Sylow permutability in generalized soluble groups

A finite group G is a soluble PST-group if and only if γ∞⁢(G) is abelian of odd order, |γ∞⁢(G)| and |G:γ∞(G)| are relatively prime and elements of G induce power automorphisms in γ∞⁢(G).

Let G be a finitely generated nilpotent group.

1. If π is an infinite set of primes, then ⋂p∈πGp is finite.

2. For each prime p there is an integer kp>0 such that Gpkp∩Gp=1.

Proof.

(i) Clearly we may assume that G is torsion-free. It follows from the theory of basic commutators – see [6, Theorem 12.3.1] – that Gp consists of all pth powers of elements of G, provided that p is greater than c, the nilpotent class of G. Thus Gp∩Z⁢(G)=Z⁢(G)p since G/Z⁢(G) is torsion-free. Writing I=⋂p∈π,p>cGp, we have

and consequently I=1.

(ii) Let 1≠g∈Gp. Since G is residually finite, there exists Lg◁G which is maximal with respect to g∉Lg and G/Lg being finite. Now G/Lg cannot have two non-trivial primary components and g⁢Lg is a p-element. Hence G/Lg is a p-group. Define M⁢(p)=⋂1≠g∈GpLg and note that G/M⁢(p) is a finite p-group. Hence there exists kp>0 such that Gpkp≤M⁢(p). Finally,

A group G is of infinite dihedral type if and only if it has a normal subgroup A such that:

1. A is a finitely generated, infinite abelian group containing no involutions,

2. G/A is a finite 2 -group and |G:CG(A)|=2,

3. elements in G\CG⁢(A) induce inversion in A.

Proof.

Assume that G is of infinite dihedral type, so the hypercentre H of G is a finite 2-group and G/H=Dih⁢(B) where B is a finitely generated, infinite abelian group without involutions. There is a nilpotent normal subgroup C such that H≤C<G, where C/H=B, G=〈t,C〉, t inverts in C/H and t2∈H. Next C2k∩H=1 for some k>0 by Lemma 1, so we have

Hence t inverts in A=C2k. Also [A,C]≤A∩C′≤A∩H=1, so we have |G:CG(A)|=2. Thus G has the stated property.

Conversely, assume that there is a finitely generated, infinite abelian normal subgroup A without involutions such that the quotient G/A is a finite 2-group, |G:C=CG(A)|=2, and t∈G\C induces inversion in A. We can write A in the form F×D, where F is free abelian and D has odd order. Then A≤Z⁢(C), so C/Z⁢(C) is a finite 2-group, as is C′. Therefore C′≤C2=H, say. By Lemma 1 there is a k>0 such that C2k∩H=1 and also C2k≤A. Hence we have

Therefore t inverts in B=C/H. Observe that t is a 2-element and t2∈C, so t2∈H. This shows that G/H≃Dih⁢(B). Notice that [H,rG]≤A∩H=1 for some r>0, so H is contained in the hypercentre of G; indeed H coincides with the hypercentre since Z⁢(Dih⁢(B))=1. Therefore G is of infinite dihedral type. ∎

If G is a group of infinite dihedral type, then all its finite quotients are PST-groups.

Proof.

By Lemma 2 there is a finitely generated, infinite abelian normal subgroup A◁G, without involutions, such that G/A is a finite 2-group, |G:CG(A)|=2 and t∈G\CG⁢(A) induces inversion in A. Let G/M be a finite quotient; then Am≤M for some m>0 and G¯=G/Am is finite. By Proposition 1 it is enough to prove that G¯ is a PST-group. Let B/Am=O2′⁢(A/Am). Then G/B is a finite 2-group and γ∞⁢(G¯)≤B¯. Since t inverts in A and |B¯| is odd, [B¯,t¯]=B¯, so B¯=γ∞⁢(G¯) and by Proposition 1 we conclude that G¯ is a PST-group. ∎

If G is a group of infinite dihedral type, then G is not a PST-group.

Proof.

There is a finitely generated, infinite abelian normal subgroup A without involutions such that G/A is a finite 2-group, |G:CG(A)|=2 and t∈G\CG⁢(A) induces inversion in A. Write A=F×D, where F is free abelian of positive rank and D has odd order. Let a∈F\F2; then (t⁢a)2=t2, so t and ta are 2-elements of equal order. Let P be a Sylow 2-subgroup containing ta and put |P|=2k. Set H=〈t,A2k+1〉, noting that H is subnormal in 〈t,A〉 and hence in G.

Now assume that G is a PST-group; then, since H is subnormal in G, it is Sylow-permutable and H⁢P=P⁢H=J, say. Notice that a∈J since t∈H and t⁢a∈P. Hence J=H⁢P=〈t〉⁢〈a〉⁢A2k+1⁢P. Now

while on the other hand

Now 〈t〉∩A=1 since t is a 2-element. Hence

since a∈F\F2. Therefore |J:H|≥|〈a〉:〈a2k+1〉|=2k+1, a contradiction which shows that G is not a PST-group. ∎

Let G be a polycyclic group. If every finite quotient of G is a PST-group, then G is finite soluble PST or nilpotent or a group of infinite dihedral type.

