Two Classes of Finite Semigroups and Monoids Involving Lucas Numbers K. Ahmadidelir, C. M. Campbell and H. Doostie Communicated by M. V. Volkov n
n
Abstract The class of finitely presented groups ha, b | an = bn , abab 2 c bb 2 c =1i is an extension of triangle groups studied recently. These groups are finite and their orders depend on the Lucas numbers. In this paper, by considering the three presentations n
n
π1 = ha, b | an = bn , abab 2 c bb 2 c = 1i, n
n
π2 = ha, b | an = bn , a2 bab 2 c bb 2 c = ai and
n
n
π3 = ha, b | an = bn , a2 bab 2 c bb 2 c+1 = abi, we study M on(πi ), i = 1, 2, 3, and Sg(πi ), i = 2, 3, for their finiteness. In this investigation, we find their relationship with Gp(πi ), where M on(π), Sg(π) and Gp(π) are used for the monoid, the semigroup and the group presented by the presentation π, respectively. Keywords: Group presentation, monoid presentation, semigroup presentation, Lucas number K. Ahmadidelir Department of Mathematics, Science and Research Branch, Islamic Azad University, P.O. Box 14515/1775, Tehran, Iran e-mail:
[email protected] C. M. Campbell School of Mathematics and Statistics, University of St. Andrews, North Hough, St Andrews, Fife KY16 9SS, Scotland e-mail:
[email protected] H. Doostie Mathematics Department, Tarbiat Moallem University, 49 Mofateh Ave., Tehran 15614, Iran e-mail:
[email protected]
1
Introduction The class of deficiency zero groups presented by n
n
π1 = ha, b | an = bn , abab 2 c bb 2 c = 1i has been studied in [11] where the be finite of order n(n+2) n (1 + 3 2 ), 2 2n(n + 1)gn+1 , n(n + 1)gn+1 ,
corresponding group has been proved to n ≡ 0 or ± 2 (mod 6), n≡3 (mod 6), n ≡ ±1 (mod 6),
for every integer n ≥ 2, where btc denotes the integer part of a real t and {gi }∞ i=1 is the sequence of Lucas numbers g1 = 2, g2 = 1, gi+2 = gi+1 + gi , i ≥ 1. In [11], it has been proved that all of these groups are metabelian and that if n ≡ 0 (mod 4) or n ≡ ±1 (mod 6) they are metacyclic. For every integer n > 2 consider the presentations: n
n
π2 = ha, b | an = bn , a2 bab 2 c bb 2 c = ai, n
n
π3 = ha, b | an = bn , a2 bab 2 c bb 2 c = abi, of semigroups and/or monoids. Our notation is standard and follows [3, 4, 15]. For more information on group, semigroup and monoid presentations and the related algorithms one may consult [3, 4, 12, 14, 15]. To avoid confusion we denote a semigroup presentation by Sg(π), a monoid presentation by M on(π) and a group presentation by Gp(π). Note that we will refer to [1, 3–5, 13–15] for certain results which are necessary for our calculations. Our main results in this paper are the following theorems: Theorem A. For every n > 2, M on(π1 ) is a group and M on(π1 ) ∼ = Gp(π1 ). Theorem B. For every n > 2, |Sg(π2 )| = |Gp(π2 )| + n − 1. Theorem C. For every n > 2, |Sg(π3 )| = |Gp(π3 )| + n2 . Let i ∈ {2, 3}. As a result of Proposition 1.12 of [13], we have M on(πi ) ∼ = Sg(πi ) ∪ {1}. Hence, |M on(πi )| = |Sg(πi )| + 1. Also by Lemma 2.3 of [4] we deduce that Sg(πi ) does not define a group, since there is no word w ∈ {a, b}+ such that (b = w) is a relator. 2
1
Proof of Theorem A
Some preliminaries are necessary. For an alphabet A let A+ be the free semigroup over A, and let A∗ = A+ ∪ {1} be the free monoid over A. For a subset R of A+ ×A+ (respectively of A∗ ×A∗ ) let ρ be the congruence relation generated by R. Then the semigroup S = A+ /ρ (the monoid M = A∗ /ρ) will be denoted by hA | Ri which is called a semigroup presentation for S (a monoid presentation for M ). To lessen the likelihood of confusion, for w1 , w2 ∈ A+ we write w1 ≡ w2 if w1 and w2 are identical words, and w1 = w2 if they represent the same element of S (i.e. if (w1 , w2 ) ∈ ρ). Thus, for example, if A = {a, b} and R is {ab = ba}, then aba = a2 b but aba 6≡ a2 b. Proof of Theorem A. Let n > 2 and m = b n2 c. Let M = M on(π1 ). First we show that M has a unique minimal left ideal and a unique minimal right ideal. We claim that M an−1 is the unique minimal left ideal and an−1 M is the unique minimal right ideal of M . We have to show that: ∀w ∈ {a, b}+ , ∃w1 ∈ {a, b}+ ; w1 w = an−1 , +
+
n−1
∀w ∈ {a, b} , ∃w1 ∈ {a, b} ; a
= ww1 .
