W. A. Bogley and N. D. Gilbert

New York J. Math.6(2000)55–71.

The Homology of Peiffer Products of Groups

W. A. Bogley and N. D. Gilbert

Abstract. The Peiffer product of groups first arose in work of J.H.C. White- head on the structure of relative homotopy groups, and is closely related to problems of asphericity for two-complexes. We develop algebraic methods for computing the second integral homology of a Peiffer product. We show that a Peiffer product of superperfect groups is superperfect, and determine when a Peiffer product of cyclic groups has trivial second homology. We also introduce a double wreath product as a Peiffer product.

Contents

Introduction 55

1. The low-dimensional homology of products of subgroups 57

2. Twisted bilinear relations 60

3. The structure ofS_G∗H 61

4. Computations 63

References 70

Introduction

Given two groups acting on each other by automorphisms, it is natural to ask whether these groups can be embedded in an overgroup in such a way that the given actions are realized by conjugation. If the actions are trivial, this can be done simply by forming the direct product of the two groups. In general, the question has a negative answer.

One is led to the following construction. Let Gand H be groups and suppose we are given fixed actions (g, h)7→g^h and (h, g)7→h^g of each group on the other.

Received October 1, 1999.

Mathematics Subject Classification. 20J05, 20E22, 20F05.

Key words and phrases. homology, Peiffer product, asphericity, two-complex, double wreath product.

W. A. Bogley would like to thank the Department of Mathematics at Heriot-Watt University, Edinburgh, for its gracious hospitality while this work was in progress. Bogley was supported by a U. K. Engineering and Physical Sciences Research Council Visiting Fellowship (GR/L49932).

ISSN 1076-9803/00

55

(These are assumed to be right actions, so that (g^h)^h0 =g^hh0, for example.) Let Π denote the normal closure in the free productG∗H of all elements

g⁻¹h⁻¹gh^g, h⁻¹g⁻¹hg^h :g∈G, h∈H.

The quotient G ./ H := (G∗H)/Π is the Peiffer product of G and H with the given actions. For example, whenGand H act trivially on each other, the Peiffer product is just the direct product G ./ H ∼= G×H, and Π is the Cartesian subgroup Π =GH = ker(G∗H →G×H). WhenG andH are infinite cyclic with nontrivial actions, it is shown in [8] that the Peiffer product is the quaternion group of order eight: Z./Z∼=Q8. The question that was alluded to in the opening paragraph can now be posed as the following

Embedding Question. When are the natural mapsG→G ./ H←H injective?

The Peiffer product of groups (so named in [8]) first arose in a topological setting in the work of J.H.C. Whitehead [14] on the structure of relative homotopy groups.

Suppose that a connected two-complexZ is a union of connected subcomplexesX and Y that intersect in the common one-skeleton X∩Y =Z¹. One consequence of Whithehead’s work in [14] is that the relative homotopy groupπ2(Z, Z¹) can be decomposed as the Peiffer product of the relative groupsπ2(X, Z¹) andπ2(Y, Z¹):

π₂(Z, Z¹)∼=π₂(X, Z¹)./ π₂(Y, Z¹).

(1)

(The actions arise via the homotopy action ofπ₁(Z¹).) As an application, White- head proposed his notorious

Asphericity Question. Are subcomplexes of aspherical two-complexes themselves aspherical?

The point here is that by comparing the homotopy sequences of the pairs (X, Z¹) and (Z, Z¹), one sees that ifZ is aspherical (i.e.,π2(Z) = 0), thenX is aspherical if and only if π₂(X, Z¹) embeds inπ₂(Z, Z¹). The longstanding unresolved status of Whitehead’s Asphericity Question therefore stands as testimony to the subtlety of the Embedding Question for Peiffer products.

The Peiffer product has been applied to algebraic problems (see for example its implicit role in [10]) and to the topological setting in which it first appeared, the calculation of low dimensional homotopy and homology groups [3, 5, 7, 9]. The following theorem of M. N. Dyer [6] connects the vanishing of the second homotopy group of a two-complex to the vanishing of the second homology group of its second relative homotopy group.

Theorem. [6] Let Xbe a connected two-complex with one-skeleton X¹. IfX does not have the homotopy type of the two-sphere, then X is aspherical if and only if H₂(π₂(X, X¹)) = 0.

In this paper we consider the purely algebraic problem of determining the second integral homologyH₂(G ./ H) of a Peiffer productG ./ H in terms of information about the factorsGandH. As in [5], we exploit the description of a Peiffer product via semidirect products. By way of general results, we show that any Peiffer product of superperfect groups is superperfect (Corollary2.3) and we give a very short proof of the K¨unneth formula for the second homology of direct products (§4.1).

Our main results support a systematic approach to the problem of H2 calcu- lations for Peiffer products. We draw the reader’s attention to Theorem 3.2 and

Corollary 3.3, which are definitive technical results within this context. We illus- trate the effectivenes of our approach by showing how to determine whether a Peiffer product of cyclic groups has trivial second homology (Theorem4.4). This in turn is related to the Embedding Question for these Peiffer products. (See Corollary4.5 and the succeeding examples.) We also introduce (§4.4) a double wreath product construction as a Peiffer product and we investigate its second homology.

