ : S˜(MM4) → S¯ be the corresponding homomorphism. Then it is
easy to see that ψ (w1) = ψ (w2). Notice that in S¯, every non-zero element has finite
number of divisors. Indeed, it is true for S˜(MM2) (see Remark 3.10) and the number of
different elements of S¯ of the form v(w) is finite. Hence we can again use Lemma 3.11.
The function ρ(n) for the semigroup S˜(MM4) is at least as large as the following
function (n) associated with the machine MM4: (n) is the smallest number such
that for every non-accepted input configuration of M of length ≤ n, the machine MM4
halts after at most (n) steps (we call this function the co-time function of MM4).
Indeed let c be an input configuration of length at most n such that MM4 halts after
exactly (n) steps starting at c. Suppose that the word w(c) in S˜(MM4) corresponding
to the configuration c can be separated from 0 in a homomorphic image E of S˜(MM4)
with at most (n) − 1 elements. Then the images of a3, a4 in that semigroup satisfy
z D = z2D for some D < T (n). Note that
• the halting computation has > D steps,
• the letter a3 does not occur in w(c),
• every old command of MM4 adds one coin in glass 3,
Therefore there exists a word W which is equal to w(c) in S˜(MM4) and which has the
form
Modulo relations corresponding to the commands (4), this word is equal to
q j a1l1 a2l2 A1 A2a3D A3 A4.
q j a1l1 a2l2 A1 A2a32D A3a4D A4.
The image of the latter word in E is equal to
q j a1l1 a2l2 A1 A2a3D A3a4D A4
q j a1l1 a2l2 A1 A2 A3 A4.
which, again modulo the relations corresponding to the commands (4), is equal to
Here j > 1 by our assumption that command number 1 cannot be used in the middle of
a computation consisting of old commands. Hence the latter word is equal to 0 by the
relations corresponding to the commands (5) and the relation q0 = 0, a contradiction.
Note that the co-time function of a Turing machine recognizing a recursive set can
be larger than any given recursive function. Indeed, we can assume that the Turing
machine has a history tape as in Sect. 2.2, so if the machine does not accept the
input, the last configuration in the halting computation has the history of computation
written on the tape. Then after the Turing machine halts without accepting, we can
make it compute some large recursive function, taking as a variable the word written
on the history tape. It remains to note that the co-time function of a Minsky machine
simulating that Turing machine cannot be smaller.
In the next section, we shall use the following properties of the semigroups studied
in this section. Let Sˇ be the semigroup given by all non-Minsky defining relations of
S(MMk ) and let Sˇ¯ be the semigroup given by all non-Minsky defining relations of the
semigroup S¯ from the proof of Theorem 3.12.
Lemma 3.15 (a) The growth function of Sˇ is polynomial of degree k.
(b) The semigroup Sˇ satisfies the following property:
where α, α , β j , β j ∈ {0, 1} and w = w in Sˇ, then qiα = qiα (i.e., either
α = α = 0 or i = i and α = α ), m j = m j , β j = β j , j = 1, . . . , k.
(c) The semigroup Sˇ¯ satisfies the following two properties
(Q1) If
where i, i = 0, α, α , β j , β j ∈ {0, 1}, and w = w in Sˇ¯, then qiα = qiα , β j =
β j , j = 1, . . . , k.
(Q2) For every word w the equality w Ai = 0 in Sˇ¯ implies wai Ai = 0 in Sˇ¯.
Proof (a) Indeed every non-zero element of length ≤ n of Sˇ is represented (modulo
the commutativity relations) by a word of the form qi a1m1 . . . akmk A11 . . . Akk where
m j ≤ n, j ∈ {0, 1}.
To prove Properties (P) and (Q1) notice that the exponents of qi , a j , A j do not
change when we apply the defining relations of these semigroups to w.
To prove (Q2) notice that word the w = qiαa1m1 . . . akmk A1β1 . . . Akβk is equal to 0 in
Sˇ¯ if and only if i = 0 and then apply (Q1).
4 Simulation of Minsky machines in solvable groups
Recall that a variety of algebraic structures is a class of all algebraic structures of a
given signature satisfying a given set of identities (also called laws). Equivalently, by
a theorem of Birkhoff [
35
] a variety is a class of algebraic structures closed under
taking cartesian products, homomorphic images and substructures. Every variety
contains free objects (called relatively free algebraic structures). One can define algebraic
structures that are finitely presented in a variety as factor-structures by congruence
relations generated by finite number of equalities. Every finitely presented algebraic
structure which belongs to a variety V is finitely presented inside V but the converse is
very rarely true. See [
32
] for a survey of algorithmic problems for varieties of different
algebraic structures (mostly semigroups, groups, associative and Lie algebras). In this
section we concentrate on varieties of groups (see [
43
]). The most well known varieties
are the variety of Abelian groups A given by the identity [x , y] = 1, the variety of
nilpotent groups of class c, Nc given by the identity [. . . [x1, x2], . . . , xc+1] = 1, etc.
The class of Abelian groups of finite exponent d, Ad , is also a variety, given by two
identities [x , y] = 1, x d = 1.
If U and V are two varieties of groups then the class of groups consisting of
extensions of groups from U by groups from V is again a variety (the product of U and V)
denoted by U V. The product of varieties is associative [
43
]. For example the variety
of all solvable groups of class c is the product of c copies of the variety A. If V is a
variety of groups, then ZV is the variety consisting of all central extensions of groups
from V. For example N2 = ZA and, more generally, Nc+1 = ZNc for every c ≥ 1.
4.1 The construction
Let MMk be a Minsky machine with k glasses and N + 1 commands (numbered
0, . . . , N ). We are going to construct a group G(MMk ) simulating MMk . The group
G(MMk ) will be very close to the semigroup S(MMk ) constructed above. The main
idea will be to replace the product in S(MMk ) by another, derived, operation and make
sure that with respect to the new operation the semigroup S(MMk ) “embeds” into our
group, in such a way that the “image” of S(MMk ) is “almost the whole” group.
The group will be generated by the x -letters which will be related to the letters
qi from S(MMk ), and also a-letters a1, . . . , ak , A-letters A1, . . . , Ak and some other
a- and A-letters that help us impose the necessary commutativity relations that, in
particular, make the group solvable, and contain “very few” extra elements. The group
we are going to construct will be a semidirect product of an elementary Abelian normal
subgroup generated (as a normal subgroup) by the x -letters by a semidirect product
of an Abelian subgroup generated (as a normal subgroup) by A-letters and an Abelian
subgroup generated by a-letters.
