C om mun. Fac. Sci. U niv. A nk. Ser. A 1 M ath. Stat. Volum e 68, N umb er 2, Pages 1895–1908 (2019) D O I: 10.31801/cfsuasm as.472024
ISSN 1303–5991 E-ISSN 2618-6470
http://com munications.science.ankara.edu.tr/index.php?series= A 1
AUTOMATIC STRUCTURE FOR GENERALIZED
BRUCK-REILLY -EXTENSION OF A MONOID
EYLEM GÜZEL KARPUZ
Abstract. In the present paper, we study the automaticity of generalized Bruck-Reilly -extension of a monoid. Under some certain situations, we prove that the automaticity of the monoid implies the automaticity of the generalized Bruck-Reilly -extension of this monoid.
1. Introduction and Preliminaries
One of the most popular areas of computational algebra has recently been the theory of automatic groups. The description of a group by an automatic struc-ture allows one e¢ ciently to perform various computations involving the group, which may be hard or impossible given only a presentation. Groups which admit automatic structure also share a number of interesting structural and geometric properties [8]. Recently, many authors have followed a suggestion of Hudson [12] by considering a natural generalization to the broader class of monoids or, even more generally, of semigroups, and a coherent theory has begun to develop from the point of geometric aspects [21], computational and decidebility aspects [17, 18, 19], other notions of automaticity for semigroups [9, 10].
Many results about automatic semigroups concern automaticity of semigroup constructions. For instance, in [5] free product of semigroups, in [4] direct product of semigroups, in [7] Rees matrix semigroups, in [1, 3] Bruck-Reilly extension of monoids and wreath product of semigroups were studied. In [6], the author showed that a Bruck-Reilly extension BR(S; ) of an automatic monoid S is itself automatic
if S is …nite (Theorem 5:1),
if the mapping : S ! S sends every element of S to 1S (Theorem 5:2),
if : S ! S is the identity mapping (Theorem 5:3),
if S is a …nite geometric type automatic monoid and S is …nite (Theorem 5:4).
Received by the editors: October 18, 2018; Accepted: February 28, 2019.
2010 Mathematics Subject Classi…cation. Primary 20M05; Secondary 20M35, 68Q45. Key words and phrases. Automatic semigroup, generalized Bruck-Reilly -extension, presentation.
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These results and their proofs are reproduced in a survey article by Andrade et al. [1]. In the present paper, by considering the results given in [6], we study on generalized Bruck-Reilly -extension of a monoid of which presentation was …rstly de…ned in [14]. A generalized Bruck-Reilly -extension was …rst introduced in [2]. Since then many research papers have been published see for example [13, 15, 16, 20]. We prove the following results:
Theorem 4If T is a …nite monoid then generalized Bruck-Reilly -extension of T is automatic.
Theorem 6 If T is an automatic monoid and ; : T ! H1; t 7! 1T then
generalized Bruck-Reilly -extension of T is automatic.
Theorem 7If T is an automatic monoid and ; are identity homomorphisms of T then generalized Bruck-Reilly -extension of T is automatic.
Theorem 10 Let T be a …nite geometric type automatic monoid and let ; : T ! H1 be homomorphisms. If T , T are …nite then generalized Bruck-Reilly
-extension of T is automatic.