Notation. When a groupH acts on a groupGon the right, [G;H] will denote the subgroup ofGgenerated by all elementsg⁻¹g^h, g∈G, h∈H.Then [G;H] is normal in G, and we denote the quotient G/[G;H] by G_H. We shall use this subscript notation for any quotient group obtained by killing an action. For example, ifH normalizes G in a common overgroup, then H acts on G by conjugation: g^h = h⁻¹ghand [G;H] = [G, H] is the subgroup generated by the commutators [g, h] = g⁻¹h⁻¹gh =g⁻¹g^h. Thus GH =G/[G, H] and if G=H, thenGG =G/[G, G] is the abelianized groupG^ab.

Now suppose that Gand H are groups acting on each other on the right. We form the following normal subgroups in the free productG∗H:

S=hhg⁻¹h⁻¹gh^g:g∈G, h∈Hii (2)

T =hhh⁻¹g⁻¹hg^h:g∈G, h∈Hii

and we set Π =ST. Note that (G∗H)/Sand (G∗H)/T are the semidirect products GnH and GoH respectively. The quotient group (G∗H)/Π = G ./ H is the Peiffer product ofGandH with the given actions. The images of the natural maps G→G ./ H←H will be denoted by ¯Gand ¯H respectively.

1. The low-dimensional homology of products of subgroups

R. Brown’s five-term exact sequence [5] for the homology of a groupP, equal to the product of two normal subgroupsM andN, is

(3) H₂(P)→H₂(P/M)⊕H₂(P/N)→(M ∩N)/[M, N]

→H1(P)→H1(P/M)⊕H1(P/N)→0.

We shall be interested in the setting in whichM ∩N = [M, N], in which case the groupP decomposes as a Peiffer product ofM andN.

Proposition 1.1. In a Peiffer productG ./ H, the subgroupsG¯ andH¯ are normal subgroups satisfying G ./ H = ¯GH¯ and G¯∩H¯ = [ ¯G,H¯]. Conversely, if P is a group with normal subgroupsM, N satisfyingP =MN andM∩N = [M, N], then P ∼=M ./ N.

Proof. It is immediate from the defining relations of the Peiffer product that ¯G and ¯H are normal subgroups ofG ./ H and that ¯GH¯ =G ./ H. Next,

(G ./ H)/G¯∼=HG and (G ./ H)/H¯ ∼=GH.

As in [3,7], we note that the quotient (G ./ H)/[ ¯G,H¯] is isomorphic toG_H×H_G and elements of ¯G∩H¯ lie in the kernel of the quotient mapG ./ H→GH×HG. Hence ¯G∩H¯ ⊆[ ¯G,H].¯

Conversely, suppose that M, N are normal subgroups of a group P with P = MN. Following Brown [5], we form the Peiffer productM ./ N using the conjuga- tion actions inP. By identifying M ./ N as a quotient of the semidirect product

MoN, Brown obtains a short exact sequence

1→(M∩N)/[M, N]→M ./ N→P →1.

Hence ifM ∩N = [M, N] thenM ./ N ∼=P.

With Proposition1.1, the Brown homology sequence (3) for the groupG ./ H and its normal subgroups ¯G and ¯H shows that the mapsGH ← G ./ H → HG

induce homomorphisms

H₁(G ./ H)→^α1H₁(G_H)⊕H₁(H_G) H₂(G ./ H)→^α2H₂(G_H)⊕H₂(H_G)

where α₁ is an isomorphism and α₂ is surjective. For later convenience, we note the following immediate corollary.

Corollary 1.2. Let GandH be perfect groups (i.e.,H1(G) =H1(H) = 0). Then for any actions of GandH on each other, the Peiffer productG ./ H is perfect.

Our next aim is to investigate the kernel ofα₂. The mapsG←G∗H →H induce a homomorphism

ΠG∗H→β [G;H]G⊕[H;G]H

that is obviously surjective.

Proposition 1.3. There is a surjective homomorphism kerα₂ → kerβ. If the mapsH₂(G)→H₂(G_H) andH₂(H)→H₂(H_G) are each injective, then kerα₂∼= kerβ.

Proof. The standard five-term exact sequences for the extensions 1→Π→G∗H→^π G ./ H→1

1→[G;H]→G→G_H →1 1→[H;G]→H →HG →1

combine into a commutative diagram involving the mapsα₁, α₂ andβ as shown in Figure 1.1. A diagram chase gives a surjection kerα₂ → kerβ. If π denotes the quotient mapG∗H →G ./ H then we obtain an extension of abelian groups

0→kerα₂∩imH₂(π)→kerα₂→kerβ →0.

Now each ofH2(G)→H2(GH) andH2(H)→H2(HG) factors throughH2(π) and it follows that ifH2(G)→H2(GH) andH2(H)→H2(HG) are each injective then

kerα2→kerβ is an isomorphism.

The converse of Proposition1.3is false, as the following example shows.

Example. Let A=ha, b|[a, b]ibe free abelian of rank 2 and letV ={1, x, y, xy}

be a Klein 4–group. We letxact onAby inversion and let y act trivially. Define an action of A on V by x^a = y, x^b =xy, y^a = xy, y^b = x. In A ./ V we have a⁻¹ya= xy and y⁻¹ay =a: hence y =xy and x= 1. Then a⁻¹xa = y implies y= 1, so that ¯V = 1 and

A ./ V = (A ./ V)/V¯ =AV ∼=V,

H₂(G)⊕H₂(H)

H2(π)

H₂(G)⊕H₂(H) H2(G ./ H)

α2 //H2(GH)⊕H2(HG)

ΠG∗H β //

[G;H]G⊕[H;G]H

H₁(G)⊕H₁(H)

H₁(G)⊕H₁(H) H₁(G ./ H) ^α1 //H₁(G_H)⊕H₁(H_G)

Figure 1.1. Five term sequences

and H2(A ./ V) = Z2, H2(AV) = Z2 and H2(VA) = 0. We see that αi is an isomorphism (i≥1), and soβ is an isomorphism: Π/[Π, A∗V]∼=Z²⊕V. However Z=H₂(A)→H₂(A_V) =Z₂ is not injective.

