Bidiagonal ranks of completely (0-)simple semigroups Текст научной статьи по специальности «Математика»

УДК 512.53

Bidiagonal Ranks of Completely (0-)simple Semigroups

Ilia V. Barkov*

National Research University of Electronic Technology Shokin square, 1, Moscow, Zelenograd, 124498

Russia

Received 12.09.2015, received in revised form 19.01.2016, accepted 24.02.2016 A bidiogonal act over a semigroup is a two-sided act, where the semigroup acts on its Cartesian power. A bidiagonal rank of a semigroup is the least power of a generating set of the bidiagonal act over this semigroup. In this paper we compute bidiagonal ranks of completely (0-)simple semigroups.

Keywords: act over a semigroup, diagonal rank, completely (0-)simple semigroup. DOI: 10.17516/1997-1397-2016-9-2-144-148.

A right act over a semigroup S is a set X with a map X x S ^ X, (x, s) ^ xs satisfying (xs)s' = x(ss') for all x £ X, s, s' £ S (see [1]). A left S-act Y over the semigroup S is defined analogously: S x Y ^ Y, (s,y) ^ sy, s(s'y) = (ss')y for all s,s' £ S, y £ Y. Let S,T be semigroups. A set Z is called an (S,T)-act (bi-act over S and T), if it is a left S-act and a right T-act at the same time and (sz)t = s(zt) for all z £ Z, s £ S, t £ T. The right S-act X, left S-act Y and (S,T)-bi-act Z may be denoted by Xs, sY, and sZT.

A generating set G of the act (S x S)s is called irreducible if none of its subsets G' c G is a generating set of this act. Clearly, any finite generating set may be reduced to an irreducible one. A generating set is called minimal if it is minimal with respect to power.

Note that a diagonal act over a semigroup is a unary algebra. Indeed, if S is a semigroup, then multiplication by s £ S may be thought as applying unary operation <s : x ^ xs, where x £ S. Therefore, the following theorem is applicable to diagonal acts.

Theorem 1 (Kartashov, [2], Theorem 1). Let A be an algebra with signature £ = [<i | i £ I}, where all operations <i are unary. If A is finitely generated, then any irreducible generating set of A is minimal.

Let S be a semigroup. A right diagonal rank of S (denoted by rdr S) is called the least power of generating sets of the diagonal rigth act of S, or

rdrS = min {IAI | A C S x S A AS1 = S x S} .

A bidiagonal rank bdr S of S is defined in a similar way.

Diagonal acts were used in [3,4] to study conditions of wreath products to be finitely generated. Diagonal acts themselves became a subject of study in [5-7] and others. In these papers the prime problem were conditions of finite generateness of infinite diagonal acts. The notion of diagonal rank was explicitly formulated in the paper [8]. In the paper [9] one-sided diagonal ranks of completely (O-)simple semigroups were calculated. In this paper we continue the study of diagonal ranks of semigroups and calculate bidiagonal ranks of completely (O-)simple semigroups. For completeness we cite the prime results of [9] concerned with completely (O-)simple

*zvord@b64.ru © Siberian Federal University. All rights reserved

semigroups. These theorems may be proven using reasoning similar to the one in the presented theorems.

Theorem 2 ( [9], Theorem 2). Let S be a Rees matrix semigroup with sandwich-matrix P: S = M (G,I, A,P). Then the right diagonal rank of S equals \I\2G, if A is singleton and \I\2\G\2A (A - 1) otherwise.

Theorem 3 ( [9], Theorem 3). Let S be a Rees matrix semigroup with zero: S = M0 (G, I, A, P). Let \I\ = k, \G\ = t, \A\ = l. If l > 2 then:

• if there are no zeros in P, then rdr S = k2t2 (l2 — l) + 2k;

• if there are zeros in P, but there is no column with two or more nonzero elements, then rdr S = k2t2 (l2 — l) + k2t;

• otherwise, rdr S = k2t2 (l2 — l). If l =1, then rdr S = k2t + 2k.

The following theorems are the body of this paper. As in the case of one-sided ranks, bidiagonal ranks of completely (O-)simple semigroups depend on sandwich-matrices insignificantly.

Theorem 4. Let S be a completely simple semigroup: S = M (G,I, A, P), where G is a group of t elements and f conjugacy classes, I and A are index sets of k and l elements correspondingly and P is a sandwich-matrix. Then

• if k,l = 1, then bdr S = t2l(l — 1)k(k — 1);

• if k = 1,l =1, then bdr S = tl(l — 1);

• if l = 1,k =1, then bdr S = tk(k — 1);

• if l = 1,k =1, then bdr S = f.

Proof. Let k,l > 1, M = {((x)^Xl, (y)i2x2) \ h = i2, Ai = A2}. We prove that M is an irreducible generating set of S (S x S)S. Indeed, from

((x)iiXi , (y)i2xj = ((aPxiix)iXi , (apXi2y)iXJ ,

and

((x)iiXi , (y)i2xj (b)ix = ((xPXiib)iiX , (ypx2ib)i2x) >

we see that pairs from M cannot be obtained from any other pairs using one-sided or two-sided multiplication. Therefore, M is a subset of any generating set.

