We provide short and direct proofs for some classical theorems proved by
Howie, Levi and McFadden concerning idempotent generated semigroups of
transformations on a finite set.Comment: three page

Araújo, João

Semigroup Forum

English

arXiv

João Araújo

Crossref

On idempotent generated semigroups

We provide short and direct proofs for some classical theorems proved by Howie, Levi and McFadden concerning idempotent generated
semigroups of transformations on a finite set. Mathematics subject classification: 20M20

The Repositório Aberto

On Idempotent Generated Semigroups
Joa˜o Arau´jo
Abstract
We provide short and direct proofs for some classical theorems
proved by Howie, Levi and McFadden concerning idempotent gener-
ated semigroups of transformations on a finite set.
Mathematics subject classification: 20M20.
Let n be a natural number and [n] = {1, . . . , n}. Let Tn, Sn be, respec-
tively, the transformation semigroup and the symmetric group on [n]. Let
P = (Ai)i2[k] be a partition of [n] and let C = {x1, . . . , xk} be a cross-
section of P (say xi 2 Ai). Then we represent Ai by [xi]P and the pair
(P,C) induces an idempotent mapping defined by [xi]P e = {xi}. Con-
versely, every idempotent can be so constructed. To save space instead of
e =
 
[x1]P . . . [xk]P
x1 . . . xk
!
we write e = ([x1]P , . . . , [xk]P ). This notation ex-
tends to e = ([x1, y]P , [x2]P , . . . , [xk]P ) when y 2 [x1]P and [xi]P e = {xi}. By
([x1], . . . , [xi, y], . . . , [xk]) we denote the set of all idempotents e with image
{x1, . . . , xn} and such that the Ker(e)-class of xi contains (at least) two ele-
ments: xi and y, where the underlined element (in this case xi) is the image
of the class under e. For a 2 Tn, we denote the image of a by ra.
Lemma 1 Let a 2 In = Tn \ Sn, rank(a) = k, and (xy) 2 Sym(X).
Then a(xy) = ab, where b = 1 or b = e1e2e3, with e2i = ei and rank(ei) = k.
Proof. Let ra = {ai | i 2 [k]}. If x, y 62 ra, then a(xy) = a. If, say,
x = a1 and y 62 ra, then ae1 = a(a1y), for all e1 2 ([a1, y], [a2], . . . , [ak]).
Finally, if x, y 2 ra, without loss of generality, we can assume that x = a1
and y = a2. Then a(a1 a2) = ae2e3e4, for e2 2 ([a1], [a2, u], [a3], . . . , [ak]),
e3 2 ([u], [a1, a2], [a3], . . . , [ak]) and e4 2 ([u, a1], [a2], . . . , [ak]).
1
Theorem 2 [1] Every ideal of In is generated by its own idempotents.
