On normal closures of involutions in the group of limited permutations Текст научной статьи по специальности «Математика»

УДК 512.54

On Normal Closures of Involutions in the Group of Limited Permutations

Yuri S. Tarasov*

Institute of Mathematics and Computer Science Siberian Federal University Svobodny, 79, Krasnoyarsk, 660041

Russia

Received 10.03.2016, received in revised form 08.05.2016, accepted 20.06.2016 We study the group G = Lim(N) of limited permutations of a set N of all natural numbers. Found the link between the dispersion subsets of a set N and normal subgroups of G.

Keywords: group, limited permutation, dispersion set, normal subgroup, involution. DOI: 10.17516/1997-1397-2016-9-3-393-400.

Introduction

Let N be the set of all natural numbers, Z be the set of all integers, M is any of these sets. The S(M) will denote the group of all permutations of the set M.

Definition 1. Permutation g € S(M) is called limited if

wiq) = max \a — a9\ < oo.

If g, h are limited permutations, so the same is about the permutations g-1 and gh, as

w(g-1) = w(g), w(gh) < w(g) + w(h). Thus set

Lim(M) = {x | x € S(M),w(x) < to}

form a group, which is a natural extension of a locally finite group Fin(M) of all finitary permutations of the set M, i.e. such permutation y € S(M), for which the set {a | a € M,ay = a} is finite.

In the work of N. M. Suchkov [1] an example of the mixed group H = AB was first built, where A, B is periodic (and even locally-finite) subgroups. Then in [2,3] it was found that

H = {g Ig € Lim(Z), g < to),

any countable free group and Aleshin 2-group isomorphically embeddable into the group H and

Lim(Z) = H X {d),

where d-shift, ad = a + 1 for any a € Z.

* gigtorus@yandex.ru © Siberian Federal University. All rights reserved

In their work [4] N. M. Suchkov and N. G. Suchkova proved the factorization of the whole group Lim(N) by two locally-finite subgroups and it is shown that the group Lim(M) is generated by the permutations x G S(M), for which w(x) = 1.

These generators are either involutions, in which decomposition into independent cycles only transpositions of the (a a + 1), a G M or M = Z and x G {d, d-1}. The relation between groups Lim(N) and Lim(Z) is found in [5]. Assuming that permutations of the group S(N) have identical influence on the set Z \ N, we get a natural embedding of S(N) < S(Z). Denote by t the involution of the groups S(Z) for which a* = -a(a G Z). It is proved that

H = Fin(Z)(Lim(N) x (Lim(N))*).

From this congruence it follows that in the study of normal structure of the group Lim(Z) is the defining description of normal subgroups of the group Lim(N).

The first result in this direction is obtained in [5]. To formulate it is necessary to provide some definitions being introduced in this work. Let

L = {Vi, V2, ..., Vn, ■■■ }

is an infinite subset of N, where v1 < v2 < • • • < vn < ...; m is a fixed natural number. By definition, elements of vi and Vj are equivalent, if i = j, or when i < j (j < i) all inequations are fulfilled vk+1 — Vk ^ m; i < k < j — 1(j < k < i — 1). It is easy to see that this relation is indeed an equivalence relation, therefore it induces a decomposition of the set L into equivalence classes. This partition is called m-partition. Let Bm(L)be the set of all equivalence classes of elements of the set L.

Definition 2. The set L is called m-dispersion, if all classes of the set Bm(L) are finite and are completely m-dispersion if

cm = max < to.

AeBm(L)

The set L is called (completely) dispersion, if it is (completely) m-dispersion for every natural m.

The example of completely dispersion set is the set L, for elements of which the following inequations are used

V2 — Vi < V3 — V2 < • • • < Vn — Vn-1 < Vn+1 — Vn < ... .

Let Vn + 1 < Vn+1(n = 1, 2,...) and

a = (vI Vi + 1)(V2 V2 + 1)... (Vn Vn + 1)...

is the decomposition of an involution a into independent transpositions. The main result of [5] is the theorem according to which the normal closure of the involution of a in the group Lim(N) if and only if locally finite, when L is an completely dispersion set.

