1 Introduction
Let the group G=AB be the product of two subgroups A and B, i.e. G is of the form G={ab∣a∈A,b∈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]).
In connection with Itô’s theorem a natural question is whether every group G=AB with abelian-by-finite subgroups A and B is metabelian-by-finite (see [1, Question 3]) or at least soluble-by-finite. However, this seemingly simple question is very difficult to attack and only partial results in this direction are known. A positive answer was given for linear groups G by the second author in [8] (see also [9]) and for residually finite groups G by J. Wilson [1, Theorem 2.3.4]. Furthermore, N. S. Chernikov proved that every group G=AB with central-by-finite subgroups A and B is soluble-by-finite (see [1, Theorem 2.2.5]).
It is natural to consider first groups G=AB where the two factors A and B have abelian subgroups with small index. There are a few known results in the case when both factors A and B have an abelian subgroup of index at most 2. It was shown in [3] that G is soluble and metacyclic-by-finite if A and B have cyclic subgroups of index at most 2, and it is proved in [2] that G is soluble if A and B are periodic locally dihedral subgroups. A more general result that G=AB is soluble if each of the factors A and B is either abelian or generalized dihedral was obtained in [4] by another approach. Here a group is called generalized dihedral if it contains an abelian subgroup of index 2 and an involution which inverts the elements of this subgroup. Clearly dihedral groups and locally dihedral groups, i.e. groups with a local system of dihedral subgroups, are generalized dihedral.
We recall that a group is called quasicyclic (or a Prüfer group) if it is an infinite locally cyclic p-group for some prime p. It is well known that quasicyclic subgroups of abelian groups are their direct factors. Furthermore, it seems to be known and will be shown below that every non-abelian group having a quasicyclic subgroup of index 2 is either an infinite locally dihedral or a locally quaternion group. It should be noted that for each prime p, up to isomorphism, there exists a unique locally dihedral group whose quasicyclic subgroup is a p-group, and there is only one locally quaternion group. These and other details about such groups can be found in [6, pp. 45–50].
Theorem 1.1.
Let the group G=AB be the product of two subgroups A and B each of which is either abelian or has a quasicyclic subgroup of index 2. Then G is soluble with derived length at most 3. Moreover, if the subgroup B is non-abelian and X is its quasicyclic subgroup, then AX=XA is a metabelian subgroup of index 2 in G.
As a direct consequence of this theorem, we have an affirmative answer to Question 18.95 of the “Kourovka notebook” [7] posed by A. I. Sozutov.
Corollary 1.2.
If a group G=AB is the product of an abelian subgroup A and a locally quaternion subgroup B, then G is soluble.
It is also easy to see that if each of the factors A and B in Theorem 1.1 has a quasicyclic subgroup of index 2, then their quasicyclic subgroups are permutable. As a result of this the following holds.
Corollary 1.3.
Let the group G=A1A2⋯An be the product of pairwise permutable subgroups A1,…,An each of which contains a quasicyclic subgroup of index 2. Then the derived subgroup G′ is a direct product of the quasicyclic subgroups and the factor group G/G′ is elementary abelian of order 2m for some positive integer m≤n.
The notation is standard. If H is a subgroup of a group G and g∈G, then the normal closure of H in G is the normal subgroup of G generated by all conjugates of H in G, and gG is the conjugacy class of G containing g, respectively.
2 Preliminary lemmas
Our first lemma lists some simple facts concerning groups with quasicyclic subgroups of index 2 which will be used without further explanation.
Lemma 2.1.
Let G be a non-abelian group containing a quasicyclic p-subgroup X of index 2 and y∈G∖X. Then y2∈X and the following statements hold:
Every subgroup of X is characteristic in G.
The group G is either locally dihedral or locally quaternion.
The derived subgroup G′ coincides with X.
Every proper normal subgroup of G is contained in X.
If G is locally quaternion, then p=2,
y4=1,
xy=x-1 for all x∈X,
the center Z(G) coincides with 〈y2〉 and is contained in every non-trivial subgroup of G,
the coset yX coincides with the conjugacy class yG=yX.
