Computably Enumerable Semigroups, Algebras, and
Groups
Bakhadyr Khoussainov
The University of Auckland
New Zealand
Research is partially supported by Marsden
Fund of New Zealand Royal Society.
Plan
• Equational presentations.
• The specification problem.
• A semigroup example.
• Algorithmically finite universal algebras.
• An algebra example.
• A group example.
Definition: A universal algebra is a tuple
A=(A; f1,f2,…., fk, c1,…,cr),
where A is the non-empty set (domain),
each fj is a total function on A, and
each ck is a constant.
Definition (terms):
Variables and constants are terms.
Suppose that t1,…,tm are terms, f is a function
symbol, then the expression
f(t1,…,tm)
is a term.
A ground term is a term with no variables.
The set of ground terms is a universal algebra,
called the term algebra. Notation:
T = the term algebra
Fact 1: T is finitely generated and computable.
Fact 2: Every universal algebra generated by
the constants is a homomorphic image of T.
An equational presentation is a finite set S
of formulas of the type
t=q.
A quasiequational presentation is a finite set
S of formulas of the type
t1=q1…… tn=qn→ t=q.
Here t,q are terms that may contain variables.
Definition: (Specified Universal Algebras)
Let E(S) be the congruence relation generated by S.
The universal algebra TS = T/ E(S) is called specified by S.
TS is equationally presented if S is an equational presentation.
TS is quasi-equationally presented if S is a quasi-equational presentation.
Examples
• Finitely presented groups and semigroups.
• Finitely presented algebras.
• Finitely presented semigroups with left-cancellation properties.
Properties of TS:
1. TS satisfies S and is finitely generated.
2. The equality relation E(S) of TS is computable enumerable.
3. TS is universal.
4. TS is unique.
The Specification Problem:
Let A be a universal algebra.
(1) Is A equationally presented?
(2) Is A quasi-equationally presented?
Clearly, we need to assume the following:
(a) A is finitely generated.
(b)The equality relation in A is c.e.
Definition (c.e. universal algebra)A c.e. universal algebra is one of the form
T/E, where
(1) T is the term algebra,
(2) E is a c.e. congruence relation on T.
Examples: finitely presented groups, semi-
groups, and universal algebras.
Example:
Consider the universal algebra
( ω; x+1, 2x ).
(Bergstra and Tucker):
The universal algebra ( ω; x+1, 2x ) does
not have an equational presentation.
Important Observation
Consider the expansion:
(ω, x+1, 2x , +, x, 0).
The expanded universal algebra is now
finitely presented.
Definition: (Computable Algebra)
A universal algebra
A=(A; f1,f2,…., fk, c1,…,cr)
is computable if the set A and all functions
fj are computable.
Definition: An expansion of
A=(A; f1,f2,…., fk, c1,…,cr)
is
A’ = (A; f1,f2,…., fk, g1,…,gr, c1,…,cr),
where g1,…,gr are new functions.
Theorem (Bergstra Tucker, ≈1980).
Every computable universal algebra can be
equationally presented in an expansion.
The question of Goncharov: (early 1980s, Goncharov (also Bersgtra and Tucker))
Let A be a finitely generated computably
enumerable universal algebra:
1. Does A have an equationally presented expansion?
2. Does A have a quasi-equationally presented expansion?
Resemblance to Higman’s theorem
Theorem
(Kassymov, 1988; Khoussainov, 1994)
There exists a finitely generated computably
enumerable universal algebra no expansion
of which is equationally presented.
Theorem (Khoussainov, 2006)
There exists a finitely generated computably enumerable universal algebra no expansion of which is quasi-equationally presented.
Important comments
• The counter-examples constructed are universal algebras built specifically.
• The counter-examples do not belong to natural classes of structures such as the classes of groups, semi-groups, rings, etc.
Main Question:
Can such examples be found among the standard algebraic structures:
(1) Semigroups,
(2) Algebras (these are rings that form vector spaces over fields), and
(3) Groups?
Theorem A (with Hirschfeldt, 2011)
There exists a finitely generated computably
enumerable semigroup no expansion of
which is equationally presented.