Let A be an alphabet. We denote by A+ the free semigroup generated by A
consisting of …nite sequences of elements of A, which we call words, under the concatenation; and by A the free monoid generated by A consisting of A+ with
the empty word , the identity in A . For a word w 2 A , we denote the length of w by jwj. Let S be a semigroup and : A ! S a mapping. We say that A is a …nite generating set for S with respect to if the unique extension of to a semigroup homomorphism : A+ ! S is surjective. For u; v 2 A+ we
write u v to mean that u and v are equal as words and u = v to mean that u and v represent the same element in the semigroup. In other words u = v . We say that a subset L of A , usually called a language, is regular if there is a …nite state automaton accepting L ([5]). To be able to deal with automata that accept pairs of words and to de…ne automatic semigroups we need to de…ne the set A(2; $) = ((A [ f$g) (A [ f$g)) f($; $)g where $ is a symbol not in A (called the padding symbol) and the function A: A A ! A(2; $) de…ned by
(a1 am; b1 bn) A= 8 > > < > > : if 0 = m = n (a1; b1) (am; bm) if 0 < m = n (a1; b1) (am; bm)($; bm+1) ($; bn) if 0 m < n (a1; b1) (an; bn)(an+1; $) (am; $) if m > n 0:
Let S be a semigroup and A a …nite generating set for S with respect to : A+! S. The pair (A; L) is an automatic structure for S (with respect to ) if
L is a regular subset of A+ and L = S,
L== f( ; ) : ; 2 L; = g A is a regular in A(2; $)+, and
La = f( ; ) : ; 2 L; a = g A is a regular in A(2; $)+ for each a 2 A.
We say that the pair (A; L) is an automatic structure with uniqueness (with respect to ) for a semigroup S, if it is an automatic structure and each element in S is represented by an unique word in L (the restriction of to L is a bijection).
2. Generalized Bruck-Reilly -Extension
Let T be a monoid with H1 and H1as the H - and H- class which contains the
identity 1T of T , respectively. Assume that and are morphisms from T into H1.
Let u be an element in H1and let ube the inner automorphism of H1 de…ned by
x 7! uxu 1such that
u= . Now we can consider N0 N0 T N0 N0 into
a semigroup by de…ning multiplication (m; n; v; p; q)(m0; n0; v0; p0; q0) = 8 < : (m; n p + t; (v t p)(v0t n0); p0 n0+ t); q0) if q = m0 (m; n; v(((u n0 (v0p0 ) q m0 1 ) p); p; q0 m0+ q) if q > m0 (m q + m0; n0 n(v )up) m0 q 1 ) n0)v0; p0; q0) if q < m0;
where t = max(p; n0) and 0; 0 are interpreted as the identity map of T and u0 is interpreted as the identity 1T of T . The monoid N0 N0 T N0 N0
constructed above is called generalized Bruck-Reilly -extension of T determined by the morphisms ; and the element u. This monoid is denoted by GBR (T ; ; ; u) and the identity of it is the element (0; 0; 1T; 0; 0) ([20]). For some information
concerning semigroup theory such as H - and H-Green relations, see [11]. In [14], the authors have obtained the following results.
Lemma 1. Suppose that X is a generating set for the monoid T . Then f(0; 0; x; 0; 0) : x 2 Xg [ f(1; 0; 1T; 0; 0) [ (0; 1; 1T; 0; 0) [ (0; 0; 1T; 1; 0)
[(0; 0; 1T; 0; 1)g
is a generating set for the monoid GBR (T ; ; ; u).
Theorem 2. Let T be a monoid de…ned by the presentation < X; R >, and let ; be morphisms from T into H1. Therefore the monoid GBR (T ; ; ; u) is de…ned by the presentation
< X; y; z; a; b ; R; yz = 1; ba = 1;
yx = (x )y; xz = z(x ); bx = (x )b; xa = a(x ) (x 2 X); yb = uy; ya = u 1y; bz = zu; az = zu 1> :
As a consequence of Theorem 2, we have the following result. Corollary 3. Let v be an arbitrary word in X . The relations
ymv = (v m)ym; vzm= zm(v m); bnv = (v n)bn; van= an(v n); ymbn= (u m 1)nym; yman= (u 1 m 1)nym;
bnzm= zm(u m 1)n; anzm= zm(u 1 m 1)n
hold in GBR (T ; ; ; u) for all m; n 2 N0. As a consequence, every word w 2
(X [ fy; z; a; bg) is equal in GBR (T ; ; ; u) to a word of the form zmanvbpyq for
some v 2 X and m; n; p; q 2 N0.
3. Main Results We give the …rst result of this paper.
Theorem 4. If T is a …nite monoid then any generalized Bruck-Reilly -extension of T is automatic.
