arXiv:1310.5023v1 [math.GR] 18 Oct 2013
A proof of Zhil’tsov’s theorem on decidability
of equational theory of epigroups
Inna Mikhailova∗
October 21, 2013
Abstract
Epigroups are semigroups equipped with an additional unary operation called pseudoinversion. Each finite semigroup can be considered
as epigroup. We prove the following theorem announced by Zhil’tsov
in 2000: the equational theory of the class of all epigroups coincides
with the equational theory of the class of all finite epigroups and is
decidable. We show that the theory is not finitely based but provide
a transparent infinite basis for it.
1
Introduction
A semigroup S is called an epigroup if, for every element x ∈ S, some power
of x belongs to a subgroup of S. The class E of all epigroups includes all
periodic semigroups (i.e. semigroups in which each element has an idempotent
power), all completely regular semigroups (i.e. unions of groups), and many
other important classes of semigroups. See [7, 8] and the survey [9] for more
examples and an introduction into the structure theory of epigroups.
It is known that, for every element x of epigroup S, there exists a unique
maximal subgroup Gx that contains all but finitely many powers of x. Let
ex stand for the identity element of Gx . Then it is known that xex = ex x
and that the product belongs to Gx . The latter fact allows one to consider
the inverse of xex in the group Gx ; we denote this inverse by x. This defines
∗
Ural Federal University, Ekaterinburg, Russia.
1
the unary operation x 7→ x on each epigroup; we call this operation pseudoinversion. Thus, epigroups can be treated as unary semigroups, that is,
as algebras with two operations: multiplication and pseudoinversion, and we
shall adopt this meaning of the term ‘epigroup’ throughout.
A systematic study of epigroups as unary semigroups was initiated in
[7, 8], see also [9]. In particular, there it was suggested to investigate the
collection of all unary semigroup identities holding in all epigroups, that is,
the equational theory of the class E treated as a class of unary semigroups1 .
The most fundamental questions about this theory are whether or not it is
decidable and whether or not it is finitely axiomatizable.
Yet another motivation for studying unary identities of epigroups comes
from the theory of finite semigroups. Every finite semigroup can be treated
as an epigroup, and the unary operation of pseudoinversion is an implicit
operation in the sense of [6], that is, it commutes with homomorphisms between finite semigroups. Therefore, for each collection Σ of unary identities
of epigroups the class of all finite semigroups satisfying Σ is a pseudovariety.
In fact, many important pseudovarieties of finite semigroups can be defined
by identities involving the operation of pseudoinversion (which within the
realm of finite semigroups is usually denoted by x 7→ xω−1 ) or the operation
x 7→ xω := xx; see the monograph [1] for plentiful examples.
We denote by Ef in the class of all finite epigroups. Along with E and
Ef in , among important classes of epigroups is the class Af in consisting of
finite combinatorial (or aperiodic) semigroups. Remind that a semigroup is
combinatorial if all its maximal subgroups are one-element.
At the end of the 1990s Ilya Zhil’tsov began to consider this problem
along with a more general question. First results he obtained in [10] and
the full solution of the problem for Af in (received independently to [5]) was
announced in the extended abstract [11]. [11] also contained similar result
for the pseudovariety Ef in .
Theorem 1. The equational theories of the classes E and Ef in coincide and
1
In order to prevent any chance of confusion, we mention here that the class E does
not form a variety of unary semigroups; of course, this is not an obstacle for considering
the equational theory of E.
2
are decidable. The following identities form its basis
(xy)z ≈ x(yz),
(xy)x ≈ x(yx),
x2 x ≈ x,
x2 x ≈ x,
xx ≈ xx.
xp ≈ xp for each prime p.
Very unfortunately, soon after that Zhil’tsov died in a tragic accident and
left no implementation of the statements indicated in [11]. It took us considerable effort to reconstruct all necessary steps of the proof. In Sections 2
and 3 we follow his plan quite closely while in Section 4 we choose some
different way.
In Section 1 we introduce Zhil’tsov’s concept of Z-unary word. He suggested to consider words with not just one but countably many additional
unary operations. In Section 2 a normal form of a Z-unary word is considered. We prove two Zhil’tsov’s propositions. First, it is possible algorithmically construct the normal form. Second, there is an algorithm that, for
two Z-words in normal form, returns their longest common prefix [suffix]. In
particular, this means that there is an algorithm that decides whether normal
forms of given Z-unary words coincide. In Section 3 we consider epigroup
terms as Z-unary words and show that given epigroup identity holds in each
epigroup if and only if the normal forms of corresponding words coincide.
We denote by En the pseudovariety consisting of all epigroups where n-th
power of any element belongs to a subgroup. In Section 3 we show that
normal forms of Z-unary words act almost like ordinary words and, hence,
it is likely to apply purely combinatorial methods similar to those used in [4]
to prove the analogue of Theorem 1 not only for pseudovariety Af in but for
pseudovarieties En .
2
Z-unary words
Let A be a non-empty set of letters that refers to as an alphabet. As usual
by A+ we denote free semigroup over A. Consider free algebra Z(A) over
alphabet A with one binary associative and countably many unary operations. The unary operations map arbitrary element x to the element written
3
as xω+q where q runs through the set N. The algebra Z(A) is called Zunary semigroup and an element σ ∈ Z(A) is a Z-unary word. For example,
(xω−4 yxω+30 )ω−1 xy ω is a Z-unary word over 2-letter alphabet. Note that any
ordinary word over alphabet A is also a Z-unary word.