The structure ofH₂(G ./ H) is summarised in the Hasse diagram in Figure1.2.

H₂(G ./ H)

H₂(G_H)⊕H₂(H_G)

ker (ΠG∗H→β [G;H]G⊕[H;G]H) kerα2

kerα₂∩imH₂(π)

Figure 1.2. Structure ofH₂(G ./ H)

2. Twisted bilinear relations

Recall that Π =ST whereS denotes the normal closure inG∗H of all elements g⁻¹h⁻¹gh^g,g∈G, h∈H andT is the normal closure of all elementsh⁻¹g⁻¹hg^h, g∈G,h∈H. The inclusions ofS andT in Π induce a surjective homomorphism

SG∗H⊕TG∗H→ΠG∗H.

In this and the following section, we completely describe the structure ofS_G∗H, with analogous remarks holding forTG∗H. The image inSG∗H of the normal generator g⁻¹h⁻¹gh^g forS will be denotedhg, hi. Thus,

hg, hi=g⁻¹h⁻¹gh^g[S, G∗H]∈S_G∗H =S/[S, G∗H].

Lemma 2.1. The abelian group S_G∗H is generated by the elements hg, hi (g ∈ G , h∈H) and the following relations hold:

(i) hgg⁰, hi=hg⁰, h^gihg, hi, (ii) hg, hh⁰i=hg, hihg, h⁰i,

These relations imply thathg,1i=h1, hi= 1inS_G∗Hand thathg, hi⁻¹=hg, h⁻¹i= hg⁻¹, h^gi. Furthermore, if x (resp. y) is a generating set for G (resp. H), then S_G∗H is generated by the elementshx, yi,x∈x,y∈y.

Proof. The elementshg, higenerateS_G∗H sinceS is the normal closure inG∗H of all the elements g⁻¹h⁻¹gh^g. The validity of the relations (i) and (ii) may be checked directly: for example, working modulo [S, G∗H],

hgg⁰, hi = g⁰⁻¹g⁻¹h⁻¹gg⁰h^gg0

= g⁰⁻¹h^−gg⁰h^gg0h^−gg0g⁰⁻¹h^gg⁻¹h⁻¹gg⁰h^gg0

= hg⁰, h^gih^−gg0g⁰⁻¹h^gg⁻¹h⁻¹gh^gh^−gg⁰h^gg0

= hg⁰, h^gihg, hi.

The remaining assertions of the lemma follow directly from the relations (i) and

(ii).

Denote the augmentation ideal inZGbygand writeH₁(H) =H^ab=H/[H, H].

Here is a restatement of Lemma2.1. The elementary proof is left to the reader.

Corollary 2.2. There is a surjective homomorphismσ:H^ab⊗_Gg→S_G∗H given byσ(h[H, H]⊗(g−1)) =hg, hi.

Corollary 2.3. If G and H are perfect (i.e., G^ab =H^ab = 0), then the natural mapH₂(G)⊕H₂(H)→H₂(G ./ H)is surjective. IfGandH are both superperfect (i.e.,H1=H2= 0), then so isG ./ H.

Proof. WhenH is perfect we haveSG∗H = 0 by Corollary2.2. When bothGand H are perfect, this implies that ΠG∗H= 0. The first statement follows from the five term homology sequence for G ./ H = (G∗H)/Π. (See Figure 1.1.) The second

statement follows immediately.

3. The structure of S

In this section we show that the surjectionσ:H^ab⊗Gg→SG∗H is actually an isomorphism. Of course, the discussion also applies to show thatTG∗H ∼=G^ab⊗Hh, where h denotes the augmentation ideal inZH. Following Brown [5], we rely on the fact that the Peiffer product G ./ H is a quotient of the semi-direct product GnH = (G∗H)/S. We begin by building a standard two-complex with fundamental groupGnH.

We have an action ofGby automorphisms onH. Let K (resp. L) be the two- complex modeled on a presentation P =hx :ri(resp. Q =hy : si) for G(resp.

H). For each (x, y)∈x×y, we can choose a reduced word v_x,y in the free group ony so that the relationy^x=v_x,y holds in GnH. If we set t_x,y =x⁻¹y⁻¹xv_x,y andt={t_x,y: (x, y)∈x×y}, then

R=hx,y:r,s,ti

is a presentation for the split extension GnH. Let M denote the two-complex modeled on R and let p : Mf → M be the universal covering projection. The complex M contains the one-point union K ∨L as a subcomplex and the pre- image p⁻¹(K∨L) = K∨L has fundamental group S = ker(G∗H → GnH).

The homology sequence for the pair (M, Kf ∨L) determines an exact sequence of Z(GnH)-modules

π₂(M)→H₂(M, Kf ∨L)→S^ab→0 (4)

in which the second term is the free Z(GnH)-module with basis elements ˜ex,y, (x, y)∈x×y, corresponding to the two-cells ofM −(K∨L). The basis element

˜

e_x,y ∈H₂(M, Kf ∨L) is mapped to the cosett_x,y[S, S]∈S^ab, wheret_x,y is viewed inG∗H.

