Cyclic quasinormal subgroups of arbitrary groups

Cyclic Quasinormal Subgroups of Arbitrary Groups (*).

STEWARTE. STONEHEWER(**) - GIOVANNIZACHER(***)

To Professor Guido Zappa on his 90th birthday

ABSTRACT - In recent years several papers have appeared showing how cyclic quasinormal subgroups are embedded in finite groups and many structure theorems have been proved. The purpose of the present work is twofold. First we show that, without exception, all of these theorems remain valid for finite cyclic quasinormal subgroups ofinfinitegroups. Secondly we obtain analogous results for infinite cyclic quasinormal subgroups, where the statements turn out to be even stronger.

1. Introduction and statement of results.

LetAbe a cyclicquasinormalsubgroup of a groupG. Thus for every subgroupXofG,AXˆXAˆ hA;Xi. WhenGis finite, then the structure of the normal closureA^GofAinGis quite well understood. IfAhas odd order, then [A;G] is abelian andAacts on it by conjugation as a group of power automorphisms ([2]). WhenAhas even order, then [A;G] is nilpo- tent of class at most 2. When dihedral actions are excluded in certain subgroups of productsAX, whereXis cyclic (see below), then [A;G]⁰has order at most 2 andAacts on [A;G]=[A;G]⁰as power automorphisms ([3]).

But always [A;G;A] is abelian and A acts on it again as power auto- morphisms ([4]).

Many of the arguments used in obtaining these results involved in- duction on group orders; and at the time, infinite groups did not appear to be the natural setting for the work. However, it has now transpired thatall

(*) 2000Mathematics Subject Classification: Primary: 20E15, 20E34.

(**) Indirizzo dell'A.: Mathematics Institute - University of Warwick, Coventry CV4 7AL (Gran Bretagna).

(***) Indirizzo dell'A.: Seminario Matematico - UniversitaÁ degli Studi di Padova, Via Belzoni, 7 - 35131 Padova.

the major results in [2], [3] and [4] are true without the hypothesis thatGis finite. A large part of the present work, therefore, is devoted to extending those earlier theorems to the case whereGis infinite. Many of the original proofs can be repeated almost verbatim, modulo some initial lemmas, and we shall endeavour to avoid tedious repetition. But of course when Gis infinite we have the new situation allowing the cyclic quasinormal sub- groupAto be infinite also. The arguments in this case are new and we give them in full. As one might intuitively expect, the structure ofA^Gis simpler whenAis infinite than whenAis finite.

Our work therefore divides naturally into three parts, dealing with the two cases when A is finite, corresponding to jAj odd and jAj arbitrary (analogous to the contents of [2], [3] and [4]); followed by the situation when Ais infinite. In general

all groups G are assumed to be arbitrary,

i.e. finite or infinite. For any setpof primes,Apdenotes thep-component of a cyclic groupA. Recall that all subgroups of a cyclic quasinormal subgroup are also quasinormal (see Lemma 2.3 below). Our main results are as follows.

THEOREM A. Let A be a finite cyclic quasinormal subgroup of odd order in a group G. Then[A;G]is abelian and A acts on it as a group of power automorphisms.

THEOREMB. Let A be a finite cyclic quasinormal subgroup of a group G. Suppose that each cyclic subgroup X of G is quasinormal in A2X, and let Nˆ[A;G]⁰. Then

(i) jNj42and N4A;

(ii) A acts on[A;G]=N as a group of power automorphisms;and (iii) [A;G]is abelian if and only if A acts on [A;G] as a group of power automorphisms.

In Example 5.2 of [3], a finite groupGof order 2¹⁷is constructed, with a cyclic quasinormal subgroupAof order 2⁷, in which [A;G]⁰has order 2 and the hypotheses of Theorem Bare satisfied.

THEOREMC. Let A be a finite cyclic quasinormal subgroup of a group G. Then

(i) [A;G]ˆ[A2;G][A2⁰;G];

(ii) [A2⁰;G]is an abelian 2⁰-group on which A acts as a group of power automorphisms;

(iii) [A2;G]is a2-group of class at most2;

(iv) [A2;G;A]is normal in G and lies in Z([A2;G]);and (v) A acts on[A2;G;A]as a group of power automorphisms.

In Theorems A, Band C, the power automorphisms areuniversal, i.e.

each element maps to thesamepower. Also one can easily check that if we form the product ofAwith any of the abelian subgroups on which we claim that Aacts as a group of power automorphisms, then that product is a modular group, i.e. its subgroups form a modular lattice.

THEOREM D. Let A be an infinite cyclic quasinormal subgroup of a group G. Then

(i) [A;G]is abelian and A acts on it as a group of power auto- morphisms;

(ii) [A;G]is periodic if and only if A\[A;G]ˆ1;and

(iii) when A\[A;G]6ˆ1, i.e. when[A;G]is not periodic, then A^Gis abelian.

We shall see in Section 4 that the power automorphisms in Theorem D (i) are not universal in general. It follows from Lemma 2.5 below that in all the above theorems, the normal closureA^GofAinGis locally nilpotent and so the periodic elements ofA^Gform a subgroupT, say. In [1], Theorem 2.2, Busetto shows that with the hypotheses of Theorem D, A^GˆAT, T is abelian andT4Z_v(G). Busetto also shows thatAacts onTas a group of power automorphisms and A^Gis a modular group. Clearly A^GˆA[A;G]

and when A\[A;G]ˆ1, then [A;G]ˆT. But in general [A;G]6ˆT. In- deed Example 5 in Section 5 shows thatT64[A;G] in general.

Sections 2, 3 and 4 of our paper will establish the above results se- quentially. The final Section 5 contains examples relevant to the many questions that arise naturally throughout our discussion. The notation is standard. ThusA^GandAGdenote, respectively, the normal closure and the core of a subgroup Ain a group G. When A is a quasinormal sub- group of G, then we shall write A qn G. Cyclic groups of order n and infinity will be denoted byC_nandC₁, respectively, and Q₈ denotes the quaternion group of order 8. The centre ofGisZ(G) and, for any ordinal a, thea-th term of the upper central series isZa(G). Ifpis prime andAis ap-group, thenV(A) denotes the subgroup generated by the elements of orderp. Whenpis a set of primes, thenp⁰is the complementary set. Also G_p is a maximal p-subgroup of G and G_p⁰ a maximal p⁰-subgroup. The intersection of the normalisers ofallthe subgroups of a groupGis the

norm, writtennorm(G). Finally the subgroup lattice ofGis denoted by L(G) and the lattice of all subgroups of Gcontaining a subgroupH by L(G:H).