Now we consider two important characteristics of Z-unary words. First,
we define a height of an arbitrary Z-unary word by the following rule:
1) the height of any word from A+ or empty word is equal to 0;
2) for h ≥ 0, by Z-unary word of height h + 1 we mean
σ = π0 τ1ω+q1 π1 . . . τnω+qn πn ,
where n ≥ 1, each Z-unary word τi is of height h and the height of Z-unary
words πi do not exceed h. We call this record the height representation of σ.
The fact that σ ∈ Z(A) has the height h we denote by h(σ) = h.
Next, we define the length of a Z-unary word as a polynomial with integer
coefficients. We denote by Z[ω] the ring of polynomials over Z. Consider a
map | | : Z(A) 7→ Z[ω] such that,
1) for any letter a ∈ A, we have |a| = 1;
2) for any two Z-unary words σ, τ and an integer q, we have that |στ | =
|σ|
+
|τ |
and
ω+q
|σ | = (ω + q)|σ|.
We say that the image |σ| = |σ|(ω) is the length of the Z-unary word σ. Note
that due to this definition the height of the Z-unary word σ is the degree of
the polynomial |σ|.
In this article we will try to apply to Z-unary words the same combinatorial methods as for ordinary words. In particular, for two Z-unary words
σ, τ , we have |σ| ≤ |τ | if the leader coefficient of polynomial |τ | − |σ| is
non-negative.
We say that a Z-unary word τ is a prefix (suffix ) of σ ∈ Z(A) if, for some
ρ ∈ Z(A), σ = τ ρ (resp. σ = ρτ ). Similarly, a Z-unary word τ is a subword
for σ if, for some ρ1 , ρ2 ∈ Z(A), σ = ρ1 τ ρ1 .
4
2.1
Singular words
Consider the fully invariant congruence S on Z-unary semigroup Z(A) generated by the pairs
(xω+q , xxω+q−1 ),
(xω+q , xω+q−1 x),
(x(yx)ω+q , (xy)ω+q x),
for each q ∈ N.
By singular word or shortly sword we mean S-class. The class corresponding to an arbitrary Z-unary word σ is denoted by σ S . Note that the
class including ordinary word consists of unique element the word itself. The
factor-semigroup Z(A)/S of all swords is denoted by Sing(A).
Example 1 (Zhil’tsov). Let us describe some important relations between
members of the same S-class. Consider the word x(xyz)ω−5 xy(zx)ω+4 xx of
height 1. We can picture it in the following way:
✛✘
✛
✓✏
☛
❑ x 4 z
−5
z ✚✙
x
✲ ✲ ✒✑
✲ ✲ ✲
y
x(xyz)ω−5 xy(zx)ω+4 xx
=
x
x
y
x
x
.
Here we can see two circles xyz and zx with circle powers −5 and 4
respectively. Consider three more words that belong to the same S-class.
x(xyz)ω−5 xyzx(zx)ω+3 xx
xx(yzx)ω−5 yzx(zx)ω+3 xx
=
=
y
✛✘
☛
−5 ❑x
z ✲
✲ ✲
✚✙
x
x
x
y
z
✲
z
✛✘
☛
−5 ❑y
x
✲ ✚✙
✲ ✲ ✲
x
5
y
z
✛
✓✏
x 3 z
✲ ✲ ✲
✒✑
x x x
,
✛
✓✏
x 3 z
✲ ✲ ✲
✒✑
x x x
,
x(xyz)ω−4 xyzx(zx)ω+1 xx
=
x
z
✛✘
✛
✓✏
☛
❑ z 1 x
−4
x ✲ y✲ ✲ ✲ ✲
✲ ✚✙
✒✑
x
z
x
z
x
x
✲
x
✲
.
It is easy to see that to obtain one word from another we can:
• ”wind off ” a copy of a circle to the left or the right side of the circle
simultaneously decreasing the circle power;
• ”wind on” (if it is possible) a copy of a circle from the left or the right
side of the circle simultaneously increasing the circle power;
• ”roll” the circle to the left (right) side if some suffix (prefix) of the
circle is written before (after) the circle.
Thus, any sword can be considered as a word with some ”singularities”
that can move to the left or to the right. This explains the notion of singular
word.
Note that due to the definition of the congruence S the following conditions hold for all words σ, τ in the same S-class:
• σ and τ have the same heights;
• σ and τ have the same lengths.
Thus, we can define height and length of a sword as the height and the
length of an arbitrary member of the class.
We say that a sword τ S is a prefix [suffix, subsword ] of the sword σ S
if, for some words τ ′ , σ ′ such that (τ ′ )S = τ S , (σ ′ )S = σ S , the word τ ′ is a
prefix [respectively suffix, subword] of σ ′ . Then the unary semigroup Sing(A)
satisfies the following properties:
Lemma 1.
1. Sing(A) is a J -trivial cancelative semigroup;
2. The set of all prefixes [suffixes] of a certain sword is linearly ordered by
the relation of left [resp. right] division.
6
Proof. Let us start with proof of cancelation property. Suppose that σ S τ1S =
σ S τ2S . This implies (στ1 )S = (στ2 )S , hence τ1 Sτ2 . The right cancelative
property can be proved in the same way.
Assume now that, for some swords α, β, the corresponding J classes are
equal. Therefore, α = σ1 βτ1 and β = σ2 ατ2 . Hence, α = σ2 σ1 ατ1 τ2 that
implies the lengths of all swords σi and τi , where i = 1, 2, are equal to zero
and the swords are empty. Therefore, Sing(A) is J -trivial.