If G is locally dihedral, then y2=1,
xy=x-1 for all x∈X,
Z(G)=1 and the coset yX coincides with the conjugacy class yG=yX for p>2 and Z(G) is the subgroup
of order
2
in X for p=2.
The factor group G/Z(G) is locally dihedral.
Proof.
In fact, only statement (2) needs an explanation. Clearly G=X〈y〉 for some y∈G with y2∈X and each cyclic subgroup 〈x〉 of X is normal in G. Therefore for p>2 we have y2=1 and either xy=x or xy=x-1. Since X contains a unique cyclic subgroup of order pn for each n≥1, the equality xy=x for some x≠1 holds for all x∈X, contrary to the hypothesis that G is non-abelian. Therefore xy=x-1 for all x∈X and hence the group G is locally dihedral. In the case p=2 each subgroup 〈x〉 of X properly containing the subgroup 〈y2〉 has index 2 in the subgroup 〈x,y〉. If x is of order 2n for some n>3, then the element y can be chosen such that either y4=1 and 〈x,y〉 is a generalized quaternion group with xy=x-1 or y2=1 and 〈x,y〉 is one of the following groups: dihedral with xy=x-1, semidihedral with xy=x-1+2n-2 or a group with xy=x1+2n-2 (see [5, Theorem 5.4.3]). It is easy to see that from this list only generalized quaternion and dihedral subgroups can form an infinite ascending series of subgroups, so that the 2-group G can be either locally quaternion or locally dihedral, as claimed.
∎
Lemma 2.2.
Let G be a group and M an abelian minimal normal p-subgroup of G for some prime p. Then the factor group G/CG(M) has no non-trivial finite normal p-subgroup.
Proof.
Indeed, if N/CG(M) is a finite normal p-subgroup of G/CG(M) and x is an element of order p in M, then the p-subgroup K=〈xN〉 is finite and N acts on K as a finite p-group of automorphisms. Therefore the centralizer CK(N) of N in K is non-trivial and hence CM(N) is a non-trivial normal subgroup of G properly contained in M, contradicting the minimality of M.
∎
We will say that a subset S of G is normal in G if Sg=S for each g∈G which means that sg∈S for every s∈S.
Lemma 2.3.
Let G be a group, and let A and B be subgroups of G. If a normal subset S of G is contained in the set AB and S-1=S, then the normal subgroup of G generated by S is also contained in AB. In particular, if i is an involution with iG⊆AB and N is the normal closure of the subgroup 〈i〉 in G, then AN∩BN=A1B1 with A1=A∩BN and B1=AN∩B.
Proof.
If s,t∈S, then t=ab and (s-1)a=cd for some elements a,c∈A and b,d∈B. Therefore s-1t=s-1ab=a(s-1)ab=(ac)(db)∈AB and hence the subgroup 〈s∣s∈S〉 is contained in AB and normal in G. Moreover, if N is a normal subgroup of G and N⊆AB, it is easy to see that AN∩BN=(AN∩B)N=(AN∩B)N=(A∩BN)(AN∩B) (for details see [1, Lemma 1.1.4]).
∎
The following slight generalization of Itô’s theorem was proved in [8] (see also [9, Lemma 9]).
Lemma 2.4.
Let G be a group and let A,B be abelian subgroups of G. If H is a subgroup of G contained in the set AB, then H is metabelian.
3 The product of an abelian group and a group containing a quasicyclic subgroup of index 2
In this section we consider groups of the form G=AB with an abelian subgroup A and a subgroup B=X〈y〉 in which X is a quasicyclic p-subgroup of index 2 and y∈B∖X.
Lemma 3.1.
Let the group G=AB be the product of an abelian subgroup A and a non-abelian subgroup B with a quasicyclic p-subgroup X of index 2. If G has non-trivial abelian normal subgroups, then one of these is contained in the set AX.
Proof.