Proof. Let a 2 In. Then a = eg for some e = e2 2 Tn and g 2 Sn. Therefore
a = e(x1y1) . . . (xmym) and hence, applying m times Lemma 1, a can be
obtained as a product of idempotents of the same rank as a.
Let t 2 In and g 2 Sn. Denote g 1tg by tg and let Ct = {tg | g 2 Sn} and
tSn = h{t}[Sni\Sn. For f = ([x1, w]Q, [x2]Q, . . . , [xk]Q) and g 2 Sn, it is easy
to check that we have f g = ([x1g, wg]Qg, . . . , [xkg]Qg) 2 ([x1g, wg], . . . , [xkg]).
Lemma 3 Let a 2 In, rank(a) = k, let f = ([x1, w]Q, [x2]Q, . . . , [xk]Q)
and let (xy) 2 Sn. Then a(xy) = ab, where b = 1 or b = e1e2e3, with ei 2 Cf .
Proof. As in Lemma 1, one only has to show that the sets ([a1, y], [a2], . . . , [ak]),
([a1], [a2, u], . . . , [ak]), ([u], [a1, a2], . . . , [ak]) and ([a1, u], [a2], . . . , [ak]) inter-
sect Cf . Let g 2 Sn such that x1g = a1, wg = y and xig = ai (2  i  k).
Thus f g = ([x1g, wg]Qg, [x2g]Qg, . . . , [xkg]Qg) 2 ([a1, y], [a2], . . . , [ak]). The
proof that Cf intersects the remaining three sets is similar.
Repeating the arguments of Theorem 1 together with Lemma 3 we have
Corollary 4 Let eg 2 In (for some e = e2, g 2 Sn) and let f 2 = f with
rank(eg) = rank(f). Then eg 2 ehCfi and, in particular, eg 2 hCei.
From now on let a = eg 2 In (for some e = e2, g 2 Sn).
Corollary 5 hCei = eSn = (eg)Sn = aSn.
Proof. The only non-trivial inclusion is eSn ✓ hCei. As eSn is generated by
{geh | g, h 2 Sn}, let g, h 2 Sn. By Corollary 4, he = (heh 1)h 2 hCheh 1i
and eg 2 hCei. Since heh 1 2 hCei, it follows that hCheh 1i  hCei. Thus
heg = (he)(eg) 2 hCei. It is proved that eSn ✓ hCei.
Theorem 6 [2] hCai = hCei = aSn and hence is idempotent generated.
Proof. Since a = eg, then e = ag 1 and it is easy to check that we have
(ag 1)n = aagag2 . . . agn 1g n. But for some natural m we have g m = 1.
Thus (ag 1)m = aagag2 . . . agm 1 2 hCai so that e = em = (ag 1)m 2 hCai
and hence, by Corollary 5, it follows that aSn = hCei  hCai  aSn . (Cf.
[3]).
2
References
[1] J. M. Howie, R. B. McFadden Idempotent rank in finite full transformation semi-
groups, Proc. R. Soc. Edinburgh, Sect. A 114 (1990), 161–167.
[2] I. Levi and R. McFadden Sn-normal semigroups, Proc. Edinburgh Math. Soc., 37
(1994), 471-476.
[3] D.B. McAlister Semigroups generated by a group and an idempotent, Comm. Algebra,
26 (1998), 515-547.
3