Thus we should have a way to ensure that in a subgroup generated by two sets of
letters Z ∪ Y , the normal subgroup generated by Z is Abelian. This is done with the
help of the following lemma due to Baumslag [3] and Remeslennikov [51]. In that
lemma we denote ua = a−1ua and ua+b = ua ub (note that although ua+b is not
necessarily equal to ub+a , the equality will hold if the normal subgroup generated by
u is Abelian, which is going to be the case every time we apply this lemma).
Lemma 4.1 ([3,51]). Suppose that a group H is generated by three sets X, F = {ai |
i = 1, . . . , m}, F = {ai | i = 1, . . . , m} such that
(1) The subgroup generated by F ∪ F is Abelian;
(2) For every a ∈ F and every x ∈ X we have x f (a) = x a for some monic polynomial
f of a which has at least two terms (everywhere below f (t ) = t − 1 and hence
x f (a) = a−1x ax −1 = [a, x −1], so we will not mention f again);
(3) [x1a1α1 ...amαm , x2] = 1, for every x1, x2 ∈ X , and every α1, . . . , αm ∈ {0, 1, −1}.
Then the normal subgroup generated by X in the group H = X ∪ F ∪ F is
Abelian, and H is metabelian.
If the elements ai and ai and the set X satisfy the conditions of Lemma 4.1 we will
call ai , BR-conjoints to ai with respect to X (and the polynomial f ), i = 1, . . . , m.
Consider the free commutative monoid generated by letters A0, . . . , Ak . Let U be
the set of all divisors of the element A0 A1 . . . Ak in that monoid, and U be the set of
all symbols q j u, u ∈ U , j = 0, . . . , N . Also fix a prime p (say, p = 2).
The generating set of our group G = G(MMk ) will consist of three subsets:
L0 = {x (u) | u ∈ U, i = 0, . . . , N };
L1 = { Ai | i = 0, . . . , k, };
L2 = {ai , ai , a˜i , a˜i | i = 1, . . . , k}.
We introduce notation for some subgroups of the group G. Denote Hi = Li , i =
0, 1, 2. Denote also
M0 = {a˜i , a˜i , A0 | i = 1, . . . , k}, Mi = {ai , ai , Ai }, i = 1, . . . , k.
The group G(MMk ) has the following finite set of defining relations:
(G1) Relations saying that H0 and H1 are Abelian groups of exponent p, and H2 is
an Abelian group.
(G2) Any y ∈ Mi , z ∈ M j , i = j ∈ {0, . . . , k}, commute.
(G3) For every i = 1, . . . , k, (ai )−1 is a BR-conjoint to ai−1 with respect to { Ai }.
(G4) The elements of the set {(a˜i )−1 | i = 1, . . . , k} are BR-conjoints to elements of
the set {a˜i−1 | i = 1, . . . , k} with respect to { A0}.
(G5) a) If u ∈ U does not contain Ai for some i = 0, . . . , k, then [x (u), Ai ] =
x (u Ai ).
b) For every i = 1, . . . , k, if u does not contain Ai , then x (u)ai −1 = x (u)ai
(where x a−1 = xa x −1),
c) For every i = 0, . . . , k, if u contains Ai , z ∈ Mi , then [x (u), z] = 1.
(G6) x (q j )ai = x (q j )a˜i , x (q j )ai = x (q j )a˜i , j = 0, . . . , N , i = 1, . . . , k.
(G7) [x (u)z , x (v)] = 1, where z = a1α1 . . . akαk , αi ∈ {−1, 0, 1}, u, v ∈ U
Remark 4.2 Relations (G7) together with (G1) and (G5) b) imply that for every subset
I ⊆ {1, . . . , k} the letters {ai | i ∈ I } are BR-conjoints of {ai | i ∈ I } with respect to
the set of all x (u)’s where u does not contain letters Ai , i ∈ I .
(G8) Relations constructed from the program of the machine MMk . For every f ∈ G
denote
f ∗ ai = f −1 f ai f −ai−1 f (ai )−1 , i = 1, . . . , k,
also let
f ∗ Ai = [ f, Ai ], i = 0, . . . , k.
We denote (. . . (t1 ∗ t2) ∗ . . .) ∗ tm by t1 ∗ . . . ∗ tm , and t1 ∗ t2 ∗ . . . ∗ t2 by t1 ∗ t2(n).
i, n1 > 0, . . . , nm > 0 → Sub(n1, . . . , nm ); j x(qi A0) ∗ an1 ∗ . . . ∗ anm = x(q j A0)
x (qi A0) ∗ a1(m1) ∗ . . . ak(mk ) ∗ A(1α1)
∗ . . . ∗ A(kαk ) = x (q j A0) ∗ a1(n1) ∗ . . . ∗ ak(nk) ∗ A(1β1) ∗ . . . A(kβk )
where αi , βi ∈ {0, 1} is true in G(MMk ) if and only if the equality
qi a1m1 . . . akmk A1α1 . . . Akαk = q j a1n1 . . . akmk A1β1 . . . Akβk
is true in the semigroup S(MMk ) (in particular, αi = βi for every i ).
(c) The equality
x (qi ) ∗ a1(m1) ∗ . . . ak(mk ) ∗ A(1α1)
∗ . . . ∗ A(kαk ) = x (q j ) ∗ a1(n1) ∗ . . . ∗ ak(nk ) ∗ A(1β1) ∗ . . . A(kβk )
where αi , βi ∈ {0, 1} is true in G(MMk ) if and only if the equality
qi a1m1 . . . akmk A1α1 . . . Akαk = q j a1n1 . . . akmk A1β1 . . . Akβk
is true in the semigroup Sˆ (i.e., these words coincide, see Lemma 3.15 (b)).
Proof The proof of Part (a) is divided into several lemmas.
Lemma 4.4 The subgroup H1 ∪ H2 of G is metabelian and a semidirect product of
the Abelian normal subgroup H1H2 of exponent p, and H2.
Proof Indeed by relations (G2),
k
i=1
Mi , i = 0, . . . , k =
ai , ai , Ai × a˜i , a˜i , A0, i = 1, . . . , k .
Using relations (G1), (G3), (G4), we can apply Lemma 4.1 to each of the factors
in that direct product and conclude that each of them is metabelian and a semidirect
product of the Abelian normal subgroup of exponent p generated by the intersection of
{ Ai | i = 0, . . . , k} with that factor, and the Abelian group generated by the a-letters
from that factor.
Lemma 4.5 The normal subgroup T of G generated as a normal subgroup by all the
elements x (u), u ∈ U, is Abelian of exponent p.