Suppose the contrary and let 𝒩 be the set of all non-trivial normal subgroups of G contained in the derived subgroup G′. Then AG=1 and ANX≠AX for each N∈𝒩. Since G=AB=AX∪AXy and AX∩AXy=∅, for every N∈𝒩 the intersection NX∩AXy is non-empty and so G=ANX. Moreover, as X=B′≤G′ by Lemma 2.1, it follows that G′=DNX with D=A∩G′. It is also clear that G=〈A,X〉, because otherwise 〈A,X〉=AX is a normal subgroup of index 2 in G. In particular, A∩X=1.
For each N∈𝒩 we put AN=A∩BN and BN=AN∩B. Then ANN=BNN=ANBN by [1, Lemma 1.1.4], and the subgroup BN is not contained in X, because otherwise N is contained in the set AX, contrary to the assumption. Let XN=BN∩X and CN=AN∩NXN. Then XN is a subgroup of index 2 in BN and CNN=NXN is a normal subgroup of G, because (NXN)X=NXN and (NCN)A=NCN. Put M=⋂N∈𝒩N.
Since G=ANX for each N∈𝒩, the factor group G/N is metabelian by Lemma 2.4. Therefore also the factor group G/M is metabelian and so its derived subgroup G′/M is abelian. Clearly if M=1, then D=A∩G′≤AG=1 and hence G′=⋂N∈𝒩NX=X, contrary to the assumption. Thus M is the unique abelian minimal normal subgroup of G. We show first that the centralizer CG(M) of M in G does not contain the subgroup X.
Indeed, otherwise the group G=A(MX) is metabelian by Itô’s theorem and so the derived subgroup G′=DMX is abelian. Since G=AG′, it follows that D=A∩G′≤AG=1 and so G′=MX. If M contains elements of order p, then it is an elementary abelian p-subgroup and hence X is the finite residual of G′. In the other case M has no element of order p, so that X is the maximal p-subgroup of G′. Therefore in both cases X is a characteristic subgroup of G′ and so normal in G, contrary to the assumption. Thus X≦̸CG(M) which implies in particular that the subgroup M is infinite and the centralizer CX(M) is finite.
Now, if M is a p-subgroup, then the factor group G¯=G/CG(M) has no non-trivial finite normal p-subgroup by Lemma 2.2. On the other hand, G=AMX and G′=DMX with D=A∩G′, so that G=AG′. Using bars for images under the homomorphism G→G¯, we derive that the group G¯=A¯X¯=A¯G¯′ is metabelian and the derived subgroup G¯′=D¯X¯ is abelian. Therefore the intersection A¯∩G¯′ is a central subgroup of G¯ and hence it has no subgroup of order p. Since D¯≤A¯∩G¯′, it follows that X¯ is the maximal p-subgroup of G¯′ and so normal in G¯. As X¯ is the union of its finite p-subgroups each of which is also normal in G¯, this implies X¯=1 and thus X≤CG(M), contrary to the above.
Suppose next that M is not a p-subgroup and so M has no element of order p. As was shown above, the subgroup MXM=CMM is normal in G. If XM=X, then G=A(MX)=A(CMM)=AM and so G′=M. But then X≤M which is not the case. Therefore the subgroup XM is finite of order pk for some k≥0. If the subgroup MXM is non-abelian, then its center is trivial, because the subgroup M is minimal normal in G. In particular, CM∩M=1 and hence the subgroups AM and BM are finite of order 2pk. As AMM=BMM=AMBM, the subgroup M is also finite which contradicts what has been proved above.
Thus the subgroup MXM is abelian and hence XM is its maximal p-subgroup. Therefore XM is normal in G and so XM=1 by assumption. Then AM∩X=1 and the subgroup BM=AM∩B is of order 2, because BM∩X=1 and M≦̸A by assumption. Therefore AM=ABM and BM=〈y〉 with y2=1. Since the subgroup B is non-abelian by the hypothesis of the lemma, it follows from Lemma 2.1 that B=X⋊〈y〉 is locally dihedral and so xy=x-1 for each x∈X. Furthermore, the index of A in AM is equal to 2 and so A∩M is a subgroup of index 2 in M. As M is abelian and minimal normal in G, it follows that M is an elementary abelian 2-subgroup. It is also clear that the subgroup AM is nilpotent and the intersection A∩M is centralized by y.
