Shunkov groups with the minimal condition for noncomplemented abelian subgroups

Chernikov Nikolai S.

УДК 519.41/47

Shunkov Groups with the Minimal Condition for Noncomplemented Abelian Subgroups

Nikolai S. Chernikov*

Institute of Mathematics National Academy of Sciences of Ukraine Tereschenkivska, 3, Kyiv-4, 01601

Ukraine

Received 10.09.2015, received in revised form 21.09.2015, accepted 02.11.2015 In the present paper, we give a complete exhaustive description of the pointed out Shunkov groups.

Keywords: Shunkov, periodic, locally finite, completely factorizable, Chernikov group, minimal conditions,

complemented, abelian subgroups.

DOI: 10.17516/1997-1397-2015-8-4-377-384

Introduction

A great many deep and bright results are connected with groups, satisfying various minimal conditions, and with groups, having wide systems of complemented subgroups (see, for instance, [1-7]).

The present paper is devoted to the Shunkov groups with the minimal condition above.

Below p and q are always primes; min — ab, min — abc, min — p and min — p' are the minimal conditions respectively for abelian, abelian noncomplemented, for p- and p'-subgroups. All other notations are standard.

Remind that the group G is called Shunkov, if for any its finite subgroup K, every subgroup of the factor group Nq(K)/K, generated by two conjugate elements of prime order, is finite (V. D. Mazurov, 1998). The class of Shunkov groups is wide and includes, for instance, binary finite groups, 2-groups. The known Suchkova-Shunkov Theorem [8] (see also [4, Theorem 4.5.1]) asserts: The Shunkov group with min — ab is Chernikov.

Further, remind that the subgroup H of the group G is called complemented in G, if for some subgroup K of G, G = HK and H n K = 1; K is called a complement of H in G. The group G is called completely factorizable, if every its subgroup is complemented in it (N. V. Chernikova [9]). The fundamental N. V. Chernikova's Theorem [9,10] (see also, for instance, [1, Theorem 7.2]) gives an exhaustive description of completely factorizable groups and asserts: The group G =1 is completely factorizable iff G = AX B where A is a direct product of normal subgroups of prime orders of G and B is a direct product of subgroups of prime orders or B = 1; in particular, the p-group G is completely factorizable iff it is elementary abelian. The known Kargapolov [11]-Gorchakov [12] Theorem asserts: The group is completely factorizable iff all its abelian subgroups are complemented.

It is natural to consider groups, having infinite abelian subgroups, in which all such subgroups are complemented. B. I. Mishchenko [13] has described the infinite solvable and the infinite radical in the sense of B.I.Plotkin groups with complemented infinite abelian subgroups (see Theorem 1 [13] and Corollary [13, p. 158]). Since all such groups are locally finite, it is natural to consider the locally finite groups with min — abc. N. S. Chernikov [14,15] has described these locally finite groups (see Theorem [15] and Corollary 3.5 [15]). N. S. Chernikov [14,16] has established that binary finite groups with min — abc are locally finite (see Theorem 3 [16]).

1. The main result and some corollaries

The author succeeded in proving the following general theorem, which is the main result of the present paper.

Theorem. For the Shunkov group G the following statements are equivalent:

(i) G satisfies the minimal condition for abelian noncomplemented subgroups.

(ii) G is a Chernikov group or a non-Chernikov group with complemented infinite abelian sub-

groups.

(iii) G is a Chernikov group or G is a completely factorizable group, or G = A X B where A is infinite and A is a direct product of normal in G subgroups of prime orders, B = C x D is finite, C is a direct product of subgroups of prime orders or C = 1, D is cyclic = 1 and for every p G n(D), p2||D|, and also for every g G D \ {1}, C^(g) is finite.

(In view of O.Yu. Shmidt's Theorem (see, for instance, [20, Theorem 1.45]), in (iii) G is locally finite.)

Theorem is equivalent to the author's Theorem [17]. Theorem implies the following proposition.

Proposition ([17]). The Shunkov p-group G (in particular, the 2-group G) satisfies the minimal condition for abelian noncomplemented subgroups iff it is Chernikov or elementary abelian.

Note that Theorem [17] and Proposition [17] are exactly all results of [17].

The following new author's assertions are the immediate consequences of Proposition.

Corollary 1. For the 2-group G the following statements are equivalent:

(i) G satisfies the minimal condition for abelian noncomplemented subgroups.