Proof.

Since finite soluble PST-groups are supersoluble by Agrawal’s theorem (see Proposition 1), every finite quotient of G is supersoluble, and hence G is supersoluble by a theorem of Baer [2]. We assume that G is infinite and argue by induction on the Hirsch number. Thus G has an infinite cyclic normal subgroup N=〈z〉 and by induction hypothesis G/N is finite or nilpotent or a group of infinite dihedral type. Each of the three possibilities will be handled separately.

(i) Case: G/N is finite. We show that G is either nilpotent or of infinite dihedral type. Suppose first that N≤Z⁢(G). If G is not nilpotent, γ∞⁢(G/N)≠1; let p be a prime dividing |γ∞⁢(G/N)|. Now G¯=G/Np is a finite PST-group and p divides |γ∞⁢(G¯)|, so it cannot divide |G¯:γ∞(G¯)| by Proposition 1. Therefore N¯≤γ∞⁢(G¯). However, N¯≤Z⁢(G¯) and γ∞⁢(G¯)∩Z⁢(G¯)=1 by Proposition 1, so we have a contradiction. Hence G is nilpotent in this case. Therefore we may assume that N≰Z⁢(G).

Set C=CG⁢(N), so that |G:C|=2 and G=〈t,C〉, where t induces inversion in N. Let p be an odd prime and consider G¯=G/Np, which is a finite soluble PST-group. Since t inverts in N¯, we have 1≠N¯≤γ∞⁢(G¯). Therefore G⁢(p)¯≤γ∞⁢(G¯), where G⁢(p)/N is a Sylow p-subgroup of G/N. By Proposition 1, G⁢(p)◁G and G⁢(p)¯ is abelian. Therefore A/N=O2′⁢(G/N) is abelian and G/A is a finite 2-group. Clearly A≤C. Next C/A induces a 2-group of power automorphisms in each G⁢(p)¯, but it also centralizes N¯. Therefore C centralizes G⁢(p)¯ and hence A/N. Thus [A,C]≤N. But C′ is finite since N≤Z⁢(C), so [A,C]=1. It follows that CG⁢(A)=C. Thus A is abelian and clearly it does not contain involutions. Also t induces a power automorphism of order 1 or 2 in each G⁢(p)¯, since t inverts in N¯. Therefore it must invert in A/N, as well as in N. A simple argument now shows that t inverts in A. By Lemma 2 the group G is of infinite dihedral type.

(ii) Case: G/N is nilpotent. Here we may assume that N≰Z⁢(G) since otherwise G is nilpotent. Put C=CG⁢(N), so that |G:C|=2 and C is nilpotent. Also G=〈t,C〉 where t inverts in N.

Let p be an odd prime and set G¯=G/Cp, which is a finite PST-group. Since G/N is nilpotent, γ∞⁢(G¯)≤N¯≤C¯. Suppose that γ∞⁢(G¯)≠1; then p divides |γ∞⁢(G¯)|, so it does not divide |G¯/γ∞⁢(G¯)|, which implies that γ∞⁢(G¯)=N¯=C¯, i.e., C=N⁢Cp. Since C/N is finitely generated and nilpotent, it follows that C/N, and hence G/N, is finite. By case (i) either G is nilpotent or it is a group of infinite dihedral type. Thus we may assume that γ∞⁢(G¯)=1, i.e., G/Cp is nilpotent for all p>2. Since t inverts in N, it follows that N¯=1 and hence N≤⋂p>2Cp. Lemma 1 gives the contradiction that N is finite.

(iii) Case: G/N is a group of infinite dihedral type. In the first place G is not nilpotent, as a group of infinite dihedral type cannot be nilpotent. By Lemma 2 there exists L◁G such that G/L is a finite 2-group, L/N is finitely generated, infinite abelian without involutions, |G:CG(L/N)|=2 and G=〈t,CG⁢(L/N)〉 where t induces inversion in L/N. Write K=CL⁢(N); then we have |L:K|≤2 and K/N has the same structure as L/N. Note also that G/K is a 2-group and CG⁢(L/N)=CG⁢(K/N). Thus we can replace L by K without loss: from this point on we assume that N≤Z⁢(L); thus L is nilpotent of class ≤2.

Suppose that L is non-abelian. If ℓ1,ℓ2∈L, then ℓit=ℓi-1⁢ai for some ai∈N and

Thus [L′,t]=1 and, since 1≠L′≤N, it follows that [N,t]=1. Let p be an odd prime and put G¯=G/Lp; then γ∞⁢(G¯)≤L¯ since G/L is a 2-group. Now γ∞⁢(G¯) cannot be trivial: for if it were, since t¯ inverts in L¯/N¯, we would have L¯=N¯, i.e., L=Lp⁢N. But this implies that L/N is finite. Therefore γ∞⁢(G¯)≠1 and p divides |γ∞⁢(G¯)|, so it cannot divide |G¯:γ∞(G¯)|. It follows that γ∞⁢(G¯)=L¯, so L¯ is abelian by Proposition 1. Thus L′≤⋂p>2Lp, which is finite. But L′≤N, so it follows that L′=1 and L is abelian.

