Products of locally cyclic groups

Abstract

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 where A and B are torsion-free, this gives a complete classification of all groups that are the product of two locally cyclic subgroups. As an application, it is shown that the Prüfer rank of a periodic product of two locally cyclic subgroups does not exceed 3, and this bound is sharp. It is also proved that a product of a finite number of pairwise permutable periodic locally cyclic subgroups is a locally supersoluble group. This generalizes a well-known theorem of B. Huppert for finite groups.

1 Introduction

Let the group $G=AB$ be the product of two subgroups A and B, i.e. $G=\{ab\mid a\in A, b\in B\}$. It was proved by N. Itô that the group G is metabelian if the subgroups A and B are abelian (see [1, Theorem 2.1.1]). This result laid the foundation for a systematic study of groups of the form $G = AB$ with various conditions on the subgroups A and B. In particular, it follows directly from Itô’s result that every periodic group $G=AB$ with abelian subgroups A and B is locally finite. It is also well-known that the group $G=AB$ with cyclic subgroups A and B is supersoluble and abelian-by-finite ([1, Lemma 7.4.6]). Furthermore, a detailed description of the structure of the group $G=AB$ with torsion-free locally cyclic subgroups A and B was obtained by the second author in [9].

The aim of this paper is to describe the structure of groups which are products of two locally cyclic subgroups in the periodic and the mixed case. Altogether this gives a complete answer to [1, Question 15].

Theorem 1.1

Let the periodic group $G=AB$ be the product of two locally cyclic subgroups A and B. Then G contains uniquely determined locally cyclic normal subgroups S and T and a locally nilpotent subgroup $H=A^*B^*$ with $A^*\le A$ and $B^*\le B$ such that

$$\begin{aligned} G=(S\times T)\rtimes H=(S\times A^*)(T\times B^*) \end{aligned}$$

where $\pi (S)\cap \pi (A^*)=\pi (T)\cap \pi (B^*)= \emptyset$, $S=[S,B^*],$ and $T=[T,A^*]$.

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 the product $G=AB$ of two periodic locally cyclic subgroups A and B.

Corollary 1.2

If $G=AB$ is a periodic group with locally cyclic subgroups A and B, then the Prüfer rank of G does not exceed 3.

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 [2, Question 1].

Theorem 1.3

Let the group $G=A_1A_2\cdots A_n$ be the product of finitely many pairwise permutable periodic locally cyclic subgroups $A_1,\ldots ,A_n$. Then G is a periodic locally supersoluble group.

2 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 hold:

• (1) $G=A\times B$ with B cyclic or quasicyclic;

• (2) $p=2$ and $G=A\rtimes \langle b\rangle$ for some element $b \in G$ with $a^b=a^{-1}$ for all $a\in A$;

• (3) $p=2$ and $G=A\langle b\rangle$ for some element $b \in G$ with $b^{2^n}=1$ for some $n>1$, $b^{2^{n-1}}\in A$ and $a^b=a^{-1}$ for all $a\in A$.

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 non-central 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 $B=\langle b\rangle$ with $b^{2^n}=1$ for some $n\ge 1$ and b induces on A an automorphism of order 2 that inverts the elements of A. In particular, if $A\cap B=1$, we obtain statement (2). In the second case, $A\cap B=\langle b^{2^{n-1}}\rangle$ and hence statement (3) holds, as claimed. $\square$

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. $\square$

Lemma 2.3

If the group $G=AB$ is periodic and H is a finite subgroup of G, then H is contained in a finite subgroup E of G such that $E=(A\cap E)(B\cap E)$. In particular, the group G is locally supersoluble.

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 $A_0=\langle C\rangle$, $B_0=\langle D\rangle$, and $\langle C,D\rangle$ are finite, because the group G is locally finite. Furthermore, it follows from [1, Lemma 1.2.3] that the normalizer $N_G(\langle A_0, B_0\rangle )$ contains a finite subgroup E such that $\langle A_0, B_0\rangle \le E=(A\cap E)(B\cap E)$. Since the subgroup E is supersoluble by [1, Lemma 7.4.6] and $H\subseteq CD\subseteq \langle C,D\rangle = \langle A_0, B_0\rangle$, the lemma is proved. $\square$

As a direct consequence of this lemma, we have

Corollary 2.4

If the group $G=AB$ is periodic, then there exists an ascending series of finite subgroups $1=G_0<G_1<\cdots<G_n<\cdots G$ such that $G_n=(A\cap G_n)(B\cap G_n)$ for each $n>0$ and $G=\bigcup _{n=1}^\infty G_n$.