(ii) G satisfies the minimal condition for noncomplemented subgroups.

(iii) G is Chernikov or elementary abelian.

Corollary 2. For the Shunkov p-group G the following statements are equivalent:

(i) G satisfies the minimal condition for abelian noncomplemented subgroups.

(ii) G satisfies the minimal condition for noncomplemented subgroups.

(iii) G is Chernikov or elementary abelian.

In connection with the results above, note that for every p > 665, there exists the non-solvable group of exponent p containing an infinite abelian subgroup, in which every abelian subgroup of order > p is complemented (N. S. Chernikov [18]). Thus the above requirements: "G is a 2-group", "G is Shunkov" are essential.

2. Proof of the main result

A. Show that (i) implies (iii).

Let (i) hold. The subsequent proof will be accomplished in a series of steps.

(1) G is periodic.

Proof. Let G have some element g of infinite order. Then some subgroup < g2 > of the infinite chain < g2 >D< g4 >D ... D< g2 >D< g2+1 >D ... has a complement D in G. But 1 < \Dn < g > \ < to, which is a contradiction. □

(2) If G has a normal infinite locally finite subgroup H, then the statement (iii) is valid.

Proof. First, let H be Chernikov. Now remind the following S. N. Chernikov's Proposition

(see, for instance, [1, Proposition 1.13, p. 62]): A periodic group of automorphisms of the group, which is a direct product of finitely many quasicyclic subgroups, is finite. Further, H contains the characteristic subgroups R of finite index, which is such product. Since G is periodic (see (1)), in view of the last Proposition, \G : CG(R)\ < to. In accordance with Lemma 1.1 [15], an abelian group with min — abc is precisely Chernikov or a direct product of groups of prime orders. Every maximal abelian subgroup of CG (R) satisfies min — abc and is not such product and so is Chernikov. Hence follows: G satisfies min — ab. Therefore in virtue of Suchkova-Shunkov Theorem [8] (see above), G is Chernikov and, at the same time, (iii) is valid.

Now let H be non-Chernikov. Remind the following N. S. Chernikov's Theorem (see [15, Theorem]): The locally finite group with min — abc is the same as in (iii). Consequently, with regard to N. V. Chernikova's Theorem (see, above), H = K X L, where K is a direct product of normal in H subgroups of prime orders, L is abelian without quasicyclic subgroups. Let F be the Fitting subgroup of H. Then F is locally nilpotent and F = K X (F n L) < G. Since H is solvable, in view of Proposition 5.4.4 (ii) [19, (see p. 144)] , CH(F) = Z(F). Therefore because of H is infinite, F is infinite too. Obviously, F is non-Chernikov. Further, every mentioned direct multiplier of K belongs to Z(F) (for instance, in view of Proposition 1.16 [1, (see p. 70)]). So F is abelian. In accordance with Lemma 1.9 [15], the group, satisfying min — abc and having a normal abelian non-Chernikov subgroup, is the same as in (iii). Thus (iii) is valid. □

(3) Either the statement (iii) is valid, or the product L of all normal locally finite subgroups of G is finite and also G includes some normal infinite subgroup M, which does not satisfy min — ab and has no subnormal locally finite subgroups = 1.

Proof. Assume that (iii) is not valid. Then G is infinite. In consequence of O.Yu. Shmidt's Theorem (see, for instance, [20, Theorem 1.45]), L is locally finite. By virtue of the assertion (2), L is finite. So \G : CG(L)\ < to. Again by virtue of (2), CG(L) is not locally finite. Therefore, with regard to Suchkova-Shunkov Theorem [8] (see above), CG(L) does not satisfy min — ab. So some maximal abelian subgroup A of CG(L) is not Artinian. Clearly, L n CG(L) C Z(CG(L)) and so L n Cg(L) C A. Further, A has some infinite descending series

A = Ao D Ai D A2 D ... D n^=1An D L n Cg(L) D 1.

Some An has a complement D in G. Put M =< (D n A)G >. In view of Chunikhin's Lemma (see, for instance, [21, Lemma 1.36]), M C D. Also M C Cg(L) and D n L n Cg(L) = 1. So M n L C (D n CG (L)) n L = 1. In consequence of Theorem 1.1 in §2 of Chapter 5 [22] (see [22, p. 345]), every subnormal locally finite subgroup of M belongs to L. Consequently, M has no subnormal locally nontrivial subgroups. Also with regard to Suchkova-Shunkov Theorem, M does not satisfy min — ab. □

(4) If G is a p-group, then (iii) is valid and, at the same time, G is Chernikov or elementary abelian.