Proof Relations (G5) a) of the group G imply that every element x (u), u ∈ U, is a
product of elements x (q j )z , z ∈ H1, i = 0, . . . , N . Therefore, it is enough to show
that
x (qk )x (qt )z = x (qt )z x (qk )
for any z ∈ H1, H2 and any k, t . To simplify these equalities notice that z =
z0z1 . . . zk where zi ∈ Mi by (G2). Therefore equalities (7) are equivalent to
x (qk )z0 x (qt )z1...zk = x (qt )z1...zk x (qk )z0 .
(7)
(8)
We can represent element x (q j )zi , i ≥ 1, as a product of elements of the form
x (q j )aip(ai )q and x (q j Ai )a˜ip(a˜i )q . Indeed we have the following sequence of
equalities deduced using (G2), (G5), (G6):
=(G=5) =a=),=c), =(G=6) x (q j )air1+r2 (ai )s1+s2 Ait2 ...airk (ai )sk Aitk (x (q j Ai )t1 )a˜ir1 (a˜i )s1
(9)
Repeating this argument k times, one proves that x (q j )z1z2...zk can be represented as a
product of elements of the form x (u)y where u ∈ U , y ∈ H2. A similar proof (using
also (G4)) gives that x (q j )z0 is a product of elements of that form. It remains to note
that elements of the form x (u)y , u ∈ U, y ∈ H2 commute by Remark 4.2 and Lemma
4.1.
Remark 4.6 Note that equalities (9) and similar equalities when x (q j ) is replaced by
r
x (u), u ∈ U , imply the following: if y is a product of elements of the form ai l (ai )sl Ai
and l rl = l sl = 0, then [x (u), y] is equal to 1 if u contains Ai or is equal to
a product of conjugates of elements x (u Ai ) by elements from a˜i × a˜i otherwise.
Similarly, suppose that y is a product of elements from M0, each factor containing
A0, and the total exponent of every a˜i (resp. a˜i ) is 0. Then [x (u), y] = 1 provided
u contains A0 and is a product of conjugates of x (u A0) by elements from ai , ai
provided u does not contain A0.
By construction, the group G is a semidirect product of T and the metabelian group
H1H2 H2. By Lemma 4.4, G is solvable of class 3 and, moreover, belongs to A2pA.
Remark 4.7 The proof of Lemma 4.5 shows that T is generated (as an Abelian group)
by elements of the form x (u)y where u ∈ U and y ∈ H2.
Lemma 4.8 The quotient of G(MMk ) over the center satisfies the identity
[[x1, y1], [x2, y2], . . . , [xk+2, yk+2]] = 1.
This means that G belongs to the variety ZNk+1A.
Proof Let P be the derived subgroup of G(MMk ). By Lemma 4.5, every element of
P is a product of an element of T and an element of H1H2 . It also follows from Lemma
4.5 that [ P, P] ⊆ T , hence by Remark 4.7, it is generated by elements of the form
x (u)y , u ∈ U, y ∈ H2, the word u contains at least one Ai , i = 0, . . . , k. Since T is
Abelian, the subgroup [ P, P, . . . , P] is generated by the commutators
k+2
x (u)y , h1h,21,1 , . . . , h1h,2k,k
for some h1,i ∈ H1, y, h2,i ∈ H2. An easy induction shows that every such commutator
is a conjugate of
where y ∈ H2.
Let h ∈ H1, u ∈ U, y ∈ H2. Suppose that h = Ai11 · . . . · Aits where ti = 0. Consider
t
s
[x (u), h y ]. Then Remark 4.6 implies that [x (u), h y ] is a product of elements of the
form x (u )y where u ∈ U contains letters Ai1 , . . . , Ais and it may not be equal to 1
only if one of the letters Ai j does not occur in u. Therefore the commutator (10) is
either equal to 1 or is a product of elements of the form x (u )y where the word u ∈ U
contains all letters A0, A1, . . . , Ak , y ∈ H2. But every such x (u ) is in the center of
G(MMk ) by (G5) c). Hence [ P, . . . , P] is contained in the center of G(MMk ).
(10)
We now prove Parts (b) and (c) of Theorem 4.3. For every configuration c =
(i ; m1, . . . , mk ) of MMk let wG (c) = x (qi A0) ∗ a1(m1) ∗ . . . ak(mk ) ∗ A(1α1) ∗ . . . ∗ A(kαk ).
To prove (b), as we mentioned before Property 3.1, we need to prove Properties 3.1
and 3.2. Property 3.1 for G(MMk ) is proved in the same way as for the semigroup
S(MMk ) (see [
32,55
]),since the only property of S(MMk ) used there was that the
word w = qi a1l1 . . . aklk A1α1 . . . Akαk is equal in S(MMi ) to any word obtained from w
by permuting ai with a j , Ai with A j and ai with A j (i = j ). The same is true for
words of the form wG (c) in G(MMk ) by the definition of the operation ∗, relations
(G1), (G2) and Lemma 4.5.
In order to prove Property 3.2 we will define a new group G that is an image of G
under some homomorphism that is injective on the elements from Parts (b) and (c).
Let Sˇ be the semigroup with the same generating set as S(MMk ) subject all the
relations of S(MMk ) except the Minsky relations (3). That semigroup does not depend
on MMk . Thus non-zero elements in Sˇ have the form
qiα1 a1l1 . . . aklk A1α1 . . . Akαk
where l j ∈ N, α j ∈ {0, 1}. Let W be the set of all non-zero elements of Sˇ containing
a q-letter.
Let ψ be the natural homomorphism from Sˇ onto S(MMk ). Let W0 = ψ (W ).
We will need the set of vectors {1, 2, 3}k ⊂ Zk with coordinates 1, 2, 3. Its elements
will be denoted by i . The j -th coordinate of the vector i will be denoted by i j , the
standard unit vectors (0, . . . , 1, . . . , 0) are denoted by e j . Consider the set of symbols
{z(i , u) | i ∈ {1, 2, 3}k , u ∈ W ∪ W0} and the multiplicative Abelian group T1 of
exponent p freely generated by this set.
For each letter from L1 ∪ L2, we define an automorphism of T1. The group G will
be the semidirect product of T1 and the group generated by these automorphisms. For
simplicity we will denote automorphisms corresponding to letters from L1 ∪ L2 by
the same letters (the automorphisms are just conjugations by these letters).
Let us start with automorphisms a j , a j . We have to define z(i , w)a j and z(i , w)a j
for every i and every w ∈ W ∪ W0. This definition does not depend on whether w
belongs to W or W0. First suppose that w does not contain A j . Then
⎧
⎪⎪⎨ z(i , w)z(i + e j , w)z(i + 2e j , w)z(i , wa j )
z(i , w)a j = ⎪ z(i , w)z(i − e j , w)−1
⎪⎩
z(i − 2e j , w)
z(i , w)a j = z(i , w)−1z(i , w)a j .
if i j = 1;
if i j = 2;
if i j = 3.