Clearly L contains no involutions. The next step is to prove that CG⁢(L/N) is equal to CG⁢(L); let x∈CG⁢(L/N) and let p be an odd prime. Consider the finite PST-group G¯=G/Lp. Now G¯ cannot be nilpotent since t inverts in L/N, and we have seen that L¯=N¯ is impossible. Hence γ∞⁢(G¯)≠1 and p divides |γ∞⁢(G¯)|, so it does not divide |G¯:γ∞(G¯)|. As γ∞⁢(G¯)≤L¯, we obtain γ∞⁢(G¯)=L¯ and consequently x¯ induces a power automorphism of p′-order in L¯. Commuting with x¯ yields an endomorphism θ of L¯ with the property Im⁢(θ)≤N¯<L¯. Hence we have 1≠Ker⁢(θ)=CL¯⁢(x¯). Therefore [L¯,x¯]=1 and [L,x]≤Lp for all p>2, showing that [L,x]≤⋂p>2Lp, which is finite. But [L,x]≤N, so [L,x]=1 and x∈CG⁢(L). It follows that CG⁢(L)=CG⁢(L/N) and G=〈t,CG⁢(L)〉.

We know that t inverts in L/N; it will be shown next that it also inverts in N. Suppose this is false, so that zt=z. Let p be an odd prime not in π⁢(L/N) and consider the group G¯=G/Lp. Now z∉Lp since p∉π⁢(L/N). Hence we have 1≠N¯≤Z⁢(G¯). Next, if γ∞⁢(G¯)=1, then G/Lp is nilpotent and thus L=Lp⁢N since t inverts in L/N. This is impossible, so γ∞⁢(G¯)≠1. Hence γ∞⁢(G¯)=L¯ and 1≠N¯≤γ∞⁢(G¯)∩Z⁢(G¯)=1, a contradiction. Therefore t must invert in N and consequently L(t+1)2=1.

To complete the proof observe that some power t2k with k>0 belongs to L, and hence to N, from which it follows that t2k=1. Also t2 centralizes L/N and N. Hence

and since [L,t2]≤N, it follows that [L,t2]=1. Since L(t+1)2=1=Lt2-1, we conclude that L2⁢(t+1)=1 and Lt+1=1, i.e., t inverts in L. It follows via Lemma 2 that G is a group of infinite dihedral type. ∎

Proof of the theorem.

Let G be a finitely generated, hyperabelian group whose finite quotients are PST-groups. By Proposition 3 it is enough to prove that G is polycyclic, so assume this is false. Then, since polycyclic groups are finitely presented, there is a largest M◁G such that G/M is not polycyclic, and we can assume that M=1, so that G is just non-polycyclic, i.e., G is not polycyclic, but every proper quotient of G is polycyclic. By Baer’s theorem G is even just non-supersoluble. Since G is hyperabelian, there is an infinite abelian normal subgroup N and G/N is supersoluble; hence G is soluble.

At this point we invoke the theory of soluble just non-supersoluble groups given in [15, Sections 3.4, 2.4]. According to this there is a normal subgroup A≃ℚπ, the additive group of π-adic rationals with π a finite set of primes, and a free abelian subgroup X of finite rank such that |G:XA| finite and X∩A=1. Without loss we may assume that G=X⁢A, so G is metabelian. Let {x1,x2,…,xr} be a basis of X and, with p be an odd prime not in π, define

Since A/Ap≃ℤp, we have [A,Y]≤Ap and Y⁢Ap◁G. Consider G¯=G/Y⁢Ap, which is a finite PST-group, and note that γ∞⁢(G¯)≤A¯. Suppose that γ∞⁢(G¯)≠1; then p divides |γ∞⁢(G¯)|, so it cannot divide |G¯:γ∞(G¯)|. However, G¯/γ∞⁢(G¯) maps homomorphically onto X/Y, and hence onto ℤp. By this contradiction γ∞⁢(G¯)=1 and, since |A¯|=p, it follows that [A¯,G¯]=1. Hence

and G is nilpotent. But then A is finitely generated, which is a contradiction.

The converse statement follows from Lemma 3. ∎

There is an algorithm which, when a finite presentation of a polycyclic group G is given, decides if all the finite quotients of G are PST-groups.

Proof.

By the theorem it is enough to decide if G is finite or nilpotent or a group of infinite dihedral type. Algorithms to decide finiteness and nilpotence were given in [4]. Thus it remains to decide if G is of infinite dihedral type and for this purpose the original definition of these groups is convenient. Using other algorithms in [4], we can find the hypercentre F of G and check to see if it is a 2-group. If this is the case, pass to the group G¯=G/F and compute its Fitting subgroup F¯. Next determine whether F¯ is an infinite abelian group without involutions. If this is the case, check to see if |G¯:F¯|=2. If so, choose an element t in G¯\F¯ and determine whether t inverts each generator of F¯. If this happens, then G is a group of infinite dihedral type. ∎