(1) (2)
The proof of (1) is by induction on |w| (the length of the word w). If |w| = 1 then w ≡ a or w ≡ b. If w ≡ a, we set w1 ≡ bam bm an−1 and then, w1 w = bam bm an−1 · a = bam bm an = an bam bm (an is central in M ) m m n−1 =a aba | {z b } = an−1 · 1 = an−1 .
If w ≡ b, we put w1 ≡ an bam bm−1 , n ≥ 2, and then, w1 w = an bam bm−1 · b = an bam bm m m n−1 · 1 = an−1 . = an−1 aba | {z b } = a Now, suppose that the assertion holds for all w such that |w| < k + 1, k ∈ N, and let |w| = k + 1. If w ≡ w0 a, then since |w0 | = k < k + 1, by the induction hypothesis we get: ∃w10 ∈ {a, b}+ ;
w10 w0 = an−1 .
Letting w1 ≡ bam bm w10 gives us w1 w = bam bm w10 · w0 a = bam bm · an−1 · a = an bam bm | {z } m m = an−1 aba = an−1 · 1 = an−1 . | {z b } 3
If w ≡ w00 b, then since |w00 | = k < k + 1, ∃w100 ∈ {a, b}+ ;
w100 w00 = an−1 .
So, by considering w1 ≡ bam bm−1 aw100 we deduce that w1 w = bam bm−1 a w100 · w00 b = bam bm−1 a · an−1 · b = an bam bm = an−1 . | {z } This completes the proof of (1), showing that M an−1 is the unique minimal left ideal of M . To prove (2) we may use a similar method. If I is the minimal (two-sided) ideal of M then by Theorem 1 in [3], n−1 a M = I = M an−1 , and then by Theorem 2 in [3], an−1 M ∩ M an−1 = I is a group. Since 1 = a(bam bm ), it follows that aM = M and so an−1 M = M = I, whence M is a group. But, every monoid (semigroup) presentation for a group G is also a group presentation for G. So M ∼ = Gp(π1 ). This completes the proof.