Lemma 3.1. Given two-complexes K andL modeled on presentations (x:r)for Gand(y:s) forH and with M constructed as above, there is an exact sequence

π2(M)→H2(M, K∨L)→SG∗H→0 (5)

of abelian groups. Here, the second term is the free abelian group with basis con- sisting of the two-cells ex,y, (x, y) ∈ x×y, of M −(K∨L) and the basis ele- ment ex,y is mapped to hx, yi ∈ SG∗H. The first map in the sequence factors as π2(M)→^h H2(M)→H2(M, K∨L)wherehis the Hurewicz homomorphism.

Proof. The result follows upon killing the action ofGnH (i.e., of G∗H) in the

sequence (4).

Module generators for π₂(M) have been described by Y. G. Baik and S. J.

Pride [1] (see [2]). In practice it is a simple matter to determine the images of these generators under the mapπ₂(M)→H₂(M, L∨K) and thus to work out a presentation forS_G∗H in terms of the generatorshx, yi, x∈x,y ∈ y. We briefly describe the Baik-Prideπ₂ generators for the reader’s convenience.

Recall thatK (resp.L) is modeled on a presentationP = (x:r) (resp.Q= (y: s)) forG(resp.H) and thatM is is modeled on a presentationR= (x,y:r,s,t) where t consists of relations of the form tx,y = x⁻¹y⁻¹xvx,y, x∈ x, y ∈ y, that realize the action of G on H. Baik and Pride describe generators for π2(M) in terms of spherical pictures. (See [4] for a general treatment of spherical pictures.)

In addition to the collection of all spherical pictures overP andQ, two additional families of spherical pictures are sufficient to generateπ2(M) as aZ(GnH)-module.

Given (x, s)∈x×s, construct a disc pictureAx,sover the presentation (x,y:s,t) consisting of a single positively orienteds-disc surrounded by a series oftx,y-discs according to the occurences of the letters y^±1 in the relator s. The boundary of eachtx,y-disc has two oppositely oriented occurences of arcs labeled by x; thesex- arcs are joined so as to form an annulus oft_x,y-discs surrounding the originals-disc.

Since the generatorx, when viewed as an element ofGdetermines an automorphism ofH, the boundary ofA_x,ssupports a word iny∪y⁻¹that determines the identity element of H. Thus we can choose a disc picture B_x,s over Q = (y : s) whose boundary word is the same as that ofA_x,s. Gluing these two pictures together along their common boundary, we obtain a spherical pictureP_x,s overR. Constructing one such picture for each (x, s) ∈ x×s, we refer to the resulting family as the pictures of Type I.

Given (r, y)∈r×y, choose a disc picture Cr,y over (x,y:s,t) with boundary wordr⁻¹y⁻¹ry. This is possible becauseGacts onH and so racts trivially ony.

Now attach two oppositely orientedr-discs to match the occurences ofr and r⁻¹ in the boundary ofCr,yand then close up the remaining oppositely orientedy-arcs to obtain a spherical pictureQr,y over R. Constructing one such picture for each (r, y)∈r×y, we refer to the resulting family as the pictures of Type II.

Each element ofπ2(M) can be represented by a spherical picture over R. Baik and Pride showed thatπ₂(M) is generated as aZ(GnH)-module by the homotopy elements represented by spherical pictures overP andQ, together with the selected pictures of TypeI and TypeII. The image of a homotopy element under the map π₂(M) → H₂(M, K∨L) is computed by simply counting with multiplicity the occurences oft_x,y-discs in a representative picture. In particular, all pictures over P and Q have trivial image. This process and the construction of Type I and TypeIIpictures will be illustrated in the proof of Theorem3.2below.

Theorem 3.2. The mapσ:H^ab⊗_Gg→S_G∗H is an isomorphism. (And similarly, there is an isomorphism G^ab⊗_Hh∼=T_G∗H.)

Proof. Use the multiplication tables to construct presentations P for G and Q for H. Thus the presentation P = (x : r) has generatorsx ={(g) :g ∈ G} and defining relations r={(g)(g⁰)(gg⁰)⁻¹ :g, g⁰ ∈G}. The presentationQ= (y:s) is constructed in the same way. We can then buildK,L, andM as described above.

The free abelian homology groupH₂(M, K∨L) has basis consisting of the two-cells e_(g),(h),g∈G,h∈H with boundary word readingt_(g),(h)= (g)⁻¹(h)⁻¹(g)(h^g). So e_(g),(h)∈H2(M, K∨L) maps to hg, hi ∈ SG∗H. In order to determine the image ofπ₂(M)→H₂(M, K∨L), letg, g⁰ ∈Gandh, h⁰ ∈H. It suffices to examine the occurences oft-discs in the picturesA_(g),s andC_r,(h) wherer= (g)(g⁰)(gg⁰)⁻¹∈r ands= (h)(h⁰)(hh⁰)⁻¹∈s. These pictures are displayed in Figure3.1.

Examining the black discs in Figure3.1, we find that the TypeI pictureP_(g),s has imagee_(g),(h)+e_(g),(h⁰₎−e_(g),(hh⁰₎∈H2(M, K∨L) and that the TypeIIpicture Q_r,(h)has imagee_(g),(h)+e_(g⁰_),(h^g₎−e_(gg⁰_),(h)∈H2(M, K∨L). Passing toSG∗H, this means that the relations (i) and (ii) from Lemma2.1are actually defining relations for the generatorshg, hiof the abelian groupSG∗H. From this it is straightforward to show that the assignmenthg, hi 7→ h[H, H]⊗(g−1) defines an inverse to the

mapσ.