If w contains letter A j , then let z(i , w)a j = z(i , w)a j = z(i , w).
It is easy to prove that a j is an automorphism by constructing the automorphism
a −j1 (view (11) as a triangular system of linear equations and solve it by backward
substitution). For example:
⎧
⎪⎪⎪ z(i − e j , w)−1z(i , w)−1z(i , wa j )−1,
⎪⎨
z(i , w)a −j1 = ⎪ z(i , w)z(i + e j , w), if i j = 2
⎪⎪⎪⎩
z(i + 2e j , w),
if i j = 1.
if i j = 3
⎧
⎪⎪⎨ z(i , w)z(i + e j , w)z(i + 2e j , w)z(i , va j A j ),
z(i , w)a˜ j = ⎪ z(i , w)z(i − e j , w)−1,
⎩⎪
z(i − 2e j , w),
z(i , w)a˜ j = z(i , w)−1z(i , w)a˜ j .
The automorphisms corresponding to A j , j = 1, . . . , k, are defined as follows:
if w ∈ W ∪ W0 does not contain A j and
if w contains A j .
Finally the automorphism corresponding to A0 is defined as follows:
A j
A0
= z i , w z i , w A j
= z i , w z i , ψ (w)
if i j = 1;
if i j = 2;
if i j = 3.
if w ∈ W and
if w ∈ W0.
The following lemma is obtained by a straightforward application of the definition
of the automorphisms above and the definition of the operation ∗. This lemma implies
that G satisfies (G8) if we replace x (u) by z(1, u) where 1 is the vector (1, 1, . . . , 1)
(since the corresponding relations hold in S(MMk )).
If w ∈ W ∪ W0, d ∈ {a j , A j | j = 1, . . . , k} then by wd we mean the product of
w and d in S(MMk ) provided w ∈ W0, or in Sˆ provided w ∈ W .
The proof of the following lemma is by inspection of various cases and mostly left
to the reader.
Lemma 4.9 The following relations hold in G. For every d ∈ {a j , A j | j = 1, . . . , k},
w ∈ W ∪ W0
z(1, w) ∗ d = z(1, wd)
where ∗ is defined in (G8). Here we set z(1, 0) = 1 (where 0 is the zero element in
S(MMk ) or Sˆ, 1 in the right hand side is the identity element in G(MMk )).
Proof For example, if d = A1 then z(1, w) ∗ d = [z(1, w), A1]. It is equal to 1 if w
contains A1, and it is equal to z(1, w)−1z(1, w)A1 = z(1, w)−1(z(1, w)z(1, w A j )) =
z(1, w A j ) if w does not contain A1. Thus in both cases z(1, w) ∗ d = z(1, wd).
By definition, G is the semidirect product of T1 and the subgroup of Aut(T1)
generated by the automorphisms corresponding to the elements from L1 ∪ L2. From
the definition of the automorphisms and Lemma 4.9, it follows that G is generated
by the elements z(1, u), u ∈ U , and the automorphisms corresponding to elements of
L1 ∪ L2. It is easy to check that all the relations (G1)-(G8) hold in G, therefore
Lemma 4.10 The map that sends every a- or A-letter to itself, every x (u) with u ∈ U
containing A0, u = v A0, to z(1, ψ (v)) and every x (u) with u not containing A0 to
z(1, u) extends to a homomorphism γ from G to G.
Lemma 4.11 The homomorphism γ is surjective and the preimage of T1 is T .
Proof We only need to define pre-images of elements z(i , w) ∈ G¯ , w ∈ W ∪ W0. By
the definition of γ , we have γ (x (u)) = z(1, u) for every u ∈ U which does not contain
A0, and γ (x (u A0)) = z(1, ψ (u )) so x (u), u ∈ U , are preimages of all z(1, u). The
preimage x (i , w ) of z(i , w) for every i and w is defined by induction on the length
of w and the sum of i j (the base of induction, where i = 1 is obvious).
x (i + e j , w) = x (i , w)−(a j )−1 ,
x (i + 2e j , w) = x (i , w)a −j1 ,
x (i , wa j ) = x (i , w) ∗ a j .
We also set x (i , w A j ) = x (i , w) ∗ A j for any i . This covers the case when w contains
A j and i j = 2 or 3.
It is easy to see that for every i and w ∈ W ∪ W0, we have γ (x (i , w)) = z(i , w).
This proves the lemma.
In G(MMk ), consider the set P0 of elements
where αi ∈ {0, 1} and the set P of elements
z(1, ψ (qi )) ∗ a1(m1) ∗ . . . ak(mk ) ∗ A(1α1) ∗ . . . ∗ A(kαk )
z(1, qi ) ∗ a1(m1) ∗ . . . ak(mk ) ∗ A(1α1) ∗ . . . ∗ A(kαk )
By construction P ∩ P0 = ∅, elements (12) are different if and only if elements
qi a1m1 . . . akmk A1α1 . . . Akαk
from S(MMk ) are different, and elements (13) are different if and only if the
corresponding elements qi a1m1 . . . akmk A1α1 . . . Aαk of Sˇ are different. By Lemma 4.9 the
k
element x (qi A0) ∗ a1(m1) ∗ . . . ak(mk ) ∗ A(1α1) ∗ . . . ∗ Ak(αk ) is equal to the element (12)
and the element x (qi ) ∗ a1(m1) ∗ . . . ak(mk ) ∗ A(1α1) ∗ . . . ∗ A(kαk ) is equal to the element
(13). This completes the proof of Property 3.2 and Theorem 4.3 (b), (c).
We shall need a few more properties of the group G(MMk ).
Lemma 4.12 Let elements x (i , w), w ∈ W ∪ W0, from G(MMk ) be defined as in the
proof of Lemma 4.11. Let y ∈ L1 ∪ L2, w ∈ W ∪ W0. Then x (i , u)y is a product of
one or several elements of the form x (i , w ) such that every letter a j occurs in w at
least as many times as in w (in particular if for some R > 0, w belongs to the ideal
VR defined in Lemma 3.7, then w ∈ VR .
Proof Indeed it is easy to check that for every i ∈ {1, 2, 3}{1,...,k}, x (i , w) satisfies the
same equalities as elements z(i , w) from the definition of automorphism of G with z
replaced by x everywhere.
Then the statement of the lemma for y ∈ ∪ j≥1 M j , follows from the way x (i , u)
are constructed. For y ∈ M0, one needs to use (G2), (G5) c), and (G6).
Lemma 4.13 The normal subgroup T of G = G(MMk ) generated by the elements
x (u), u ∈ U is the direct product of cyclic subgroups generated by the elements
x i , w , i ∈ {1, 2, 3}{1,...,k}, w ∈ W ∪ W0.