Three hypotheses about normal subgroups of the group Lim(N) are provided ibid. In this article one of these hipotheses is proved; namely, the following theorem is the main result of the present paper.

Theorem 1. An involution a if and only if contained in a proper normal subgroup of the group Lim(N) when L is a dispersion set. If L is dispersion, but is not completely dispersion set, then (a9 lg G Lim(N)) is a mixed group.

All the designations used in this work are either discussed, or standard [6].

1. Preliminary results

Let j, e be integers and j < e. Let us call the set

Uj = {3\3 G Z,j < 3 < e}

a segment of integers; j is the left end of the segment, e is the right. In particular, UY = {7}. For each m G N, a G L let be

vm = ua+m n n, Em = u vm.

aeL

Lemma 1. If the set L is dispersion, then the set Em is 1-dispersion for every natural m.

Proof. If the Lemma is wrong, then such integers m and 7 will be found that will prove that 1-decomposition of the Em set contains an infinite class UY = {3 \ 3 G N, 3 > 7}. Let h > 7, then the union V^U Km+i includes a segment of integers with h, №+1 endpoints. Therefore, 2m is a decomposition of the set L contains an infinite equivalence class with representative Hi We have come to a contradiction with the dispersion of the set L. The Lemma is proved. □ For brevity let us define G = Lim(N) and for each of dispersion set of L we define the subgroup Q = Q(L). As to Lemma 1 each set Em is split into segments

Wmi, Wm2, ..., Wmn, . . .

of integers and if 3mn is the right end of the segment Wmn, and amn+1 is the left end of the segment Wmn+1, then amn+1 > 3mn + 1(n = 1, 2,...); each segment is Wmn is included into some interval Wm+1 s. Let

Qm = {x \ x G G; WXn = Wmn (n =1, 2,...); 3x = 3(3 G N \ Em)}.

Obviously, Qm is a subgroup of G and Qm < Qm+1, m = 1, 2,... . Let, finally,

Q = Q(L) = (J Qm

m£N

Lemma 2. Q is proper normal subgroupin the group G.

Proof. Let 1 = h G Q; g G G and w(g) = k. From the definition of the group Q, it follows that there are such natural m that the element h contains the subgroup Qm. We claim that hg G Qt, where t = m + k +1. Indeed, consider the decomposition of permutation h into independent cycles. Since h leaves untouched the segments of Wmn (n = 1, 2,...) and acts identically on the integers which are not contained in these segments, then all the cycles are finite. If x = (j1 ...js) is one of these cycles is (s > 1), then j1, ..., js are contained in some interval Wmn, which coincides with the union of several segments

Vm ym V m

Mq , Mq + 1 , . . . , Me .

We fix any number of ji of set {j1, ..., ys}. Then ji G V^ for some index j, q < j < e; and since yX = if and \ji — jf \ < w(g) = k, then jf G Vt.. Next, segment Wmn is part of the

segment

VM IV4 II I l\

Me

vM V I L.I V .

Mq^J Mq + 1 ^ \J Me

In turn, this segment is contained in some segment Wtd, d G N. Therefore, all elements of the cycle x9 = (j9 ... y9) belong to Wtd, where due to the definition of the group Qt we deduce that x9 G Qt and h9 G Qt. Thus, Q is a normal subgroup of G. It remains to show that Q is proper subgroup of G. In fact, in the course the corroboration we have seen that any permutation of the subgroup Q is decomposed into finite independent cycles. Therefore an infinite cycle

y =(... 2n ... 4213 ... 2n — 1 ...)

is not contained in Q, but since w(y) = 2, then y G G. Thus, Q = G. The Lemma is proved. □

Lemma 3. If z is involution and f is a triple of the cycle of the alternating group A4, zzf zf = 1.