The relations of left and right division are partial orders on Sing(A). It
is obviously reflexive and transitive, the antisymmetric property follows from
J -triviality of the semigroup Sing(A).
Thus, Sing(A) satisfies almost the same properties as free semigroup.
The difference is that each sword of height more than zero has infinitely
many subswords. Moreover, for ordinary word w, one can find its prefix of
length 0 ≤ t ≤ |w| which is not true for arbitrary sword. For example, the
sword (x2 )ω has no prefix of length ω.
3
Normal swords
In this section we introduce a new form of a Z-unary word that has the key
role in decidability question.
Let us denote by R the fully invariant stabile relation on Z-unary semigroup Z(A) generated by the pairs
(xω+q xω+p , xω+q+p ),
((xn )ω+q , xω+nq ),
((xω+p )ω+q , xω+pq ),
for all integers p, q and n ≥ 2.
Recall that the factor-relation R/S on factor-semigroup Sing(A) = Z(A)/S
is defined by the following condition: α R/S β if and only if there exist Zunary words σ and τ such that αS = σ S , β S = τ S and σRτ .
A normal sword (or R/S-irreducible sword ) is a sword that is not R/Srelated to any other sword. By normal form of a sword we call an (R/S)∗ related normal sword, where (R/S)∗ is denoted the reflexive and transitive
closure of R/S. A sword α is called fully normal if, for all n ≥ 2, its power
αn is also fully normal.
7
This definition imply that the normal form of a sword does not contain
any subsword of the form xω+p xω+q , (xn )ω+q , (xω+p )ω+q . Note also that if the
sword is fully normal then it is obviously normal and can not have the form
xω+m or xω+p yxω+q . Note that all ordinary words are obviously fully normal.
The main goal of this section is to prove the two following propositions.
Proposition 1. Let σ be a sword of height h. Then there exists an algorithm
that constructs a normal form of the sword σ.
Proposition 2. Let σ and δ be two normal swords of height h. Then there
exists an algorithm that constructs their longest common prefix.
We are going to prove a number of statements simultaneously by induction
on height h assuming that for all swords of height h−1 the propositions hold.
Note that for ordinary words (that is, the swords of height zero) these two
propositions are trivial.
3.1
Periodic normal swords
We will need some auxiliary facts similar to those that can be obtained to
ordinary words.
We say that a sword σ a power of the sword τ if, for some n ≥ 2, it is
equal to τ n . Likewise, the sword τ ω+k , where k is an integer, is called an
ω-power of the sword τ . A sword σ is called τ -periodic if it is a power or an
ω-power of the sword τ . The following lemma being applied to the words of
height 0 is well-known.
Lemma 2. Let σ, τ be two normal swords of height no more than h and
τ σ = στ . Then there exists a fully normal sword π such that both τ and σ
are π-periodic. Suppose τ and σ are fully normal. Then h(σ) = h(τ ) = h(π)
and, for some positive integers n1 , n2 , σ = π n1 , τ = π n2 .
Proof. Assume that |σ| ≥ |τ |. Observe that since στ = τ σ then τ is a prefix
(and also a suffix) of σ. Therefore, there exists a sword σ ′ such that σ = τ σ ′
and στ = σ ′ τ 2 = τ σ = τ σ ′ τ . Then σ ′ τ = τ σ ′ and if σ ′ is not empty we can
similarly divide the longest sword into two parts again.
Let us consider the case h(σ) > h(τ ). Then τ is a prefix and suffix of σ ′
and there exists a sword σ ′′ of height h(σ) such that σ ′′ τ = τ σ ′′ and so on.
Therefore, it is easy to establish that, for any positive integer m, the sword
8
τ m is a proper prefix and suffix of the sword σ. Note that since σ is normal
then any degree of τ is normal. This implies the sword τ to be fully normal.
It is obvious that σ has some ω-powers of τ as a prefix and a suffix.
Note that if σ = τ ω+k then the lemma is proved. Therefore, we may assume
that σ = τ ω+k1 σ1 τ ω+k2 for some integers k1 , k2 . Suppose that h(σ1 ) = h(σ).
Observe that σ1 τ = τ σ1 and we can repeat the same steps as before to
obtain σ1 = τ ω+k3 σ2 τ ω+k4 but this means that σ is not normal. Hence,
h(σ1 ) < h(σ) and we can use Lemma 2 for swords σ1 , τ of height less than h.
Then there exists a fully normal sword π such that σ1 and τ are π-periodic.
Since σ = τ ω+k1 σ1 τ ω+k2 this implies σ is not normal. Therefore, σ = τ ω+k
what was required.
Assume now that h(σ) = h(τ ) then |σ| = ℓ1 ω h + g1 (w), |τ | = ℓ2 ω h + g2(ω)
where deg(gi ) < h. Let us prove the lemma by induction on maximum of
ℓ1 , ℓ2 .
Suppose that ℓ1 = ℓ2 . Then the sword σ ′ such that σ = τ σ ′ = σ ′ τ has the
height less than h which leads us to the case above. We receive that there is
a fully normal sword π such that, for some integers k, m, τ = π ω+k , σ ′ = π m
and, thus, σ = π ω+k+m . If ℓ1 6= ℓ2 then we can apply induction hypothesis
to the pair σ ′ , τ to obtain the required property.
The last statement trivially follows from the definition of the fully normal
sword.
