Автоморфизмы полугрупп-степеней циклических групп простого порядка Текст научной статьи по специальности «Математика»

НАУЧНОЕ ИЗДАНИЕ МГТУ ИМ. Н. Э. БАУМАНА

НАУКА и ОБРАЗОВАНИЕ

Эл № ФС77 - 48211. Государственная регистрация №0421200025. ISSN 1994-0408

электронный научно-технический журнал

Автоморфизмы полугрупп-степеней циклических групп

простого порядка

# 11, ноябрь 2012

DOI: 10.7463/1112.0495704

Степанов Д. А.

УДК 512.53

Россия, МГТУ им. Н.Э. Баумана [email protected]

Полугруппой-степенью некоторой группы называется множество всех непустых подмножеств данной группы с естественной операцией умножения, индуцированной операцией группы. В статье описаны автоморфизмы полугруппы-степени циклической группы простого порядка. Показано, что для всех простых чисел за исключением 3 и 5 все автоморфизмы полугруппы-степени индуцированы автоморфизмами исходной группы.

Список литературы

1. Mazorchuk V.S. All automorphisms of FP+(Sn) are inner // Semigroup Forum. 2000. V. 60, no. 3. P. 486-490.

2. Артамонов В.А., Салий В.Н., Скорняков Л.А. и др. Общая алгебра. В 2 т. Т. 2 / под ред. Л.А. Скорнякова. М.: Наука, 1991. 480 с. (Справочная математическая библиотека).

SCIENTIFIC PERIODICAL OF THE BAUMAN MSTU

SCIENCE and EDUCATION

EL № FS77 - 48211. №0421200025. ISSN 1994-0408

electronic scientific and technical journal

Automorphisms of global semigroups of cyclic groups of prime order # 11, November 2012 DOI: 10.7463/1112.0495704 Stepanov D. A.

Russia, Bauman Moscow State Technical University

[email protected]

1. Introduction

We use terminology from [2] throughout. The global semigroup or the semigroup-power of a semigroup S is the set P + (S) of all nonempty subsets of S with a natural associative operation AB = {ab | a e A, b e B}. In the general case, the description of the group of automorphisms of P+(S) is not known. In the present work we solve this problem in a particular case when S is a cyclic group Cp of prime order p.

Let G be a group. An element A from P+(G) is called a k-element, if A as a subset of G consists of k elements. We distinguish two subgroups in the group Aut P+(G) of all automorphisms of the global semigroup:

a) the group of induced automorphisms Ind P + (G). Each automorphism ( of the group G extends naturally to an automorphism (p of the semigroup P + (G): ((A) = {((a) | a e A}. The group Ind P+(G) consists of all automorphisms of P + (G) that are induced in this way by the automorphisms of G;

b) the group of proper automorphisms Own P + (G). We say that an automorphism of P+(G) is proper, if it fixes all the 1-elements of P + (G).

Proposition 1. If G is a group, then

Aut P+(G) ~ Own P +(G) * Ind P +(G) (semidirect product with Own P + (G) normal).

< Let ( be an automorphism of P+(G). Its restriction onto 1-elements gives an automorphism of the group G. Consider the induced automorphism (|G of P+(G) and also the proper automorphism

^ = ( O ((G)-1. We have ( = ^ o thus AutP +(G) = OwnP+(G)inf P+(G). Also OwnP+(G) n IndP+(G) = {e}, OwnP +(G) < AutP+(G). ►

2. Auxiliary lemmas

Let e be the unit, and a a generator of the group Cp. In the sequel the words set, subset mean a subset of Cp, i. e. an element of P+(Cp). The class of the Green's relation (all Green's relations on P+(Cp) coincide, since P + (Cp) is commutative) that contains a subset (gi,... , gk} will be denoted [(g1,..., gk}]. Throughout this section we assume p ^ 5. Lemma 1.