Proof By Lemma 4.12 elements x (i , w) span T . We defined elements x (i , w), w ∈
W ∪ W0 in such a way that they are pre-images of the corresponding elements z(1, w)
in G under γ . Thus the elements x (i , w), i ∈ {1, 2, 3}{1,...,k}, w ∈ W ∪ W0 are
linearly independent since their images under γ are linearly independent in T1 (here
we temporarily view T , T1 as vector spaces over the field with p elements).
The proof of Lemma 4.12 actually gives the following two facts (the proof of the
first of them also employs Lemma 3.15 (a)).
Lemma 4.14 If v is a word in a- and A-letters (i.e. over L1 ∪ L2), u ∈ U , then x (i , u)v
is a product in G of elements x ( j , w) as in Lemma 4.12 where the length of each w
does not exceed |v| + O(1) hence the total number of different x ( j , w) occurring in
this product is polynomial in |v| of degree at most k.
Lemma 4.15 Let v, w ∈ W . Suppose v = w is a Minsky relation of the
semigroup S(MMk ). Then the corresponding relation from (G8) has the form
x (1, v)(x (1, w))−1 = 1. The conjugate of x (1, v)x (1, w)−1 by an element g ∈
H1, H2 is a product
m
(x (i , vm )x (i , wm )−1),
where for each m, vm = vum , wm = wum for some word um whose length does not
exceed the length of g.
Remark 4.16 Instead of semigroup S(MMk ) we could use the semigroup S˜(MMk ) and
construct a group G˜ (MMk ) which is a quotient of G(MMk ) by the subgroup spanned
by x (i , w) where w ∈ W0 ∪ W contains q0. It is easy to see that the construction of
G(MMk ), Lemmas 4.4, 4.5, 4.8–4.15 (with notation modified in the appropriate way
by “killing” q0) and Theorem 4.18 hold for the group G˜ (MMk ) as well.
Moreover let S¯ be any homomorphic image of S obtained by adding defining
relations to S(MMk ). Let Sˇ¯ be the semigroup given by the same presentation as S¯
excluding the Minsky relations. Suppose that Properties (Q1), (Q2) from Lemma 3.15
hold in Sˇ¯. Then we can replace S(MMk ) by S¯, and Sˇ by Sˇ¯ in the construction of
G(MMk ). The resulting group will satisfy all the lemmas and the theorem mentioned
in the previous paragraph (with S(MMk ) replaced by S¯, and Sˇ replaced by Sˇ¯). To
verify this statement, one just needs to routinely check the proofs of these lemmas and
the theorem.
4.1.1 A finitely presented solvable group with undecidable word problem
By Theorem 2.7, there exists a 2-glass Minsky machine which computes a
nonrecursive partial function. The corresponding group G(MMk ) has undecidable word
2
problem and belongs to the variety A pA ∩ ZN 3A by Theorem 4.3. Hence we obtain
the following:
Theorem 4.17 See ([28]). There exists a finitely presented group with undecidable
2
word problem that belongs to the variety A pA ∩ ZN 3A.
The proof of this theorem presented in this paper is simpler than the original proof
in [
28
].
4.1.2 Residually finite finitely presented groups
Theorem 4.18 If a Minsky machine MMk is sym-universally halting then the group
G(MMk ) is residually finite. The word problem in G(MMk ) and the configuration equivalence problem for MMk are polynomially reducible to each other.
Proof Let MMk be a sym-universally halting Minsky machine. Let g = 1 ∈ G(MMk ).
We use the notation from the definition of G(MMk ). There exists a natural
homomorphism ζ from G(MMk ) to the metabelian group H1H2 H2 with kernel T . Since every
finitely generated metabelian group is residually finite, we can assume that ζ (g) = 1.
Hence g ∈ T . By Lemma 4.13, g is a product of elements of the form
x (i , w), i ∈ {1, 2, 3}{1,...,k}, u ∈ W ∪ W0.
(14)
Hence g = g0g1 where g0 (resp. g1) is a product of elements (14) with w ∈ W0 (resp.
w ∈ W ). Suppose that g1 is not 1. Let T be the subgroup of G(MMk ) generated by
elements (14) with w ∈ W0. Then T is a normal subgroup of G(MMk ) by Lemma
4.12. Let G (MMk ) = G(MMk )/ T . This group is a semidirect product of T / T and
the metabelian group H1H2 H1. Let D be the sum of lengths of words w ∈ W that
appear in the factors of g1. Let YD be the set of all words in Sˇ where at least one a-letter
appears at least D times, and 0. Then YD is an ideal in Sˇ, and the image of the set
of elements (14) with w ∈ YD in G (MMk ) form a normal subgroup N of G (MMk )
of finite index (because T is an Abelian group of finite exponent p). That normal
subgroup does not contain g by Theorem 4.3 (c). Then G (MMk )/N is a semidirect
product of a finite group and the metabelian group H1H2 H2. Hence G (MMk )/N
is residually finite and g can be separated from 1 by a homomorphism from G(MMk )
onto a finite group.
Finally suppose that g1 = 1. Let w1, . . . , wl be the elements from W0 that appear
in the representation of g as a product of elements (14). Let E be the set of words
that is equal to one of the w j in S(MMk ). Since MMk is sym-universally halting,
E is finite. Let D be the maximal length of a word in E . Let, as above, YD be the
ideal in Sˇ consisting of 0 and all elements where one of the a-letters appears at least
D times. Let Z D be the set of non-zero elements of S(MMk ) that are images of
words from YD under the natural homomorphism Sˇ → S(MMk ). Then Z D does not
contain w1, . . . , wl . Consider the subgroup F of T spanned by all elements (14) with
w ∈ Z D ∪ YD. From Lemma 4.12, it follows that F is a normal subgroup of G(MMk )
of finite index in T . Since Z D does not contain w1, . . . , wl , the subgroup F does not
contain g. The factor-group G(MMk )/F is a semidirect product of a finite group and
the metabelian group H1H2 H2, and we can complete the proof as above.