Lemma 3. Let x and y be fully normal swords such that |x| ≥ |y|. Suppose
that x2 is a subword of some y-periodic sword. Then there exists a fully
normal sword z such that x and y are powers of z. In particular, if some
ω-powers of x and y are subwords of some normal sword then x = y.
Proof. First, assume that h(y) < h(x) then, for some integer q, x2 is a
subword of the an ω-power of y. Hence, there are swords y1 y2 = y such that
the sword x2 is equal to (y1 y2 )ω+t1 y1 y2 (y1 y2 )ω+t2 y1 where x = (y1 y2 )ω+t1 y1 =
y2 (y1 y2 )ω+t2 y1 . Then x is not fully normal. Thus, h(x) = h(y).
Let x = (y1 y2 )n y1 and x = y2 (y1 y2 )m y1 where y1 y2 = y. It is easy to see
that x has the prefix y1 y2 and y2 y1 of the same length, therefore, y1 y2 = y2 y1 .
By Lemma 2 the swords y1 and y2 are z-periodic. Since y is fully normal we
have that x = z k1 , y = z k2 .
The last statement of the lemma follows trivially from the definition of
normal sword.
9
3.2
Normalization
It clear that the production of two normal swords could be not normal. For
example, if σ1 = (xz ω+2 )ω , σ2 = z ω−5 then it is necessary to make just
one R/S-transition in the sword (xz ω+2 )ω z ω−5 to obtain its normal form
(xz ω+2 )ω−1 xz ω−3 .
The algorithm that constructs the normal form of an arbitrary sword will
have two procedures. Recall that each sword of height h can be presented as
π0 ρ1ω+q1 π1 . . . ρnω+qn πn . Each procedure will construct R/S-reduced form for
subswords of the sword. First, we study how to build a normal form for the
product of two normal swords.
Lemma 4. Let σ1 , σ2 be two normal swords and the sword σ1 σ2 is not
normal. Then there exists an algorithm that decides wether there is a fully
normal sword y such that a suffix of the sword σ1 and a prefix of the sword σ2
are some ω-powers of y and constructs the normal form of the sword σ1 σ2 .
If there exists another fully normal sword z such that a suffix of σ1 and a
prefix of σ2 are ω-powers of z then z = y.
Proof. First we prove that if the production σ1 σ2 is not normal then it has
the unique subsword of the form y ω+q y ω+p .
Suppose the statement of the lemma does not hold and σ1 σ2 has subsword
of the other form. This means that there exists a sword z such that, for some
z-periodic sword x, the sword xω+n is a subsword of σ1 σ2 . But this implies
that some ω-power of x is a subsword of one of the swords σ1 , σ2 and, thus,
that sword is not normal.
Therefore, if the production of σ1 , σ2 is not normal then σ1 σ2 has a subsword of the form y ω+q y ω+p . Assume now that y is not unique, that is, there
exists another fully normal sword z such that σ1 has a suffix z ω+k . Assume
that |y| ≤ |z| then by Lemma 3 y = z.
Let σ1 = σ1′ ρ1ω+q1 π1 and σ2 = π2 ρ2ω+q2 σ2′ where h(ρ1 ) = h(ρ2 ) = h − 1,
h(π1 ), h(π2 ) are no more than h − 1. First, we apply Lemma 4 for the swords
ρ21 π1 and π2 ρ22 of height h − 1.
Assume that there is a fully normal sword y satisfying the conditions of
the lemma for this pair then ρ21 π1 = σ1′′ y ω+p1 , π2 ρ22 = y ω+p2 σ2′′ . Note that
σ1′′ , σ2′′ are not empty, otherwise the swords ρ1 , ρ2 are y-periodic which contradicts to Lemma 3. Then the sword σ1′ σ1′′ y ω+p1 +p2 σ2′′ σ2′ is R/S-related to
σ1 σ2 and clearly normal.
10
Otherwise, the required ω-powers of the sword y either do not exist or have
the height h. In the last case we should check if ρ1 = π1 ρ′1 , ρ2 = ρ′2 π2 (thus,
σ1 = σ1′ π1 (ρ′1 π1 )ω+q1 and σ2 = (π2 ρ′2 )ω+q2 π2 σ2′ ) and, finally, ρ′1 π1 = π2 ρ′2 = ρ.
Note that we certainly can compare all the pairs using the Proposition 2
for swords of height h − 1. Then the required normal form is the sword
σ1′ π1 ρω+q1 +q2 π2 σ2′ .
Now we study how to construct a normal form for the ω-power of the
normal sword.
Lemma 5. Let ρ be a normal sword of height h − 1. Then there is an
algorithm that, for the sword ρω+q , reduces it to a sword α1 β ω+t α2 where
h(β) ≤ h − 1 and the sword β ω+t is normal.
Proof. Assume that ρω+q is not normal. By Lemmas 4 and 6 applied to the
normal sword ρ of height h − 1 we can find a fully normal sword x such that
1) ρ = xω+m1 x′ xω+m2 for some integers m1 , m2 ;
2) ρ = xn where n is a positive integer;
3) ρ = xω+m where m is an integer.
Note that Lemma 3 provides that only one of three cases holds .
Assume that, for some fully normal sword x0 , ρ = x0ω+m1 x′0 x0ω+m2 . Lemma 4
implies that the sword x satisfying this property is unique. Consider the
sword ρω+q = (x0ω+m1 x′0 x0ω+m2 )ω+q . It is obvious that it can be reduced to
the sword x0ω+m1 x′0 (x0ω+m1 +m2 x′0 )ω+q−1 x0ω+m2 .