2
Proofs of Theorems B and C
An almost similar method to that of Theorem A will be used here. Proof of Theorem B. Let S = Sg(π2 ). We show that S 1 an−1 and an−1 S 1 are the unique left and unique right minimal ideals of S, respectively. As well as the proof of Theorem A we have to show, by induction on the length of w, that: ∀w ∈ {a, b}+ , ∃w1 ∈ {a, b}+ ; w1 w = an−1 If |w| = 1 then w ≡ a or w ≡ b. If w ≡ a, by setting w1 ≡ bam bm an−1 we get: w1 w = bam bm an−1 · a = bam bm an = an bam bm 2 m m n−2 · a = an−1 . = an−2 a | ba{z b } = a If w ≡ b, by putting w1 ≡ an bam bm−1 ,
(n > 2) we get:
2 m m n−2 w1 w = an bam bm−1 · b = an bam bm = an−2 · a · a = an−1 . | ba{z b } = a
Now, suppose that the assertion holds for all words with length < k + 1 where k ∈ N. Let w be a word of length |w| = k + 1. If w ≡ w0 a, then by the induction hypothesis, ∃w10 ∈ {a, b}+ ;
w10 w0 = an−1 . 4
So, by defining w1 ≡ bam bm w10 , we conclude that w1 w = bam bm w10 · w0 a = bam bm · an−1 · a = an bam bm = an−1 . | {z } If w ≡ w00 b then, ∃w100 ∈ {a, b}+ ;
w100 w00 = an−1 ,
and considering w1 ≡ bam bm−1 aw100 gives us w1 w = bam bm−1 a w100 · w00 b = bam bm−1 a · an−1 · b = an bam bm = an−1 . | {z } Therefore, S 1 an−1 = San−1 ∪ {an−1 } is the unique minimal left ideal of S. To prove that an−1 S 1 is the unique minimal right ideal of S it suffices to show that ∀w ∈ {a, b}+ , ∃w1 ∈ {a, b}+ ; an−1 = ww1 and the proof is almost similar to that of the above proof. Now, if we denote the minimal (two-sided) ideal of S by I, then an−1 S 1 = I = S 1 an−1 . Consequently, an−1 S 1 ∩ S 1 an−1 = I is a group, indeed, the group I ∼ = Gp(π2 ), by Theorem 4 of [3]. To calculate the order of Sg(π2 ), 2 since a bam bm = a, then a ∈ a2 S 1 and a2 S 1 = aS 1 (for, a2 ∈ aS 1 ). This yields in turn ai S 1 = aS 1 , for every positive integer i. Now, a = a2 bam bm ∈ a2 S 1 = an−1 S 1 and so a ∈ S 1 an−1 . So S 1 an = S 1 an−1 = · · · = S 1 a and then, by S 1 bn = S 1 an , we get S 1 ai = ai S 1 = I = S 1 bn
for all i = 1, . . . , n.
Thus, all the elements of S are in I ∼ = Gp(π2 ), except b, b2 , . . . , bn−1 which are pairwise distinct. So, |S| = |I| + (n − 1), i.e. |Sg(π2 )| = |Gp(π2 )| + (n − 1). This completes the proof. Remark 2.1. In Sg(π2 ), every bi (1 6 i 6 n − 1) is alone in its H-class and the number of H-classes is (n − 1) + 1 = n. (Namely, Ha ∼ = Gp(π2 ), Hb = 2 n−1 {b}, Hb2 = {b }, . . . , Hbn−1 = {b }). Trivially by the above discussion, all Green’s relations are equal in S, i.e. H = L = R = D = J . Proof of Theorem C. Let S = Sg(π3 ). We show that S 1 an−1 b and abn−1 S 1 are the unique left and the unique right minimal ideals of S, respectively. We have to prove the following assertions by induction on the length of w. ∀w ∈ {a, b}+ , ∃w1 ∈ {a, b}+ ; w1 w = an−1 b, ∀w ∈ {a, b}+ , ∃w1 ∈ {a, b}+ ; ww1 = abn−1 . 5
If |w| = 1, then w ≡ a or w ≡ b. In the first case we let w1 ≡ bam bm+1 an−1 . Then, w1 w = bam bm+1 an−1 · a = an bam bm+1 = an−2 · ab = an−1 b. Likewise if w ≡ b, we take w1 ≡ an bam bm . If w = w0 a or w = w0 b where |w0 | = k < k + 1 then, ∃w10 ∈ {a, b}+ ;
w10 w0 = an−1 b.
In the first case by choosing w1 ≡ (bam babam−1 bn−1 a)w10 we get w1 w = (bam babam−1 bn−1 a) w10 · w0 a = (bam babam−1 bn−1 a)(an−1 b)a | {z } m−1 n n = (bam babam−1 )(bn−1 a)(an−1 b)a = bam ba ba | {za b a}
2 m m+1 m−1 = (bam ba)(an−2 a ) | ba {zb } b
= (bam ba)(an bam bn )
= (bam b a)(an−2 a bbm−1 ) | {z } = an bam bm+1
= bam ban bm
= an−2 ab = an−1 b.