& %

' $

QQ

QQ QQ QQ Q

~

~

~ 6 (g)

@@R (g) (g)

(h)

(h⁰) JJ^

(hh⁰) - (h^g)

(h^0g) JJ^

(h^gh^0g) -

~

~

~ (h)

(h^g)

(h^gg0)

(h) (g)?

(g?⁰)

(gg⁰) 6

(g)?

(g?⁰)

(gg⁰) 6

Figure 3.1. A_(g),s andC_r,(h)

Corollary 3.3. There is an isomorphismS_G∗H∼=G^ab⊗H^ab if either (a) Gis a free group or

(b) Gacts trivially onH^ab.

Proof. When G is a free group, the augmentation ideal g (resp. abelianization G^ab) is the free ZG-module (resp. free abelian group) with basis in one-to-one correspondence with a basis for G. The result (a) follows easily. The result (b)

follows from the fact thatG^ab∼=g/g².

The setting of Corollary3.3(a) is of interest in low dimensional homotopy theory.

When a two-complexZ is a union of aspherical subcomplexesX andY with inter- sectionX∩Y =Z¹, Whitehead’s result (1) shows that the relative homotopy group π2(Z, Z¹) is the Peiffer product G ./ H whereG=π2(X, Z¹) and H =π2(Y, Z¹).

SinceX andY are aspherical, the groupsGandH are free.

4. Computations

4.1. Trivial actions: The K¨unneth formula. WhenGandH act trivially on each other, the Peiffer product is simply the direct product: G ./ H ∼= G×H. Inspection of the normal generators for S and T reveals that S = T = Π is the Cartesian subgroup Π =GH = ker(G∗H →G×H). Corollary3.3(b) implies that ΠG∗H∼=G^ab⊗H^ab. This result was first proved by MacHenry [11]. When we examine the five-term homology sequence forG×H = (G∗H)/Π (see Figure1.1), it is clear that the map H1(G)⊕H1(H)→H1(G×H) is an isomorphism and so the sequence

H₂(G)⊕H₂(H)→H₂(G×H)→G^ab⊗H^ab→0

is exact. Finally, the mapH2(G)⊕H2(H)→H2(G×H) is readily seen to be split injective, so we recover the K¨unneth formula for the second homology of direct

products:

H₂(G×H)∼=H₂(G)⊕H₂(H)⊕(G^ab⊗H^ab).

4.2. Conjugation action. LetGact on itself by conjugation. As in [3,8] we have G ./ G∼=G^ab×Gwith the two canonical mapsG→G ./ Gtakingg7→(1, g) and g7→(¯g, g). There is an isomorphism

H₂(G ./ G)∼=H₂(G)⊕H₂(G^ab)⊕(G^ab⊗G^ab).

Corollary3.3(b) shows that the mapsSG∗H →[H;G]H and TG∗H →[G;H]G can each be identified with the commutator pairing G^ab⊗G^ab→[G, G]/[G,[G, G]] = γ2(G)/γ3(G). It follows from the commutative diagram in Figure1.1that

Π_G∗G ∼= (G^ab⊗G^ab)⊕γ₂(G)/γ₃(G) and thatβ : ΠG∗G→[G;G]G⊕[G;G]G is the map

(G^ab⊗G^ab)⊕γ₂(G)/γ₃(G)→(γ₂(G)/γ₃(G))^⊕2

which is the sum of the identity on the second summand and the diagonal commu- tator pairing

χ⊕χ: ¯g⊗¯h7→([g, h]γ₃(G),[g, h]γ₃(G)).

Therefore kerβ ∼= ker(χ:G^ab⊗G^ab→γ2(G)/γ3(G)).

4.3. Peiffer products of cyclic groups. Consider a Peiffer product G ./ H whereGandHare cyclic groups generated byxandyrespectively. Proposition1.3 shows that H2(G ./ H) is isomorphic to the kernel of the map β : ΠG∗H → [G;H]_G⊕[H;G]_H. Suppose that the actions are given by

x^y=x^a+1 andy^x=y^b+1. (6)

whereaandbare integers. Note that ifaorbis zero, thenG ./ His abelian. Given the orders ofGand H, it is a simple matter to work out the structure of G ./ H and ofH₂(G ./ H) in this case. We therefore assume that aandb are nonzero.

With the given actions (6), the Peiffer productG ./ H is a quotient of the group P(a, b) with presentation

P(a, b) =hx, y:x⁻¹y⁻¹xy^b+1, y⁻¹x⁻¹yx^a+1i and we begin by examining some relations inP(a, b) and its quotients.

Lemma 4.1. In the group P(a, b)and all of its quotient groups, (a) x^a= [x, y] =y^−b is central and

(b) x^a2 =x^ab=y^b2=y^ab= 1.

Proof. Working inP(a, b), we have 1 =x⁻¹y⁻¹xy^b+1=x⁻¹x^a+1y^b=x^ay^b, which proves (a). Since x^a is central we have x^a =y⁻¹x^ay = (x^a+1)^a =x^a2+a so that x^a2 = 1. In the same way we havey^b2= 1. Sincex^a andy^b are central, we further have 1 = (x^ay^b)^a =y^ab and similarlyx^ab= 1. This proves (b).

In a Peiffer product of cyclic groups with actions given by (6) whereab6= 0, we may as well assume that the factors GandH are cyclic of finite orders mand n, respectively. We introduce the groupP(a, b;m, n) with presentation

P(a, b;m, n) =hx, y:x^m, yⁿ, x⁻¹y⁻¹xy^b+1, y⁻¹x⁻¹yx^a+1i.