It was noted above that G=AMX and G′=DMX with D=A∩G′. Passing to the factor group G¯=G/M and using bars for images under the homomorphism G→G¯, we obtain that the group G¯=A¯X¯ is metabelian and so its derived subgroup G¯′=D¯X¯ is abelian. Since A is abelian, the subgroup D¯ is central in G¯ and thus the subgroup DM is normal in G. Furthermore, (DM)′≠M, because DM as a subgroup of AM is nilpotent. As M is the unique minimal normal subgroup of G, the subgroup DM must be abelian. But then D2=1, because D2=(DM)2 is a normal subgroup of G. Thus X¯ is the maximal p-subgroup of G¯′. Since p≠2 and X is quasicyclic, this means that each subgroup of X¯ is characteristic in G¯′ and so normal in G¯. Therefore for each x∈X the subgroup M〈x〉 and MX itself are normal in G. In particular, for each g∈G there exists m∈M such that 〈x〉g=〈x〉m from which it follows that gm∈NG(〈x〉) and thus G=MNG(〈x〉). It is easily seen that M∩NG(〈x〉)=1 and hence M=CG(M), because otherwise the intersection CG(M)∩NG(〈x〉) is a non-trivial normal subgroup of G which does not contain M. Moreover, as G=AB and B is contained in NG(〈x〉), it follows that NG(〈x〉)=NA(〈x〉)B and so NA(〈x〉)G=NA(〈x〉)B is a normal subgroup of G contained in NG(〈x〉). Therefore we have NA(〈x〉)=1 and we conclude that G=AB=M⋊B, the subgroup B=X⋊〈y〉 is locally dihedral and A∩B=1.
Finally, taking an element x of order p in X and considering M as an irreducible B-module, we derive from Clifford’s theorem (see [5, Theorem 4.1]) that M is decomposed in an infinite direct product M=M1×…×Mi×⋯ of finite 〈x〉-invariant subgroups Mi. Furthermore, it was proved above that M=CG(M) and A∩M has index 2 in M and is centralized by y. This gives [M,y]=〈a〉 for some involution a∈A∩M. Let N be one of the subgroups Mi which does not contain a. Then the subgroup Ny is also 〈x〉-invariant and (A∩N)y=A∩N≠1. Therefore Ny=N and so [N,y]≤〈a〉∩N=1. But then we have 1=[N,y]x=[N,yx]=[N,yx2]=[N,x2] and so the centralizer CM(x) contains N. Since M〈x〉 is a normal subgroup of G, so is CM(x) and thus CM(x)=M. This final contradiction completes the proof.
∎
It should be noted that if in Lemma 3.1 the subgroup B is locally dihedral, then the group G=AB is soluble by [4, Theorem 1.1]. Therefore the following assertion is an easy consequence of this lemma.
Corollary 3.2.
If the group G=AB is the product of an abelian subgroup A and a locally dihedral subgroup B containing a quasicyclic subgroup X of index 2, then AX=XA is a metabelian subgroup of index 2 in G.
Proof.
Indeed, let H be a maximal normal subgroup of G with respect to the condition H⊆AX. If X≤H, then AH=AX is a metabelian subgroup of index 2 in G by Itô’s theorem. In the other case the intersection H∩X is finite and hence HX/H is the quasicyclic subgroup of index 2 in BH/H. Since G/H=(AH/H)(BH/H) is the product of the abelian subgroup AH/H and the locally dihedral subgroup BH/H, the set (AH/H)(HX/H) contains a non-trivial normal subgroup F/H of G/H by Lemma 3.1. But then F is a normal subgroup of G which is contained in the set AX and properly contains H. This contradiction completes the proof.
∎
In the following lemma G=AB is a group with an abelian subgroup A and a locally quaternion subgroup B=X〈y〉 in which X is the quasicyclic 2-subgroup of index 2 and y is an element of order 4, so that xy=x-1 for each x∈X and z=y2 is the unique involution of B. It turns out that in this case the conjugacy class zG of z in G is contained in the set AX.