Proof. Let T = ft1; t2; ; tlg and let T = ft1; t2; ; tlg be an alphabet in
bijec-tion with T . We de…ne the alphabet A = fy; z; a; bg [ T and the regular language L = fzmantbpyq : m; n; p; q 0; t 2 T g
on A. De…ning the homomorphism
: A+ ! GBR (T ; ; ; u); t 7! (0; 0; t; 0; 0); y 7! (0; 0; 1T; 0; 1); z 7! (1; 0; 1T; 0; 0); a 7! (0; 1; 1T; 0; 0); b 7! (0; 0; 1T; 1; 0);
it is clear that A is a generating set for GBR (T ; ; ; u) with respect to and, in fact, given an element (m; n; t; p; q) 2 N0 N0 T N0 N0the unique word in L
representing it is zmantbpyq.
In order to prove that (A; L) is an automatic structure with uniqueness for GBR (T ; ; ; u) we have to prove that, for each generator k 2 A the language Lk
is regular. To prove that Ly, Lz, La and Lb are regular we observe that
(zmantibpyq)y = zmantibpyq+1; (zmantibpyq)z = zmant ibpyq 1 if q 1; zm+1(t i ) if q = 0; (zmantibpyq)a = 8 < : zman(t i((u 1 q 1) p))bpyq if q 1; zmantibp 1 if q = 0, p 1; zman+1(ti ) if q = p = 0, (zmantibpyq)b = zman(t i((u q 1) p))bpyq if q 1; zmant ibp+1 if q = 0,
and so we can write Ly =
l
[
i=1
=
l
[
i=1
(f(z; z)g f(a; a)g f(ti; ti)g f(b; b)g f(y; y)g f($; y)g)
which is a regular language. We have Lz = l [ i=1 f(zmantibpyq; zmantibpyq 1) A: m; n; p 2 N0; q 1g [ l [ i=1 f(zmti; zm+1(ti )) A: m 2 N0g = l [ i=1
(f(z; z)g f(a; a)g f(ti; ti)g f(b; b)g f(y; y)g f(y; $)g)
[
l
[
i=1
(f(z; z)g f(ti; z)($; ti )g);
and we conclude that Lz is a regular language. Now we consider the language La
La = l [ i=1 f(zmantibpyq; zman(ti((u 1 q 1) p))bpyq) A: m; n; p 2 N0; q 1g [ l [ i=1 f(zmantibp; zmantibp 1) A: m; n 2 N0; p 1g [ l [ i=1 f(zmanti; zman+1(ti )) A: m; n 2 N0g:
Since T is …nite the set H1 is …nite as well. So f(u 1 q 1) p; (u q 1) p
: p; q 2 N0; q 1g is …nite. Then we get
La = l
[
i=1
(f(z; z)g f(a; a)g f(ti; ti)($; ((u 1 q 1) p))g f(b; b)g f(y; y)g+)
[ l [ i=1 (f(z; z)g f(a; a)g f(ti; ti)g f(b; b)g f(b; $)g) [ l [ i=1
(f(z; z)g f(a; a)g f(ti; a)($; ti )g)
which is a …nite union of regular languages and so is regular. Lb =
l
[
i=1
[ l [ i=1 f(zmantibp; zmantibp+1) A: m; n; p 2 N0g = l [ i=1
(f(z; z)g f(a; a)g f(ti; ti)($; ((u q 1) p))g f(b; b)g f(y; y)g+)
[
l
[
i=1
(f(z; z)g f(a; a)g f(ti; ti)g f(b; b)g f($; b)g);
and we conclude that Lb is a regular language as well.