(a) VA e P+(Cp), A = Cp, Vg, h e Cp, gA = hA ^ g = h;

(b) if 2 ^ |A| <p, 2 ^ |B| < p, then |AB| > |B|;

(c) B e [A] ^ B = gA, g e Cp;

(d) for A = Cp the class [A] has p elements: A, aA, ..., ap-1A; the class [Cp] has only one element Cp, which is zero of P+(Cp);

(e) there are (p — 1)/2 Green's classes of 2-elements: [{e, a}], [(e, a2}],..., [{e, a(p-1)/2}];

(f) VA e P +(Cp), A = Cp, Vg e Cp one has: g(Cp \ A) = Cp \ gA.

< Let us prove (c) and (e).

(c). B e [A] means that there exist D1, D2 e P + (Cp) such that A = D1B, B = D2A, or A = D1D2A. If A = Cp, then by (b) we have |D1D2| = 1, |D2| = 1, that is D2 = (g}, g e Cp. If A = Cp, the statement does not need a proof.

(e). In view of (d) it remains only to prove that the classes listed in (e) are different. Suppose that [{e, ak}] = [{e, a1}], k = l, 1 ^ k, l ^ (p — 1)/2. Then it follows from (c) that

(e, ak} = am{e, a1} = {am, am+1}

for some m, 0 ^ m ^ p — 1. But k = l, thus m = 0, am = e. Then it must be a1+m = e, l + m = p, m = k. But this implies l = p — k ^ (p + 1)/2 > (p — 1)/2, which is a contradiction. ►

Note that the degree of nilpotency of each 2-element equals p — 1. It follows from Lemma 1, 2, that for | A| ^ 3 the degree of nilpotency is not greater than p — 2; 1-elements are not nilpotent. Therefore any automorphism maps the 2-elements to 2-elements.

In this section we speak only about proper automorphisms of P +(Cp), so the word proper will usually be omitted.

Lemma 2. Let ^ be a proper automorphism of P +(Cp). If for any 2-element A it holds that <^(A) e [A], then for all 2-elements <^(A) = A. That is, if every class of 2-elements is invariant under the automorphism then ^ acts on 2-elements identically.

< Let <^({e, a}) = g{e, a}. Take arbitrary k, 2 ^ k ^ (p — 1)/2, and let <^({e, ak}) = h{e, ak}, g, h e Cp. We have:

^({e, a}p-2) = gp-2{e, a}p-2 = gp-2(Cp \ {a-1}), ^({e,afc}p-2) = hp-2(Cp \{a-k}).

But Cp \ {a-k} = a1-k(Cp \ {a-1}), thus

<^({e,afc }p-2) = hp-2{e,ak }p-2 = <^(a1-fc {e,a}p-2) = gp-2{e,ak }p-2.

It follows that gp-2 = hp-2, g = h. Thus if A is a 2-element, then <^(A) = gA, in particular

^({e,a2}) = g{e,a2}. Then

<^({e, a}3) = g3{e, a}3 = g3{e,a,a2,a3}, but {e, a, a2, a3} = {e, a}{e, a2}, that gives

^({e,a}3) = ^({e,a}{e,a2}) = g2{e,a}3.

This is possible only for g3 = g2, that is for g = e. Therefore <^(A) = A for every 2-element A. ►

Lemma 3. Let ^ be a proper automorphism of P + (Cp). If for some 2-element {g1,g2} ^({g1,g2}) e [{g1,g2}], then ^ acts on all 2 elements identically.

M Without loss of generality we may assume that the class [{e, a}] is invariant (we can take a different generator of Cp if necessary). So, let <^({e, a}) = g{e, a} for some g e Cp.

Let us take arbitrary k, 2 ^ k ^ (p — 1)/2, and show that the class [{e, ak}] is also invariant under Suppose that ^({e,afc}) = h{e,am}, h e Cp, 2 ^ m ^ (p — 1)/2. We have:

{e, a}k-1{e, ak} = {e, a, a2,..., a2k-1} = {e, a}2k-1 = Cp,

because 2k — 1 ^ p — 2. Applying the automorphism ^ to this relation we get

gk-1h{e, a}k-1{e, am} = g2k-1{e, a,..., a2k-1},

or

gkh-1{e, a,..., a2k-1} = {e, a,..., ak-1, am,..., am+k-1}.