Proof. Let G be a p-group. It is easy to see, with regard to N. V. Chernikova's Theorem above: G is Chernikov or elementary abelian iff (iii) is valid.

Assume that (iii) is not valid. Now define the finite subgroup H of G in the following way.

First, if G has an element g of order p2, then put H =< g >.

Suppose that G is of exponent p. Then G is non-abelian. If for some g, h e G, [g,gh] = 1, then we put H =< g,gh >. Since G is Shunkov, H is finite. Further, assume that also for every g e G and h e G, [g, gh] = 1. Take a, b e G such that [a, b] = 1. Since < ah : h e G > and <bh : h e G > are normal abelian subgroups of G, the subgroup < ah : h e G ><bh : h e G >is metabelian and non-abelian. Further, the known S. N. Chernikov's Theorem (see, for instance, [1, Proposition 1.1]) asserts: Periodic locally solvable groups are locally finite. Then < a,b > is finite non-abelian. Now put H =< a, b >.

Let A be any abelian subgroup of Cq(H). Then AH is a nilpotent non-(elementary abelian) group with min — abc. In accordance with Lemmas 2.2 [15] and 1.1 [15]: Every non-Chernikov locally nilpotent p-group with min — abc is elementary abelian. Thus, AH is Chernikov. So CG(H) satisfies min — ab. Now remind Shunkov's Theorem [23]: The 2-group with min — ab is Chernikov. Remind An. Ostilovskiy's Theorem [24] (see also [4, Theorem 4.4.1]): The Shunkov 2'-group with min — ab is Chernikov. In view of these theorems, CG(H) is Chernikov.

Let F be a subgroup of maximal order among all X < H, for which CG(X) is non-Chernikov. Take u e H \ F such that up e F and also uF e Z(H/F). Since F < H and also | F | > \F |, the CG(F) is Chernikov.

Put T = Cg(F ). If | \ = p, then up = 1 e Z (T). If | \ = p, then H and, at the same time, F are cyclic. Therefore in this case we have: up e F C Z(CG(F)). Consequently, [up,T] = [up, Cg(F )] = 1, i.e. up e Z (T).

In view of S. N. Chernikov's Lemma (see, for instance, [1, Lemma 3.7, p. 151]), CT(u) = (Ct (u) n Cg(F )). Then Ct (u) : Ct (u) n Cg(F )| < to. Since Ct (u) n Cg(F ) C Cg(F) and CG( F) is Chernikov (see above), the subgroup CT(u) n CG(F) is Chernikov too. Therefore CT(u) is also Chernikov.

Further, it is easy to see: the statement (iii) of Theorem with T in the character of G is not valid. Therefore in view of the assertion (3), T contains some normal subgroup M that does not satisfy min — ab and has no normal locally finite subgroups = 1.

Let K be a normal subgroup of T, having some abelian non-Chernikov subgroup B. In view of Lemma 1.2 [15], K contains some subgroup L<T with infinite B/L n B and non-Chernikov L n B. Taking this into account it is easy to see: M has some infinite descending series

M = Mo D Ml D M2 D ... D Ma D Ma+1 D ... D M7 = na<7Ma D 1

of normal subgroups of T such that all Ma, a < 7, do not satisfy min — ab and MY satisfies min — ab. In view of mentioned Shunkov's and An. Ostilovskiy's Theorems, MY is Chernikov. So MY = 1. Further, since CM (u) is Chernikov, for some ß such that 0 < ß < 7, we have: CMß(u) = 1. Take v e Mß \ {1}. Then, because of up e Z(T), we have: u, v e CG(up) and n < uv >=< up >. Since G is Shunkov, <,< uv >> is a finite p-group. So <, < uv >> nMß has some element = 1 centralizing u, which is a contradiction.

Thus (iii) is valid. □

(5) If for some element g e G of prime order and for some infinite normal subgroup H of G we have: H n CG(g) = 1, then (iii) is valid.