Proof. We have A4 = ((z) x (u)) X (f), where |u| = 2. Thus f transitively permutes the involution z, u, zu whose product equals 1. Therefore, the Lemma is correct. □

In further calculations we will use the well-known and easily verifiable Assertion 1. Let g, h are permutation of some set. If

g = ... (... ai a2 ...) ...

is decomposition of g into independent cycles, then

gh = h-1gh = ... (...ah ah ...) ....

Lemma 4. Let y = (e1 e2)(e3 e4)(e5 e6) is decomposition of the permutation y into independent transpositions, f = (e3 e4 e5). Then yyfyf = (e1 e2).

Proof. The elements z = (e3 e4)(e5 e6), f generate a group isomorphic to alternating grope A4, all elements of which commute with the transposition (e1 e2). Therefore, in view of Lemma 3 we have

y = (ei e2)z, yf = (ei e2)zf, yf = (ei e2)zf ,

yyf yf = (ei e2)3zzf zf = (ei e2).

The Lemma is proved. □

Lemma 5. Let

C = (№i + 1)(fol32 + 1) ... (PnPn + 1) ...

is the decomposition of permutation c of the group G into independent transpositions If all of the inequalities are fulfilled

6 < lln+i — In < m (n =1, 2,...),

where m is some fixed natural number, then the normal closure of B(c) = (c9 lg G G) the involution of c in the group G contains a group Fin(N) of all finitary permutations of the set N.

Proof. Since the group Fin(N) coincides with the normal closure of any of its transposition, in order to prove the Lemma it is enough to show that B(c) contains a transposition (li li + 1) from the decomposition of permutation c. In fact, since ln+i — ln > 6 for all natural n, the transpositions of the permutation's decomposition

l = (li li + 2)(li + 1 li + 3)(l2 I2 + 2)(l2 + 112 +33)... (In In + 2)(ln + 1 In + 3) ...

is independent, and in the process w(l) = 2, in particular l e G. Therefore, the group B(c) contains an involution ce. Using the Assertion 1 we get

ci = ce = (3i + 2 3i + 3)3 + 2 /32 + 3)... (¡3n + 22 /3n + 3)....

Now let

S = (3i +2 32)(3i + 3 32 + 1)3 + 2 3s)(32 + 3 33 + 1)... (3n + 2 3n+l)(3n + 3 3n+1 + 1)....

By condition of the Lemma 3n+1 — 3n ^ m (n = 1, 2,...), therefore, w(s) < m. Thus, c\ e B(c) and we get

ci = (32 32 + 1)(33 33 + 1) ... (3n+1 3n+1 + 1) ... . Thus, cci = (3i 3i + 1) is a contained in B(c). The Lemma is proved. □

Assertion 2( [5], Lemma 7). Let {a1 ,...,ak} is a subset of set N and a + 1 < ai+i, 1 < i < k — 1. If

b = (ai ai + 1)(a.2 a.2 + 1) ... (a.k-i ak-i + 1)(a.k ak + 1) is the decomposition involutive permutation b e Fin(N) into independent of transpositions,

u = (ai a2 ... ak-i ak ak + 1 ak-i + 1 . .. a2 + 1 ai + 1) is cycle, then the permutation bbu has the order k.

2. Proof of the Theorem 1

Let us proceed to the direct corroboration of the theorem formulated at the end of the introduction. Initially, it will prove the first part.

Suppose that L is a dispersion set. Then from the constrution in Section 1 of the group

Q = Q(L) = U Qn

n

and Lemma 2 it follows that the involution a belong to the subgroup of Qi, which is contained in proper normal subgroup Q of G.

Conversely, suppose that a is contained in proper normal in G = Lim(N) subgroup. Obviously, is equivalent to, subgroup B(a) = (a9 | g e G) is a proper subgroup of G. Suppose that the set L is not dispersion. This means that there is such a natural integer m0 that the set Bm0 (L) contains an infinite class of A. Then if ¡iY is the minimal number of the set A, then from definition it follows that — in < m0 for all i > 7. Hence we deduce that Bm(L) consists of a single class {L}, if m> ). Fix mi > m. Thus,

in+i — in < mi (n =1, 2,...) .