To prove that the configuration equivalence problem in MMk polynomially reduces
to the word problem in G(MMk ) we notice that the length of a word w(c) corresponding
to an input configuration c of MMk in the semigroup S(MMk ) is |c| + O(1). If w(c) is
the word of length n in S(MMk ), then the corresponding element in G(MMk ) according
to relations (G8), can be represented as a similar word wg(c) with respect to the ∗
operation. Every time when we evaluate the ∗ operation, we rewrite the word as a
product of x (q j A0)u , u = u1u2 . . . uk , ui = aiki Ai or ui = ai i , j = 0, . . . , k where u
k
is a subword of w(c). Using commutativity relations from Lemma 4.5, we can collect
all x (q j A0)u with the same u. Elements x (q j A0)u have order p. Therefore, we have a
product of conjugates (x (q j A0)u )r , where 1 ≤ r ≤ p − 1, and k1 + k2 + k3 ≤ n. The
number of such elements is at most O(n3), each of them can be written as a group
word of length at most 2n + 1. Therefore for w(c) of length n, the corresponding
element in G(MMk ) can be represented as a word of length at most O(n4). Now
two configurations c and c of MMk are equivalent if and only if w(c) = w(c ) in
S(MMk ) and if and only if wg(c) = wg(c ) in G(MMk ) by Properties 3.1 and 3.2.
Thus checking whether c ≡MMk c can be done in polynomial time in terms of |c|+|c |
with using the oracle responsible for the word problem in G(MMk ) only once.
To get a polynomial reduction in the other direction we consider any element g of
represented by a word w of length ≤n in G(MMk ). We need to check if g = 1. First
check if g is in the normal subgroup T . By construction G(MMk ) is the semidirect
product of T and a finitely generated metabelian group (generated by a- and A-letters).
Since metabelian groups embed into finite direct products of linear groups over fields
[62], the membership g ∈ T can be checked in polynomial time. If the answer is “no”,
then g = 1.
Now suppose that g is in T , then we represent w in G(MMk ) as a product of a fewer
than n conjugates x (i , u)v where u ∈ U , and v is a word in a-letters and A-letters whose
length is bounded by n. By Lemmas 4.14 and 4.11 then w is a product in G(MMk ) of
elements of the form (12) and (13) whose number is bounded by a polynomial in n and
whose lengths (in terms of the operation ∗) are bounded by n. Since different words of
that form are linearly independent (by Lemma 4.11) in order to check whether g = 1,
we need to verify equalities of words of the form (12) and (13) which by Theorem
4.3 is equivalent to verifying equalities of corresponding words in S(MMk ), which, in
turn, reduces to verifying equivalence of the corresponding configurations of MMk by
Properties 3.1 and 3.2 which hold for S(MMk ). Thus the word problem in G(MMk )
polynomially reduces to the configuration equivalence problem for MMk .
Remark 4.19 Note that if instead of the semigroup S(MMk ) we could start with any
semigroup S¯ satisfying Properties (Q1), (Q2) of Lemma 3.15. By Remark 4.16, the
resulting group G¯ satisfies all the properties mentioned in that remark. The proof of
Theorem 4.18 shows that if in S¯ every non-zero element has finitely many divisors,
then the group G¯ is residually finite. Moreover if an element w ∈ W0 has finite number
of divisors in S¯, then there exists a homomorphism φ from G¯ to a finite group with
φ (x (1, w)) = 1.
Lemma 4.20 Let d(n) be the Dehn function of G(MMk ) and t (n) be the time function
of MMk . Then
t (n)
Proof Let Gˇ be the group given by the presentation of G(MMk ) except the defining
relations (G8). Let g be an element of Gˇ which is equal to 1 in G(MMk ). Let w be
a word in generators of Gˇ that represents g in Gˇ , |w| = n. Then w is equal in Gˇ
to a product of conjugates of relators (G8). This representation can be obtained
as follows. Consider a minimal area van Kampen diagram over the presentation of
G(MMk ) with boundary label w. We can read off of this diagram a representation of w
as a product of conjugates of all relations (G1)–(G8). Now remove from that product
all conjugates of relators (G1)–(G7). The remaining product is .
Let t (n) be the time function of MMk . Let c be an accepted configuration of MMk
such that |c| = n and the length of the computation connecting c with the accept
configuration s0 of MMk is t (n). Let w(c) and wg(c) be the corresponding elements
in S(MMk ) and Gˇ respectively. Let Tˇ be the normal subgroup of Gˇ generated (as a
normal subgroup) by the x -letters. Since the configuration c is accepted by MMk , we
have that wg(c)−1wg(s0) = 1 modulo the relations (G8). Therefore wg(c)−1wg(s0)
is a product of conjugates of the relations (G8). Since the length of wg(c) does not
exceed |c|4 for some uniform constant (see the proof of Theorem 4.18), the number
of factors in the product does not exceed d( n4).
By Lemma 4.15, wg(c) = x (1, uc), wg(c ) = x (1, uc ) for some words uc, uc in the
generators of S(MMk ), and we can rewrite the product and obtain a product of
elements of the form x (i , u)x (i , v)−1 where v is obtained from u by applying a Minsky
relator of S(MMk ) once. This product is also equal to x (1, uc)x (1, uc )−1. Since
elements x (i , u), u ∈ W ∪ W0 form a basis of T viewed as a vector space of the field
with p elements (Lemma 4.13), this implies that there exists an accepting computation
for the configuration c of length at most d( n4) and inequality (15) follows.
Theorem 4.21 For every recursive set of natural numbers X and every recursive
function g(n) there exists a finitely presented residually finite solvable of class 3
group G such that the word problem in G is as hard as the membership problem in X
and polynomially reduces to it; the Dehn function G is bigger than g(n).
Proof By Theorems 2.4 and 2.7 there exists a sym-universally halting 2-glass Minsky
machine MM2 whose configuration equivalence problem polynomially reduces to the
membership problem for X . By Lemma 3.5, the time complexity of the problem of
recognizing equality to 0 in the semigroup S(MM2) is as large as f (n) and the word
problem in S(MM2) polynomially reduces to the membership problem for X . The first
statement now follows from Theorems 4.18 and 3.9. For the Dehn function part of the
theorem we use Lemma 4.20. Thus we just need to modify the Minsky machine MM4
so that the new machine MMm , m > 4, has the same complexity of the configuration
equivalence problem but time function larger than g(n4).
This can be done in the following straightforward way. After the machine MM4 is
supposed to stop, we make MMm compute a recursive function that is greater than
g(n4), then stops. We leave it as an exercise for the reader to determine the exact value
of m and the program of MMm .
4.2 A residually finite finitely presented group with large depth function
Theorem 4.22 For every recursive function f and a recursive set X of natural num
bers, one can construct two residually finite finitely presented solvable of class 3
groups G1, G2. Both groups have depth functions greater than f . The group G1 has
word problem as hard as the membership problem for X . The group G2 has the word
problem decidable in polynomial time.