Lemma 3.3.
If G=AB and A∩B=1, then the intersection zA∩AXy is empty.
Proof.
Suppose the contrary and let za=bxy for some elements a,b∈A and x∈X. Then b-1z=(xy)a-1 and from the equality (xy)2=z it follows that (b-1z)4=1 and b-1zb-1z=za-1. Therefore we have b-1zab-1=zza and hence bzab=zaz. As za=bxy, we have b(bxy)b=(bxy)z and so bxyb=xyz. Thus (xy)-1b(xy)=zb-1. Furthermore, we have bxyb-1=(zb-1)a, so that bzb-1=((zb-1)2)a=(xy)-ab2(xy)a, i.e. the elements z and b2 are conjugate in G by the element g=b-1(xy)-a. Since g=cd for some c∈B and d∈A, we have b2=zg=zd and so z=(b2)d-1=b2, contrary to the hypothesis of the lemma. Thus zA∩AXy=∅, as desired.
∎
Theorem 3.4.
Let the group G=AB be the product of an abelian subgroup A and a locally quaternion subgroup B. If X is the quasicyclic subgroup of B, then AX=XA is a metabelian subgroup of index 2 in G. In particular, G is soluble of derived length at most 3.
Proof.
Let Z be the center of B, N the normal closure of Z in G and X=B′, so that X is the quasicyclic subgroup of index 2 in B. If A∩B≠1, then Z is contained in A∩B by statement (4) of Lemma 2.1 and so N=Z. Otherwise it follows from Lemma 2.3 that N=ZG=ZA is contained in the set AX. Then N is a metabelian normal subgroup of G by Lemma 2.4 and the factor group BN/N is locally dihedral by statement (7) of Lemma 2.1. Since the factor group G/N=(AN/N)(BN/N) is the product of an abelian subgroup AN/N and the locally dihedral subgroup BN/N, it is soluble by [4, Theorem 1.1], and so the group G is soluble.
Now if X≤N, then AN=AX is a metabelian subgroup of index 2 in G and so the derived length of G does not exceed 3. In the other case the intersection N∩X is finite and hence NX/N is the quasicyclic subgroup of index 2 in BN/N. Therefore AX=XA by Corollary 3.2 and this completes the proof.
∎
4 The product of groups each of which is locally quaternion or generalized dihedral
Since the groups of the form G=AB with two generalized dihedral subgroups A and B are soluble by [4, Theorem 1.1], in this section we consider the remaining cases in which the subgroup A is locally quaternion and B is either generalized dihedral or locally quaternion. The main part is devoted to the proof that every group G of this form has a non-trivial abelian normal subgroup.
In what follows up to Theorem 4.5G=AB is a group in which A=Q〈c〉 with a quasicyclic 2-subgroup Q of index 2 and an element c of order 4 such that ac=a-1 for each a∈Q and B=X⋊〈y〉 with an abelian subgroup X and an involution y such that xy=x-1 for each x∈X.
Let d=c2 denote the involution of A. The following assertion is concerned with the structure of the centralizer CG(d) of d in G. It follows from statement (4) of Lemma 2.1 that the normalizer of every non-trivial normal subgroup of A is contained in CG(d).
Lemma 4.1.
The centralizer CG(d) is soluble.
Proof.
If Z=〈d〉, then the factor group CG(d)/Z=(A/Z)(CB(d)Z/Z) is a product of the generalized dihedral subgroup A/Z and the subgroup CB(d)Z/Z which is either abelian or generalized dihedral. Therefore CG(d)/Z and thus CG(d) is a soluble group by [4, Theorem 1.1], as claimed.
∎
The following lemma shows that if G has no non-trivial abelian normal subgroup, then the index of A in CG(d) does not exceed 2.
Lemma 4.2.
If CB(d)≠1, then either CX(d)=1 or G contains a non-trivial abelian normal subgroup.
Proof.
If X1=CX(d), then X1 is a normal subgroup of B and CG(d)=ACB(d). Therefore the normal closure N=X1G is contained in CG(d), because X1G=X1BA=X1A. Since CG(d) and so N is a soluble subgroup by Lemma 4.1, this completes the proof.