Consider the obtained sword σ1 = x0ω+m1 +m2 x′0 . Suppose that ρω+q−1
is
1
ω+ℓ2
1 ′
not normal. Therefore, ρ1 has the form xω+ℓ
x
x
(otherwise
we
obtain
1
1 1
a contradiction with ρ being normal). If h(x0 ) = h(x1 ) then by Lemma 3
that implies ρ = x0ω+m1 x′0 x0ω+m2
x0 = x1 and the sword x′0 has a suffix xω+p
1
1
not to be normal. If h(x0 ) > h(x1 ) then x0 has a prefix xω+p
and x′0 has
1
ω+p2
a suffix x1
and ρ is again not normal. Thus, h(x0 ) < h(x1 ). So, we can
proceed constructing swords ρ1 , ρ2 , . . . , ρk , where k ≤ h. We finally obtain
′
the word β = ρk and reduce the sword ρω+q to the sword α1 β ω+q α2 where
′
β ω+q is normal.
Suppose no that ρ is x-periodic. Then ρω+q can be reduced to xω+nq ,
where n is the suitable power. It is easy to check that x has none of forms
listed above. Therefore, the sword xω+nq is normal.
11
Now we are ready to prove Proposition 1.
Let π0 ρ1ω+q1 π1 . . . ρnω+qn πn be the height representation of the sword σ. Remind that by inductional hypothesis on height we can suppose Proposition 1
and 2 hold for all swords of height no more than h−1. Thus, we can consider
the swords ρi and πi to be normal.
By Lemma 5 every sword (ρi )ω+qi can be reduced to the sword of the
form αi1 βiω+pi αi2 where the sword βiω+pi is normal of height h and αi1 , αi2
have height no more than h − 1. Thus, we can consider the swords ρiω+qi to
be normal too.
Take the normal prefix π0 and apply Lemma 4 a number times for swords
πi and ρiω+qi to obtain the normal sword σ ′ . Therefore, the sword σ ′ R/Sσ
and it is the required normal form of the sword σ.
3.3
Longest common prefix
For two normal swords σ, δ we denote by LCP (σ, δ) their longest common
prefix. Here we prove Proposition 2.
Proof. We prove the proposition by induction of maximal number of ”rings”
in height representations of the swords σ and δ.
Let σ = πρω+q , δ = αβ ω+r . Without loss of generation we may assume
that |α| ≥ |π|.
First, we should find using Proposition 2 for the swords of less height the
sword τ1 = LCP (α, π). Suppose τ1 6= α then clearly τ1 = LCP (σ, δ).
Otherwise, τ1 = α Let π = τ π ′ then we should find the longest common
prefix τ2 for β ω+r and π ′ ρω+q . Suppose τ2 has the height h then obviously
it has ω-period β and by Lemma 3 π ′ ρ = βπ ′ . The last we can check using
Corollary 1. If the swords are equal then clearly τ2 is the longest between
π ′ ρω+q and β ω+r .
Suppose now that τ2 has height at least h − 1. In view of Lemma 3 the
prefix τ2 can not be very long. First, assume that |π ′| ≤ |β|. Then we may
find τ2 as CLP (β ℓ1 , π ′ ρℓ2 ) where positive integers ℓ1 , ℓ2 are large enough and
lengths of both swords exceed max(|ρ2 |, |β 2|). Finally, LCP (σ, δ) = τ1 τ2 .
Now consider the following representations of the swords σ, δ:
σ = πρω+q σ ′ , δ = αβ ω+r δ ′ ,
12
where h(ρ) = h(β) = h − 1, h(α), h(π) ≤ h − 1.
Let τ1 be the LCP (πρω+q , αβ ω+r ). If h(τ1 ) < h then clearly LCP (σ, δ) =
τ1 . Otherwise, we may assume that τ1 = πρω+q and αβ ω+r = τ1 δ ′′ . It is
easy to see that h(δ ′′ ) < h then the swords σ ′ , δ ′′ δ ′ satisfies the induction
hypothesis and we can find τ2 = LCP (σ, δ ′′ δ ′ ). Again, LCP (σ, δ) = τ1 τ2 .
Corollary 1. There is an algorithm that checks whether two given normal
swords coincide.
And we finish the algorithmic part by the following
Lemma 6. Let σ be a normal form of height h. Then there exists an
algorithm that decides whether there is a fully normal sword x such that σ is
x-periodic.
Proof. The proof of the lemma is based on the following observation. If the
sword σ is x-periodic then we can test by Proposition 2 the swords xω+m (or
respectively xm ) and σ. For that it is necessary to find the suitable prefix x
and number m.
Note that if σ = xω+m then |σ| = (ω + m)|x| and the number −m should
be chosen between all integer root of the polynomial |σ|. In the other case
we should take as m a common divisor for the coefficients of |σ|. Having
obtained a number m we receive the length of the possible period x. As we
noticed above it is not necessary that sword has a prefix of a certain length
but if the sword periodic it has one.
Let π0 ρ1ω+q1 . . . ρnω+qn πn be the height representation of the sword σ. Suppose that h(x) ≤ h − 1 then we can choose a positive integer k such that
|π0 ρk1 | ≥ |x| and try to search the prefix of the sword of smaller height.
h
Now suppose that
Pk h(x) = h, |x| = Kω + gh−1 (ω) and consider the polynomials fk (ω) = i=1 |ρi |. There exists k0 such that the leader coefficient
ak0 of fk0 is not less than K while ak0 +1 > K.