This is possible only if m = k or if m = p — k. We agreed to choose m ^ (p — 1)/2, k ^ (p — 1)/2, hence m = k, <^({e, ak}) = h{e, ak}, i. e., ^ does not move the class [{e, ak}]. Now our lemma follows from Lemma 2. ►

We can conclude that an automorphism is either identity on 2-elements, or moves all the 2-elements.

Lemma 4. Each proper automorphism of P+(Cp) acts identically on (p — 1)-elements.

M If an automorphism acts identically on 2-elements, it acts also identically on p — 1 elements, since {e, a}p-2 = {e, a,..., ap-2} = Cp \ {a-1}.

If an automorphism ^ is not identity on 2-elements, consider one of the cycles of the permutation induced by ^ on the classes of 2-elements:

^({e,afc1}) = g1{e,ak2}, ..., ^({e,afcn}) = g„{e,afcl},

g G Cp, 1 ^ i ^ n. p — 1-elements constitute one class of the Green's relation on P + (Cp); every 2-element in degree p — 2 lies in this class, in particular

{e, aki}p-2 = Cp \ {a-ki},

{e, akn }p-2 = Cp \{a-kn }.

Thus (Lemma 1, 5) the following relations hold:

a'

kl-k2 {e,akl }p-2 = {e,ak2}

k2 \ p-2

kn — 1 kn

{e, akn-1}p-2 = {e, akn}p-2.

Applying to these relations the automorphism we get

gp-V1-k2 (e,ak2 }p-2 = gp-2(e,ak3 }p-2 = gp-V2-ks (e,ak2 }p-2,

gn^2iakn-1-fcn (e,akn }p-2 = gr2{e,akl }p-2 = g^-2«"1 (e,«kn }p-2,

from where we get

gp-2akl-k2 = gp-2ak2-k3,

p-2 kn_i—kn _ rP-2nkn-ki

gn— la = g« a .

On the other hand, we have

({e, akl}) = <^-1(gi{e, ak2}) = gi... g„{e, akl},

that is the automorphism leaves invariant the class [{e, akl}]. But then Lemma 3 implies that fixes all the 2-elements, that is g1... gn = e. Hence the elements gi, ..., gn of Cp satisfy the system

gp-Vl-k2 = g2k2-k3 ak2-fc3 ,

Let us express gp

,p-2

gp-2 akn-i-kn — yp-2akn-ki

gn— 1a = gn a ,

g« 2 through gp 2:

glg2 . . . gn = e. ,p-2.

gp-2 = aki-2k2+k3 yp-2,

gp-2 = aki-k2-k3+k4 gp-2

gP-2 = aki-k2-kn-i+kn gP-2 gn- 1 = a g1 ,

g

,p-2 = a2ki-k2-kn gp-2gp-2

Exponentiate the last equation of the system to the degree p — 2 and substitute for gf 2, ..., g their expressions through gf-2. We obtain gp(p-2)ara(kl-k2) = e. Since n < p, gp-2akl-k2 = e, gf-2 = ak2-kl. It follows that

7p-2 n

^({e,ak}p-2 = ak2-kl (Cf \ {a-k2}) = Cf \ {a-kl} = {e,akl}p-2.

This means that the set {e,akl} is fixed by In the same way one proves that all the other 2-elements are fixed by ►

Lemma 5. Assume that p ^ 7. Then each proper automorphism of P + (Cp) leaves invariant all the 2-elements.