This group decomposes as a Peiffer productP(a, b;m, n) =Z/m ./Z/nwhenever the congruence relations

(a+ 1)ⁿ ≡ 1 modm (7)

(b+ 1)^m ≡ 1 modn

are satisfied. (These are necessary and sufficient ensure that the factorsG=Z/m andH =Z/nact on each other via (6).) For convenience, we introduce the following notation:

µ= gcd(a, m) and ν = gcd(b, n).

In the presence of the congruence relations (7), we can examine the structure of P(a, b;m, n) using Proposition1.1as follows:

P(a, b;m, n)/hxi ∼=Z/ν (8)

hxi/(hxi ∩ hyi)∼=P(a, b;m, n)/hyi ∼=Z/µ (9)

hxi ∩ hyi= [hxi,hyi] =h[x, y]i=hx^ai (10)

Lemma 4.2. Assume that the parameters a, b, m, and n satisfy the congruence relations (7). IfH2(P(a, b;m, n)) = 0, then the subgroup hx^aiof P(a, b;m, n) has ordergcd(µ, ν)and the groupP(a, b;m, n) has orderµνgcd(µ, ν).

Proof. LetP=P(a, b;m, n) and consider the quotient groupP/hx^ai ∼=Z/µ⊕Z/ν.

The five-term sequence associated to the central extension 0→ hx^ai →P→Z/µ⊕Z/ν→1 takes the form

0 =H₂(P)→Z/µ⊗Z/ν→ hx^ai →H₁(P)→^∼= Z/µ⊕Z/ν→0.

From this we conclude that the subgrouphx^aiofP has order gcd(µ, ν). Computa- tion of the order ofP is enabled by (8), (9), and (10).

As an example, note thatP(−2,−2)∼=P(−2,−2; 4,4) is the Peiffer productZ./

Zwith nontrivial actions by the infinite cyclic factors. SinceP(−2,−2) is a finite group with a balanced presentation, Lemmas4.1and4.2show thatx²=y²= [x, y]

is a central element of order two and that Z./ Zhas order eight. As seen in [8], Z./Zis the quaternion group of order eight.

We now examineH₂(P(a, b;m, n)) under the assumption that the parametersa, b, m, andnsatisfy the congruence relations (7). Using Lemma4.1, note that

x^an/ν =y^bm/µ= 1

in P(a, b;m, n). The group P(a, b;m, n) is therefore unchanged if we replace m andnby gcd(a², ab, m, an/ν) and gcd(b², ab, n, bm/µ), respectively. The values of µ and ν are unchanged and one can use the binomial theorem to show that the congruence relations (7) are still satisfied. These observations show that we can restrict our attention to the case where P = G ./ H where G =Z/m=hxi and H =Z/n=hyi, and where the parametersm, n, a, bsatisfy the divisor relations

m|a², ab,an

ν and n|b², ab,bm µ . (11)

Recall thatS(respectivelyT) is the normal closure ofx⁻¹y⁻¹xy^b+1(respectively y⁻¹x⁻¹yx^a+1) in the free productG∗H, and that Π =ST. Let Σ (resp. Θ) denote

the image ofSG∗H (resp. TG∗H) in ΠG∗H. Thus ΠG∗H = Σ + Θ. Note that the intersection Σ∩Θ is contained in the kernel of β : ΠG∗H → [G;H]G⊕[H;G]H. We obtain explicit information about the kernel of β by studying the sequence of surjections

SG∗H→Σ→Σ/(Σ∩Θ)→[H;G]H.

Lemma 4.3. Assume that the parameters a, b,m, andn satisfy the divisor rela- tions (11).

(a) The group S_G∗H is cyclic of ordergcd(m, n), generated by hx, yi.

(b) The order of the cyclic groupΣis a common divisor ofa,b,m, andn. Thus the order of Σdivides gcd(µ, ν).

(c) The order of the cyclic groupΣ/(Σ∩Θ) is the least common multiple ofm/µ andn/ν.

(d) The group [H;G]H is cyclic of ordern/ν.

Analogous results hold for the groups in the sequence of surjections T_G∗H→Θ→Θ/(Σ∩Θ)→[G;H]_G.

Proof. (a) Lemma2.1shows thatSG∗His cyclic, generated byhx, yi.The structure of SG∗H can be worked out from Theorem3.2. For example, using the fact that the augmentation idealginZGis given byg∼=ZG/hP_m−1

i=0 xⁱi, the groupS_G∗H∼= H^ab⊗_Ggis cyclic of order gcd(n,1 + (b+ 1)· · ·+ (b+ 1)^m−1). Using the binomial theorem and the divisor relations (11), one can show that 1+(b+1)· · ·+(b+1)^m−1 is congruent tommodulonand so SG∗H is cyclic of order gcd(m, n).

(b) Considerhx, yi^b=hx, y^bi=x⁻¹y^−bx(y^b)^x[S, G∗H]. We have x⁻¹y^−bx(y^b)^x=x⁻¹y^−bxy^b2+b=x⁻¹y^−bxy^b= [x, y^b].

On the other hand, since x^ay^b ∈ ST, we have [x, y^b] = [x, x^ay^b] ∈ [ST, G∗H].