If ak0 < K then the prefix of length |x| does not exists and the sword is
not x-periodic. Then we have to check another possible periods. Otherwise,
ω+q
we denote the prefix π0 ρ1ω+q1 . . . ρk0 k0 by σ1 and the corresponding suffix by
σ2 . We can again reduce the problem to the case when height of both swords
is no greater than h − 1 by the following way. For |σ1 | < |x|, we search the
prefix τ of σ2 of length |x| − |σ1 | then x = σ1 τ . Otherwise, the suffix σ1 = xτ
where |τ | = |x| − |σ1 |.
13
4
4.1
Proof of the theorem
Normal swords and epigroup identities
Remind that any epigroup can be considered as a unary semigroup with the
following unary operation: x 7→ x, where xx = xx is the idempotent power of
the element x. Then any unary semigroup term with operation x replaced by
xω−1 is clearly a Z-unary word. It is known [9] that the following identities
hold in every epigroup:
(xy)x ≈ x(yx),
(1)
x2 x ≈ x,
(2)
x2 x ≈ x,
(3)
xx ≈ xx,
(4)
xp ≈ xp
for each prime p.
(5)
Inversely, any Z-unary word can be treated as epigroup term such that,
for any integer q, the result of applying the unary operation
xxq+1 , if q is non-negative;
ω+q
x
=
x(−q) ,
otherwise.
Proposition 3. Let σ, σ ′ be two Z-unary words belonging to the same Sclass then the identity σ ≈ σ ′ holds in E. Therefore, if E satisfies the identity
σ ≈ τ then, for all σ ′ ∈ σ S , τ ′ ∈ τ S , the identity σ ′ ≈ τ ′ also holds in every
epigroup.
Proof. Since the singularity relation S is a fully invariant congruence then it
is enough to prove that, for any integer q, each epigroup satisfies the identities
(xy)ω+q x ≈ x(yx)ω+q , xxω+q ≈ xω+q+1 , xω+q x ≈ xω+q+1
generating the relation S.
Suppose q ≥ 0 then xω+q = xxq+1 . Note that since xx = xx then
xxq+1 = xq+1 x. Therefore, the last two identities are clearly hold in E since
xxω+q = xxq+1 x = xq+2 x = xω+q+1 .
Consider the first identity. We have
(xy)ω+q x = xy(xy)q+1x = xyx(yx)q+1 ≈ xyx(yx)q+1 = x(yx)ω+q .
14
For a negative integer we have xω+q = x(−q) . Thus,
(xy)ω+q x = xy (−q) x = xy (−q−1) xyx ≈ xy (−q−1) xyx ≈ . . . ≈ xyx(−q) = x(yx)ω+q .
When q = −1 the second identity has the form xxω−1 = xx = xω . For
q < −1 using 2 we obtain
xxω+q = xx(−q) = xx2 x(−q−2) ≈ xx(−q−2) = xω+q+1 .
Thus, instead of identities of Z-unary words we will consider for short the
identities of corresponding swords.
Proposition 4. Let σ1 and σ2 be two swords, α1 and α2 be their normal
forms. Then the equality σ1 ≈ σ2 holds in E if and only if the equality α1 ≈ α2
holds in E.
Proof. We prove again that, for every sword σ and its normal form α, the
identity σ ≈ α holds in E. This clearly imply the statement of the proposition.
Assume now σR/Sα and prove that each epigroup satisfies the identity
σ ≈ α. It is clear that we should prove that, for all integers p, q, n ≥ 2, the
following identities hold in E:
(xn )ω+q ≈ xω+nq , xω+p xω+q ≈ xω+p+q , (xω+p )ω+q ≈ xω+pq .
First, note that the identity xn ≈ xn holds for any positive integer n. To
establish this it is enough to apply the identity 5 to each prime divisor of
number n.
Let q ≥ 0 then (xn )ω+q = xn (xn )q+1 ≈ xn xnq+n . We apply the identity 2
(n − 1) times to obtain xn xnq+n ≈ xn−1 xnq+n−1 ≈ . . . xxnq+1 = xω+nq . For the
(−q)
negative integer q the prove is trivial: (xn )ω+q = xn
≈ x(−nq) = xω+nq .
ω+q ω+p
ω+p+q
Next identity, x x
≈x
. For positive integers p and q we have
xxp+1 xxq+1 ≈ x2 xp+q+2 . The identity 2 implies x2 xp+q+2 ≈ xxp+q+1 = xω+p+q .
For both negative p and q, the prove is straightforward since xω+p xω+q =
x−p x−q = x(−p−q) = xω+p+q .
Assume now that p and q have different signs, say, p ≥ 0 and q < 0. Then
xω+p xω+q = xxp+1 x(−q) = x(−q+1) xp+1 .
15
Let m be the minimum of absolute values of q and p. It is easy to understand
that applying the identity 2 m times we obtain xxp−m+1 = xω+p+q , when
m = −q, and x−q−m = xω+p+q , when m = p.
Now we show that xω+p ≈ xω−p for every integer p. First, let p be a
negative integer. Consider xω+p = x(−p) . Applying the identity 5 we obtain
(−p)
x(−p) ≈ x
. Since x ≈ xx2 (identity 3) we have xω+p ≈(xx2 )−p = x−p x−2p .
Using the identity 2 several times we obtain xω+p ≈ xx−p+1 = xω−p .
Suppose now that p ≥ 0 then xω+p = xxp+1 . Let us apply the identity 2
p times and then identities 5 and 3 to receive
p
p
xxp+1 ≈ x2p xp = (x2 x)p ≈ x2 x ≈ x ≈ xp = xω−p .