< Let <^({e, a}) = g{e, ak}, g G Cp, 2 ^ k ^ (p - 2)/2. From the relation

{e, a}{e, a, a3} = {e, a, a2, a3, a4} = {e, a}4

we get

g4{e,ak}4 = g{e,akM{e, a, a3}). (1)

First note that |^({e, a, a3})| = 3. Indeed, since |{e, ak}4| = 5, it must be 3 ^ |^({e, a, a3}) ^ 4. But from |^({e, a, a3})| = 4 and (1) it follows that

|^({e, a, a3}) n ak^({e,a,a3})| = 3,

which is possible only if ^({e,a,a3}) G [{e,ak}3]. But this class is the image of the class

[{e, a}3] = [{e, a, a3}]. Thus, if we denote g-3^({e, a, a3}) = {ax, ay, az}, where 0 ^ x, y, z ^ p — 1, then (1) takes the form

{ax, ay, az}{e, ak} = {e, ak, a2k, a3k, a4k},

or

{ax, ay, az, ax+k, ay+k, az+k} = {e, ak, a2k, a3k, a4k}. Since the set on the right has 5 elements, one of the following congruences must hold:

x = (y + k), x = (z + k), y = (x + k)(mod p), y = (z + k), z = (x + k), z = (y + k)(mod p).

Let us suppose that z = (x + k)(modp) holds. Then we have

{ax, ay, ax+k, ay+k, ax+2k} = {e, ak, a2k, a3k, a4k}.

Considering 5 cases 1) ax = e, x = 0, 2) ay = e, y = 0, 3) ax+fc = e, x = — k(mod p), 4) ay+fc = e, y + k = — k(mod p), 5) ax+2fc = e, x = —2k(modp), one checks that only

1) [{e, a, a3}] [{e, ak, a3k}], or

2) [{e, a, a3}] [{e,a2k, a3k}]

are possible. The argument below depends on the residue of p modulo 3, so we have to consider 4 cases.

But for the beginning let us prove 2 relations. Let <^({e, a, a3}) = g{e, ak, a3k}, <^({e, a}) = h{e, ak}, g, h G Cp. First, applying to the relation

{e, a}{e, a, a3} = {e,a}4 = Cp

the automorphism we get

g = h3. (2)

Also, {e, a}p-2 = Cp \ {a-1}, thus Cp \ {a-1} hp-2 (Cp \ {a-1}). But by Lemma 4 Cp \ {a-1} Cp \ {a-1}, therefore hp-2a-k = a-1,

hp-2 = ak-1. (3)

Relations (2) and (3) hold also when <^({e, a, a3}) = g{e, a2k, a3k} and can be proved in the same way.

I. Let p = 3/ + 1, {e, a, a3}1 = {e, a,..., a31-2, a31} = Cp \ {a31-1}. Consider the case 1): {e, a, a3} g{e,ak, a3k}, ^({e,a}) = h{e,ak}, g, h G Cp, 1 ^ k ^ (p - 1)/2.

We have (2)

Cp \ {ap-2} = Cp \ {a31-1} = {e, a, a3}1 ^ g1 (Cp \ {a31-1}) = hp-1(Cp \ {a(p-2)k}). Lemma 4 implies that

hp-1a(p-2)fc = aP-2, h-1 = a(P-2)(fc-1), h = a(P-2)(fc-1).

Then, by (3), (p - 2)2(k - 1) = (k - 1)(modp). If 2 ^ k ^ (p - 1)/2, k - 1 iscoprimetop, thus (p - 2)2 = 1(modp), but this may hold only for p = 3, whereas we have p ^ 7. Hence k = 1, h = e, <^({e, a}) = {e, a}, and by Lemma 3 all the 2-elements are invariant. Consider the case 2):

{e, a, a3} g{e, a2k, a3k}, {e, a} h{e, ak}.

We have Cp \ {ap-2} = {e,a,a3} h31 {e,a2k, a3k} = hp-1(Cp \ {ak}), and by Lemma 4, h-1ak = ap-2, h = ak+2. Now by relation (3) a(p-2)(fc+2) = ak-1, a-2k-4+1-k = e, -3(k + 1) = 0(modp). Under our restrictions on k the last congruence does not hold, thus variant 2) is impossible.