Repositório Aberto da Universidade Aberta

On Idempotent Generated SemigroupsJoa˜o Arau´joAbstractWe provide short and direct proofs for some classical theoremsproved by Howie, Levi and McFadden concerning idempotent gener-ated semigroups of transformations on a finite set.Mathematics subject classification: 20M20.Let n be a natural number and [n] = {1, . . . , n}. Let Tn, Sn be, respec-tively, the transformation semigroup and the symmetric group on [n]. LetP = (Ai)i2[k] be a partition of [n] and let C = {x1, . . . , xk} be a cross-section of P (say xi 2 Ai). Then we represent Ai by [xi]P and the pair(P,C) induces an idempotent mapping defined by [xi]P e = {xi}. Con-versely, every idempotent can be so constructed. To save space instead ofe = [x1]P . . . [xk]Px1 . . . xk!we write e = ([x1]P , . . . , [xk]P ). This notation ex-tends to e = ([x1, y]P , [x2]P , . . . , [xk]P ) when y 2 [x1]P and [xi]P e = {xi}. By([x1], . . . , [xi, y], . . . , [xk]) we denote the set of all idempotents e with image{x1, . . . , xn} and such that the Ker(e)-class of xi contains (at least) two ele-ments: xi and y, where the underlined element (in this case xi) is the imageof the class under e. For a 2 Tn, we denote the image of a by ra.Lemma 1 Let a 2 In = Tn \ Sn, rank(a) = k, and (xy) 2 Sym(X).Then a(xy) = ab, where b = 1 or b = e1e2e3, with e2i = ei and rank(ei) = k.Proof. Let ra = {ai | i 2 [k]}. If x, y 62 ra, then a(xy) = a. If, say,x = a1 and y 62 ra, then ae1 = a(a1y), for all e1 2 ([a1, y], [a2], . . . , [ak]).Finally, if x, y 2 ra, without loss of generality, we can assume that x = a1and y = a2. Then a(a1 a2) = ae2e3e4, for e2 2 ([a1], [a2, u], [a3], . . . , [ak]),e3 2 ([u], [a1, a2], [a3], . . . , [ak]) and e4 2 ([u, a1], [a2], . . . , [ak]).1Theorem 2 [1] Every ideal of In is generated by its own idempotents.Proof. Let a 2 In. Then a = eg for some e = e2 2 Tn and g 2 Sn. Thereforea = e(x1y1) . . . (xmym) and hence, applying m times Lemma 1, a can beobtained as a product of idempotents of the same rank as a.Let t 2 In and g 2 Sn. Denote g 1tg by tg and let Ct = {tg | g 2 Sn} andtSn = h{t}[Sni\Sn. For f = ([x1, w]Q, [x2]Q, . . . , [xk]Q) and g 2 Sn, it is easyto check that we have f g = ([x1g, wg]Qg, . . . , [xkg]Qg) 2 ([x1g, wg], . . . , [xkg]).Lemma 3 Let a 2 In, rank(a) = k, let f = ([x1, w]Q, [x2]Q, . . . , [xk]Q)and let (xy) 2 Sn. Then a(xy) = ab, where b = 1 or b = e1e2e3, with ei 2 Cf .Proof. As in Lemma 1, one only has to show that the sets ([a1, y], [a2], . . . , [ak]),([a1], [a2, u], . . . , [ak]), ([u], [a1, a2], . . . , [ak]) and ([a1, u], [a2], . . . , [ak]) inter-sect Cf . Let g 2 Sn such that x1g = a1, wg = y and xig = ai (2  i  k).Thus f g = ([x1g, wg]Qg, [x2g]Qg, . . . , [xkg]Qg) 2 ([a1, y], [a2], . . . , [ak]). Theproof that Cf intersects the remaining three sets is similar.Repeating the arguments of Theorem 1 together with Lemma 3 we haveCorollary 4 Let eg 2 In (for some e = e2, g 2 Sn) and let f 2 = f withrank(eg) = rank(f). Then eg 2 ehCfi and, in particular, eg 2 hCei.From now on let a = eg 2 In (for some e = e2, g 2 Sn).Corollary 5 hCei = eSn = (eg)Sn = aSn.Proof. The only non-trivial inclusion is eSn ✓ hCei. As eSn is generated by{geh | g, h 2 Sn}, let g, h 2 Sn. By Corollary 4, he = (heh 1)h 2 hCheh 1iand eg 2 hCei. Since heh 1 2 hCei, it follows that hCheh 1i  hCei. Thusheg = (he)(eg) 2 hCei. It is proved that eSn ✓ hCei.Theorem 6 [2] hCai = hCei = aSn and hence is idempotent generated.Proof. Since a = eg, then e = ag 1 and it is easy to check that we have(ag 1)n = aagag2 . . . agn 1g n. But for some natural m we have g m = 1.Thus (ag 1)m = aagag2 . . . agm 1 2 hCai so that e = em = (ag 1)m 2 hCaiand hence, by Corollary 5, it follows that aSn = hCei  hCai  aSn . (Cf.[3]).2References[1] J. M. Howie, R. B. McFadden Idempotent rank in finite full transformation semi-groups, Proc. R. Soc. Edinburgh, Sect. A 114 (1990), 161–167.[2] I. Levi and R. McFadden Sn-normal semigroups, Proc. Edinburgh Math. Soc., 37(1994), 471-476.[3] D.B. McAlister Semigroups generated by a group and an idempotent, Comm. Algebra,26 (1998), 515-547.3

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On idempotent generated semigroups

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