And in the second case, we let w1 ≡ bam bm−1 aw10 and get, w1 w = bam bm−1 a w10 · w0 b = bam bm−1 a · an−1 b · b = bam bm−1 an b2 | {z } m+1 = an−2 ab = an−1 b. = an−2 |a2 bam {zb } Therefore, S 1 an−1 b is the unique minimal left ideal of S. For w ∈ {a, b}+ , the proof of the existence of w1 , where ww1 = abn−1 , is by induction: w ≡ a ⇒ w1 ≡ bn−1 ⇒ ww1 = abn−1 w ≡ b ⇒ w ≡ bn−1 a2 bam bm−1 ⇒ ww = b · bn−1 a2 bam bm−1 1 1 | {z } |w| = 1 ⇒ = a2 bam bm−1 bn = (a2 bam bm+1 )bn−2 = ab · bn−2 = abn−1 . Let |w| ≥ 1. So w ≡ aw0 or w ≡ bw0 and ∃w10 ∈ {a, b}+ ;
w0 w10 = abn−1 .
For the case w ≡ aw0 , we let w1 ≡ w10 b2 am bm−1 . Then, ww1 = (aw0 )(w10 b2 am bm−1 ) = a(w0 w10 )b2 am bm−1 = a(abn−1 )b2 am bm−1 2 m m+1 n−2 = a2 bam bm−1 bn =a = ab · bn−2 = abn−1 . | ba {zb } b 6
And for the case w ≡ bw0 we let w1 ≡ w10 ban−1 am (bm+1 am )n−2 bm−1 . It is necessary to consider two subcases for n. If n ≥ 3 is even then, ww1 = (bw0 )(w10 ban−1 am (bm+1 am )n−2 bm−1 ) = b(w0 w10 )ban−1 am (bm+1 am )n−2 bm−1 = b (abn−1 )ban−1 am (bm+1 am )n−2 bm−1 | {z } = an bam bn (bm+1 am )n−2 bm−1 m+1 m−1 m+1 m n−2 m−1 = an−2 |a2 bam (b a ) b {zb } b = an−2 · ab · bm−1 (bm+1 am )n−2 bm−1 = an−1 bm (bm+1 am )(bm+1 am )n−3 bm−1 = an−1 bam bn (bm+1 am )n−3 bm−1 m+1 m−1 m+1 m n−3 m−1 = an−3 |a2 bam (b a ) b {zb } b
= an−3 · ab · bm−1 bm+1 am (bm+1 am )n−4 bm−1 m+1 m−1 m+1 m n−4 m−1 = an−4 |a2 bam (b a ) b {zb } b = an−4 · ab · bm−1 bm+1 am (bm+1 am )n−5 bm−1 m+1 m−1 m+1 m n−5 m−1 = an−5 |a2 bam (b a ) b {zb } b = an−5 · ab · bm−1 (bm+1 am )n−5 bm−1 = an−4 bm (bm+1 am )n−5 bm−1 . = .. = abm bm−1 = abn−1 . And if n is odd, in the 5th row we must substitute bn = bm+1 bm instead of bn = bm+1 bm−1 and continue the calculations. Also in the case n = 2 we may let w1 ≡ w10 ba2 . This yields ww1 = (bw0 )(w10 ba2 ) = b(w0 w10 )ba2 = b · abba2 = a2 bab2 = ab. Consequently, abn−1 S 1 is the unique minimal right ideal of S. Denoting the minimal two-sided ideal of S by I and using Theorem 1 of [3] yields abn−1 S 1 = I = S 1 an−1 b. Hence, abn−1 S 1 ∩ S 1 an−1 b = I is a group and I ∼ = Gp(π3 )(∼ = Gp(π1 ), by Theorem 4 of [3]). On the other hand, it is easy to prove that the set T of all elements of S which are not in I is formed by the pairwise distinct elements aj , with 1 6 j 6 n, bi , with 1 6 i 6 n − 1, and bi aj , with 1 6 i, j 6 n − 1. Counting the elements gives us: |T | = n + (n − 1) + (n − 1)2 = 2n − 1 + n2 − 2n + 1 = n2 . Thus, |Sg(π3 )| = |Gp(π3 )|+n2 = |Gp(π1 )|+n2 . This completes the proof.