Proof. Consider the free semigroup
A=({0,1}★;).
Let X be a subset of {0,1}★. We say that a
string u realizes X if u contains a substring
in X. Otherwise, we say that u avoids X.
Define: R(X)={u | u realizes X}.
Clearly, R(X) is a subset of {0,1}★.
Define the following relation ≈X on {0,1}★.
u ≈X v if either u=v or u and v both realize X.
Lemma 1.
The relation ≈X is a congruence relation on
the free semigroup A=({0,1}★;).
Set:
A(X) = A/≈X .
Lemma 2. (Miller)
If for all k there are at most k many strings of
length ≤ k+4 in X, then R(X) is co-infinite.
Lemma 3.
There is a c.e. set X such that R(X) is simple.
Proof. Let W0, W1, …. be a standard list ofc.e. subsets of {0,1}★.
Put string y into X if for some i the string y is
the first string of length ≥i+5 appeared Wi.
The set X is a desired c.e. set.
Consider the semigroup
A(X) = A/≈X .
It is finitely generated, c.e., and infinite.
Let h1, …, hn be computable functions
compatible with ≈X. Consider the expansion
A’(X)= ( A(X); h1, …, hn)
Lemma 4 (Kassymov).
Any c.e. universal algebra whose equality relation
coincides with ≈X is residually finite. In particular,
A’(X) is residually finite.
Lemma 5 (Malcev). If a universal algebra A is
finitely presented and residually finite then the
word problem in A is decidable.
Hence, A’(X) is not equationally presented.☐
Definition (Kassymov, Khoussainov, 1986)
A finitely generated infinite c.e. universal
algebra
A = F/E
is effectively infinite if there is an infinite
c.e. sequence u0, u1, u2 ….listing pair-wise
distinct elements of A.
If A is not effectively infinite then we call
A algorithmically finite (Miasnikov).
Example.
The semi-group A(X) constructed above is
algorithmically finite.
Let A = T/E
be an algorithmically finite universal algebra.
Property 1. Each expansion of A is algorithmically finite.
Property 2.Each finitely generated subalgebra is algorithmically finite.
Property 3.For every term t(x) the trace
a, t(a), tt(a),…is eventually periodic. In particular, if A is a semigroup then every element of A is of finite order.
Property 4.All infinite homomorphic images of A are also algorithmically finite.
Property 5. If A=T/E is residually finite
then for all distinct elements x, y of A there
exists a subset S(x,y) of T such that:
(1) S(x,y) is E-closed and computable.
(2) x belongs S(x,y).
(3) y does belong to S(x,y).
Lemma. If A is residually finite, then all
expansions of A are also residually finite.
Proof. Let A’ be an expansion of A. Take
two distinct elements x,y of A’. Select the
separator set S(x,y) from Property 5.
Define the following binary relation ≈(x,y)
on A’:
a ≈(x,y) b
if and only if
no elements in S(x,y) and in its complement
are identified by the congruence relation on
A’ generated by the pair (a,b).
Properties of ≈(x,y) :
(1) ≈(x,y) is a congruence relation on A’.
(2) ≈(x,y) is a co-c.e. relation.
(3) In the quotient algebra A’/ ≈(x,y) the images of x and y are distinct.
Since A’ is algorithmically finite, A’/ ≈(x,y)
must be finite.
Theorem B.
Let A be an algorithmically finite universal
algebra.If A is residually finite then no
expansion of A has equational presentation.
Proof. If A’ is an equationally presented
expansion of A, then A’ is residually finite.
By Malcev’s lemma the word problem in A’
is decidable. Contradiction.
Question (Khoussainov):
Are there algorithmically finite groups?
Motivation of the question:
Algorithmically finite groups are candidates
that have no equationally presented expansions.
Theorem (Miasnikov, Osin, 2011)
There exists an algorithmically finite group.
Miasnikov motivates the theorem from a generic
complexity view point. Algorithmically finite
groups are called Dehn monsters.
Miasnikov and Osin ask if there are
residually finite Dehn monsters.