2. Finite cyclic quasinormal subgroups of odd order.

Whenp is an odd prime, Lemma 2.3 of [2] considers a finitep-group GˆAX, whereAandXare cyclic subgroups. Here we need the following generalisation.

LEMMA2.1. Let GˆAX be a nilpotent group, where Aˆ haiis a cyclic p-subgroup, with p an odd prime, and Xˆ hxiis also cyclic (possibly in- finite). Then

(i) G is metacyclic and G⁰is a p-group;

(ii) every subgroup of G is quasinormal;

(iii) G⁰ˆ h[a;x]i;

(iv) for each integer i,h[aⁱ;x]i ˆ h[a;xⁱ]i ˆ h[a;x]ⁱi; and (v) each element of G⁰has the form[a;g], for some g2X.

PROOF. (i) IfX is finite, thenXˆXpXp⁰. SoGˆ(AXp)Xp⁰ and AXp is metacyclic, by [2], Lemma 2.3 (i). Thus G is metacyclic. Clearly G⁰4AX_p and so G⁰ is a p-group. If X is infinite, then A/G and both statements are immediate.

(ii) By (i) and Iwasawa's Theorem (see [11], Theorems 2.4.4 and 2.4.11), G is a modular group. Since G is also nilpotent, all its subgroups are subnormal and therefore quasinormal ([14], Proposition 1.7).

(iii) The easy argument is the same as in [2], Lemma 2.3 (iii).

(iv) WhenXis finite, then we argue as in (i) above and [2], Lemma 2.3 (iv). WhenXis infinite, thenA/GandX₁ˆX=C_X(A)C_pⁿ, for somen.

Again the result follows from [2], Lemma 2.3 (iv) applied to the split ex- tensionA^jX1.

(v) The argument of [2], Lemma 2.3 (v) applies, making use of (i), (iii)

and (iv) above. p

We shall establish Theorem A in a somewhat circuitous way, as in [2]. It turns out that each element of [A;G] has a special form and we deduce this by considering first the caseA\[A;G]ˆ1. In fact with this hypothesis we can prove the strongest statement of all our results for a very general si- tuation.

THEOREM2.2. Let A be any quasinormal subgroup of a group G and suppose that A\[A;G]ˆ1. Then[A;G]is abelian and A acts on it as a group of power automorphisms.

PROOF. LetXbe a subgroup of [A;G]. Then [A;X]4AX\[A;G]ˆX.

SoAand hence alsoA^Gnormalise all subgroups of [A;G]. Therefore [A;G]

is a Dedekind group. If [A;G] is not abelian, we factorGby the 2⁰-compo- nent of [A;G]. Then

[A;G]Q₈C₂C₂

and there is a cyclic subgroupYof order 4 in [A;G]. IfAdoes not centralise Y, thenA=CA(Y) has order 2 andAY=CA(Y) is the dihedral group of order 8, withA=C_A(Y) a non-central subgroup of order 2, contradicting the quasi- normality ofA. ThereforeAmust centralise [A;G] and thus so also doesA^G.

But then [A;G] is abelian, a contradiction. p

Next we considera cyclic quasinormal subgroup A of odd prime-power order in a group G satisfying the hypotheses of Theorem 2.2, i.e.

A\[A;G]ˆ1. LetAˆ hai. We will show that [A;G]ˆ f[a;g]jg2Gg:

…1†

In Lemma 2.4 of [2], this is proved for the case in which Gis a finitep- group, with p an odd prime, using the following two results (see [11], Lemma 5.2.11 and [1], respectively).

LEMMA2.3. If A is a cyclic quasinormal subgroup of a group G, then every subgroup of A is quasinormal in G.

LEMMA 2.4. Let A be a quasinormal subgroup of prime order p in a group G. Then A^Gis an elementary abelian p-group. Moreover, if A is not normal in G, then A^G4Z₂(G).

It is convenient to recall two more well-known results (see [9], [3] and [13], Lemma 2.1).

LEMMA 2.5. A quasinormal subgroup of a finite group is subnormal.

More generally, a quasinormal subgroup of any group is ascendant.

LEMMA2.6. Let A be a quasinormal subgroup of a group G and let X be an infinite cyclic subgroup such that A\Xˆ1. Then X normalises A.

The argument used to establish (1) for finitep-groupsGin Lemma 2.4 of [2] proceeds by induction on jGj. When G is infinite we can use the following (see [8], Lemma 7.1.9).

LEMMA 2.7. Suppose that a group Gˆ hA;g₁; :::;g_ni, where A qn G.

ThenjA^G:Ajis finite.

Thus in order to prove (1) whenGis infinite, clearly we may assume that Gis finitely generated. ThenjA^G:Ajis finite and soA^Gis finite. Therefore suppose thatGis a counter-example withjA^Gjminimal. LetA1ˆA_G(<A), A₂=A₁ˆV(A=A₁) andA₂ˆ ha₂i. Then by Lemmas 2.3 and 2.4

[A₂;G]A₁=A₁4Z(G=A₁):

Hence [A2;G;G]4A1\[A;G]ˆ1 and so [A2;G]4Z(G). Therefore there is an elementxinGsuch that [a2;x]2Z(G) and

[a₂;x]6ˆ1:

…2†

We would like to write [a₂;x] in the form [a;g], whereAˆ hai. In fact this can be achieved by generalising a result due to Busetto (see [1]; also [11], Theorem 5.2.12).

LEMMA2.8. Let A be a cyclic quasinormal subgroup of order pⁿ (p a prime) in a group G and let jA_Gj ˆp^m, where 04m<n. Then A4Z2n m(G).

