Products of locally cyclic groups

We consider groups of the form $G=AB$ with two locally cyclic subgroups $A$ and $B$. The structure of these groups is determined in the cases when $A$ and $B$ are both periodic or when one of them is periodic and the other is not. Together with a previous study of the case when $A$ and $B$ are torsion-free, this gives a complete classification of all groups that are the product of two locally cyclic groups. It is also shown that the product of a finite number of pairwise permutable periodic locally cyclic subgroups is a locally supersoluble group.

The aim of this paper is to describe the structure of groups that are products of two locally cyclic subgroups in the periodic and in the mixed case. Altogether this gives a complete answer to Question 15 in the book [1]. We recall that a group G has finite Prüfer rank r = r(G) (special rank in the sense of Mal'cev in Russian terminology) if each finitely generated subgroup of G can be generated by r elements and r is the least positive integer with this property. Clearly a group is of rank 1 if and only if it is locally cyclic. It is also known that every finite p -group of the form G = AB with cyclic subgroups A and B has rank at most 2 for p odd ( [5], Satz 8) and at most 3 for p = 2 ([6], Theorem 5.1). The following consequence of Theorem 1.1 gives an exact upper bound for the Prüfer rank of a product G = AB of two periodic locally cyclic subgroups A and B . It should be noted that this result is also new for arbitrary finite groups of the form G = AB with cyclic subgroups A and B .
Finally, the following theorem extends a well-known result of B. Huppert on the supersolubility of finite groups which are products of pairwise permutable cyclic subgroups (see [5], Satz 34 or [4], Satz VI. 10.3). This gives, in particular, an affirmative answer to Question 1 of the article [2]. Theorem 1.3. Let the group G = A 1 A 2 ...A n be the product of finitely many pairwise permutable periodic locally cyclic subgroups A 1 , ..., A n . Then G is a periodic locally supersoluble group.

Preliminaries
In what follows G = AB is a group with locally cyclic subgroups A and B .
Lemma 2.1. If G = AB is an infinite p -group, then up to a permutation of the factors A and B the subgroup A is quasicyclic and one of the following statements holds: Proof. Clearly without loss of generality we may assume that the subgroup A is infinite and so quasicyclic. Then the subgroup B is either cyclic or quasicyclic. Since in the latter case the group G is abelian by [1], Lemma 7.4.4, the subgroup A is complemented in G and hence statement 1) holds. Let the group G be non-abelian. Then the subgroup B is cyclic and so A as a quasicyclic p -subgroup of finite index in G must be normal and noncentral in G . In particular, B induces on A a non-trivial cyclic p -group of automorphisms. On the other hand, since quasicyclic p -groups have no automorphisms of order p > 2 , it follows that p = 2 . But then B = b with b 2 n = 1 for some n ≥ 1 and b induces on A an automorphism of order 2 that inverts the elements of A . In particular, if A ∩ B = 1 , we obtain statement 2). In the second case, A ∩ B = b 2 n−1 and hence statement 3) holds, as claimed.
Corollary 2.2. If G = AB is a p -group and C, D are subgroups of A and B , respectively, then CD = DC and so CD is a subgroup of G .
Proof. This is known if G is finite (see [5], Satz 3), and follows from Lemma 2.1 in the general case. Proof. Since the group H is finite, there exist finite subsets C of A and D of B such that H is contained in the set CD . Then the subgroups A 0 = C , B 0 = D and C, D are finite, because the group G is locally finite. Furthermore, it follows from [1], Lemma 1.2.3, that the normalizer Since the subgroup E is supersoluble by [1,Lemma 7.4.6] and H ⊆ CD ⊆ C, D = A 0 , B 0 , the lemma is proved.
As a direct consequence of this lemma, we have If G is a periodic group and π is a set of primes, then a subgroup H of G is called a π -subgroup, provided that all prime divisors of the order of any element of H are contained in π . By a Sylow π -subgroup of G we mean simply a maximal π -subgroup G π of G which will be denoted by G p if π = {p} . Lemma 2.5. Let G = AB be a periodic group and π a set of primes. Then the following statements hold. 1) If A π and B π are Sylow π -subgroups of A and B , respectively, then G π = A π B π is a Sylow π -subgroup of G and 2) If p, q are primes with p > q , then a Sylow p -subgroup G p is normalized by a Sylow q -subgroup G q . In particular, G {p,q} = G p G q for any primes p and q .
Proof. 1) It follows from [1], Lemma 1.3.2, that in the notation of Corollary 2.4 for each n ≥ 1 the set 2) If π = {p, q} for some primes p > q , then . As G and so its subgroup G {p,q} are locally supersoluble by Lemma 2.3, the Sylow p -subgroup A Sylow basis of a periodic group G is defined to be a complete set S = {G p } of Sylow p -subgroups of G , one for each prime p , such that G p G q = G q G p for all pairs p, q of primes, and G π = G p | p ∈ π is a Sylow π -subgroup of G for each set π of primes. As is well known (see [3],Lemma 2.1), every countable periodic locally soluble group possesses Sylow bases.
Proof. Indeed, by Lemma 2.5 the set S = {G p } forms a Sylow basis of G and N G (G p ) = N A (G p )N B (G p ) for every p by [1], Lemma 1.2.2. Therefore, The following lemma is a direct consequence of a well-known result of L. Kovacs (see [7], Theorem 2). Lemma 2.7. Let G be a finite soluble group, π a set of primes and H a Hall π -subgroup of G . If for each p ∈ π the Prüfer rank of a Sylow p -subgroup of G does not exceed r , then H is a subgroup of rank at most r + 1 .
Proof. Indeed, it is obvious that if K is a subgroup of H , then every Sylow subgroup of K is generated by r elements. Therefore, K can be generated by r + 1 elements by the result of Kovacs cited above. Thus every subgroup of H is generated by r + 1 elements and so H has rank at most r + 1 .