7
Remark 2.2. In the proof of Theorem C each of the elements of T is alone in its H-class because, if S 1 bi aj = S 1 bk al , bi aj S 1 = bk al S 1
(0 ≤ i, j, k, l ≤ n − 1)
(if i = 0 then we assume that S 1 bi aj = S 1 aj and so on) then, ∃w1 , w2 ∈ {a, b}+ ;
w1 bi aj = bk al , bi aj = w2 bk al ,
∃w3 , w4 ∈ {a, b}+ ;
bi aj w3 = bk al , bi aj = bk al w4 .
and These yield the equations w1 w2 bk al = bk al and bk al w3 w4 = bk al which are obvious contradictions to the existing relators of the semigroup. So, there are exactly n2 + 1 H-classes and all Green’s relations on S coincide. Remark 2.3. For i = 2, 3, it is clear, by the previous proofs, that Sg(πi ) has a unique idempotent (up to isomorphism, the identity of the group Gp(πi )), which is precisely e = abam bm = am bm ab. Also, we can show that e = bam bm a. By the way, this is also (trivially) the case for M on(π1 ).
3
Anti-isomorphic structures
Together with the presentations π1 , π2 and π3 we consider here the following presentations: π4 π5 π6 π7
= ha, b | = ha, b | = ha, b | = ha, b |
an an an an
= bn , = bn , = bn , = bn ,
abam bm+1 = b am bm ab = 1 am+1 bm ab = a am bm ab2 = b
i, i, i, i,
where 2 6 n ∈ N and m = b n2 c. We have the following proposition. Proposition 3.1. (i) Gp(πi ) ∼ = Gp(π1 ), for every i = 4, 5, 6, 7; (ii) M on(π5 ) is anti-isomorphic to M on(π1 ); (iii) Sg(π4 ) ∼ = Sg(π2 ); (iv) Sg(π7 ) and Sg(π6 ) are anti-isomorphic to Sg(π2 ) and Sg(π4 ), respectively. Proof. Considering π4 , π5 , π6 and π7 as group presentations and using Tietze transformations gives (i) at once. For example, Gp(π5 ) = ha, b | a−n = b−n , b−1 a−1 b−m a−m = 1i ∼ = ha−1 , b−1 | b−n = a−n , b−1 a−1 b−m a−m = 1i ∼ = hc, d | cn = dn , cdcm dm = 1i ∼ = Gp(π1 ). 8
To prove (ii), let S = M on(π1 ) and T = M on(π5 ) where, π1 = ha, b | an = bn , abam bm = 1i, π5 = hc, d | cn = dn , cm dm cd = 1i. By defining f : S → T as f (a) = d , f (b) = c and f (1S ) = 1T we get that f (ab) = cd = f (b)f (a). Then, for every x, y ∈ S, f (xy) = f (y)f (x) and, for every x, y ∈ S we have f (xy) = f (y)f (x). Also for every relation that holds in S, certainly a relation holds in the opposite direction in T and thus S and T are anti-isomorphic. Repeating the proof of Theorem B for Sg(π4 ) yields Sg(π4 ) ∼ = Sg(π2 ) (in fact, the two-sided minimal ideal of Sg(π4 ) is isomorphic to Gp(π4 ) ∼ = Gp(π2 ) and |Sg(π4 )| = |Gp(π2 )| + (n − 1)). We now may use the above method to prove (iv).
