International Journal of Scientific & Engineering Research, Volume 4, Issue 7, July-2013 73
ISSN 2229-5518
Rushadije Ramani Halili(1) and Merime Mustafi (2) Department of Mathematics
State University of Tetova
Avenue “Blagoja Toska” Bllok 82-A1/6
Tetova 1200, Fyrom Makedonia
Abstract. The paper [1] represents construction of free left distributive semigroups - by J. Jezek and T. Kepka. In that paper Jezek and Kepka has also shown, that proceeding similarly, one can construct free left distributive idempotent semigroups.
In this paper it is presented the construction of free left distributive idempotent semigroups.
Key words: Free distributive semigroups, idempotent, permutation.
0.INTRODUCTION
A semigroup S is called left(right) distributive semigroup if , xyz = xyxz ( yzx = yxzx )where
x, y, z belong to S .
Semigroup S is called distributive if it is both left and right distributive. A semigroup is called
- medial if it satisfies the equation xuvy = xvuy ;
-left (right) semimedial if it satisfies the identity
x2 yz = xyxz( zyx2 = zxyx) ;
-semimedial if it is both left and right semimedial;
-middle semigroup if it satisfies the identity xyzx = xzyx .
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International Journal of Scientific & Engineering Research, Volume 4, Issue 7, July-2013 74
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The element x ∈ S is an idempotent if
x2 = x . Also we define the set of idempotents in S to
be Id (S ) = {x ∈ S xx = x}
Let A be a finite alphabet,
F [ A] denote the set of nonempty words
a1a2 ....an over A . The
binary operation "⋅" an
F [ A] is defined
(a1a2 ...an )(b1b2 ...bn ) = a1a2 ...anb1b2 ...bn
...(*)
The set
F [ A] with a binary multiplicative operations defined by (*) is called free
semigroups over A .
1. EXAMPLES
Some examples of left distributive idempotent semigroups
a b
a a a b a b
a b a a b
b a b
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b a b b c a b c
2. CONSTRUCTION OF FREE LEFT DISTRIBUTIVE IDEMPOTENT SEMIGROUPS Let X be a nonempty set. Denote by F the free semigroup over X .
Denote by F the union of the following two pairwise disjoint subsets
A, B of F:
A = {x1 x2 ...xn ;
x1 , x2 ,..., xn ∈ X , n ≥ 1
pairwise different }
B = {x1 x2 ...xn xk ;
x1 , x2 ,..., xn ∈ X , n ≥ 2,1 ≤ k ≤ n
pairwise different } .
For every element u of F, expressed as u = x1...xn
where
n ≥ 1; xi ∈ X ,1 ≤ i ≤ n
and x1 ≠ x2 ≠ x3 ≠ ... ≠ xn ,
we define an element
f (u) of F as follows:
(i) If
n = 1,
let
f (u) = x1
(ii) If
n = 2,
let
f (u) = x1 x2
(iii) If n ≥ 3 and
xn ∉{x1 , x2 ,..., xn −1} let
f (u) = x1 y1...ym xn and (by induction on i )
yi is the
fist member of
x1 ,..., xn −1 not contained in {x1 , y1 ,..., yi −1} .
(iv)
If n ≥ 3 and
xn ∈{x1 ,..., xn−2 } , let
f (u) = x1 y1...yn xn
and (by induction on i )
yi is the
first member of
x1 ,..., xn −1 not contained in {x1 , y1 ,..., yi −1} .
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It is easy to see that
f (u) ∈ F in any case. Also, it is easy to see that
f (u) = u
for u ∈ F . Let
us define a binary operation * on F in this way: u * v =
f (u) for any u, v ∈ F . We are going
to prove that
F (*)
is a free left distributive idempotent semigroup over X .
Proposition 1: Let u, v ∈ F and u ≠ v . Then there is an left distributive idempotent
semigroup not satisfying u = v .
Proof: Suppose that u = v is satisfied in all left distributive idempotent semigroup. Since every left zero semigroup is left distributive the words u, v hare the same first letters.
Similarly right zero semi group is left distributive and hence
u, v have the same last letters.
Furthermore, every semilattice is distributive and we conclude that the set of letters accurring in u coincides with the set of letters occurring in v . Now, we distinguish the following
cases.
Case1: u = x1 x2 ...xn−1 xn ∈ A and
v = xp (1) xp ( 2) ...xp ( n −1) xp ( n ) ∈ A
for a permutation p of {1,..., n}
with
p(1) = 1 and
p(n) = n . If n ≥ 4 , then every left distributive idempotent semigroup
satisfying u = v
is medial.
Hewever, semigroup in example
a b c
a a a a b b b b
c a b c
is a non- medial left distributive idempotent semigroup. Consequently n = 3 .
Case 2: u = x1 x2 ...xn−1 xn ∈ A and
v = xp (1) xp ( 2) ...xp ( n −1) xp ( n ) xp ( k ) ∈ B for a permutation p of
{1,..., n} with
p(1) = 1 and
p(k ) = n . One can easily check that every left distributive
idempotent semigroup satisfying u = v is distributive. However, semigroup in example is
not distributive, a contradiction.
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c a b c
Case 3: u = x1 x2 ...xn−1 xk ∈ B
and
v = xp (1) xp ( 2) ...xp ( n −1) xp ( n ) xp (l ) ∈ B
for a permutation p of
{1,..., n} with
p(1) = 1 and
p(l ) = k .
Since the semigroup
a b c
a a a a b b b b
c a b c
is not middle semimedial, we have
p(2) = 2,..., p(n) = n .
However, the left distributive idempotent
semigroup in example does not satisfy axa = a
Thus, u = v , a contradiction. #
Theorem: For a nonempty set X , the grupoid
idempotent semigroup over X .
F (*)
constructed in 2 is a free left distributive
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ISSN 2229-5518
Proof: Denote by the set of the ordered pairs (u, v) of elements of F such that the equation u = v is satisfied in all left distributive idempotent semigroups. So, is a congruence of F and F / is a free left distributive idempotent semigroup over X .
It is obvious that
f (u) = u
for any F so that (by proposition 1) u v if and only if
f (u) =
f (v) for any u, v ∈F and is just the kernel of f .
Now, f is a homomorphism of F onto F (*) : if u, v ∈F, then both
f (u) * f (v) belong to F and are congruent modulo with uv .
The result follows from the homomorphism theorem.
f (uv) and
[1]Jezhek J., Kepka T. and Nemec P., Distributive grupoids, Rozpravy CSAV 91/3 (1981). [2] Markovski S., Za distributivnite polugrupi, God. Zbor. (Skopje) 30 (1979), 15-27.
[3]Zejnullahu A., Free left distributive semigroups, Acta Univ. Carolinae Math. Phys. 30 (1989), n0.1,
29-32.
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