Now for t2 T we have (zmantibpyq)t =
zmant
i((t q) p)bpyq if q 1;
zmant
i(t p)bp if q = 0, p 1:
Since T is …nite the sets f(t q) p : p; q 2 N0; q 1g and ft p : p 2 N0g are …nite as well. Thus we have
Lt = l [ i=1 f(zmantibpyq; zmanti((t q) p)bpyq) A: m; n; p 2 N0; q 1g [ l [ i=1 f(zmantibp; zmanti(t p)bp) A: m; n 2 N0; p 1g = l [ i=1
(f(z; z)g f(a; a)g f(ti; ti)($; ((t q) p))g f(b; b)g f(y; y)g+)
[
l
[
i=1
(f(z; z)g f(a; a)g f(ti; ti)($; t p)g f(b; b)g+)
which is a …nite union of regular languages and so is regular. Hence the result.
Now on we assume that T is an automatic monoid and we …x an automatic struc-ture (X; K) with uniqueness for T , where X = fx1; ; xng is a set of semigroup
generators for T with respect to the homomorphism : X+! T: We de…ne the alphabet
A = fy; z; a; bg [ X (1)
to be a set of semigroup generators for GBR (T ; ; ; u) with respect to the homo-morphism
: A+ ! GBR (T ; ; ; u); xi 7! (0; 0; xi ; 0; 0);
z 7! (1; 0; 1T; 0; 0);
a 7! (0; 1; 1T; 0; 0);
b 7! (0; 0; 1T; 1; 0);
and the regular language
L = fzmanwbpyq : w 2 K; m; n; p; q 2 N0g (2) on A+, which is a set of unique normal forms for GBR (T ; ; ; u), since we have (zmanwbpyq) = (m; n; w ; p; q) for w 2 K; m; n; p; q 2 N0. As usual, to simplify notation, we will avoid explicit use of the homomorphisms and , associated with the generating sets, and it will be clear from the context whenever a word w 2 X+
is being identi…ed with an element of T , with an element of GBR (T ; ; ; u) or considered as a word. In particular, for a word w 2 X+ we write w instead of
(w ) , seeing also as a homomorphism : X+ ! T , and we will often write
(m; n; w; p; q) instead of (m; n; w ; p; q) for m; n; p; q 2 N0.
To show that GBR (T ; ; ; u) has automatic structure (A; L), the languages Ly = f(zmanwbpyq; zmanwbpyq+1) A: w 2 K; m; n; p; q 2 N0g; Lz = f(zmanwbpyq; zmanwbpyq 1) A: w 2 K; m; n; p 2 N0; q 1g [f(zmw1; zm+1w2) A: w1; w22 K; m 2 N0; w2= w1 g; La = f(zmanw1bpyq; zmanw2bpyq) A: w1; w22 K; m; n; p 2 N0; q 1; w2= w1((u 1 q 1) p)g [f(zmanwbp; zmanwbp 1) A: w 2 K; m; n 2 N0; p 1g [f(zmanw1; zman+1w2) A: w1; w22 K; m; n 2 N0; w2= w1 g; Lb = f(zmanw1bpyq; zmanw2bpyq) A: w1; w22 K; m; n; p 2 N0; q 1; w2= w1((u q 1) p)g [f(zmanwbp; zmanwbp+1) A: w 2 K; m; n; p 2 N0g; Lxr = f(z manw 1bpyq; zmanw2bpyq) A: (w1; w2) X 2 K(xr q) p; m; n; p; q 2 N0; (xr2 X)g;
must be regular. We note that the language Ly is regular, since we have
Ly= f(z; z)g f(a; a)g f(w; w) X: w 2 Kg f(b; b)g f(y; y)g f($; y)g;
but there is no obvious reason why the languages Lz, La, Lband Lxr should also be
regular. Hence we will consider particular situations where (A; L) is an automatic structure for GBR (T ; ; ; u). We will use the notion of padded product of lan-guages and the following result. The proof of the following result can be found in [6]. Now we …x an alphabet A, and take two regular languages M; N in (A A ) . Then the padded product of languages M and N is
M N = f(w1w 0 1; w2w 0 2) : (w1; w2) 2 M; (w 0 1; w 0 2) 2 Ng:
Lemma 5. Let A be an alphabet and let M; N be regular languages on (A A ) . If there exists a constant C such that for any two words w1; w22 A we have
(w1; w2) 2 M ) jjw1j jw2jj C;
then the language M N is regular. Now we give our result.