If G is a periodic group and $\pi$ is a set of primes, then a subgroup H of G is called a $\pi$-subgroup provided that all prime divisors of the order of any element of H are contained in $\pi$. By a Sylow $\pi$-subgroup of G we simply mean a maximal $\pi$-subgroup $G_{\pi }$ of G which will be denoted by $G_p$ if $\pi =\{p\}$.

Lemma 2.5

Let $G=AB$ be a periodic group and $\pi$ a set of primes. Then the following statements hold.

• (1) If $A_{\pi }$ and $B_{\pi }$ are Sylow $\pi$-subgroups of A and B, respectively, then $G_{\pi }=A_{\pi }B_{\pi }$ is a Sylow $\pi$-subgroup of G and

  $$N_G(G_{\pi })=N_A(G_{\pi })N_B(G_{\pi }).$$

• (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, $G_{\{p, q\}}=G_pG_q$ 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 $n\ge 1$ the set $(A_{\pi }\cap G_n)(B_{\pi }\cap G_n)$ is a Hall ${\pi }$-subgroup of $G_n$. Therefore $G_{\pi }=\bigcup _{n=1}^\infty (A_{\pi }\cap G_n)(B_{\pi }\cap G_n)$ is a Sylow $\pi$-subgroup of G. Since $A_{\pi }=\bigcup _{n=1}^\infty (A_{\pi }\cap G_n)$ and $B_{\pi }=\bigcup _{n=1}^\infty (B_{\pi }\cap G_n)$, this implies $G_{\pi }=A_{\pi }B_{\pi }$. In addition, applying [1, Lemma 1.2.2], we have $N_G(G_{\pi })= N_A(G_{\pi })N_B(G_{\pi })$.

(2) If ${\pi }=\{p, q\}$ for some primes $p>q$, then $A_{\{p,q\}}=A_p\times A_q$, $B_{\{p,q\}}=B_p\times B_q$, and $G_{\{p,q\}}=(A_p\times A_q)(B_p\times B_q)$. As G and so its subgroup $G_{\{p,q\}}$ is locally supersoluble by Lemma 2.3, the Sylow p-subgroup $G_p=A_pB_p$ is normal in $G_{\{p,q\}}$. Therefore, $G_{\{p,q\}}=(A_p\times A_q)(B_p\times B_q)=G_pA_qB_q=G_pG_q$, as claimed. $\square$

A Sylow basis of a periodic group G is defined to be a complete set $\mathbf{S} =\{G_p\}$ of Sylow p-subgroups of G, one for each prime p, such that $G_pG_q=G_qG_p$ for all pairs p, q of primes, and $G_{\pi }=\langle G_p \mid p\in {\pi }\rangle$ is a Sylow $\pi$-subgroup of G for each set $\pi$ of primes. As is well-known (see [3, Lemma 2.1]), every countable periodic locally soluble group possesses Sylow bases. The basis normalizer $N_G(\mathbf{S})$ of a Sylow basis $\mathbf{S}$ of G is by definition the intersection $N_G(\mathbf{S})=\bigcap _p N_G(G_p)$ of the normalizers $N_G(G_p)$ of the Sylow p-subgroups $G_p$ of $\mathbf{S}$ for all p.

Lemma 2.6

Let the group $G=AB$ be periodic and $G_p=A_pB_p$ for each prime p. Then $\mathbf{S}=\{G_p\}$ is a Sylow basis of G. Moreover, if $A^*=\bigcap _p N_A(G_p)$ and $B^*=\bigcap _p N_B(G_p)$, then $N_G(\mathbf{S})=A^*B^*$.

Proof

Indeed, by Lemma 2.5, the set $\mathbf{S}=\{G_p\}$ 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, $N_G(\mathbf{S})=\bigcap _p N_G(G_p)=\bigcap _p N_A(G_p)N_B(G_p)$ and it is easy to check that $\bigcap _p N_A(G_p)N_B(G_p)=(\bigcap _p N_A(G_p))(\bigcap _p N_B(G_p))=A^*B^*$ (see [1, Lemma 1.1.2]). Therefore $N_G(\mathbf{S})=A^*B^*$. $\square$

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, $\pi$ a set of primes and H a Hall $\pi$-subgroup of G. If for each $p\in \pi$ 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 $r+1$. $\square$

3 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 it is abelian. It was proved by Hartley [3, Theorem 1] that in this case $G=R\rtimes H$, 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, $G'=R\times H'$ and so $H'$ is a normal subgroup of G. Since R is abelian and $N_G(H)= N_R(H)\times H$, it follows that $N_R(H)\le Z(G)\le H$. Therefore, $N_R(H)=1$ and hence $H=N_G(H)$. We now show that the subgroup $H=A^*B^*$ commutes with both subgroups A and B.