Proof. First, give the following Popov-Sozutov-Shunkov Theorem (see Lemmas 2.7, 5.24 [25], Theorem 5.11 [25], Lemma 5.20 [25]): Let X = UX <v> be an infinite group with \ < v > \ = p, CX(v) =< v > and \ < v,vu > \ < to, u G U. Then: X is periodic; all divisible abelian subgroups of U belongs to Z(U); every finite subgroup of U, normalized by v belongs to some infinite locally finite subgroup of U, normalized by v. Further, if for some u G U, all subgroups Un < u, fv > with f G U are abelian, then the normal closure < uX > of u in X is abelian.

Now give some comments. Since < v > is obviously a Sylow p-subgroup of < v,vu > and < v,vu >= (Un <v,vu >) < v >, for some w G Un < v,vu > we have: < v >u=< v >w, i.e. u = w and u G Un < v,vu >. Obviously, for some a G< uv > and x G Un < v,vu >, \ < a > \ = p and < a >=< v >x. So < uv >=< v >x. Thus, (X \ U) U {1} = UueU < vu >. Hence follows: for y, z G X \ U, \ <y, z > \ < to.

Now return directly to the present assertion (5). Since G is Shunkov, for any x, y G G, we have: \ < gx,gy > \ < to.

If H contains a quasicyclic subgroup, then in view of Popov-Sozutov-Shunkov Theorem above, Z(H) contains all such subgroups. Then every maximal abelian subgroup of H contains a quasicyclic subgroup. Consequently in view of Lemma 1.1 [15], all maximal abelian subgroups of H are Chernikov and so H satisfies min — ab. Therefore in view of Suchkova-Shunkov Theorem mentioned above, H is Chernikov. So in accordance with the assertion (2), the statement (iii) is valid.

Now let H have no quasicyclic subgroups. Take u, f G H. For some h G H, fg = gh (see comments above). Also Hn < gh,ghu > is a finite subgroup, normalized by gh, and u G Hn < gh,ghu > (see comments above). Then Hn < u,fg >C Hn < gh,ghu >. Further, in view of Popov-Sozutov-Shunkov Theorem above, Hn < gh, ghu > belongs to some infinite locally finite subgroup R of H, normalized by gh. By virtue of J.G.Thompson Theorem [26], R is locally nilpotent. Since R has no quasicyclic subgroups, R is also non-Chernikov. Therefore in view of Lemma 2.2 [15], R is abelian. At the same, Hn < u, fg > is abelian. Consequently, in view of Popov-Sozutov-Shunkov Theorem above, < uH<g> > is abelian. Thus H is the product of normal locally finite subgroups < uH<g> > , taking by all u G H. Then in consequence of O. Yu. Shmidt's Theorem, H is locally finite. Therefore (iii) is valid (see (2)). □

(6) If for g G G of prime order the centralizer CG(g) satisfies min — ab, then (iii) is valid.

Proof. Let CG(g) satisfy min — ab. In view of Suchkova-Shunkov Theorem, CG(g) is

Chernikov. Assume that (iii) is not valid. Let M be such as in (3). Then M has some descending series

M = M0 D M1 D M2 D ... D MY = na<YMa

such that MY < G and MY satisfies min — ab, and for a < y, Ma < G and Ma does not satisfy min — ab (see above the proof of the assertion (4)). In view of Suchkova-Shunkov Theorem [8], MY is Chernikov. Consequently MY = 1. Therefore because of CG(g) is Chernikov, for some ¡3 < Y we have: CG (g) n Mp = 1. But then, with regard to (5), (iii) is valid, which is a contradiction. □

Remind: the group with a normal abelian subgroup of finite index is called almost abelian.

(7) If for g G G of prime order the CG(g) is almost abelian, then (iii) is valid.

Proof. First, (iii) is valid, if CG(g) is Chernikov (see (6)). Let CG(g) be almost abelian non-Chernikov and A be its abelian subgroup of finite index. Since A is non-Chernikov, it is a direct product of groups of prime orders (see Lemma 1.1 [15]). Therefore, obviously, A has an infinite chain A1 D A2 D ... D An D An+1 D ... with factors of prime orders. Since G satisfies

min — abc, the set of all complemented in G terms of the chain is infinite. Let Dn complements some An in G. Then A = An x (A n Dn) (by S. N. Chernikov's Lemma). In view of Chunikhin's Lemma (see, for instance, [21, Lemma 1.36]), < (A n Dn)G >C Dn. Since also Dn n CG(g) is finite, < (AnDn)G > nCG(g) is finite too. Therefore the centralizer of g in < g >< (AnDn)G > is finite. Then in view of the assertion (6), the statement (iii) with < g >< (A n Dn)G > in the character of G is valid. At the same time, < (AnDn)G > is locally finite. Then in consequence of O. Yu. Shmidt's Theorem, the product of subgroups < (A n Dn)G >, taken by all complemented in G subgroups An, is an infinite normal locally finite subgroup of G. Therefore in view of assertion (2), the statement (iii) is valid. □

(8) For g e G and n = n(< g >) and H =< gG >, all n'-subgroups of CH(g) are Chernikov.