PROOF. Busetto's theorem is the casemˆ0. Suppose that the result is false and letG be a counter-example with jAj minimal. Thenm51. So BˆV(A)/Gand by choice ofG

A=B4Z2(n 1) (m 1)(G):

…3†

Letg2G. Ifjgj ˆ 1orjgj ˆq^{`</sup>, whereqis a prime different fromp, then A<sup>g</sup>ˆA. The first case follows from Lemma 2.6 and the second from Lemma 2.5. Thus [A;g]4A<sup>p</sup>, by (3), and therefore [B;g]ˆ1. On the other hand, if jgj ˆp<sup>`}, thenB/hB;giimplies [B;g]ˆ1. It follows thatB4Z(G). Then

A4Z2n m(G), again by (3). p

Returning to (2) above, sinceA\[A;G]ˆ1,x does not normaliseA.

Then Lemma 2.8 shows thathA;xiis nilpotent. Therefore, by Lemma 2.1 (v), there is an elementy2 hxisuch that [a2;x]ˆ[a;y]. The argument of Lemma 2.4 in [2] now suffices to complete the proof of the following.

LEMMA 2.9. Let Aˆ hai be a cyclic quasinormal subgroup of odd prime-power order in a group G. Suppose that A\[A;G]ˆ1. Then each element of[A;G]has the form[a;g];for some element gin G.

We should point out that the hypothesis here thatAhasoddorder is necessary, even whenGis finite. This is illustrated in Example 2 in Sec- tion 5.

The next step is to remove the hypothesisA\[A;G]ˆ1 in Lemma 2.9.

LEMMA 2.10. Let Aˆ hai be a cyclic quasinormal subgroup of odd prime-power order in a group G and suppose that A is not normal in G.

Then each element of[A;G]has the form[a;g];for some g2G.

We have to exclude A/Ghere, as can be seen from the non-abelian group of order 6. In [2], Theorem 2.5, Lemma 2.10 is proved in the case whenGis ap-group. We wish to use the same argument here. SinceAis not normal in G, it follows from Lemma 2.8 thatA4Zm(G) for somem.

Then it is convenient for us to prove the following result.

Let Aˆ hai be a cyclic quasinormal subgroup of odd prime-power order in a group G and suppose that A4Z_m(G), for some finite m. Then each element of[A;G]has the form[a;g];for some element gin G.

Now we can prove this stronger form of Lemma 2.10 by means of the argument in Theorem 2.5 of [2]. There a counter-example was chosen with jAjminimal. Here we assume the above statement is false and so there is a finitely generated counter-example G. Then jA^Gj is finite and we can chooseGwithjA^Gjminimal. LetjAj ˆpⁿ. Then we need to know that the statement holds withAreplaced byA^p. But this must be the case, because j(A^p)^Gj<jA^Gj is an immediate consequence of the following elementary result.

LEMMA2.11. Let A be a cyclic quasinormal subgroup of finite order k in a group G. Then A^Ghas exponent k.

PROOF. Ifkˆ1, the Lemma is trivial. Thus we proceed by induction on kand assume the usual induction hypothesis. Letpbe a prime divisor ofk and let Bbe the subgroup of order p in A. Then B qn G and B^G is ele- mentary abelian, by Lemma 2.4. By inductionA^G=B^Ghas exponentk=pand

henceA^Ghas exponentk. p

Thus our argument above can continue as in Theorem 2.5 of [2] and so Lemma 2.10 follows. As a consequence we obtain

THEOREM2.12. Let A be a cyclic quasinormal subgroup of odd prime- power order pⁿin a group G. Then[A;G]is an abelian p-group of exponent dividingjAjand A acts on[A;G]as a group of power automorphisms.

The proof is exactly the same as that of Theorem 2.1 in [2], using Lemmas 2.10, 2.1 (iii) and 2.11 above. The power automorphisms here are universal, by [10], 13.4.3 (ii).

We can now extend Theorem 2.12 to the case whereAhas finite odd order, i.e. Theorem A. If AˆP₁P₂ P_s is the primary decom- position ofA, then [A;G]ˆ[P1;G][P2;G] [Ps;G] and eachPiis quasinormal inG. Theorem A follows from Theorem 2.12 by the easy di- vision algorithm argument which was used to establish Theorem 1.1 from Theorem 2.1 in [2]. The only difference here is thatGmay be infinite, but each step in the proof remains valid using the lemmas above.

3. Arbitrary finite cyclic quasinormal subgroups.

It suffices to prove Theorem Bfor the case whereAis a 2-group. For, if AˆPQ, withPthe 2-component andQthe 2⁰-component ofA, then P andQare quasinormal inG. Also [A;G]ˆ[P;G][Q;G]. By Theorem A, [Q;G] is abelian andQacts on it as a group of power automorphisms. By Lemma 2.5,Pcentralises [Q;G] andQcentralises [P;G]. Thus if Theorem Bholds for quasinormal 2-subgroups (Phere), then it holds generally. The statements referring to power automorphisms follow as in Theorem A.

Thus we may assume thatAin Theorem Bis a 2-group and we have the hypothesis

() X is quasinormal in AX for all cyclic subgroups X of G.

The key to making progress is again the understanding of the structure of the productsAX, whereXis a cyclic subgroup ofG. We need the ana- logue of Lemma 2.1 for the casepˆ2.

LEMMA 3.1. Let GˆAX, where Aˆ hai is a cyclic quasinormal 2- subgroup of G and Xˆ hxiis a cyclic subgroup. Then

(i) G is metacyclic and G⁰is a2-group;

(ii) G⁰ˆ h[a;x]i;

(iii) for each integer i,h[aⁱ;x]i ˆ h[a;x]ⁱi;

(iv) a conjugates[a;x]to a power congruent to1modulo4; and (v) each element of G⁰has the form[aⁱ;x];for some integer i.

If in addition()holds, then

(iii)⁰ for each integer i,h[a;xⁱ]i ˆ h[a;x]ⁱi;

(iv)⁰ x conjugates[a;x]to a power congruent to1modulo4; and (v)⁰ each element of G⁰has the form[a;xⁱ];for some integer i.