Theorem 6. If T is an automatic monoid and ; : T ! H1; t 7! 1T then
GBR (T ; ; ; u) is automatic.
Proof. To show that the pair (A; L) de…ned by (1) and (2) is an automatic structure for GBR (T ; ; ; u), we have to prove that the languages Lz, La, Lb and Lx
(x 2 X) are regular. But now we denote by w1T the unique word in K representing
1T. Then we have
Lz = f(zmanwbpyq; zmanwbpyq 1) A: w 2 K; m; n; p 2 N0; q 1g
[f(zmw; zm+1w1T) A: w 2 K; m 2 N
0
g
= (f(z; z)g f(a; a)g f(w; w) X : w 2 Kg f(b; b)g f(y; y)g f(y; $)g)
[(f(z; z)g (K fw1Tg) X) and La = f(zmanwbpyq; zmanwbpyq) A: w 2 K; m; n; p 2 N0; q 1g [f(zmanwbp; zmanwbp 1) A: w 2 K; m; n 2 N0; p 1g [f(zmanw; zmanw1T) A: w 2 K; m; n 2 N 0 g
= (f(z; z)g f(a; a)g f(w; w) X: w 2 Kg f(b; b)g f(y; y)g+)
[f(z; z)g f(a; a)g f(w; w) X: w 2 Kg f(b; b)g f(b; $)g
[((f(z; z)g f(a; a)g ) (K fw1Tg) X);
which are regular languages by Lemma 5. Now we consider the language Lb and
then we have
Lb = f(zmanwbpyq; zmanwbpyq) A: w 2 K; m; n; p 2 N0; q 1g
[f(zmanwbp; zmanwbp+1) A: w 2 K; m; n; p 2 N0g
= (f(z; z)g f(a; a)g f(w; w) X: w 2 Kg f(b; b)g f(y; y)g+)
[(f(z; z)g f(a; a)g f(w; w) X: w 2 Kg f(b; b)g f($; b)g);
which is a regular language. Since, for any zmanwbpyq2 L with q 1, we have
(zmanwbpyq)x = zmanwbpyq; and for zmanwbp2 L with p 1, we have
(zmanwbp)x = zmanwbp; and for zmanw 2 L we have
we get
Lx = f(zmanwbpyq; zmanwbpyq) A: w 2 K; m; n; p 2 N0; q 1g
[f(zmanwbp; zmanwbp) A: w 2 K; m; n 2 N0; p 1g
[f(zmanw1; zmanw2) A: (w1; w2) X 2 Kx; m; n 2 N0g
= (f(z; z)g f(a; a)g f(w; w) X: w 2 Kg f(b; b)g f(y; y)g+)
[(f(z; z)g f(a; a)g f(w; w) X: w 2 Kg f(b; b)g+)
[(f(z; z)g f(a; a)g Kx):
Hence Lxis a regular language and so GBR (T ; ; ; u) is automatic.
Theorem 7. If T is an automatic monoid and ; are identity homomorphisms of T then GBR (T ; ; ; u) is automatic.