Proof. Assume that CH(g) has some non-Chernikov n'-subgroup. Then in view of Suchkova-

Shunkov Theorem (mentioned above), this subgroup has some infinite chain A D A1 D A2 D ... D An D An+1 D ... of abelian subgroups. Some An has a complement D in G. Then, with regard to S.N. Chernikov's Lemma, we have:

Ax <g>= An x (D n Ax < g >) = An x (D n A) x (Dn < g >) = A x (Dn < g >).

Therefore, clearly, < g >= Dn < g >, i.e. < g >C D. Since also G = (Ax < g >)D, by virtue of Chunikhin's Lemma (see, for instance, [21, Lemma 1.36]), H C D. But A C H and A ^ D, which is a contradiction. □

(9) If G satisfies min — p' for some p, then (iii) is valid and also G is Chernikov or contains a normal elementary abelian p-subgroup of finite index.

Proof. Assume that (iii) is not valid. Let M be from the assertion (3). In view of the assertion (4), every p-subgroup of G is abelian or Chernikov. Consequently, M has an element g of prime order q = p. Put H =< gM >. In view of the assertion (8), in CH(g) all q'-subgroup are Chernikov. Consequently CH (g) satisfies min — p. Also CH (g) satisfies min — p'.

Further, every abelian subgroup of CH(g) is a direct product of a p-subgroup and a p'-subgroup. Thus it is a direct product of two Artinian subgroups, and so it is Artinian. Thus, CH (g) satisfies min — ab. Then in view of the assertion (6), the statement (iii) with H in the character of G is valid. Therefore H is a normal locally finite subgroup of M, which is a contradiction. Thus, (iii) is valid.

Now let G be non-Chernikov. Then, with regard to N. V.Chernikova's Theorem [9,10] (see also Introduction), G = U X V, U and V are abelian, U is a direct product of normal in G subgroups of prime orders and G has no quasicyclic subgroups. So U = Up x Up', V = Vp x Vp', where Up and Vp are p-subgroups, Up' and Vp' are p'-subgroups. Since Up' and Vp' are Artinian abelian, by Kurosh'es Theorem (see, for instance, [19, Proposition 4.2.11, p. 101]), Up' and Vp' are Chernikov. Since G has no quasicyclic subgroups, Up' and Vp' are finite. Therefore G : Up X V^ < to. Since Up is obviously a direct product of normal in G subgroups of order p, if Up = 1, and Vp is a p-subgroup, Up X Vp = Up x Vp. In consequence of Lemma 1.1 [15], Up x Vp is elementary abelian. □

(10) The statement (iii) is necessarily valid.

Proof. Assume that (iii) is not valid. Let M be from (3). Further, let g be an element of some prime order p of M. Put H =< gM >. Then CH(g) satisfies min — p' (see (8)). Therefore in view of the assertion (9) (with CH (g) instead of G), CH (g) is almost abelian. Therefore by virtue of the assertion (7), the statement (iii) with H in the character of G is valid. At the same time, H is locally finite, which is a contradiction. □

B. Show that (iii) implies (ii).

Put A* = Cg(A). In view of S. N. Chernikov's Lemma, A* = A X (A* n B). Then because of A and A* n B are abelian, A* is abelian too. Obviously, A* = CG(A*). Further, clearly, D n A* = 1. Since C is a direct product of groups of prime orders or C = 1, we have for some subgroup C* C C: B = (A* nB) x (D x C*). Then B = D x (A* nB) x C*. So for B* = D x C* we have: G = A*B = A* X (D x C*) = A* X B*. Therefore in view of Proposition 2 [27], every infinite abelian subgroup of G is complemented in it.

Of course, (ii) implies (i).

Theorem is proven.

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Шунковские группы с условием минимальности для недополняемых абелевых погрупп

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