∎
Consider now the normalizers in A of non-trivial normal subgroups of B.
Lemma 4.3.
Let G have no non-trivial abelian normal subgroup. If U is a non-trivial normal subgroup of B, then NA(U)=1. In particular, A∩B=1.
Proof.
If NA(U)≠1, then d∈NA(U) and so the normal closure 〈d〉G=〈d〉B is contained in the normalizer NG(U)=NA(U)B. Since NA(U)≠A, the subgroup NA(U) is either finite or quasicyclic, so that NG(U) and thus 〈d〉G is soluble. This contradiction completes the proof.
∎
Lemma 4.4.
If CX(d)=1, then G contains a non-trivial abelian normal subgroup.
Proof.
Since G=AB, for each x∈B there exist elements a∈A and b∈B such that dx=ab. If b∉X, then b=a-1dx is an element of order 2 and so dxadx=a-1. As a2k=d for some k≥0, it follows that dxddx=d and hence ab=dx=(dx)d=(ab)d=abd. Therefore bd=b and so b∈CB(d). In particular, if CB(d)=1, then b∈X, so that in this case the conjugacy class dG=dB is contained in the set AX.
Assume that CB(d)≠1 and the group G has no non-trivial normal subgroup. Then CX(d)=1 by Lemma 4.2 and without loss of generality CB(d)=〈y〉. Then G=(A〈y〉)X and so the quasicyclic subgroup Q of A is normalized by y. In particular, dy=d and the subgroup Q〈y〉 can be either abelian or locally dihedral. We consider first the case when y centralizes Q and show that in this case the conjugacy class dG is also contained in the set AX.
Indeed, otherwise there exist elements a∈A and b,x∈B such that dx=ab and b∉X. Then b∈CB(d)=〈y〉 by what was proved above, so that b=y and dx=ay. As dB=d〈y〉X=dX, we may suppose that x∈X. But then dx-1=(dx)y=ay=dx and hence dx2=d. Therefore we have x2∈〈y〉 and so x2=1. In particular, if X has no involution, then dG=dX⊆AX. We show next that the case with an involution x∈X cannot appear.
Clearly in this case x is a central involution in B and so the subgroup D=〈d,x〉 generated by the involutions d and x is dihedral. It is easy to see that d and x cannot be conjugate in G and the center of D is trivial, because otherwise the centralizer CG(x) properly contains B, contradicting Lemma 4.3. Thus dx is an element of infinite order and so D=〈dx〉⋊〈x〉 has no automorphism of finite order more than 2. On the other hand, if u∈A, v∈B and uv normalizes D, then 〈d,xu〉=Du=Dv-1=〈dv-1,x〉 and so D≤Du. Since u is an element of finite order, it follows that D=Du=Dv-1 and thus NG(D)=NA(D)NB(D). Therefore NA(D)=〈d〉 and hence z=(dx)2 is an element of infinite order in NB(D). But then z∈X and so 〈z〉 is a normal subgroup of B normalized by d, again contradicting Lemma 4.3. Thus X has no involution, as claimed.
Finally, if N is the normal closure of the subgroup 〈d〉 in G, then AN=NX=A1X1 with A1=A∩NX and X1=AN∩X by Lemma 2.3. Therefore the subgroup A1X1 is soluble by Theorem 3.4, so that N and hence G has a non-trivial abelian normal subgroup, contrary to our assumption.
Thus the subgroup Q〈y〉 is locally dihedral and so y inverts the elements of Q. Since A=Q〈c〉 with ac=a-1 for all a∈A, the element cy centralizes Q and hence the subgroup Q〈cy〉 is abelian. But then the group G=(Q〈cy〉)B as the product of an abelian and a generalized dihedral subgroup is soluble by [4, Theorem 1.1]. This final contradiction completes the proof.
∎
Theorem 4.5.
Let the group G=AB be the product of a locally quaternion subgroup A and a generalized dihedral subgroup B. Then G is soluble. Moreover, if B has a quasicyclic subgroup of index 2, then G is metabelian.