Proof. To show that the pair (A; L) de…ned by (1) and (2) is an automatic structure for GBR (T ; ; ; u) we have to prove that the languages Lz, La, Lband Lx(x 2 X)
are regular. To do that we have
Lz = f(zmanwbpyq; zmanwbpyq 1) A: w 2 K; m; n; p 2 N0; q 1g
[f(zmw; zm+1w) A: w 2 K; m 2 N0g
= (f(z; z)g f(a; a)g f(w; w) X : w 2 Kg f(b; b)g f(y; y)g f(y; $)g)
[((f(z; z)g f($; y)g) f(w; w) X: w 2 Kg); and La = f(zmanwbpyq; zmanwu 1bpyq) A: w; u 12 K; m; n; p 2 N0; q 1g [f(zmanwbp; zmanwbp 1) A: w 2 K; m; n 2 N0; p 1g [f(zmanw; zman+1w) A: w 2 K; m; n 2 N0g = (f(z; z)g f(a; a)g f(w; w) X: w 2 Kg f($; u 1) X : u 12 Kg f(b; b)g f(y; y)g+) [(f(z; z)g f(a; a)g f(w; w) X: w 2 Kg f(b; b)g f(b; $)g)
[((f(z; z)g f(a; a)g f($; a)g) f(w; w) X : w 2 Kg);
which are regular languages by Lemma 5. We have
Lb = f(zmanwbpyq; zmanwubpyq) A: w; u 2 K; m; n; p 2 N0; q 1g
[f(zmanwbp; zmanwbp+1) A: w 2 K; m; n; p 2 N0g
= (f(z; z)g f(a; a)g f(w; w) X : w 2 Kg f($; u) X : u 2 Kg
f(b; b)g f(y; y)g+)
[(f(z; z)g f(a; a)g f(w; w) X : w 2 Kg f(b; b)g f($; b)g);
which is a regular language. Also we have
m; n; p; q 2 N0; (x 2 X)g
= f(z; z)g f(a; a)g Kx f(b; b)g f(y; y)g
which is a regular language. So (A; L) is an automatic structure for GBR (T ; ; ; u).
A semigroup T is called of …nite geometric type (fgt) (see [21]) if for every t12 T ,
there exists k 2 N such that the equation xt1= t2has at most k solutions for every
t22 T .
To prove the next theorem we need the following two lemmas which were proved in [6].
Lemma 8. Let T be a …nite geometric type monoid with an automatic structure with uniqueness (X; K). Then for every w 2 X+ there is a constant C such that
(w1; w2) X 2 Kw implies jjw1j jw2jj < C.
Lemma 9. Let S be a …nite semigroup, X be a …nite set and : X+ ! S be a
surjective homomorphism. For any s 2 S the set s 1 is a regular language. Theorem 10. Let T be a …nite geometric type automatic monoid and let ; : T ! H1 be homomorphisms. If T , T are …nite then GBR (T ; ; ; u) is automatic. Proof. We will prove that the pair (A; L) de…ned by (1) and (2) is an automatic structure for GBR (T ; ; ; u). To do that we have
Lz = f(zmanwbpyq; zmanwbpyq 1) A: w 2 K; m; n; p 2 N0; q 1g
[f(zmw1; zm+1w2) A: w1; w22 K; m 2 N0; w2= w1 g:
It is seen that the language
f(zmanwbpyq; zmanwbpyq 1) A: w 2 K; m; n; p 2 N0; q 1g =
f(z; z)g f(a; a)g f(w; w) X: w 2 Kg f(b; b)g f(y; y)g f(y; $)g
is regular. Thus we just have to prove that the language
M = f(zmw1; zm+1w2) A: w1; w22 K; m 2 N0; w2= w1 g
is also regular. For any t 2 T , let wtbe the unique word in K representing t. Let
N = f(w1; w2) X : w1; w22 K; w2= w1 g = [ t2T f(w1; w2) X: w1; w22 K; w2= w1 = tg = [ t2T f(w1; wt) X : w12 K; w12 (t 1) 1g = [ t2T (((t 1) 1\ K) fwtg) X:
We can de…ne : X+! T ; w 7! w and, since T is …nite, for any t 2 T , we
can apply Lemma 9 and conclude that (t 1) 1= t 1 is regular. Therefore, N
is a regular language. By Lemma 5, the language
M = f(zmw1; zm+1w2) A: (w1; w2) X 2 N; m 2 N0g