Let K be a finite field. Consider the algebra
F=K<x1, x2,..., xm>
of polynomials in non-commuting variables.
We can represent F as the direct sum
∑Fn
where Fn is the vector space spanned over
monomials of degree n.
Let H be a set of homogeneous polynomials,
I be the ideal generated by H.
Theorem (Golod Shafarevich).
Let rn be the number of polynomials in H of
degree n, and ε be such that 0< ε < m/2 and
rn ≤ ε2(m-2 ε)n-2.
Then the algebra
A=F / I
is infinite dimensional.
Let H be a subset of {x,y}★ constructed in
Lemma 3 above. Consider the ideal I=<H>.
Theorem C. The algebra A=F/I satisfies the
following properties:
(1) A is effectively infinite.
(2) All expansions of A are residually finite.
(3) A has no equationally presented expansions.
Proof. It is clear that the algebra is infinite.
Write any polynomials f of F in the form
a+ h,
where a is a sum of monomials not from KH
and h is a sum of monomials in KH. So:
f=(a1+.... an)+(h1+....+hk).
Identify this sum with the set
{a1,.... an,h1,....+hk},
and call the set {a1,.... an}
the true representative of a.
Since H is simple but not hypersimple, H
has a strong array of finite sets for the
complement of H.
The identification of polynomials with finite
subsets implies that the algebra A is
effectively infinite.
Claim 1. The collection of all true
representatives is an immune set.
Claim 2. For all distinct elements x, y of A
there exists a subset S(x,y) of F such that:
(1) S(x,y) is I-closed and computable.
(2) x belongs S(x,y).
(3) y does belong to S(x,y).
Claim 3.
All expansions of A are residually finite.
Proof.
Take two distinct elements x,y of an
expansion A’. Select the separator set
S(x,y) from Claim 2.
Define the following binary relation ≈(x,y) :
f ≈(x,y) g
if and only if
no elements in S(x,y) and in its complement
are identified by the congruence relation on
A’ generated by the pair (f,g).
Properties of ≈(x,y) :
(1) ≈(x,y) is a congruence relation on A’.
(2) ≈(x,y) is a co-c.e. relation.
(3) In the quotient algebra A’/ ≈(x,y) the images of x and y are distinct.
Claim 4. The collection of all true
representatives that belong to distinct
≈(x,y) -equivalence classes is a c.e. set.
Thus, the A’/ ≈(x,y) must be finite.
Hence, A’ is c.e., infinite, residually finite.
Therefore, the algebra can not be
equationally specified by Theorem B.
Theorem C. There exists an algorithmically
finite and residually finite algebra.
Proof. Consider F=K<x1, x2,..., xm>.
Construct a set H of homogeneous
polynomials by stages as follows.
Let W0, W1, …. be a list of c.e. subsets of F.
For each i, let f and g the first polynomials
occurring in Wi such that:
(1) f=f1+f2, g=g1+ g2, f1= g1, and(2) the degrees of homogeneous polynomials
occurring in both f2 and g2 are greater than i+64.
Put all homogeneous polynomials occurring
in both f2 and g2 into H.
For each n >2, in H there are at most 2n
homogeneous polynomials of degree n.
Hence, for a small ε we have 0< ε < m/2
and
rn ≤ ε2(m-2 ε)n-2
for all n.
By Golod Shafarevich theorem, we have
that the algebra A=F / I is infinite. By
construction, it is algorithmically finite.
It is well-known that Golod-Shafarevich
algebras are residually finite (Golod).
So, A is algorithmically finite and residually
finite. By Theorem B no expansion of A is
equationally presented.
Theorem D. There exists an algorithmically
finite and residually finite group G.
Proof. Consider the algebra A constructed
above. The semigroup G=G(A) generated
by the elements (1+x)/I and (1+y)/I of the
algebra A forms a group under the product
operation.
The group G is the one desired.
As a corollary we obtain the following
theorem.
Theorem E. There exists a group that has
no equationally presented expansion.
Open Question:
Are there semigroups, algebras and groups
whose all expansions are not quasi-
equationally presented?