Now we prove that M itself is a generating set. A pair of a kind (iu)ji^, (v)j2^), where j1 = j2, may be obtained from M in the following way. Take A1 = A2 e A and any i e I. Then

((1)jixi , (vu-1pxiip-1i)j2x2) (P-1iU in = {(u)jiH, , (v)j2^ ■

In a similar way we may get pairs of a kind ((u)ixi , (v)ix2), where A1 = A2. Pairs of a kind ((u)jn , (v)j^ are obtained by multiplying the pair

i2 x2 '

(p-k vp-31iJ i3x3 )

by elements (1jxi and (1)i4M. Here i2 = i3, and A2 = A3. These pairs are in M and M is a generating set. Hence, bdrS = \M\ = t2l(l — 1)k(k — 1).

Let l = 1, k > 1. Matrix P is a vector. Using Lemma 3.6 from [10] and the remark after we can say that P consists of 1's. Let M = j , (t j | i = .

Take a pair x = a)i, (b j such that i = j. Since

((1)i, (ba-1)^ (a)i =((a)i, (bj , then x e MS. Pairs ((a)i, (b)i) we get via

(1)i ((1)i, (ba-1)j) (a)i = ((a)i, (b)i) .

Hence S x S C S1MS and M is a generating set. Now we prove that M is irreducible. It is sufficient to show that no pairs from M are products of an another pair and elements of S1. Let ^(1)i, (a)j j = s' ^(1)p, (b)J s for some s, s' e S1, where i = j, p = q. Let s = (c)r. Then

((1)i, (a)j) = ((1)p , (b) J (c)r = ((c)p , (bc)q) . Hence p = i, q = j, c =1 and ((1)p , (b) J =

^(1)i, (a)^j. It means that M is irreducible, so bdr S = IM| = tk(k — 1). Let k =1, l = 1. In a similar way one can prove that bdrS = tl(l — 1). Let k =1, l = 1. Now S ~ G. Using Lemma 4.1 from [5] we get bdr G = f. So bdr S = f. □

Theorem 5. Let S be a completely (0-)simple semigroup: S = M0 (G,I, A, P), where the group G consists of t elements and f conjugacy classes, I and A are index sets of k and l and elements correspondingly and P is a sandwich matrix

• Let k,l > 1. If a row or a column of P has two or more non-zero elements we say that it is good. Consider the following cases.

1. There are no zeros in P. Then bdrP = t2k(k — 1)l(l — 1) + 2.

2. There are zeros in P. Moreover, there is a good row and a good column in P. Then bdr P = t2k(k — 1)l(l — 1).

3. There is a good row in P, but no good column. Then bdrP = t2k(k — 1)l(l — 1) + tk(k — 1).

4. There is a good column in P, but no good row. Then bdr P = t2k(k — 1)l(l — 1)+tl(l — 1).

5. There are no good rows or columns in P. Then bdr P = t2k(k — 1)l(l — 1) + tk(k — 1)+ tl(l — 1) + f.

• If k = 1, l> 1, then bdr S = tl(l — 1) + 2.

• If l =1, l> 1 then bdr S = tk(k — 1) + 2.

• If k = l = 1, then bdr S = f + 2.

Proof. Throughout the following text consider the indices denoted by different symbols to be different.

Let k,l > 1. Action in the biact S (S x S)S is defined by the following expressions:

(a)i\ ((x)i1A1 , (V)i2\2) = ((«PAii x)iX! , (aP\i2 V)iX2) , {(x)i1X1 , (v)i2\2) (b)iX = ((xPAiib)iiA , (yp\2ib)i2 J . Divide all pairs from S x S cnto the following classes:

1 (H'iw , H2J ; 2. (H>I , H'J ; 3. (Hi M , (v)j2 J ; 4. ((u j, Hm) ; 5. ((u j, 0); 6. i0, (v)jj ; 7. (0,0).

Pairs of the class 1 do not belong to A = S1 ■ (S x S) ■ S1. So we add them to the generating set. There are t2k(k — l)l(l — 1) such pairs.

Consider the class 2. Depending on P we have two cases.

• There is a good row in P. Then pairs of the class 2 are obtainable via left multiplication of the class 1 pairs.

• There are no good rows in P. Then the class 2 pairs are not obtainable from the class 1 pairs. To fix that we add to the generating set the pairs of a kind ((1)qAl , (v)q\2^J, where index q is arbitrary, but fixed. There are tl(l — 1) such pairs.

Consider pairs of the class 3. Depending of P we have the following cases.

• There is a good column in P. Pairs of the class 3 are obtainable from the class 1 pairs via right multiplication.

• There is no good columns in P. Then the class 2 pairs are not obtainable from the class 1 pairs. To fix that we add to the generating set the pairs of a kind ((1)ilM, (v)i2^J, where index n is arbitrary, but fixed. There are tk(k — 1) such pairs.