Now let us prove that B(a) = G. Then we get a contradiction to our assumption B(a) = G and the first part of the theorem will be proved. Firstly it should be noted, that in the group B(a) we can find a permutation

c =(3i 3i + 1)(32 32 + 1) ... (3n 3n + 1) ... ,

that for some natural m all the inequations are fulfilled

6 < 3n+1 — In < m (n =1, 2,...). (1)

Indeed, we will split the transpositions from decomposition a into triples:

a = (^1 H1 + 1)(H2 H2 + 1)(H3 H3 + 1),... (H3k+1 H3k+1 + 1)(^3k+2 H3k+2 + 1) (H3k+3 H3k+3 + 1)... . Let

t =(H2 H2 + 1 H3) ... (,H3k+2 H3k+2 + 1 H3k+3) ....

Since Hn+1 — Hn ^ m1, therefore w(t) ^ m1, and it means that, t G G. Therefore, if c = a a4 a*2, then c G B(a) and by Lemma 4

c = (P1 H1 + 1)... (H3k+1 H3k+1 + 1)... .

Thus from the inequation 2 < Hn+1 — Hn ^ m1 it easily implies that 6 < H3k+4 — H3k+1 ^ 3m1 = m. Assuming 31 = ¡i1, 32 = , ..., 3n = H3n-2,..., we get that the permutation c is the sought fore.

Let us note that the inclusion c G B(a) immediately implies that B(c) ^ B(a), and therefore for the proof of the part 1 of the Theorem it is enough to establish the congruence B(c) = G. It was noted in the introduction that the group G is generated by involutions, in decomposition into independent transpositions of which only transpositions of the (a a +1) form take part. Since Fin(N) < B(c) by Lemma 5, to prove the congruence B(c) = G it is enough to show that if

x = (j1 J1 + 1)... (jn in + 1)...,

where jn+1 > jn + 1 (n = 1,2,...), then x G B(c). Since B(c) contains any finitary permutation, xn = (j1 j1 + 1)... (jn jn + 1), then without loss of generality we can assume that j1 > 31.

Denote Lx = {jn\ n G N} and consider the case when the inequations are fulfilled for the elements of this set

jn+1 — in > 5m (n =1, 2,...). (2)

Let us split the set of N \ {1,2,... ,31} into the segments of the integers is = U£+V ^ An = An+1 = ....

In virtue of the inequation (1)

\An\ = 32n+1 — 32n-1 = (32n+1 — 32n) + (32n — 32n-1) < 2m . From inequations (2) and j1 > 31 this implies that

Lx C y An,

neN

the intersection of An n Lx for every n is either empty or contains max one element; ji, jj is not contained in the adjacent segments for every i = j. Thus, there is such a sequence j1,j2, ...,jn,..., that jn+1 — jn > 1(n =1,2,...) and

j1 G Aji, j2 G Aj2, ...,jn G j, ....

Let us define the permutation 0 e S(N) as follows: for n = 1,2,... assume

it = 3j, j = Yn, (Yn + 1)t = 32jn + 1, (32jn + 1)t = Yn + 1; Yt = Y, if Y e U ({Yn, Yn + 1} U {32jn, 32+ 1}) .

neN

Since the elements Yn, 32jn belong to the segment An and |An| < 2m, then w(0) < 2m, i.e.

0 e G. As to Assertion 1 we have

xt = (32j! 32j! + 1) ... (32jn 3j + 1) ... .

Now let

ai, a2, . . . , an, . . .

be elements of the set {3i,..., 3n,... }\ {32ji,..., 32jn,... }, arranged in ascending order. From the above it follows that if ai = 3k, then ai+i is element of the set {3k+i, 3k+2}, and therefore ai+i — a.i < 3k+2 — 3k ^ 2m. Here i is any natural number, k = k(i). It's easy to deduce that the permutation

f = (ai a2 a2 + 1) ... (a2n-i a2n a2n + 1) ...

is an element of the group G. Applying Lemmas 3, 4 we get the congruence ccf cf = xt from which it immediately follows that x e B(c).