Proof.
If A∩X≠1, then the centralizer CG(d) is of index at most 2 in G and so G is soluble by Lemma 4.1. Let N be a normal subgroup of G maximal with respect to the condition A∩NX=1. Then BN=(A∩BN)B and the subgroup A∩BN is of order at most 2. Therefore the subgroup N is soluble and the factor group G/N=(AN/N)(BN/N) is the product of the locally quaternion subgroup AN/N and the subgroup BN/N which is either abelian or generalized dihedral. Hence it follows from Theorem 3.4 and Lemmas 4.2 and 4.4 that G/N has a non-trivial abelian normal subgroup M/N. Put L=MQ∩MX, Q1=Q∩MX and X1=MQ∩X. We have L=MQ1=MX1 and Q1≠1, because A∩MX≠1 by the choice of M. It is also clear that L is a soluble normal subgroup of G, because (MQ1)A=MQ1 and (MX1)B=MX1. Therefore the factor group G/L and so the group G is soluble if AL/L is of order 2. In the other case AL/L is locally dihedral and BL/L is abelian or generalized dihedral. Since G/L=(AL/L)(BL/L), it follows that G/L and so G is soluble by [4, Theorem 1.1]. Moreover, if the subgroup X is quasicyclic, then the subgroups Q and X centralize each other by [1, Corollary 3.2.10], so that QX is an abelian normal subgroup of index 2 or 4 in G and thus G is metabelian.
∎
Our last theorem describes the structure of groups which are products of two locally quaternion subgroups.
Theorem 4.6.
Let the group G=AB be the product of two locally quaternion subgroups A and B. If X and Y are quasicyclic subgroup of A and B, respectively, then XY=YX is an abelian subgroup of index 2 or 4 in G. In particular, G is metabelian.
Proof.
Let x and y be the unique involution of A and B, respectively. If G is soluble, then XY=YX by [1, Corollary 3.2.10]. We show now that the group G satisfies this condition.
Indeed, if A∩B≠1, then x=y is a central involution of G and the factor group G/〈x〉=(A/〈x〉)(B/〈x〉) is the product of two locally dihedral subgroups A/〈x〉 and B/〈x〉. Therefore G is soluble by [4, Theorem 1.1].
Let A∩B=1 and D=〈x,y〉. Then D is a dihedral subgroup of G and the normalizer NG(D) can be written in the form NG(D)=NA(D)NB(D) by [1, Lemma 1.2.2 (i)]. It is easy to see that NA(D)∩D=〈x〉 and NB(D)∩D=〈y〉, so that each of the factor groups NA(D)D/D and NB(D)D/D is either abelian or locally dihedral. Since NG(D)/D=(NA(D)D/D)(NB(D)D/D), the factor group NG(D)/D and so also NG(D) is soluble by [4, Theorem 1.1]. Then NG(D) is a 2-group by [1, Corollary 3.2.7], and hence D is a dihedral 2-subgroup of G. Therefore D contains a central involution z which is different from x and y. As z=ab for some a∈A and b∈B, it follows that b≠1 and x=xab=xb. But then x=xy, because y∈〈b〉, so that
D=〈x〉×〈y〉 and CG(D)=CA(y)CB(x) is a soluble 2-subgroup.
It is clear that if CG(D) is of finite index in G, then G is soluble. In the other case one of the centralizers CA(y) and CB(x), for example the second one, must be finite and thus the centralizer CG(x)=ACB(x) is soluble. But then the normal closure N=〈y〉G=〈y〉A≤ACB(x) of 〈y〉 in G is also soluble. Furthermore, in the factor group G/N=(AN/N)(BN/N the subgroup AN/N is either locally quaternion or locally dihedral and BN/N is locally dihedral. Therefore G/N and so also G is soluble by Theorem 4.5 or by [4, Theorem 1.1], as claimed.
∎
The proof of Theorem 1.1 is completed by a direct application of Corollary 3.2 and Theorems 3.4, 4.5 and 4.6.
and
Yaroslav P. Sysak