The proofs proceed by considering the cases X finite and X infinite separately. In the former,Gˆ(AX₂)X₂⁰; and in the latter,A/G. The arguments are straightforward, following the pattern established in Lemma 2.3 of [2], Lemma 2.3 of [3], Lemma 2.1 of [4] and Lemma 2.1 above, and may safely be omitted.

From (iii) we obtain

COROLLARY3.2. LetAˆ haibe a cyclic quasinormal2-subgroup of a groupG. Then[A; G]ˆ h[a; g]jg2Gi.

We have already reduced the proof of Theorem Bto the case where Aˆ haiis a 2-group. WhenGis also a finite 2-group, then Theorem Bis contained in Theorems 2.1 and 2.2 of [3]. The arguments given in Sections 2, 3 and 4 of [3], upto and including the proofs of Theorems 2.1 and 2.2, use the finiteness of G only for the purpose of applying induction. In the present context, following on from Lemma 3.1, whenever this is necessary, we may assume thatGis finitely generated. Then, as in Lemma 2.7 above, we deduce thatA^Gis finite and this suffices for the induction arguments of [3] to apply here. The presence of elements of infinite order inGis not a problem, essentially because they always normaliseA. Thus we can safely omit the remaining details of the proof of Theorem B, and thereby spare the reader much tedium.

In the same way, we can proceed as in Corollary 4.1 of [3] to obtain the following consequence of Theorem B.

COROLLARY3.3. LetA be a finite cyclic quasinormal subgroup of a groupGsatisfying(). ThenA\[A; G]4Z([A; G]).

We now pass to the proof of Theorem C. HereGis any group andAis a finite cyclic quasinormal subgroup of G. Our arguments reduce easily to the case whereAis a 2-group. Then we have the following.

THEOREM3.4. Let Aˆ haibe a finite cyclic2-subgroup, quasinormal in an arbitrary group G. Then

(i) [A;G;A]/G; and

(ii) [A;G;A]ˆ f[u;a]ju2[A;G]g.

THEOREM3.5. Let A be a finite cyclic2-subgroup, quasinormal in an arbitrary group G, and put Bˆ[A;G;A]. Then

(i) B4Z([A;G]);

(ii) A centralises[A;G]=B;

(iii) A acts by conjugation on B as a group of power automorphisms;

and

(iv) [A;G]is nilpotent of class at most2.

REMARK. As in Theorem 2.12, the power automorphisms here are universal.

WhenGis a finite 2-group, Theorem 3.5 is Theorem 1.3 of [4]. Using Lemma 3.1, Corollary 3.2 and Theorem 3.4 above, the argument of [4]

applies here, without modification, to establish Theorem 3.5.

PROOF OFTHEOREMC. We have

[A;G]ˆ[A₂A₂⁰;G]ˆ[A₂;G][A₂⁰;G];

because both factors are normal in G and are a 2-group and a 2⁰-group respectively. So (i) follows.

By Theorem A, [A2⁰;G] is abelian andA2⁰acts on it as a group of power automorphisms. Since (A₂)^G is a 2-group, A₂[A₂⁰;G]ˆA₂[A₂⁰;G].

Therefore (ii) is true. Part (iii) follows from Theorem 3.5 (iv). For part (iv), [A₂;G;A]ˆ[A₂;G;A₂]ˆB, say, sinceA₂⁰centralises [A₂;G]. By Theorem 3.4 (i),B/G; and by Theorem 3.5 (i),B4Z([A;G]).

Finally (v) follows from Theorem 3.5 (iii), again sinceA2⁰centralisesB.

p It remains to prove Theorem 3.4. When Gis a finite 2-group, this is Theorem 1.2 of [4]. For part (i) in [4], we were able to assume thatGis generated byAand at most 2 other elements. That argument did not re- quireGto be a 2-group or even finite, and so again we may assume that

Gˆ hA;x;yi:

…4†

By Lemma 2.7,A^Gis finite. Thus we suppose that Theorem 3.4 (i) is false and choose a counter-example (4) with jA^Gj minimal. Let Aˆ hai and

Bˆ[A;G;A]. ThenBGˆ1 andB6ˆ1. Two cases must be distinguished.

Case(i). Suppose that h[a;x]i \ h[a;y]i 6ˆ1. Ifa commutes with [a;x]

and [a;y], then, as in [4], it would follow thatBˆ1. (This is Lemma 4.2 below.) Thus we may assume that [a;x;a]6ˆ1. Then

V(h[a;x]i)ˆV(h[a;y]i)4B\Z(G);

contradictingB_Gˆ1.

Case (ii). Suppose that h[a;x]i \ h[a;y]i ˆ1. For each integer i50, writeA_iˆA²ⁱ. Chooseisuch that aacts non-trivially on [A_i;G] and tri- vially on [A_i‡1;G]. Then following the detailed analysis of the analogous situation in [4], we can show that a acts as a power automorphism on [A_i;G]. Thus suppose thataconjugates each element of [A_i;G] to itsn-th power and letLˆ[A_i;G]. Then

L^{n 1}4[L;A]4L^{n 1}

and so [L;A]ˆL^{n 1}/G. But 16ˆ[L;A]4B, again contradictingB_Gˆ1.

This completes the proof of Theorem 3.4 (i).

For Theorem 3.4 (ii), againAˆ haiis a finite cyclic 2-subgroup, qua- sinormal in an arbitrary groupG. We must show that

[A;G;A]ˆ f[u;a]ju2[A;G]g:

…5†

WhenGis a finite 2-group, we argued in [4] by induction onjAjand we do the same here. When jAj44, then j[a;g]j44, for all g2G, by Lemma 2.11. Thusacentralises all [a;g], by Lemma 3.1 (iv), so [A;G;A]ˆ1 and (5) is true. Therefore we suppose thatjAj58 and assume the usual induction hypothesis. This means that

[A²;G;A²]ˆ f[v;a²]jv2[A²;G]g ˆK;

say. By Theorem 3.4 (i),K/G. Also by Lemma 3.1 (ii),Anormalises each cyclic subgroup ofKand henceA^Gdoes the same. ThusK is abeliananda acts as a universal power automorphism on K. As in Theorem 3.5 (i), it follows that

[A;G;K]ˆ1:

…6†

Following the analogous argument in Theorem 1.2 (ii) of [4] (without modification), we deduce from (6) that each element of K has the form [u;a], with u2[A;G]. Recall that by Lemma 2.8, eitherA/G (in which case (5) is trivial) orA4Z_m(G), for some integerm. Thus we may assume the latter and so there is a central series of Gbetween 1 andK. Then a

simple induction on the length of this series allows us to assume that Kˆ1:

Again as in Theorem 1.2 (ii) of [4], we can now deduce from Lemma 3.1 that

[A;G;A²]ˆ h[a;g;a²]jg2Gi ˆL;

say. In turn this leads toL f[y;a]jy2[A²;G]g [A²;G;A]. But using Lemma 3.1, we can show thatLˆ[A;G;A]²/G, and so the Three Sub- group Lemma gives

[A²;G;A]4L:

Thus we have equality here and in exactly the same way as we were able to assume thatKˆ1, we may also assume that

Lˆ1:

Using Lemma 3.1 (iv), this leads to [A;G;A;[A;G]]ˆ1, from which we deduce (5). This completes the proof of Theorem 3.4 (ii). p REMARKS. IfAis aperiodic locally cyclicquasinormal subgroup of a groupG, then, for each primep, thep-componentSofAis quasinormal in G(see [1], Proposition 1.6, or [11], Lemma 6.2.16). Moreover ifS^pˆS, then S/G, as a consequence of Lemma 2.7. Then applying Theorems A, Band C to the p-components of Amakes the structure of A^G quite transparent.

Also all the subgroups ofAare quasinormal inG, by Lemma 2.3; and ifNis the join of all the normalp-components ofA, then A^G=N4Z_v(G=N), by Lemma 2.8.

4. Infinite cyclic quasinormal subgroups.

We move on now to consider an infinite cyclic quasinormal subgroupA of a groupG. As for finite cyclic quasinormal subgroups, the structure of products AX, where X is also cyclic, is of fundamental importance (see Lemmas 2.1 and 3.1 above). Once again it turns out that these subgroups are metacyclic.

LEMMA 4.1. Let GˆAX, where Aˆ haiis an infinite cyclic quasi- normal subgroup of G and Xˆ hxiis also cyclic. Then G is metacyclic and G⁰ˆ h[a;x]i.

PROOF. Clearly the second statement follows from the first. To show thatGis metacyclic, we consider five cases.

Case1.Suppose that XC₁ and A\Xˆ1. Then by Lemma 2.6, A/Gand there is nothing to prove.

Case2.Suppose that XC1 and A\X6ˆ1. We may assume that A⁶/G. LetNˆA_G. ThenN5A\X. Since xmust either invert or cen- tralise N, we can see that the latter must apply. So N4Z(G). Let p be prime and let P=N be any non-trivial p-component of A=N. Then P=N qn G=Nand hence

P=N4Zm(G=N);

for some finite integerm, by Lemma 2.8. ThereforeP4Z_m‡1(G). Since this applies for all primesp, we haveA4Z_n(G), for some finite integern, and so Gis nilpotent. The Hirsch lengthh(G)ˆ1 and the torsion subgroupTofG is finite. Moreover

L(T) L(AT:A) L(G:A) L(X:A\X);

a distributive lattice, soT is cyclic, by [9] (see also [11], Corollary 1.2.4).

Finally, sinceG=Tis torsion-free, it must be cyclic and soGis metacyclic. It is easy to see thatGis always nilpotent in this case.

Case3.Suppose that XCpⁿ, where p is an odd prime.We claim that X/ G and G is nilpotent:

…7†

To see this, we argue by induction onjXj. IfjXj ˆp, thenA/G, by Lemma 2.5. SoGis abelian and (7) is true. Therefore assume thatjXj5p²and let X1ˆV(X). Then by the same argument, AX1 is abelian and hence X₁4Z(G). Now by inductionX=X₁/G=X₁, i.e.X/G. AlsoG=X₁is nilpo- tent and soGis nilpotent, establishing (7).

REMARK. The groupGhere does not have to be a finite cyclic extension of an infinite cyclic group. For example, this is not the case whenx^aˆx^p‡1. ButGis a modular group and all its subgroups are quasinormal, by [7] (see also [11], Theorem 2.4.11).

Case4.Suppose that XC2ⁿ. Here we claim that X²/G and AX²is nilpotent:

…8†

Consider first the casejXj ˆ4. We know that all the subgroups ofAare quasinormal inG. So there are two possibilities.

(a)Suppose that axˆxⁱa, for some i. ThenX/GandGis abelian. So (8) is true.

(b)Suppose that axˆxⁱa ¹, for some i. Ifiˆ1, thenA/Ganda^xˆa ¹. Sox²2Z(G) and again (8) is true. On the other hand, ifiˆ 1, then (ax)²ˆ1.

ButGˆAhaxiand this would imply thatjG:Aj ˆ2, a contradiction.

We have shown that whenjXj ˆ4,

either GˆAX or A/G and x inverts A:

…9†

Now we suppose thatjXj58 and argue by induction onjXjto prove (8).

Again let X₁ˆV(X). Then X₁4Z(G), by (9). Therefore by induction X²=X₁/G=X₁and soX²/G. AlsoAX²=X₁ is nilpotent and soAX² is nil- potent. Thus (8) is true.

To prove thatGis metacyclic, we may assume thatjXj58, by (9). Since X²/G, we also haveX⁴/Gand so (a) or (b) above applies toG=X⁴. If (a) holds, thenX/Gand we are finished. Therefore we may assume that

axˆxⁱa ¹mod X⁴:

Hereiˆ 1. But ifiˆ 1, then moduloX⁴we get the same contradiction as in (b). Thereforeiˆ1 and so we have

axˆx^1‡4ja ¹;

for some integer j. Then x ¹axˆx^4ja ¹ and x ²ax²ˆx^4jax ^4j. Hence x^2‡4j commutes withaand so [A;X²]ˆ1. Therefore

(ax ^2j)^xˆx^4ja ¹x ^2jˆx^2ja ¹:

Thusx normaliseshax ^2jiand so doesa. Thenhax ^2ji/G andGis me- tacyclic, as required.