This shows thathx, yi^b lies in the kernel ofS_G∗H →Σ⊆Π_G∗H. In the same way, hx^a, yi=x^−ay⁻¹x^ay^xa[S, G∗H] = [x^a, y][S, G∗H] is in the kernel ofSG∗H →Σ since [x^a, y] = [y^bx^a, y] ∈ [ST, G∗H]. Using the twisted bilinear relations from Lemma 2.1and working modulo the kernel ofSG∗H →Σ, for any positive integer kwe have

hx^k, yi=hx, y^xk−1ihx^k−1, yi

=hx, y^(b+1)k−1ihx^k−1, yi

≡ hx, yi^(k−1)b+1hx^k−1, yi

≡ hx, yihx^k−1, yi.

It follows that hx^a, yi ≡ hx, yi^a modulo the kernel of SG∗H → Σ and so hx, yi^a itself lies in this kernel. Together with part (a), this shows that the order of Σ is a common divisor ofm,n,a, andb, as claimed.

(c) Information on the intersection Σ∩Θ is obtained by reducing elements ofS moduloT. For this, we first use the divisor relations (11) to show that the elements x^a and y^b are central in the semi-direct product GoH = (G∗H)/T. Working moduloT we have

y⁻¹x^ay= (x^y)^a =x^a2+a=x^a

sincem|a². In addition,

y^−bxy^b=x^(a+1)b =x^ab+1=x

sincem|gcd(a², ab). Using this, we show that [S, G∗H]⊆T. Givenu, w∈G∗H and working moduloT, we have

[w⁻¹x⁻¹y⁻¹xy^b+1w, u]≡[w⁻¹x^ay^bw, u]≡[x^ay^b, u]≡1.

This shows thatT contains a generating set for [S, G∗H].

Now Σ/(Σ∩Θ)∼= (Σ+Θ)/Θ =ST/T[S, G∗H] =ST/T embeds in (G∗H)/T ∼= GoH by the map

hx, yi+ (Σ∩Θ)7→x^ay^bT.

Once again using the fact thatx^a andy^b are central moduloT and that (G∗H)/T is the semidirect product GoH = Z/mo Z/n, we are able to conclude that hx, yi^k∈Σ∩Θ if and only ifm|akandn|bk. The result follows easily.

(d) The groupH_Gis cyclic of orderν so [H;G]_H= [H;G] is cyclic of ordern/ν,

generated byy^b.

Theorem 4.4. LetG=Z/mandH=Z/n, generated byxandyand with actions given by (6). If we assume that the parameters a, b, m, and n satisfy the divisor relations (11), thenH₂(G ./ H) = 0 if and only ifm/µ=n/ν= gcd(µ, ν).

Proof. LetP =G ./ H. We use the conclusions of Lemma4.3without reference throughout the proof of the theorem. Suppose first that m/µ=n/ν = gcd(µ, ν).

Since the order of Σ is a divisor of gcd(a, b, m, n) = gcd(µ, ν) and the order of [H;G]_H is n/ν, the fact that gcd(µ, ν) = n/ν implies that the surjection Σ → [H;G]_H is an isomorphism. In particular, Σ∩Θ = 0 so that Π_G∗H = Σ⊕Θ. In the same way, Θ→[G;H]_G is an isomorphism. Thusβ is an isomorphism and so H2(P) = 0.

Now suppose thatH2(P) = 0. Thenβ is injective, which implies that the map Σ/(Σ∩Θ)→[H;G]His an isomorphism. This in turn implies that lcm(m/µ, n/ν) = n/νso thatm/µdividesn/ν. Analogous considerations applied to the map Θ/(Σ∩

Θ)→[G;H]G show thatn/ν divides m/µ. Thusm/µ=n/ν.

The order of the subgrouphxi= ¯GofPisµgcd(µ, ν) by (9), (10) and Lemma4.2.

This implies that µgcd(µ, ν) ≤ m = µn/ν, so that gcd(µ, ν) ≤ n/ν. On the other hand, the fact that n|b² can be used to show that n/ν|gcd(µ, ν). Thus

m/µ=n/ν= gcd(µ, ν).

Corollary 4.5. With the notation and hypotheses of Theorem4.4, if H2(G ./ H) is trivial, then the factorsGandH embed in the Peiffer product G ./ H.

Proof. The order of ¯G is µgcd(µ, ν) = µm/µ = m, so that G ∼= ¯G. Similarly,

H ∼= ¯H.

Examples with H₂ = 0.Given nonzero integers a and b with g = gcd(a, b), the divisor relations (11) are satisfied if we set m = ag and n =bg. One notes that m/µ=n/ν= gcd(µ, nu) =g, so the groupP =P(a, b;ag, bg) =Z/ag ./Z/bghas orderabg,H2(P) = 0, and the factors embed in the Peiffer product.

More with H2 = 0.Suppose that a = pq, b = qr, and m = n = q² where p, q, and r are pairwise relatively prime. The divisor relations (11) are satisfied and m/µ=n/ν = gcd(µ, ν) =q, soP =P(pq, qr;q², q²) =Z/q²./Z/q² has orderq³, H2(P) = 0, and the factors embed in the Peiffer product.

Examples with H26= 0.Consider a fixed integer p≥2. The divisor relations (11) are satisfied for a =p^r, b =p^s, m =p^r+t, and n =p^r+s+t if r ≥1, s ≥0, and 0 ≤t ≤r. We havem/µ=n/ν =p^t and gcd(µ, ν) =p^r. For fixed rand s and t= 0. . . r, letGtdenote the Peiffer product

G_t=P(p^r, p^r+s;p^r+t, p^r+s+t)∼=Z/p^r+t./Z/p^r+s+t.