Indeed, put $S=R\cap \langle A,B^*\rangle$ and $T=R\cap \langle A^*,B\rangle$. It is clear that S and T are normal subgroups of G, $\langle A,B^*\rangle =S\rtimes H$, and $\langle A^*,B\rangle =T\rtimes H$. On the other hand, as $G=AB$, we have also $\langle A,B^*\rangle =AB_1$ and $\langle A^*,B\rangle =A_1B$ for some subgroups $A_1$ and $B_1$ such that $A^*\le A_1\le A$ and $B^*\le B_1\le B$. From here, we deduce $AB_1\cap A_1B=A_1B_1=(S\cap T)\rtimes H$. 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 $S\cap T$. Since $H=N_G(H)$, this implies $S\cap T=1$. Thus $A_1B_1=H=A^*B^*$ and hence $\langle A,B^*\rangle =AH=AB^*$ and $\langle A^*,B\rangle =BH=A^*B$, as asserted.

Further, taking into account the equalities $AB^*=S\rtimes H$ and $H=A^*B^*$, we conclude that the subgroup $A^*$ centralizes S, because $[A^*,S]\le H'\cap S=1$. Since in this case the normalizer $N_S(B^*)$ is contained in $N_G(H)=H$, we have $N_S(B^*)=1$. Therefore, every element $b\in B^*$ induces on S an automorphism leaving only the identity element fixed. But then every element of S can be written in the form $b^{-1}s^{-1}bs$ with $s\in S$ and hence $S=[B^*,S]$. Similarly, using the equality $A^*B=T\rtimes H$, we derive $T=[A^*,T]$.

Finally, we put $A_0=A\cap BS$ and $B_0=AT\cap B$. Clearly from the equalities $G=AB$, $AB^*=S\rtimes H$, and $A^*B=T\rtimes H$, it follows that $G=S\rtimes A^*B=T\rtimes AB^*$, $A=A^*\times A_0$, and $B=B^*\times B_0$. Therefore, $S\rtimes B=A_0B$ and $T\rtimes A=AB_0$. Furthermore, if $S_p$ is a Sylow p-subgroup of S, then $S_p\rtimes B=(A_0\cap S_pB)B$ and if $S_p\ne 1$, then $A_0\cap S_pB\ne 1$. Since the subgroup $A_0$ is locally cyclic, this implies that S is locally cyclic and $\pi (S)$ is contained in $\pi (A_0)$. Moreover, as $A^*$ and $A_0$ are subgroups of the locally cyclic subgroup A, it also follows that $\pi (A^*)\cap \pi (A_0)= \emptyset$. Similarly, using the equality $T\rtimes A=AB_0$, we obtain $\pi (T)=\pi (B_0)$ and $\pi (B^*)\cap \pi (B_0)=\emptyset$. $\square$

Proof of Corollary 1.2

By Corollary 2.4, we may restrict ourselves to the case in which the group $G=AB$ is finite. By Theorem 1.1, G then contains cyclic normal subgroups S and T and a nilpotent subgroup $H=A^*B^*$ with $A^*\le A$ and $B^*\le B$ such that

$$\begin{aligned} G=(S\times T)\rtimes H=(S\times A^*)(T\times B^*), \end{aligned}$$

where $\pi (S)\cap \pi (A^*)=\pi (T)\cap \pi (B^*)= \emptyset$, $S=[S,B^*]$, 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 $P=A_pB_p$ with $A_p=A\cap P$ and $B_p=B\cap P$, both subgroups $A_p$ and $B_p$ are non-trivial and so $p\notin \pi (S)\cup \pi (T)$. Therefore, if $G_p$ is a non-cyclic Sylow p-subgroup of G, $S_p=G_p\cap S$, and $T_p=G_p\cap T$, then up to conjugation $G_p$ coincides with one of the following subgroups of G: $P=A_pB_p$, $T_p\rtimes A_p$, $S_p\rtimes B_p$, and $S_p\rtimes T_p$. In particular, the Sylow p-subgroups 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 now show 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. It is clear that $Q=K\cap P$ for a Sylow 2-subgroup $P=A_2B_2$ of G. As the group G is metabelian, the derived subgroup $P'$ is abelian and normal in G. Moreover, $P'$ has rank at most 2 by [6, Theorems 4.2 and 4.3(e)].