II. p =3/ + 2. Then {e, a, a3}1 = Cp \ {ap-3, ap-1}, {e, a2}{e, a, a3}1 = Cp \ {ap-1}.

If [{e, a}] [{e, ak}], then [{e, a}] [{e, ak}] (this follows from the relation {e, a}{e, a2} = {e, a}3). If {e, a2} f {e,a2k}, then {e,a2}p-2 = Cp \ {a-2}^ fp-2(Cp \ {a-2k}), and by Lemma 4, fp-2a-2k = a-2,

fp-2 = a2(k-1). (4)

Consider case 1): {e,a,a3}^ h3{e,ak,a3k}. Then

Cp \ {a-1} = {e, a2}{e, a, a3}1 ^ f {e, a2k}h31(Cp \ {a(p-3)k, a(p-1)k}) = f hp-2(Cp \ {a-k}).

By Lemma 4, fhp-2 = ak-1, then by (3) f = e, and from (4) we get k = 1. It follows from Lemma 3 all the 2-elements are invariant.

Now consider case 2): {e, a, a3} h3{e, a2k, a3k}. We have:

{e, a2k, a3k} = Cp \ {ak, a-k}, {e, a2k}{e, a2k, a3k} = Cp \ {ak}.

Thus Cp \ {a-1} = {e, a2}{e, a, a3}1 fhp-2(Cp \ {ak}). By Lemma 4 and (3), fafc-1afc = a-1, / = a-2k. Then from (2) we get

(—2k(p - 2)) = 2(k - 1)(modp),

or k +1 = 0(modp), which does not hold under our restrictions on k. Hence variant 2) is impossible. In both possible cases the 2-elements are invariant. ►

Corollary 1. Elements of the subsemigroup S generated by 1- and 2-elements are invariant under each proper automorphism of the semigroup P + (Cp).

3. The main result

Theorem 1. a. Aut P+(C2) ^ {e}; Aut P + (C3) ^ C3 * C2; Aut P+(C5) ^ C2 x C4. In the second case the only nontrivial element of C2 acts nontrivially on C3; in the third the product is direct.

b. If p ^ 7, then all the automorphisms of the semigroup P+(Cp) are induced by the automorphisms of the group Cp, i.e.,

Aut P +(Cp) ~ Aut(Cp) ~ Cp-1.

M a is proved via a direct calculation. For the proof of b we use the technique of [1]. Suppose that there exist a proper automorphism ( of P + (Cp) and elements A, B e P+(Cp), A = B, such that ((A) = B. In the case |A| ^ p — 1 or |A| = 1 we immediately get a contradiction with the invariance of the elements of S, so assume that 1 < |A| < p — 1. Also assume that B £ A. Then there is a e B, a e A, and also there is b e Cp, b e A, b = a. Take D1 = {e,ab-1} and consider D1A, D1B. We have a e D1A, a e D1B, that is D1A = D1B, D1B £ D1A, |A| < |D1A| ^ p — 1. Continue this process until |D1... DnA| = p — 1, i. e. D1... DnA e S. Each Di is a 2-element, thus D = D1... Dn e S. Moreover, a e DA, but a e DB, thus DA = DB. But then ((DA) = D((A) = DB, which contradicts invariance of the elements of S. This contradiction proves the theorem. ►

The present work was done in 1999, when the author was a student at Kiev Shevchenko University, Ukraine. The author thanks to his advisor of that time Alexander Grigor'evich Ganushkin. The importance of his help appears even more evident with the years that have passed.

The work was partially supported by the Russian Grant for Leading Scientific Schools, grant no. 5139.2012.1, and RFBR, grant no. 11-01-00336-a.

References

1. Mazorchuk V. S. All automorphisms of FP +(Sn) are inner// Semigroup Forum. 2000. Vol. 60, no. 3. P. 486{490.

2. Artamonov V.A., Salii V.N., Skorniakov L.A., et al. Obshchaia algebra. V 2 t. T. 2 [General algebra. In 2 vols. Vol. 2]. Moscow, Nauka, 1991. 480 p. (Spravochnaia matematicheskaia biblioteka [Reference mathematical library]).