NOTE. In Case 4 we have shown that one of the following applies:

(i)X/G; (ii)A/G; (iii)a^xˆa ¹x^4j,x²2Z(G) andx^4j6ˆ1.

Example 5 in Section 5 shows that (iii) does occur.

Case5.Suppose that X is finite.LetjXj ˆn, so we may assume thatn is not a prime power. Ifnis odd, thenX/G, by Case 3. Therefore suppose that n is even, but not a power of 2.

We may assume thatA⁶/GandX⁶/G.

LetX2ˆ hx2ibe the 2-component ofX. By the note above, (ii) or (iii) must apply toAX2. We distinguish these possibilities.

(a)Suppose that A/AX2. Then a^x2 ˆa ¹: …10†

LetX0ˆ hx0ibe the 2⁰-component ofX. ThenX0/GandAX0is nilpotent, by Case 3. AlsoAX0 is not abelian, sinceA⁶/G. Therefore

a^x0ˆaxⁱ₀anduˆxⁱ₀6ˆ1.

As a conjugate ofA,hauiqn G. ConsiderhauiX2. Withhauithe infinite cyclic quasinormal subgroup, (i) above cannot apply, otherwiseX2/Gand thenX/G. If (iii) applies, then

(au)^x2 ˆ(au) ¹x^4j₂ ˆa ¹u,

by (10). But then we obtainx^4j₂ ˆ1, contradicting (iii). We are left with (ii).

Then (au)^x2 ˆ(au) ¹ˆa ¹u, again by (10). Sou^aˆu ¹ and this contra- dicts the nilpotency ofAX₀.

(b)Suppose that(iii)applies to A as a quasinormal subgroup of AX₂. Then for somej,

a^x2ˆa ¹x^4j₂ andx^4j₂ 6ˆ1.

If [A;X₀]ˆ1, thenGˆ hax₂^2ji ^jXandGis metacyclic. Thus suppose that [A;X₀]6ˆ1. By Case 3,

X0/AX0: …11†

So again with uˆ[a;x0], we consider hauiqnhauiX2 and the three possiblities in the note after Case 4. SinceX⁶/G, (i) cannot apply. If (ii) holds, then hauiis normalised by x₂ and hence the action is by inver- sion. So

u ¹a ¹ˆa ¹x^4j₂u,

i.e.x^4j₂ ˆ1, a contradiction. Finally, suppose that (iii) applies. Then for some integerk,

(au)^x2ˆu ¹a ¹x^4k₂ ˆa ¹x^4j₂u.

But then by (11) we must havex^4k₂ ˆx^4j₂ andu^aˆu ¹, contradicting (7).

This completes the proof of Lemma 4.1. p

REMARK. Even in a finite group GˆAX, with A and X cyclic and A qn G, it does not follow in general thatGis metacyclic. (See Example 1 in Section 5.)

For the proof of Theorem D we need one more result.

LEMMA4.2. Let Aˆ haibe a cyclic quasinormal subgroup of a group Gˆ ha;x;yi. Then

[A;G]ˆ(A\[A;G])h[a;x]ih[a;y]i:

…12†

Proof.WhenGis a finite 2-group, this is Lemma 3.3 in [3]. HereGmay be infinite, but we still need the case whereAis finite. This is a routine generalisation of Lemma 3.3 of [3], already used in the proof of Theorem 3.4 above. For convenience we include the argument here.

Thus suppose first thatAis finite. ThenA^Gis finite, by Lemma 2.7, and we can argue by induction onjA^Gj. We may assume that [A;G]6ˆ1. So let N(6ˆ1) be a normal subgroup ofGcontained inAor in [A;G]. By induction we have fromG=N

[A;G]Nˆ(A\[A;G])h[a;x]ih[a;y]iN:

…13†

Suppose thatA_G6ˆ1. Then there is a normal subgroupA₁ ofGwithjA₁j prime and A₁4A. TakeNˆA₁. If N4[A;G], thenN4A\[A;G] and (13) becomes (12). On the other hand, ifN\[A;G]ˆ1, then intersecting both sides of (13) with [A;G] again gives (12).

Now suppose thatA_Gˆ1. Then we may assume, without loss of gen- erality, thatLˆ[A₁;x]6ˆ1, whereA₁is again a subgroup of prime order in A. Since A₁qn G, we have L4Z(G), by Lemma 2.4. So we can take NˆL. But thenN4h[a;x]iand again (13) becomes (12).

Finally suppose thatAis infinite. By Lemma 2.7,jA^G:Ajis finite and soA^G=A_Gis finite. ThereforeA_G6ˆ1. Suppose thatA\[A;G]ˆ1. By the case when Ais finite, withNˆA_G we have (13). Then intersecting with [A;G] we get (12). But ifA\[A;G]6ˆ1, then there is a normal subgroupN ofGwith 16ˆN4A\[A;G]. Again (13) holds and this is (12). p PROOF OFTHEOREMD. Let Aˆ haibe an infinite cyclic quasinormal subgroup of a group G. Suppose that [A;G] is periodic. Then clearly A\[A;G]ˆ1. By Theorem 2.2, [A;G] is abelian andaacts on it as a power automorphism. Conversely, suppose that A\[A;G]ˆ1. Again [A;G] is abelian by Theorem 2.2. By Lemma 4.1,

[A;G]ˆ h[a;g]jg2Gi:

Let g2GandHˆAhgi. By Lemma 2.7,jA^H:Ajis finite and therefore j[a;g]jis finite. It follows that [A;G] is periodic. This proves (ii).

Now suppose that [A;G] is not periodic. By what we have already proved, this is equivalent to A\[A;G]6ˆ1. By Lemma 2.5,A^G is gener- ated by ascendant cyclic subgroups and so it is locally nilpotent, by [6],

Theorem 2. Thus [A;G] is locally nilpotent and so there is an elementg2G such that [a;g] has infinite order. Butanormalisesh[a;g]i, by Lemma 4.1, and soacentralises [a;g]. This applies to any commutator [a;g] of infinite order. Suppose that [a;h] has finite order. Then

[a;hg]ˆ[a;g][a;h]^g

has infinite order and soacentralises [a;hg]. PutHˆ ha;g;hi ˆ ha;g;hgi.