Put $N=P'\cap Q$. As $Q=K\cap P$, we have $N=K\cap P'$ and so N is an abelian normal subgroup of K with rank at most 2. In addition, $N\ne 1$, because otherwise the subgroup Q is embedded in the factor group $P/P'$ whose rank is equal to 2. On the other hand, since the factor group $Q/N=Q/P'\cap Q$ is isomorphic to the factor group $QP'/P'\le P/P'$, it is abelian of rank at most 2. Therefore, the Sylow 2-subgroups of the factor group K/N have rank at most 2 and hence K/N has rank at most 3 by Lemma 2.7. In particular, $d(K/N)<d(K)=4$ and so N is not contained in the Frattini subgroup $\Phi (K)$ of K, because otherwise $d(K/N)=d(K)$. Clearly, passing to the factor group $K/\Phi (K)$, we may assume that $\Phi (K)=1$. Then the normal subgroup N is complemented in K and so in Q by [4, Hilfsatz 3.4.4]. Moreover, since $\Phi (N)=1$ by [4, Hilfsatz 3.3.b], the subgroup N is elementary abelian of order at most 4.

Let L be a complement to N in K and $M=Q\cap L$. Then M is a Sylow 2-subgroup of L and $Q=N\times M$, so that M is abelian with $d(M)\le 2$. Since the subgroup K is supersoluble, its maximal subgroup U of odd order is normal in K and centralizes N. Therefore, $K=U\times Q=(U\times N)\rtimes M$ and hence the centralizer $C_N(M)$ is a non-trivial central subgroup of K. As $\Phi (K)=1$, the subgroup $C_N(M)$ is complemented in K and thus in Q. From this, it follows that $Q=M\times N$ is abelian and N is a central subgroup of K. Therefore, $K=(U\rtimes M)\times N$ and the subgroup $U\rtimes M$ is three-generated by Lemma 2.7. This means that there exist elements u, v, w of U and x, y, z of M such that $U\rtimes M=\langle ux, vy, wz\rangle$. Then M modulo U is generated by x, y, z. In particular, if $d(M)=1$, without loss of generality we may assume that $M=\langle x\rangle$ and $y=z=1$, so that $U\rtimes M=\langle ux, v, w\rangle$. In the case $d(M)=2$ we can take $M=\langle x, y\rangle$ and $z=1$. Then $U\rtimes M=\langle ux, vy, w\rangle$.

Finally, since $Q=M\times N$ is a 2-subgroup of rank 3 as noted above, only two cases are possible: either $M=\langle x\rangle$ and $N=\langle a, b\rangle$ has order 4 or $M=\langle x, y\rangle$ with $x\ne 1\ne y$ and $N=\langle a\rangle$ is of order 2. Therefore, $K=\langle ux, av, bw \rangle$ in the first case and $K=\langle ux, vy, aw \rangle$ in the second case. In both cases $d(K) < 4$ and this contradiction completes the proof. $\square$

4 Products of a periodic and a torsion-free locally 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 torsion-free rank. In this section, we describe the structure of the 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 Zaitsev [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 $A_0< A_0B_0 < G$ 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 $G/A_0B_0$ are periodic and the factor group $A_0B_0/A_0$ is isomorphic to $B_0$. Thus $r_0(G)=r_0(B)=1$, as claimed. $\square$

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 torsion-free. Then one of the following statements holds.

• (1) The subgroup A is normal in G and so $G=A\rtimes B$;

• (2) $A=A_1\langle a\rangle$ with $a^2\in A_1$, the subgroup $A_1$ is normal in G and $G=(A_1\rtimes B)\langle a\rangle$ with $b^a=b^{-1}\phi (b)$ for all $b\in B$, where $\phi : B \rightarrow A_1$ is a derivation of B into $A_1$.

Proof

It is easy to see that each periodic normal subgroup H of G is contained in A, because $AH=A(AH\cap B)$ and $AH\cap B=1$. Therefore the core $A_1=\cap _{g\in G}A^g$ 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 $B_1\ne 1$ by the theorem of 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 locally cyclic torsion-free group is of order 2. Therefore the order of $A/C_A(B_1)$ does not exceed 2 and hence either $A=C_A(B_1)$ or $A=C_A(B_1)\langle a\rangle$ with $a\in A$ and $a^2\in C_A(B_1)$.