Proof of Theorem 1.1
First of all, it follows from Lemma 2.1 that for each prime p every Sylow p -subgroup of G = AB satisfies the minimal condition for subgroups. Therefore, G satisfies the minimal condition for p -subgroups for all primes p . Since the group G is metabelian by Ito's theorem, the locally nilpotent residual R of G is contained in its derived subgroup G ′ and so is abelian. It was proved by B. Hartley in [3], Theorem 1, that in this case G = R ⋊ H , where H is any basis normalizer of G . In particular, by Lemma 2.6 we can take H = A * B * .
It is easy to see that the subgroup H is locally nilpotent and contains the center Z(G) of G . Furthermore, G ′ = R × H ′ and so H ′ is a normal subgroup of G . Since R is abelian and Moreover, passing to the factor group G/H ′ , we may restrict ourselves to the case when the subgroup H = A * B * is abelian. Then the subgroups A * and B * centralize S and T , respectively, and so the subgroup H centralizes the intersection S ∩ T . Since H = N G (H) , this implies S ∩ T = 1 . Thus Since the subgroup A 0 is locally cyclic, this implies that also S is locally cyclic and π(S) = π(A 0 ) . Moreover, as A * and A 0 are subgroups of the locally cyclic subgroup A , it also follows that π(A * ) ∩ π(A 0 ) = ∅ . Similarly, using the equality T ⋊ A = AB 0 , we obtain π(T ) = π(B 0 ) and π(B * ) ∩ π(B 0 ) = ∅ .
Proof of Corollary 1.2. By Corollary 2.4, we can restrict ourselves to the case in which the group G = AB is finite. By Theorem 1.1 G contains cyclic normal subgroups S and T and a nilpotent subgroup H = A * B * with A * ≤ A and B * ≤ B such that where π(S) ∩ π(A * ) = π(T ) ∩ π(B * ) = ∅ , S = [S, B * ] and T = [T, A * ] . In particular, if for some prime p the subgroup H contains a non-cyclic Sylow p -subgroup P , then S and T are p ′ -subgroups of G .
Since P = A p B p with A p = A ∩ P and B p = B ∩ P , both subgroups A p and B p are non-trivial and so p / ∈ π(S) ∪ π(T ) . Therefore, if G p is a non-cyclic Sylow p -subgroup of G , S p = G p ∩ S and T p = G p ∩ T , then up to conjugation G p coincides with one of the following subgroups of G : P = A p B p , A p ⋊ T p , B p ⋊ S p and S p × T p . In particular, the Sylow psubgroups of G have rank at most 2 for p > 2 (see [4], Satz III.11.5) and at most 3 for p = 2 (see [6], Theorem 5.1). We show now that G is in fact a group of rank at most 3 .
Indeed, suppose the contrary and let the group G contain a subgroup K whose minimal number of generators d(K) is at least 4 . Since the Sylow subgroups of odd orders in K have rank at most 2 by what was noted above, each Sylow 2 -subgroup Q of K must have rank 3 by Lemma 2.7. Furthermore, there exists a Sylow 2 -subgroup P = A 2 B 2 of G such that Q = K ∩ P . As the group G is metabelian, the derived subgroup P ′ is normal in G and so the subgroup N = Q ∩ P ′ is normal in K . It is easy to see that N = 1 , because otherwise the subgroup Q is embedded in the factor group P/P ′ whose rank is equal 2 . Furthermore, the factor group Q/N = Q/Q ∩ P ′ is isomorphic to the factor group QP ′ /P ′ ≤ P/P ′ and so the rank of Q/N does not exceed 2 . Thus the factor group K/N has rank at most 3 by Lemma 2.7. In particular, d(K/N) < d(K) = 4 and hence N is not contained in the Frattini subgroup Φ(K) of K , because otherwise d(K/N) = d(K) . Therefore, passing to the factor group K/Φ(K) , we may assume that Φ(K) = 1 . Then the normal subgroup N is complemented in K and so in Q by [4], Hilfsatz 3.