By Lemma 4.2,

[A;H]ˆ(A\[A;H])h[a;g]ih[a;hg]i and soacentralises [A;H]. Thereforeacentralises [a;h].

It follows that A centralises [A;G]. Thus A^G centralises [A;G] and [A;G] is abelian. Hence A^G is abelian. This completes the proof of

Theorem D. p

The power automorphisms in (i) are not always universal. For example, letHbe an abelian group of typep¹and letabe ap-adic integer such that a1mod p(a1mod4 ifpˆ2). Then the map

h7!h^a; …14†

for allh2H, defines an automorphism ofH. LetGbe the split extension of H by the cyclic group A generated by a. Then every subgroup of G is quasinormal (see [11], Theorem 2.4.11). Ifa6ˆ1, then [A;G]ˆH. Clearly the power automorphism (14) is not universal in general.

We end this Section, as we did in Section 3, with a brief discussion of how our results extend tolocally cyclicsubgroups. LetAbe a torsion-free locally cyclic group. Then the automorphism group Aut(A) of A is iso- morphic to a subgroup of the multiplicative group of the rationals. Also Aut(A) contains an element a of infinite order, if and only if we have A^pˆA, for at least one primep. In this case, no non-trivial cyclic subgroup ofAis left invariant bya. (See [5].)

THEOREM 4.3. Let A be a torsion-free locally cyclic quasinormal subgroup of a group G. Then the following conditions are equivalent.

(i) There is a non-trivial cyclic subgroup of A which is quasi- normal in G.

(ii) Every subgroup of A is quasinormal in G.

(iii) For all g2G,jhgi:C_hgi(A)jis finite.

PROOF. Assume that (iii) holds. Let a2A,a6ˆ1. Let g2Gand put Tˆ ha;gi, Lˆ hA;gi and A₁ˆA\T. If jL:Aj ˆ 1, then A/L, by

Lemma 2.6, and soA1ˆ hai^hgi/T. SinceA1is finitely generated, we have hai4A1ˆ ha1i/T and sohai/T. If jL:Aj is finite, thenhai4A1qn T and jT:A1j is finite. ThusA1 is finitely generated, hence cyclic, and so hai4A₁ˆ ha₁iqn T. Then haiqn T, by Lemma 2.3. It follows that haiqn G. Therefore (iii) implies (ii).

Now suppose that (iii) fails, i.e.jhgi:C_hgi(A)j ˆ 1, for some elementg.

ThenA^gˆA, by Lemma 2.6, and ginduces an automorphism ofAof in- finite order. But thenhai^g6ˆ haifor anya2A,a6ˆ1, as stated above. Thus

haiis not quasinormal inG, and (i) fails. p

THEOREM 4.4. Let A be a torsion-free locally cyclic quasinormal subgroup of a group G. Then

(i) [A;G]is abelian, A acts on it as a group of power automorph- isms and A^Gis a modular group;

(ii) [A;G]is periodic if and only if A\[A;G]ˆ1;

(iii) if A\[A;G]6ˆ1, then A^Gis abelian; and

(iv) for every natural number n, Aⁿis quasinormal in G.

The proofs of the above statements can safely be left as exercises, following the arguments of Lemma 4.2, Theorem D and Theorem 4.3. Of course (iv) holds for any abelian quasinormal subgroupAof any groupG, providednis odd or divisible by 4. (See [15], Theorem 1.)

5. Examples.

All the results that have been proved in the previous Sections have depended on showing that a groupGˆAX, withAandXcyclic subgroups and A quasinormal, is metacyclic, provided certain hypotheses are sa- tisfied. We begin this final Section by showing that hypotheses are re- quired here.

EXAMPLE1. LetBandX be cyclic groups of orders 7 and 9, respec- tively, and letBˆ hbi,Xˆ hxi. Form the split extension

HˆB^jX;

where b^xˆb². Now letC be a cyclic group of order 3 generated by the elementcand form the split extension

GˆH^jC;

whereb^cˆbandx^cˆx⁴. The action ofconHhas order 3 and the relations ofHare preserved. SoGexists and has order 3³:7.

SetAˆBC, a cyclic group of order 21. We claim that A is quasinormal in G:

…15†

To see this, sinceB/G, we may factor byB, i.e. we may assume thatBˆ1.

ThenGˆX^jC, which is a modular group with all subgroups quasinormal.

Therefore (15) is true. However, G is not metacyclic. For, G⁰ˆ hb;x³i C₂₁. If GˆNK, withN/GandN andK both cyclic, then G⁰4N. SinceG=G⁰C₃C₃, we must haveG⁰<N. ThusjNj ˆ3²:7 and so N₃C₉. B ut N₃ must centralise B, while C_G(B)ˆB hx³i C con- taining no cyclic subgroup of order 9.

Another key result was Lemma 2.9 showing that when Aˆ hai is a cyclic quasinormal subgroup of odd prime power order in G and A\[A;G]ˆ1, then

[A;G]ˆ f[a;g]jg2Gg:

…16†

This result is also true whenAhas 2-power order, provided condition () holds. (It follows easily from [3], Lemma 2.7.) However, if () is not sa- tisfied, then (16) can fail to hold.

EXAMPLE2. LetHandXbe cyclic groups of orders 8 and 16, respec- tively, and letHˆ hhiandXˆ hxi. We form the split extension

GˆH^jX;

with h^xˆh ¹. Letaˆhx² andAˆ hai, a cyclic group of order 8. Then C_G(A)ˆHX². AlsoA\HˆA\Xˆ1, soGˆAX and

[A;G]ˆ h[h;x]i ˆH²C₄:

Thus A\[A;G]ˆ1. Also the only elements of form [a;g] are 1 and h ²(ˆ[a;x]). Therefore

[A;G]6ˆ f[a;g]jg2Gg:

However,

A is quasinormal in G:

…17†

For, any cyclic subgroup of G of form hhⁱx^2ji commutes with A. On the other hand, hhⁱx^ji, with j odd, has order 16 (because (hⁱx^j)² ˆ

ˆhⁱx^jhⁱx ^jx^2jˆx^2j) and intersects A trivially. So jAhhⁱx^jij ˆ2⁷ and hence Ahhⁱx^ji ˆG. Thus (17) is true.