= (f(z; z)g f($; z)g) N is regular. Now we will show that the language
La = f(zmanw1bpyq; zmanw2bpyq) A: w1; w22 K; m; n; p 2 N0; q 1;
w2= w1((u 1 q 1) p)g
[f(zmanwbp; zmanwbp 1) A: w 2 K; m; n 2 N0; p 1g
[f(zmanw1; zman+1w2) A: w1; w22 K; m; n 2 N0; w2= w1 g
is regular. Since the language
f(zmanwbp; zmanwbp 1) A: w 2 K; m; n 2 N0; p 1g =
f(z; z)g f(a; a)g f(w; w) X : w 2 Kg f(b; b)g f(b; $)g
is regular, we have to prove that
M1 = f(zmanw1bpyq; zmanw2bpyq) A: w1; w22 K; m; n; p 2 N0; q 1;
w2= w1((u 1 q 1) p)g;
and
M2= f(zmanw1; zman+1w2) A: w1; w22 K; m; n 2 N0; w2= w1 g
are regular. It is seen that the language
M1 = f(zmanw1bpyq; zmanw2bpyq) A: w1; w22 K; m; n; p 2 N0; q 1;
w2= w1((u 1 q 1) p)g
= f(z; z)g f(a; a)g f(w1; w1) X: w12 Kg f($; (u 1 q 1) p)g
f(b; b)g f(y; y)g
is regular. Now for any t 2 T , let wtbe the unique word in K representing t. Let
N2 = f(w1; w2) X : w1; w22 K; w2= w1 g = [ t2T f(w1; w2) X: w1; w22 K; w2= w1 = tg = [ t2T f(w1; wt) X : w12 K; w12 (t 1) 1g = [ t2T (((t 1) 1\ K) fwtg) X:
We can de…ne 2 : X+ ! T ; w 7! w and, since T is …nite, for any t 2 T ,
N2 is a regular language. By Lemma 5, we have that the language
M2 = f(zmanw1; zman+1w2) A: (w1; w2) X2 N2; m; n 2 N0g
= (f(z; z)g f(a; a)g f($; a)g) N2;
is regular.
Now we will prove that the language
Lb = f(zmanw1bpyq; zmanw2bpyq) A: w1; w22 K; m; n; p 2 N0; q 1;
w2= w1((u q 1) p)g
[f(zmanwbp; zmanwbp+1) A: w 2 K; m; n; p 2 N0g
is regular. Since the languages
f(zmanwbp; zmanwbp+1) A: w 2 K; m; n; p 2 N0g =
f(z; z)g f(a; a)g f(w; w) X : w 2 Kg f(b; b)g f($; b)g
and
f(zmanw1bpyq; zmanw2bpyq) A: w1; w22 K; m; n; p 2 N0; q 1;
w2= w1((u q 1) p)g = f(z; z)g f(a; a)g f(w1; w1) X: w12 Kg
f($; (u q 1) p)g f(b; b)g f(y; y)g are regular, Lb is regular as well.
Now it remains to prove that the language
Lx = f(zmanw1bpyq; zmanw2bpyq) A: (w1; w2) X 2 K(x q) p;
m; n; p; q 2 N0(x 2 X)g is regular. We have
Lx= f(z; z)g f(a; a)g (K(x q) p f(b; b)g ) f(y; y)g :
Since T is …nite geometric type, by Lemma 8 there is a constant C such that (w1; w2) X 2 K(x q) pimplies jjw1j jw2jj < C, for any p; q 2 N0, and thus we can
apply Lemma 5 and we conclude that Lxis a regular language.
As known, for a given construction, natural questions are:
(1) Is the class of automatic semigroups closed under this construction? (2) If a semigroup resulting from such a construction is automatic, is the
orig-inal semigroup (or are the origorig-inal semigroups) automatic?
In this paper, we answered the …rst question “yes" under some certain situations for generalized Bruck-Reilly -extension. But the second question is still open. Acknowledgement. I would like to thank Prof. Dr. Ahmet Sinan Çevik for advising me to study on this subject and referees for their kind comments on the paper.
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Current address : Eylem Güzel Karpuz: Karamanoglu Mehmetbey University, Department of Mathematics, Kamil Özdag Science Faculty, Yunus Emre Campus, 70100, Karaman-Turkey.
E-mail address : eylem.guzel@kmu.edu.tr