4
A semigroup presentation for Gp(π1 )
Using the methods of [5] and [13] we describe a semigroup presentation for the group defined by π1 . Proposition 4.1. The group presentation Gp(π1 ) has the following semigroup presentation: π = ha, b | babam bm−1 a = a, abam bm an (abam bm−1 )n b2 am bm a = bi. Proof. Let E = abam bm and u2 = an b−n . Then, Gp(π1 ) = ha, b | E = 1, u2 = 1i. (l)
(r)
By considering a1 = a, a2 = b suppose Ei (and respectively, Ei ) denotes the cyclic permutation of E beginning (ending) with ai where (i = 1, 2) and (t) E2 is the permutation of E beginning and ending with a2 . (t) We now replace E = 1 by the relation E2 a1 = a1 and the relation (r) (l) u2 = 1 by E1 u2 a2 E1 = a2 , where u2 is obtained from u2 by replacing every occurrences of an inverse by a specific word that contains no inverses and every occurrence of the empty word by E. So we get: (r)
(l)
(t)
E1 = bam bm a, E2 = E2 = babam bm−1 , (t)
E=1 ⇒ E2 a = a ⇒ babam bm−1 a = a, (r) (l) E = abam bm ⇒ E2 = E1 = abam bm = E, (r) (l) u2 = 1 ⇒ E1 u2 bE1 = b ⇒ abam bm an (abam bm−1 )n bbam bm a = b 9
(r)
(E2 b−1 = abam bm b−1 = abam bm−1 ). This yields, Sg(π) ∼ = ha, b | babam bm−1 a = a, abam bm an (abam bm−1 )n b2 am bm a = bi. Acknowledgement. The authors wish to thank the referee for helpful comments. References 1. Ayik, H., Campbell, C. M., O’Connor, J. J., Ruskuc, N.: The semigroup efficiency of groups and monoids. Math. Proc. Royal Irish Acad. 100A, 171–176 (2000) 2. Campbell, C. M., Robertson, E. F., Thomas, R. M.: On a class of semigroups with symmetric presentations. Semigroup Forum. 46, 286–306 (1993) 3. Campbell, C. M., Robertson, E. F., Ruskuc, N., Thomas, R. M.: Semigroup and group presentations. Bull. London Math. Soc. 27, 46–50 (1995) 4. Campbell, C. M., Mitchell, J. D., Ruskuc, N.: Comparing semigroup and monoid presentations for finite monoids. Monatsh. Math. 134, 287–293 (2002) 5. Campbell, C. M., Mitchell, J. D., Ruskuc, N.: On defining groups efficiently without using inverses. Math. Proc. Cambridge Philos. Soc. 133, 31–36 (2002) 6. Campbell, C. M., Robertson, E. F., Ruskuc, N.: On semigroups defined by Coxeter type presentations. Proc. Royal Soc. Edinburgh. 125A, 1063–1075 (1995) ¨ u, Y.: 7. Campbell, C. M., Robertson, E. F., Ruskuc, N., Thomas, R. M., Unl¨ Certain one-relator products of semigroups. Comm. Algebra. 23(14), 5207– 5219 (1995) 8. Campbell, C. M., Robertson, E. F., Thomas, R. M.: Semigroup presentations and number sequences. in: Applications of Fibonacci numbers. Volume 5, eds. G. E. Bergum et al, 77–83 (1993) 9. Campbell, C. M., Robertson, E. F., Ruskuc, N., Thomas, R. M.: Fibonacci semigroups. J. Pure Appl. Algebra. 94, 49–57 (1994) 10. Campbell, C. M., Robertson, E. F., Thomas, R. M.: On subsemigroups of finitely presented semigroups. J. Algebra. 180, 1–21 (1996) 11. Doostie, H., Ahmadidelir, K.: A class of Z-metacyclic groups involving the Lucas Numbers. Novi-Sad J. Math. (to appear)
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12. Johnson, D. L.: Presentations of Groups. Cambridge Unversity Press. (1997) 13. Mitchell, J. D.: Extremal problems in combinatorial semigroup theory. Ph.D. Thesis. University of St. Andrews. (2002) ¨ u, Y.: On semigroup presentations. Proc. Edinburgh 14. Robertson, E. F., Unl¨ Math. Soc. 36, 55–68 (1993) 15. Ruskuc, N.: Semigroup presentations. Ph.D. Thesis. University of St. Andrews. (1995)
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