REMARK. Clearly the subgroupXisnotquasinormal inGand so () is not satisfied here. Also each element of [A;G] has the form [aⁱ;g], for some integeriand elementg. For many purposes this is as useful as (16) and it would be interesting to know if it isalwaysthe case.

We know (see Lemma 2.6) that ifAisanyquasinormal subgroup ofany groupGand ifXis an infinite cyclic subgroup ofGsuch thatA\Xˆ1, then X normalises A. It is natural to ask if this then implies that A is normal inG. In fact this is not the case, as is shown by the next two ex- amples.

EXAMPLE3. LetXˆ hxibe an infinite cyclic group and letYˆ hyibe a cyclic group of order 8. We form the split extension

KˆX^jY;

wherex^yˆx ¹. Then letAˆ haibe a cyclic group of order 2 and form the split extension

GˆK^jA;

wherex^aˆx,y^aˆy⁵. This extension exists, because the relationsy⁸ˆ1 andx^yˆx ¹are both preserved by thea-action (of order 2). Clearly

A is not normal in G.

However,Ais quasinormal inG. For, consider an arbitrary cyclic subgroup Hˆ hxⁱy^ja^ki, 04j47, 04k41. We have

(xⁱy^ja^k)^aˆxⁱy^5ja^k:

Case(i).Suppose that j is even. Theny^5jˆy^j andAcentralisesH.

Case(ii).Suppose that j is odd. Then [xⁱy^j;a]ˆ[y^j;a]ˆy⁴2Z(G) and hxⁱy^j;aihas nilpotency class 2. Thus (xⁱy^ja^k)⁵ˆ(xⁱy^j)⁵a^k. Also (xⁱy^j)²ˆy^2j and (xⁱy^j)⁵ˆxⁱy^5j. So (xⁱy^ja^k)⁵ˆxⁱy^5ja^kˆ(xⁱy^ja^k)^aandAnormalisesH.

ThereforeA4norm(G) and soA qn G, as claimed.

The previous group Gis isomorphic to C1^j(C8^jC2) and involves a dihedral action. We show that the same result can be achieved using odd primes instead of the prime 2.

EXAMPLE4. Letpbe an odd prime and letYbe a cyclic group of order p² generated by y. Put Y1ˆY=Y^p and y1ˆyY^p. Let RˆZY1, the in- tegral group ring ofY1and letIˆR(y1 1), the augmentation ideal ofR.

ThenIbecomes aY-module via the natural actionu^yˆuy1, allu2I. We form the split extension K ˆI^jY, a free abelian group of rank p 1 extended byC_p².

LetAbe a cyclic group of orderpgenerated byaand define an action of AonKby

u^aˆu, allu2I,y^aˆy^1‡p.

Then form the split extensionGˆK^jA. ThusAis not normal inG. B ut, as before, we claim that

A4norm(G):

…18†

For,AcentralisesIY^pAˆL, say. A typical cyclic subgroup ofG, not con- tained inL, is generated by an element of the formgˆyⁱua^j, wherei1 modpandu2I. Theng^aˆy^i(1‡p)ua^j. But we have

g^1‡pˆy^i(1‡p)ua^j: …19†

For,hyⁱu;aiis nilpotent of class 2, since its derived subgroup isY^p, which is central inG. Therefore

g^1‡pˆ(yⁱu)^1‡pa^j:

Also (yu)^1‡pˆyu(yu)^pˆyu(y^pu(1‡y₁‡y²₁‡ ‡y^p₁ ¹))ˆy^1‡pu. Thus (yⁱu)^1‡pˆy^i(1‡p)uand (19) is true. Therefore (18) holds andA is quasi- normal in G.

When proving Lemma 4.1, we considered in Case 4 a groupGˆAX, whereAˆ haiis an infinite cyclic quasinormal subgroup ofGandXˆ hxi is cyclic of order 2ⁿ. We showed that if neitherXnorAis normal inG, then

a^xˆa ¹x^4j;

withx²2Z(G) andx^4j6ˆ1. In fact this situation does occur.

EXAMPLE5. LetAˆ haibe an infinite cyclic group and letYˆ hyibe a cyclic group of order 2^{n 1}(n53). LetHˆAYand letjbe an integer.

ThenHhas an automorphism of order 2 defined by a7!a ¹y^2j, y7!y.

So there is an extensionGofHby a group of order 2 defined by Gˆ ha;xja^xˆa ¹x^4j;x²ⁿˆ1i:

(See [12], Theorem 9.7.1 (ii).) ThenGˆAX, whereXˆ hxiandax²ˆx²a.

We claim that

A is quasinormal in G:

…20†

For, a typical cyclic subgroupHofGis generated by an elementaⁱx^k. Ifkis even, thenAcentralisesH. Ifkis odd, then (aⁱx^k)²ˆx^2k‡4ij, generatingX² andX²4Z(G). SoH²ˆX²andAHˆG. Thus (20) is true. Provided 2^{n 2} does not divide j, we have a group of type described in (iii) of Case 4 of Lemma 4.1.

Our final example also concerns the case when the quasinormal sub- groupAis infinite cyclic. Situations where [A;G] are periodic or torsion- free are familiar. However, themixedcase can also occur.

EXAMPLE 6. Let Hˆ ha;yjy⁸ˆ1;y^aˆy⁵i. Then H has an auto- morphism of order 2 defined by

a7!a ¹; y7!y⁵: …21†

So we can form a split extensionGˆH^jX, whereXˆ hxiis a cyclic group of order 2 with action onHdefined by (21). We claim thatAˆ haiis qua- sinormal in G. To see this, sinceA²/G, we may factor byA², i.e. assume that a²ˆ1. Then easy calculations show thataconjugates each element ofGto its 5-th power. ThereforeA qn G. However,

[A;G]ˆ ha²i hy⁴i C1C2:

Acknowledgments. The second author is grateful to the Mathematics Institute, University of Warwick, for hospitality during the preparation of this paper.

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Manoscritto pervenuto in redazione il 15 dicembre 2005.