Proof Consider the Minsky machine MM4 constructed in the proof of Theorem 3.12,
the semigroup S˜(MMk ) and the corresponding group G˜ (MM4) (it is obtained from
G(MM4) by imposing the relation xu = 1 for every u ∈ U containing q0). Let us
prove that it is residually finite. Take an element g ∈ G(MM4) such that g = 1. As in
the proof of Theorem 4.18 we can assume g ∈ T . By Lemma 4.13, g is a product of
elements of the form 14. Hence g = g0g1 where g0 (resp. g1) is a product of elements
(14) with u ∈ W0 (resp. W ). Suppose that g1 is not 1. Let T be the subgroup of
G˜ (MM4) generated by elements (14) with w ∈ W0. Then T is a normal subgroup
of G˜ (MM4) by Lemma 4.12 (and Remark 4.16). Let G˜ (MM4) = G˜ (MM4)/ T . This
group is a semidirect product of T / T and the metabelian group H1H2 H1. Let D be
the sum of lengths of words u ∈ W that appear in the factors of g1. Let YD be the set
of all words in Sˇ where at least one a-letter appears at least D times, and 0. Then YD
is an ideal in Sˇ, and the image of the set of elements (14) with w ∈ YD in G˜ (MM4)
form a normal subgroup R of G (MN ) of finite index (because T is an Abelian group
of finite exponent p). That normal subgroup does not contain g by Theorem 4.3 (c)
(and Remark 4.16. Then G˜ (MM4)/R is a semidirect product of a finite group and
the metabelian group H1H2 H2. Hence G˜ (MM4)/R is residually finite and g can be
separated from 1 by a homomorphism from G˜ (MM4) onto a finite group.
Thus we may assume that g1 = 1. Let w1, . . . , wl be the elements from W0 that
appear in the representation of g as a product of elements (14). Let E be the set of
elements of Sˆ (i.e., words modulo the commutativity relations) which are equal to one
of the w j in S(MM4). Note that every word in E starts with a q-letter by definition of
elements (14).
For each w ∈ E let D(w) be as in the proof of Theorem 3.12. Let D be the maximum
of these numbers D(w) and, as in the proof of Theorem 3.12, let S¯ be the semigroup
S˜(MM4) with additional defining relations a3D = a32D, a4D = a42D. As we mentioned
in the proof of Theorem 3.12, every non-zero element of S¯ has finitely many divisors.
It is easy to see that S¯ satisfies Properties (Q1), (Q2) of Lemma 3.15. Therefore by
Remark 4.16, we can built a group G¯ starting with S¯ instead of S˜(MM4) and all the
statements mentioned in Remark 4.16 remain true for G¯ . The natural homomorphism
δ : S˜(MM4) → S¯ extends to a homomorphism δ¯ : G˜ (MMk ) → G¯ . By Theorem 4.3
and Lemma 4.13 the images under δ¯ of all elements (14) with w ∈ {w1, . . . , wl } form
a finite linearly independent set. Hence δ¯(g) = 1 in G¯ . Since every non-zero element
of S¯ has finitely many divisors, the group G¯ is residually finite by Remark 4.19. Hence
there exists a homomorphism of G˜ (MM4) onto a finite groups separating g from 1.
Thus G˜ (MM4) is indeed residually finite.
The fact that ρG (n) ≥ f (n) is proved the same way as in the proof of Theorem
3.12 (one only needs to replace the product by operation * everywhere in that proof).
5 Applications
In this section we present two applications of our results and methods. One application
concerns the universal theory of finite solvable groups of a given class, the second
application concerns with the membership problem in pro-finitely closed subgroups
of residually finite groups.
5.1 Universal theories of sets of finite groups
In this section we will prove the following result. For the class of all finite groups in
was proved by Slobodskoi [60] (the idea of Slobodskoi’s proof came from Gurevich’s
paper [
23
] where the same result was proved for semigroups, see also [58]).
2
Theorem 5.1 The universal theories of the class of finite groups from A pA ∩ ZN5A
and the class of all periodic groups are recursively inseparable. In particular, the
universal theory of any set of finite groups containing all finite solvable of class 3
groups is undecidable.
Proof Consider a partial function f : N → N such that f is one-to-one and
computable on its domain, and the domain is recursively enumerable but not recursive.
Let MM2 be a Minsky machine computing the function f . We can assume that for
every number not in the domain of f the machine MM2 works indefinitely long and
never stops. As in the proof of Theorem 3.12, we can assume that the start command
number 1 cannot be used in the middle of a computation of MM2.
Consider the 4-glass Minsky machine MM4 described in the proof of Theorem
3.12. Let S (MM4) be the semigroup given by the same defining relations as S˜(MM4)
except the relation q0 = 0 is substituted by the relation qi A3 A4 = 0 for every i .
Let G (MM4) be the group corresponding to S (MM4) in the same way G(MMk )
corresponds to S(MMk ). Since Properties (Q1), (Q2) of Lemma 3.15 obviously hold
2
for S (MM4) we can use Remark 4.16. Then G (MM4) belongs to A pA ∩ ZN5A
and simulates MM4 as described in Theorem 4.3. Let R be the (finite) set of defining
relations of G (MM4). Let X be the set of numbers such that MM4 accepts the
configuration ( , 0, 0, 0). Let X be the set of numbers such that MM4 works infinitely
long starting with the configuration ( , 0, 0, 0). Then X and X are recursively
inseparable by the choice of MM2 and MM4. For any configuration ( , 0, 0, 0) of MM4
consider the corresponding element
w( ) = q1 ∗ a1( ) ∗ A1 ∗ A2 ∗ a3 ∗ A3 ∗ A4.
Suppose ∈ X . Then the Minsky machine MM2 halts starting at ( , 0, 0, 0).
Since the function f is one-to-one, there are only finite number of computations
of Sym(MM2) starting at the configuration c = (1; , 0, 0, 0). Then as in the proof
of Theorem 4.22, we can find a a homomorphism φ from G˜ (MM4) to a finite group
such that φ (x (1, w)) = 1.
Hence the universal formula & R → x (1, w) = 1 does not hold in the finite group
2
H from A pA ∩ ZN5A.
Now suppose that ∈ X . Consider any periodic homomorphic image H of
G (MM4). Let t¯ be the image of t ∈ G (Mn) in H . Then there exists a number
D such that for every element x ∈ T¯ ,
x ∗ a¯ 3(D) = x ∗ a¯ 3(2D).
(16)
Since MM4 works infinitely long starting at the configuration ( , 0, 1, 0), by
Theorem 4.3 the following equality is true for some i, k1, k2:
w( ) = x¯(1, qi A0) ∗ a¯ 1(k1) ∗ a¯ 2(k2) ∗ a¯ 3(D) ∗ A¯1 ∗ A¯2 ∗ A¯3 ∗ A¯4.