Products of a periodic and a torsion-free local cyclic group
Recall that a group G has finite torsion-free rank if it has a series of finite length whose factors are either periodic or infinite cyclic. The number r 0 (G) of infinite cyclic factors in such a series is an invariant of G called its torsionfree rank. In this section, we describe the structure of a group G = AB with locally cyclic subgroups A and B , the first of which is periodic and the other non-trivial torsion-free. Clearly r 0 (B) = 1 and we note first that r 0 (G) = 1 .

Lemma 4.1. Let G = AB be a group with subgroups A and B such that
A is periodic abelian and B is non-trivial torsion-free locally cyclic. Then r 0 (G) = 1 .
Proof. It was proved by D. Zaitsev in [11], Theorem 3.7 (see also [1], Lemma 7.1.2) that there exists a non-trivial normal subgroup of G contained in A or B . Therefore G has the normal series A 0 < A 0 B 0 < G in which A 0 is the core of A in G and B 0 is the core of B in G modulo A 0 . As is easily seen, the factors A 0 and G/A 0 B 0 are periodic and the factor group A 0 B 0 /A 0 is isomorphic to B 0 . Thus r 0 (G) = r 0 (B) = 1 , as claimed.
The following lemma is a consequence of the well-known theorem of I. Schur on the finiteness of the derived subgroup of a group that is finite over its center (see [8], Corollary to Theorem 4.12).
Lemma 4.2. If a group G contains a central subgroup Z such that the factor group G/Z is locally finite, then the derived subgroup of G is locally finite. Theorem 4.3. Let the group G = AB be the product of two locally cyclic subgroups A and B such that A is periodic and B is non-trivial torsionfree. Then one of the following statements hold.

1) the subgroup A is normal in G and so
Proof. It is easy to see that each periodic normal subgroup H of G is contained in A , because AH = A(AH ∩ B) and AH ∩ B = 1 . Therefore the core A 1 = ∩ g∈G A g of A in G is the maximal periodic normal subgroup of G . Assume first that A 1 = 1 and let B 1 be the core of B in G . Then B 1 = 1 by the theorem of D. Zaitsev noted above and so the factor group G/B 1 is periodic, because it is the product of two periodic subgroups AB 1 /B 1 and B/B 1 . Moreover, since the centralizer C G (B 1 ) of B 1 in G contains B , the group G induces on B 1 a periodic group of automorphisms which is isomorphic to the factor group A/C A (B 1 ) . As is well known, a periodic group of automorphisms of any torsion-free locally cyclic group is of order 2 . Therefore the order of A/C A (B 1 ) does not exceed 2 and hence either On the other hand, since the centralizer C G (B 1 ) = C A (B 1 )B is normal in G and periodic over B 1 , its derived subgroup C G (B 1 ) ′ is periodic by Lemma 4.2 and normal in G . Therefore C G (B 1 ) ′ ≤ A 1 = 1 and hence C G (B 1 ) = C A (B 1 ) × B . But then again C A (B 1 ) is normal in G and so C A (B 1 ) = 1 . Thus in the case A 1 = 1 we have either A = 1 and G = B or A = a with a 2 = 1 and Finally, returning now to the general case, we derive that either The latter means in particular that the mapping φ : B → A 1 is a derivation of B into A 1 , as claimed.