(−q)
Thus, for a negative integer q, we receive (xω+p )ω+q ≈ xω+p
≈(xω−p )(−q) .
Finally we use the identity xω+p xω+p ≈ xω+2p several times to obtain the required xω+pq . For a non-negative q, we have
(xω+p )ω+q ≈ xω+p (xω+p )q+1 ≈ xω−p xω+pq+p ≈ xω+pq .
We are going to show that, for two normal swords σ, τ , the identity σ ≈ τ
holds in E and in Ef in if and only if σ and τ coincide.
4.2
Connections to Burnside varieties
We define by B(A, m, n) the free semigroup generated by the (finite) set A
in the Burnside variety var[xm ≈ xm+n ]. Now we recall some results of [4, 3].
Theorem 2. ([3], Theorem A) For m ≥ 3, n ≥ 1 the word problem for
B(A, m, n) is decidable
Theorem 3. For m ≥ 3, n ≥ 1, every element in the semigroup B(A, m, n)
has a finite number of prefixes and suffixes.
The next statement easily follows from the Theorem 3, we prove it here
to complete the picture.
Proposition 5. For m ≥ 3, n ≥ 1, the Burnside variety var[xm ≈ xm+n ] is
generated by its finite members.
16
Proof. For an element b ∈ B(A, m, n) consider the set Fb of all its factors. It
is easy to see that the set I(b) = B(A, m, n)\Fb is an ideal of B(A, m, n). Let
λb be the corresponding Rees congruence then in view of Theorem 3 it has
finite index. Besides, any different elements a, b ∈ B(A, m, n) are separated
by the congruence λab . Therefore, the semigroup B(A, m, n) is residually
finite for any set A and the variety var[xm ≈ xm+n ] is generated by its finite
members.
Sometimes it is convenient to consider the variety var[xk−p ≈ x2k−p ] since
xk is idempotent in it. Therefore, the identity xω ≈ xk holds in var[xk−p ≈ x2k−p ].
For a positive integer k, we denote by σ (k) the word obtained from σ
replacing ω by k. Let σ be a Z-unary word and
p(σ) = max{0, −q | the symbol ω + q is used in σ}.
Take any p > p(σ) then the identity σ ≈ σ (k) holds in var[xk−p ≈ x2k−p ].
The central notion of Guba’s works [4, 3] is the notion of reduced form
of a word relatively to variety var[xm ≈ xm+n ]. We are not going to give the
definition here since it will not be used in the following (and also given by
simultaneous induction on five another notions).
Theorem 4. (the main point of the proof of Theorem 4.1 in [4]) The identity
u ≈ v holds in var[xm ≈ xm+n ] if and only if the words u and v have the same
reduced forms relatively to the variety var[xm ≈ xm+n ].
We will shortly call the reduced form relatively to the variety var[xk−p ≈ x2k−p ]
the Bk,p -reduced form.
Consider the property of any reduced form that follows from definition of
reducible word and Lemma 2.2 from [4].
Note 1. If the word u is reducible then it contains as a factor a word of the
form v 2k−p−2.
First, consider swords of height 1.
Lemma 7.
For all k, p such that p > p(σ) and k − p − 2 > max{np + m, n}, the word
σ (k) is a Bk,p -reduced form. Two normal swords σ1 , σ2 of height 1 are equal
(k)
(k)
if and only if σ1 = σ2 for large enough k, p. Let σ be a normal sword of
height no more than 1 and length nω + m then the sword σ is periodic if and
only if σ (k) is a power of some word.
17
Proof. Indeed, the lemma obviously holds for any sword of height 0. So, let
h(σ) = 1 then
σ = u0 v1ω+q1 u1 . . . vlω+ql ul ,
where all ui , vi are ordinary words.
Suppose now σ (k) = w1 T 2k−p−2w2 and T is not a power of any other word.
Note that we have chosen the parameters p, k so that the period T can be
equal to any of vi . Thus, the word T 2k−p is the product of words of the
k+q
s1
form t1 vik+qi . . . vj j t2 where t1 is either a suffix of ui or equals to vi−1
ui ,
s2
t2 is either a prefix of uj+1 or equals to uj+1vj+1 . Let us denote by τi the
(k)
subsword of σ such that τi = T .
P
Let n1 ω + m1 be the length of τi . Since n =
|vi | then n ≥ n1 and also
n ≥ (2k − p − 2)n1 ≥ 2k − p − 2 ≥ k + n. Therefore, the word σ (k) is a
Bk,p -reduced form.
Suppose that σ1 6= σ2 and τ is their common longest prefix. Then σ1 =
′
τ ρ1 , σ2 = τ ρ′2 . It is easy to see that the first letters (by the first letter of
empty sword we mean empty word) of the swords ρ′1 and ρ′2 are distinct.
(k)
(k)
(k)
(k)
Therefore, σ1 = τ (k) ρ1 and σ2 = τ (k) ρ2 can not be equal.
Concluding the proof we assume that σ (k) is equal to T m for m ≥ 2. Then
(k)
σ = σ1 . . . σk where all σi = T and satisfy the conditions of this lemma.
Therefore, the swords σi are all equal.
Now we consider the swords of larger height.
Lemma 8. Let σ be a normal sword of height h then, for large enough k
and p, the word σ (k) is a Bk,p -reduced form. Two normal swords σ1 , σ2 are
(k)
(k)
equal if and only if σ1 = σ2 for large enough k.
The sword σ is periodic if and only if σ (k) is periodic.