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 $C_G(B_1)'\le A_1=1$ and hence $C_G(B_1)=C_A(B_1)\times B$. 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 $A=\langle a\rangle$ with $a^2=1$ and $G=B\rtimes \langle a\rangle$ with $b^a=b^{-1}$ for all $b\in B$.

Finally, returning now to the general case, we derive that either $G=A\rtimes B$ or $G=(A_1\times B)\langle a\rangle )$ with $b^a=\phi (b)b^{-1}$ for every $b\in B$ and some element $\phi (b)\in A_1$. Moreover, since $\phi (bc)(bc)^{-1}=(bc)^a =b^ac^a=(\phi (b)b^{-1})(\phi (c)c^{-1})=(\phi (b)\phi (c)^b)(bc)^{-1}$, it follows that $\phi (bc)=\phi (b)\phi (c)^b$ for any $b,c\in B$. The latter means in particular that the mapping $\phi : B \rightarrow A_1$ is a derivation of B into $A_1$, as claimed. $\square$

5 Products of finitely many periodic locally cyclic groups

A well-known theorem of Huppert cited in the introduction says that every finite group of the form $G=A_1A_2\cdots A_n$ with pairwise permuting cyclic subgroups $A_i$ for $1\le i\le n$ is supersoluble. This result was later extended to products of pairwise permutable locally cyclic Chernikov groups by Tomkinson [10]. He proved that in this case $G=A_1A_2,\ldots ,A_n$ is a locally supersoluble Chernikov 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.

Lemma 5.1

Let $G=A_1A_2\cdots A_n$ be the product of pairwise permutable periodic locally cyclic subgroups $A_i$. If the set $\pi = \bigcup _{i=1}^n\pi (A_i)$ is finite, p is the largest prime in $\pi$, $P_i$ is the Sylow p-subgroup of $A_i$, and $Q_i$ is the p-complement to $P_i$ in $A_i$ for each $1\le i\le n$, then G is a $\pi$-group, $P=P_1P_2\cdots P_n$ is a normal Sylow p-subgroup of G, and $Q=Q_1Q_2\cdots Q_n$ is a p-complement to P in G.

Proof

Since each of the $A_i$ is a subgroup of Prüfer rank 1, the group $G=A_1A_2\cdots A_n$ is hyperabelian of finite Prüfer rank by [2, Theorem 3.1]. Therefore, arguing by induction on n and applying [1, Corollary 3.2.7], and [2, Lemma 3.2], we derive that G is a $\pi$-group, $P=P_1P_2\cdots P_n$ is a Sylow p-subgroup of G and $Q=Q_1Q_2\cdots Q_n$ is a complement to P in G. Moreover, taking into account that the subgroups $A_iA_j$ are locally supersoluble by Lemma 2.3, we conclude that ${P_i}^{A_j}\le P$ for all i, j and so P is a normal subgroup of G. $\square$

Lemma 5.2

Let $G=A_1A_2\cdots A_n$ be the product of pairwise permutable locally cyclic subgroups $A_i$. If the group G is periodic and the set $\pi (G)$ is finite, then G is locally supersoluble.

Proof

Since $\pi (G)$ is finite, every locally cyclic subgroup $A_i$ is a Chernikov group, i.e. a finite extension of a direct product of finitely many quasicyclic subgroups. Therefore G is a locally supersoluble Chernikov group by [10, Theorem B]. $\square$

Proof of Theorem 1.3

Let $G=A_1A_2\cdots A_n$ 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 $\pi (G)$ is finite, then the group G is locally supersoluble hypercyclic by Lemma 5.2. In the other case the set $\pi (G)$ is infinite and thus it can be presented as a union $\pi (G)=\bigcup _{i=1}^\infty \pi _{i}$ of finite subsets $\pi _{i}$ such that $\pi _{i}\subset \pi _{i+1}$ for all $i\ge 1$. Let $P_{ij}$ be the Sylow $\pi _{i}$-subgroup of $A_j$ for $1\le j\le n$ and $G_i=P_{i1}P_{i2}\cdots P_{in}$. Then $G_i$ is a Sylow $\pi _{i}$-subgroup of G by Lemma 5.1 which is locally supersoluble as a group for each $i\ge 1$ by Lemma 5.2. Since $G=\bigcup _{i=1}^\infty G_i$, the group G is also locally supersoluble, as claimed. $\square$