Let us finally prove, that this inclusion is done in general case (without additional assumptions that for the elements of a set Lx inequations are fulfilled (2)). To do this, we fix any natural number s > 5m, and represent the permutation x as compositions of

where

Xi = (Yi Yi + 1)(Yi+s Yi+s + 1)... (Yi+ks Yi+ks + 1)... ,

1 ^ i ^ s. From the definition of the permutation of x it implies that if Lxi = {Yi+kslk = 1,2,... }, then the adjacent elements of this set an inequation is fulfilled

Yi+(k+i)s — Yi+ks > S> 5m,

which coincides with the inequation (2) for the neighbouring elements of the set Lx. But then by proved above, xi e B(c), 1 < i < s, and therefore x e B(c). The first part of the theorem is proved.

Let us prove the second part. Let L be a dispersion, but not comletely dispersion set. We need to show that the normal closure of B(a) of an involution a of a group G contains an element of infinite order. Indeed, in view of the definition for some natural number r there are pairwise disjoint sets

Ln = {ian , lan+1, . . . VPn }

n = 1,2,... of L that lLnl > n and ii+i — ii < r (an < i < 3n — 1). Let us define the permutation of the u set N by its decomposition into independent cycles un (n = 1, 2, . . . ). Let

un = (ian ian+i ... Wn Vi3n + 1 Vi3n +1 Wn-i + 1 ... ian+i + 1 ian + ^ .

Then w(u) ^ r, i.e. u e G. According to Assertion 2 the element aau e B(a) is decomposed into independent cycles which lengths is unbounded, and therefore laaul = to. The theorem is proved.

In conclusion, let us put an example of be a dispersion, but not comletely dispersion set. Let

L1 = {2, 3}, L2 = {4, 5, 6}, L3 = {8, 9,10,11}, ..., Ln = {2n, 2n + 1,..., 2n + n};

l=L1 u l2 u l3 U ••• u Ln u....

If 2n is the representative of the class A = A(n, m) G Bm (L), then from definition it follows that A contains a set Ln of the (n + 1)-th element, and if m < 2n — (2n-1 + n — 1), then A = Ln. Hence we conclude that the set L is a dispersion, but not comletely dispersion set.

References

[1] N.M.Suchkov, An example of a mixed group factorized by two periodic subgroups, Algebra i Logika, 23(1984), no. 5, 385-387 (in Russian).

[2] N.M.Suchkov, On subgroups of the product of locally finite groups, Algebra i logika, 24(1985), no. 4, 408-413 (in Russian).

[3] N.M.Suchkov, On the group of limited permutations, Konstruktsii v algebre i logike, Tver', 1990, 84-89 (in Russian).

[4] N.M.Suchkov, N.G.Suchkova, On groups of limited permutations, Journal of Siberian Federal University. Mathematics and physics, 3(2010), no. 2, 262-266 (in Russian).

[5] N.M.Suchkov, N.G.Suchkova, Normal subgroups of limited permuta tion groups, Siberskie electronnnye matematich. izvestiya, 12(2015), 344-353 (in Russian).

[6] M.I.Kargapolov, Y.I.Merzlyakov, Fundamentals of the theory of groups, New York, SpringerVerlag, 1979.

О нормальных замыканиях инволюций в группе ограниченных подстановок

Юрий С. Тарасов

Институт математики и фундаментальной информатики Сибирский федеральный университет Свободный, 79, Красноярск, 660041

Россия

Изучается группа G = Lim(N) ограниченных подстановок множества N всех натуральных чисел. Найдена связь между рассеянными подмножествами множества N и собственными нормальными подгруппами группы G.

Ключевые слова: группа, ограниченные перестановки, рассеивание, нормальная подгруппа, инволюции.