Proof. Let π0 ρ1ω+q1 π1 . . . ρnω+qn πn be the height representation of the sword
σ. Let the length |σ|(ω) be equal to nω h + gh−1 (ω). Consider integers p, k
such that p > p(σ), k − p − 2 > max{n, npk + gh−1(p)} and the sword
(k)
(k)
(k)
(k)
σ
e = π (k) (ρ1 )ω+q1 π1 . . . (ρn )ω+qn πn of height 1. It is easy to see that
σ (k) = σ
e(k) , so, we are going to prove that the conditions of Lemma 7 holds.
Indeed, it is enough to show that σ
e is normal. Suppose the contrary.
(k)
m
First, let ρi = u for some word u and integer m ≥ 2. But Proposition 4
applied to the sword of height h − 1 implies that ρi is also periodic and,
(k)
(k)
(k)
hence, not fully normal. Second, let (ρi )ω+qi πi+1 (ρi+1 )ω+qi+1 = uω+p1 uω+p2 .
18
(k) (k)
(k)
(k)
But this implies ρi πi+1 = πi+1 ρi+1 and again in view of Proposition 4 the
swords ρi πi+1 , πi+1 ρi+1 are equal and σ contains uω+p1 uω+p2 .
Therefore, we apply Lemma 7 to establish that the word σ
e(k) = σ (k) is a
Bk,p -reduced form. The proof of the last statements literally the same as in
Lemma 7.
These lemmas leads us to the important
Proposition 6. Let σ1 , σ2 be two normal swords then the equality σ1 ≈ σ2
holds in E (Ef in ) if and only if σ1 and σ2 are equal.
Proof. Remind that every Burnside semigroup is an epigroup. Also, in view
of Proposition 5 the free Burnside semigroup satisfies any identity that holds
(k)
(k)
in Ef in . Thus, the identities σ1 ≈ σ2 and, therefore, σ1 ≈ σ2 hold in the
variety var[x2k−p ≈ xk−p ] for large enough k, p. Then by Theorem 4 the words
(k)
(k)
σ1 and σ2 are equal and by Lemmas 7 and 8 the swords σ1 and σ2 coincide.
Finally, we are ready to proof Theorem 1. Let σ1 ≈ σ2 be any identity
that hold in Ef in or E. By Proposition 1 there is an algorithm that constructs
normal forms α1 , α2 of the swords σ1 , σ2 respectively and the identity α1 ≈ α2
also holds in Ef in or E. By Proposition 6 it is equivalent to the equality of
swords α1 and α2 which we can check using algorithm from Proposition 2.
The following proof was given by M. Volkov in a verbal discussion. Let us
prove that the identity basis of E is infinite. For a prime number p, consider
a non-trivial group G satisfying the identity xp ≈ 1. Let H be the semigroup
G with adjointed 0. Define a unary operation on H by the following rule:
for all x 6= 1 we set x = 0, and 1 = 1. It is clear that the unary semigroup
H satisfies the first five identities.
Let q be a prime number distinct from p. Then xq 6= 1 if and only if
x 6= 1. Therefore, 0 = xq = xq = 0. Thus, the identity xq ≈ xq holds in H.
Meanwhile, taking x ∈ G distinct from 1 we obtain 1 = xp 6= xp = 0. Thus,
the identity xp ≈ xp fails on H.
Acknowledgements
The author wants to thank M. Volkov who drew attention to this problem
and helped to improve the paper, A. Popovich for help in finishing this work
19
and opportunity to discuss the results and L. Shevrin for useful comments.
References
[1] Jorge Almeida. Finite Semigroups and Universal Algebra. World Scientific, 1995.
[2] Jorge Almeida and Benjamin Steinberg. On the decidability of iterated
semidirect products and applications to complexity. Proceedings of the
London Mathematical Society, 80:55–74, 2000.
[3] Victor Guba. The word problem for the relatively free semigroup satisfying tm = tm+n with m ≥ 3. International Journal of Algebra and
Computation, 3:335–347, 1993.
[4] Victor Guba. The word problem for the relatively free semigroup satisfying tm = tm+n with m ≥ 4 or m = 3, n = 1. International Journal of
Algebra and Computation, 3:125–140, 1993.
[5] Jon McCammond. Normal forms for free aperiodic semigroups. International Journal of Algebra and Computation, 11:581–625, 2001.
[6] Jan Reiterman. The Birkhoff theorem for finite algebras. Algebra Universalis,, 14:1–10, 1982.
[7] Lev Naumovich Shevrin. On the theory of epigroups. I. Russian Acad.
Sci. Sb. Math., 82:485–512, 1994.
[8] Lev Naumovich Shevrin. On the theory of epigroups. II. Russian Acad.
Sci. Sb. Math., 83:133–154, 1994.
[9] Lev Naumovich Shevrin. Epigroups. In V. B. Kudryavtsev and I. V.
Rosenberg, editors, Structural theory of automata, semigroups, and universal algebra, Proc. NATO Advance Study Institute Workshop at the
CRM, pages 331–380. Springer, 2005.
[10] Ilya Yu. Zhil’tsov. On ω-identities of finite semigroups. In C. L. Nehaniv
and M. Ito, editors, Algebraic Engineering, Proc. Kyoto Workshop on
Computer Systems and Formal Languages and First International Conference on Semigroups and Algebraic Engineering, pages 427–436. World
Scientific, 1999.
20
[11] Ilya Yu. Zhil’tsov. On identities of epigroups. Doklady Mathematics,
375:10–13, 2000.
21