Products of finitely many periodic local cyclic groups
A well-known theorem of B. Huppert cited in the Introduction says that every finite group of the form G = A 1 A 2 ...A n with pairwise permuting cyclic subgroups A i for 1 ≤ i ≤ n is supersoluble. This result was later extended to products of pairwise permutable locally cyclicČernikov groups by M. Tomkinson in [10]. He proved that in this case G = A 1 A 2 ...A n is a locally supersolubleČernikov group. In this section we generalize this result to products of arbitrary periodic locally cyclic groups. Recall that a group is said to be hyperabelian (respectively, hypercyclic) if it has an ascending series of normal subgroups with abelian (respectively cyclic) factors.
..A n be the product of pairwise permutable periodic locally cyclic subgroups A i . If the set π = ∪ n i=1 π(A i ) is finite, p is the largest prime in π , P i is the Sylow p -subgroup of A i and Q i is the p -complement to P i in A i for each 1 ≤ i ≤ n , then G is a π -group, P = P 1 P 2 . . . P n is a normal Sylow p -subgroup of G and Q = Q 1 Q 2 . . . Q n is a p -complement to P in G .
Proof. Since each of the A i is a subgroup of Prüfer rank 1 , the group G = A 1 A 2 . . . A n is hyperabelian of finite Prüfer rank by [2], Theorem 3.1. Therefore, arguing by induction on n and applying Corollary 3.2.7 of [1] and Lemma 3.2 of [2], we derive that G is a π -group, P = P 1 ...P n is a Sylow p -subgroup of G and Q = Q 1 Q 2 . . . Q n is a complement to P in G . Moreover, taking into account that the subgroups A i A j are locally supersoluble by Lemma 2.3, we conclude that P i A j ≤ P for all i, j and so P is a normal subgroup of G .
Lemma 5.2. Let G = A 1 A 2 ...A n be the product of pairwise permutable locally cyclic subgroups A i . If the group G is periodic and the set π(G) is finite, then G is hypercyclic.
Proof. It is easy to see that every factor group of G satisfies the hypothesis of the lemma. Therefore, in order to prove that G is hypercyclic, it suffices to show that G has a non-trivial cyclic normal subgroup.
Let p be the largest prime in π(G) , P i the Sylow p -subgroup of A i and Q i the p -complement to P i in A i for each 1 ≤ i ≤ n . Then P = P 1 P 2 . . . P n is a normal Sylow p -subgroup of G and Q = Q 1 Q 2 . . . Q n is a p -complement to P in G by Lemma 5.1. It is clear that if π(G) = {p} , then the group G = P 1 P 2 . . . P n is hypercyclic by Lemma 2.1. Therefore, arguing by induction on |π(G)| , we may assume that p > 2 and the subgroup Q = Q 1 Q 2 . . . Q n is hypercyclic. Then every non-trivial normal subgroup of Q contains a non-trivial Q -invariant cyclic subgroup. Since the centralizer C Q (P ) of P in Q is a normal subgroup of Q , every Q -invariant cyclic subgroup of C Q (P ) is normal in G . Thus, if C Q (P ) = 1 , then the group G contains a non-trivial cyclic normal subgroup. Therefore, to complete the proof of the lemma, it remains to consider the case C Q (P ) = 1 .
It is clear that if in this case the subgroup P is finite, then Q and so the group G are also finite, so that G is supersoluble and thus hypercyclic by Huppert's theorem cited above. Suppose now that the subgroup P is infinite. Then at least one of the subgroups P i , say P 1 , must be infinite and so quasicyclic. Since P A i ∩ A 1 A i = (P ∩ A 1 A i )A i = (P 1 P i )A i = P 1 A i , we conclude that P 1 A i = A i P 1 for every i . Furthermore, the subgroup P 1 P i is abelian by Lemma 2.1 and it is normal in P 1 A i , because P 1 P i = P ∩ P 1 A i . As A i = P i × Q i , it follows that P 1 Q i = P 1 and so the subgroup P 1 is normal in P 1 A i for every i . Therefore, P 1 and the subgroup of order p of P 1 are normal in G = A 1 A 2 ...A n , as claimed.
Proof of Theorem 1.3. Let G = A 1 A 2 ...A n be the product of pairwise permutable periodic locally cyclic subgroups A i . Then G is a periodic group by Lemma 5.1. If the set π(G) is finite, then the group G is hypercyclic by Lemma 5.2 and so locally supersoluble. In the other case the set π(G) is infinite and thus it can be presented as a union π(G) = ∪ ∞ i=1 π i of finite subsets π i such that π i ⊂ π i+1 for all i ≥ 1 . Let P ij be the Sylow π isubgroup of A j for 1 ≤ j ≤ n and G i = P i1 P i2 . . . P in . Then G i is a Sylow π i -subgroup of G by Lemma 2.5 which is locally supersoluble for each i ≥ 1 by Lemma 5.2. Since G = ∪ ∞ i=1 G i , the group G is also locally supersoluble, as claimed.