Takingt=r, Theorem4.4shows thatH₂(G_r) = 0. With this, Lemma4.2provides thatG_r has orderp^3r+s and Corollary4.5 shows that the elementxhas orderp^2r inG_r∼=Z/p^2r./Z/p^2r+s.

For eacht, we have a central extension

0→ hx^pr+ti →Gr→Gt→1

and the five-term sequence for this extension shows thatH₂(G_t)∼=hx^pr+tiis cyclic of orderp^r−t. Knowing the order ofG_r and the order ofxin G_r, we also conclude that G_t has order p^2r+s+t. Finally, we note that the factors embed in the Peiffer productGt∼=Z/p^r+t./Z/p^r+s+t.

4.4. A double wreath product. Given positive integersmandn, letG=Cm⁽ⁿ⁾

be the direct product ofncopies of the multiplicative cyclic groupCm of orderm and letH=Cn^(m)be the direct product ofmcopies of the cyclic groupCnof order n.

G=hx₁, . . . , x_n:x^m_i ,[x_i, x_j]i H =hy₁, . . . , y_m:yⁿ_i,[y_i, y_j]i

ThenGandH act on each other by cyclic permutation of indices.

x^y_i^j =x_i+1 (subscripts mod n) (12)

y_j^xi=y_j+1 (subscripts mod m)

The Peiffer productG ./ H is a homomorphic image of the standard wreath prod- uctsCmoCnandCnoCm, so we think ofG ./ H=Cm⁽ⁿ⁾./ Cn^(m)as adouble wreath productofCm andCn.

Lemma 4.6. For the double wreath productG ./ H=Cm⁽ⁿ⁾./ Cn^(m), the map β : Π_G∗H→[G;H]_G⊕[H;G]_H

is an isomorphism. In addition, the natural map

H2(G)⊕H2(H)→H2(G ./ H) is surjective.

Proof. We show that the kernel of the map S_G∗H → [H;G]_H is contained in the kernel of the map SG∗H → ΠG∗H. An analogous result holds for the map TG∗H →ΠG∗H and from this it follows that the mapβ is an isomorphism.

First note that [H;G]H = [H;G]∼=Cn^(m−1) is generated by the elements y⁻¹₁ y2, y⁻¹₂ y3, . . . , y_m−1⁻¹ ym.

Next, Lemma2.1shows thatSG∗H is generated by the elements hxi, yji=x⁻¹_i y_j⁻¹xiyj+1[S, G∗H].

The mapβ carrieshxi, yjito the element y_j⁻¹yj+1∈[H;G] = [H;G]H. SinceH is abelian, Theorem3.2shows thatH⊗Gg∼=SG∗H via the maph⊗(g−1)7→ hg, hi.

It follows that

hx_i, y_jiⁿ= 1 for alli= 1, . . . , nandj = 1, . . . , m. In addition,

Ym j=1

hx_i, y_ji=hx_i, y₁. . . y_mi=hx_i, y₁y^x₁ⁱ. . . y₁^xm−1i i=hx_i, y₁^{1+xi+···+xm−1i} i= 1

for alli= 1, . . . , nsince (1 +xi+· · ·+x^m−1_i )(xi−1) = 0 in the integral group ring ZG. With this we see that it suffices to show that for eachi and j, the elements hxi, yjiandhxi+1, yjihave the same image in ΠG∗H. Working inG∗H, notice that x⁻¹_i y⁻¹_j xiyj+1 = [xi, yj]y⁻¹_j yj+1 ∈ S and y_j⁻¹x⁻¹_i yjxi+1 = [yj, xi]x⁻¹_i xi+1 ∈ T so thaty⁻¹_j y_j+1x⁻¹_i x_i+1∈ST = Π. This implies that

[y_j+1, y⁻¹_j y_j+1x⁻¹_i x_i+1] = [y_j+1, x⁻¹_i x_i+1]∈[Π, G∗H].

The image of the elementhx⁻¹_i x_i+1, y_j+1iunder the mapS_G∗H →Π_G∗H is (x⁻¹_i x_i+1)⁻¹y⁻¹_j+1(x⁻¹_i x_i+1)y_j+1^{(x−1i xi+1)}[Π, G∗H] = [x⁻¹_i x_i+1, y_j+1][Π, G∗H] = 1 so thathx⁻¹_i x_i+1, y_j+1ilies in the kernel ofS_G∗H →Π_G∗H. Now, using the relations of Lemma2.1forS_G∗H, we have

hx⁻¹_i xi+1, yj+1i=hxi+1, y^x_j+1⁻¹ⁱ ihx⁻¹_i , yj+1i

=hx_i+1, y_jihx⁻¹_i , y_j^xii

=hx_i+1, y_jihx_i, y_ji⁻¹.

This shows that the elementshx_i, y_jiandhx_i+1, y_jihave the same image in Π_G∗H and completes the proof thatβ is an isomorphism.

To prove the second assertion of theLemma, note that the groupsG_H and H_G are both cyclic, so thatH₂(G_H) =H₂(H_G) = 0. Referring to Figure1.1, the map α2is the zero map. Sinceβ is injective, it follows thatH2(G ./ H)→ΠG∗H is the zero map and soH2(G∗H) =H2(G)⊕H2(H)→H2(G ./ H) is surjective.

We will not attempt to determine the kernel of the map H₂(G)⊕H₂(H) → H₂(G ./ H) here, but we can use Lemma4.6 to compute the order of the double wreath productCm⁽ⁿ⁾./ Cn^(m).

Proposition 4.7. The order of the double wreath productCm⁽ⁿ⁾./ Cn^(m)ofC_mand Cn ismngcd(m, n).