Consider pairs of the class 4. Depending on P we have the following cases.

• There is a good row and a good column in P. Then pairs of the class 4 are obtainable via two-sided action on the class 1 pairs.

• There is no good row in P, but there is a good column. Then the pairs of the class 4 are obtainable via right-sided action on the class 2 pairs. These pairs we obtain from the generating set.

• There is no good column in P, but there is a good row. Then the pairs of the class 4 are obtainable via the left-sided action on the class 3 pairs. These pairs we obtain from the generating set.

• There are no good rows or columns in P. Fix indices q, n, pick one representative gr from every conjugacy class of G and add to the generating set f pairs ((1)qM , (gr)q^j, 1 ^ r ^ f.

We show how to get an arbitrary pair ((u)jV , (v)j^ from these pairs. Choose h,a,b G G such that h-1grh = vu-1, a = h-1p-q1, b = p—1hu. Then

(a)jX ((!)„ , (gr}qß) (b)iv =((u)jv , j ■

Consider class 5 and class 6 pairs. If there are no zeroes in P we have to add two pairs to the generating set: ((x)jn , o) h ^0, (x)j^j .

The pairs (0,0) are obtainable from any pair. So we have the following cases.

1. There are no zeroes in P. Then bdr P = t2k(k — 1)l(l — 1) + 2.

2. There is a good column and a good row in P. Then bdrP = t2k(k — 1)l(l — 1).

3. There is a good row in P, but no good column. Then bdrP = t2k(k — 1)l(l — 1) + tk(k — 1).

4. There is a good column in P, but no good row. Then bdrP = t2k(k — 1)l(l — 1) + tl(l — 1).

5. There are no good rows or columns in P. Then bdrP = t2k(k — 1)l(l — 1) + tk(k — 1) + tl(l — 1) + f.

Let l = 1, k > 1. Then S is a left group with externally adjoined zero: S = A U {0} = {(g)i I g G G,i G I} U {)}, where |A| = k > 1. By Theorem 4 the act A (A x A)a is generated

by the pairs ^(1)i, (g)^j, where g G G, i,j G I. To get an irreducible generating set of the act

s (S x S)S we add the pairs (0, (1)io ) and ((1)io, 0). Hence bdr S = tk(k — 1) + 2. Analogously for the case k = 1, l > 1 we get bdr S = tl(l — 1) +2.

Let k = l = 1. Now we need only f pairs (1,gr ), 1 < r < f to get pairs (u, v) and two pairs (1,0) and (0,1) to get pairs (u, 0) and (0,u). □

References

[1] M.Kilp, U.Knauer, A.V.Mikhalev, Monoids, acts and categories, Berlin, New York, W.de Gruyter, 2000.

[2] В.К.Карташов, Independent systems of generators and the Hopf property for unary algebras, Diskretn. Mat. i Pril. 20(2009), no. 4, 79-84 (in Russian).

[3] E.F.Robertson, N.Ruskuc, M.R.Thomson, On finite generation and other finiteness conditions for wreath products semigroups, Comm. Algebra, 30(2002), no. 8, 3851-3873.

[4] M.R.Thomson, Finiteness Conditions of Wreath Products of Semigroups and Related Properties of Diagonal Acts, PhD thesis, University of St. Andrews, 2001.

[5] P.Gallagher, On the finite and non-finite generation of diagonal acts, Comm. Algebra, 34(2006), no .9, 3123-3137.

[6] P.Gallagher, N.Ruskuc, Finite generation of diagonal acts of some infinite semigroups of transformations and relations, Bull. Austral. Math. Soc., 72(2005), no. 1, 139-146.

[7] T.V.Apraksina, Diagonal acta over semigroups of isotone transformations, Chebysh. sbornik, 12(2011), no. 1. 10-16 (in Russian).

[8] T.V.Apraksina, I.V.Barkov, I.B.Kozhukhov, Diagonal ranks of semigroups, Semigroup Forum, 90(2015), no. 2, 386-400.

[9] I.V.Barkov, R.R.Shakirov, Finite semigroups with minimal diagonal rank, Mat. vestnik ped-vuzov i universitetov Volgo-Vyatskogo regiona, 16(2014), 55-63 (in Russian).

[10] A.H.Clifford, G.B.Preston, Algebraic Theory of Semigroups, Providence, R.I., American Mathematical Soc., 1967.

Бидиагональные ранги вполне (0-)простых полугрупп

Илья В. Барков

Национальный исследовательский университет МИЭТ пл. Шокина, 1, Московская обл., Зеленоград, 124498

Россия

Диагональным биполигоном над полугруппой называется полигон, в котором полугруппа действует с двух сторон на свою декартову степень. Бидиагональным 'рангом полугруппы называется

наименьшая мощность порождающего множества ее бидиагонального полигона. В данной ^работы мы вычисляем бидиагональные ранги вполне (О-)простых полугрупп.

Ключевые слова: полигон над полугруппой, диагональный ранг, вполне (О-)простая полугруппа.