Then by (16) and Theorem 4.3
= x¯1,qi A0 ∗ a¯ 1(k1) ∗ a¯ 2(k2) ∗ A¯1 ∗ A¯2 ∗ A¯3 ∗ A¯4 = 1.
w¯ ( ) = x¯1,qi A0 ∗ a¯ 1(k1) ∗ a¯ 2(k2) ∗ a¯ 3(2D) ∗ a¯ 4(D) ∗ A¯1 ∗ A¯2 ∗ A¯3 ∗ A¯4
= x¯1,qi A0 ∗ a¯ 1(k1) ∗ a¯ 2(k2) ∗ a¯ 3(D) ∗ a¯ 4(D) ∗ A¯1 ∗ A¯2 ∗ A¯3 ∗ A¯4
since qi A3 A4 = 0 in S (Mn). Hence the universal formula & R → w( ) = 1 holds
in H .
Thus the set of universal formulas &R → w( ) = 1 that do not hold in some finite
2
group from A pA ∩ ZN5A and the set of such formulas which hold in every periodic
group are recursively inseparable.
Remark 5.2 Note that the universal theory of finite metabelian groups is decidable
because every finitely generated metabelian group is residually finite (the connection
is explained in [
32
]). On the other hand, the universal theory of all finite nilpotent
groups is undecidable [
30
]. The description of all (finitely based) varieties of groups
where the universal theory of finite groups is decidable is currently out of reach. In
fact our Theorem 5.1 gives the first example of a proper variety of groups where the
universal theory of finite groups is undecidable. From Zelmanov’s solution of the
restricted Burnside problem [65,66], it immediately follows that the universal theory
of finite groups in every finitely based periodic variety of groups is decidable. That
result and simulations of Minsky machines in semigroups (as in Sect. 3) were used
by the third author [56] to obtain a complete description of all finitely based varieties
of semigroups where finite semigroups have decidable universal theory. For more
information on that problem, see [
32
].
5.2 Distortion of pro-finitely closed subgroups of finitely presented groups
Let G be a group generated by a finite set X , H ≤ G be a subgroup generated by a finite
set Y . Recall that the distortion function f H,G (n) is defined as the minimal number k
such that every element of H represented as a word w of length ≤n in the alphabet X
can be represented as a word of length ≤k in the alphabet Y [
17
]. Distortion functions
with respect to two different sets of generators for the same group are equivalent.
By [
17
] in a group G with decidable word problem, the distortion function fG,H is
recursive if and only if the membership problem in H is decidable.
Recall that H is closed in the pro-finite topology of G if H is the intersection of
some subgroups of G of finite index. If G is finitely presented and H is closed in
the pro-finite topology of G, then there exists a McKinsey-type algorithm A(G, H )
solving the membership problem for H (and thus the fG,H is recursive). For every
word w in the alphabet X , the “yes” part Ayes(G, H ) of the algorithm lists all words
in Y , rewrites them as words in X , and then applies relations of G to check whether
one of these words is equal to w. The “no” part Ano(G, H ) of the algorithm lists all
homomorphisms φ of G into finite groups and checks whether φ (w) ∈/ φ (H ). As
in Sect. 1.2, one can asks what is the complexity of the “yes” and “no” parts of that
algorithm, in particular, and of the membership problem for H in general.
The time complexity of Ayes(G, H ) can be estimated in terms of the distortion
function fG,H (n) and the time complexity of Ano(G, H ) can be estimated estimated
in terms of the relative depth function ρG,H (n) which is defined as the minimal number
r such that for every word w of length ≤n in X which does not represent an element
of H there exists a homomorphism φ from G to a finite group of order ≤r such that
φ (w) ∈/ φ (H ).
As for the word problem in residually finite finitely presented groups (discussed
above), there were no examples of finitely generated subgroups of finitely presented
groups that are closed in the pro-finite topology but have “arbitrary bad” distortion or
“arbitrary bad” relative depth function.
Mikhailova’s construction [
42
] shows that finitely generated subgroups of the
residually finite group F2 × F2 (here F2 is a free group of rank 2) could be very distorted.
In fact the set of possible distortion functions of subgroups of F2 × F2 coincides, up to
a natural equivalence, with the set of Dehn functions of finitely presented groups [47].
Finitely generated subgroups of F2 × F2 are equalizers of pairs of homomorphisms
φ : Fk → G, ψ : Fn → G (where Fk , Fn are some subgroups of F2), i.e. the
subgroups of the form {(x , y) ∈ Fk × Fn | φ (x ) = ψ (y)} (see, for example, [52] or[5]).
The equalizer subgroup is finitely generated if and only if G is finitely presented [5].
It is easy to prove (see Lemma 5.3 below) that if G is residually finite, then the
equalizer is closed in the pro-finite topology of F2 × F2. In fact we have the following
more general statement:
Lemma 5.3 Let P be a class of finite groups closed under direct products and
subgroups. Let G be a finitely generated group, let N be a normal subgroup of G, and let
φ, ψ be two homomorphisms G → G/N . If G/N is residually P, then the equalizer
E (φ, ψ ) = {(g, h ∈ G × G | φ (g) = ψ (h)} is closed in the pro-P topology on
G × G.
Proof Suppose (u, v) ∈ G × G but (u, v) ∈/ E (φ , ψ ), so φ (u) = ψ (v). Since G/N
is residually P there is a homomorphism η : G/N → K onto a finite group K ∈ P
such that ηφ (u) = ηψ (v) in K . Therefore the image of the pair (u, v) under (ηφ , ηψ )
is not in the image of the subgroup E (φ , ψ ) in K × K . Hence the subgroup E (φ , ψ )
is closed in the pro-P topology on G × G.
Lemma 5.3 and Theorems 4.22 and 4.21 immediately imply
Corollary 5.4 For every recursive function f (n) there exists a finitely generated sub
group H ≤ F2 × F2 that is closed in the pro-finite topology of F2 × F2 and whose
distortion function f F2×F2,H , the relative depth function, and the time complexity of
the membership problem are at least f (n).
Remark 5.5 Since the groups we construct are solvable of class 3, a similar corollary
is true with F2 replaced by the free solvable group of class 3 of finite rank (although
the rank is not necessarily 2 because not every free solvable group of class 3 embeds
into a 2-generated group that is solvable of class 3).
Acknowledgements The authors are grateful to Jean-Camille Birget and Friedrich Otto for pointing to the
references [13], to Ben Steinberg for pointing to the reference [
34
],to Rostislav Grigorchuk for pointing
to the references [
15,21
] and to Tim Riley for pointing to the references [
19,20
]. We are also grateful to
Markus Lohrey and Ralph Strebel for their comments. We are especially grateful to the anonymous referees
whose numerous suggestions helped us improve the paper.
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