Knit Products of Some Groups and Their Applications
FIRATATESÎ (*) - A. SINANCËEVIK(**)
ABSTRACT- L etGbe a group with subgroupsAandK(not necessarily normal) such thatGAKandA\K f1g. ThenGis isomorphic to theknit product, that is, the ``two-sided semidirect product'' ofKbyA. We note that knit products co- incide withZappa-Szep products(see [18]).
In this paper, as an application of [2, Lemma 3.16], we first define a pre- sentation for the knit productGwhereAandKare finite cyclic subgroups. Then we give an example of this presentation by considering the (extended) Hecke groups. After that, by defining the Schur multiplier ofG, we present sufficient conditions for the presentation ofGto be efficient. In the final part of this paper, by examining the knit product of a free group of ranknby an infinite cyclic group, we give necessary and sufficient conditions for this special knit product to be subgroup separable.
1. Introduction.
The structure of semidirect products is well known. In fact the semi- direct product of any two groups is a generalization of the direct product of these two groups which requires at least one of the factors to be normal in the product. In another words, if a group Gis a product ABof two sub- groups withAnormal andA\B f1g then conjugation ofAby the ele- ments ofBgives an action ofBonAby automorphisms. Moreover, ifAand Bare groups not known to be subgroups of another group and there is an action ofBonAby automorphisms then a group structure (the semidirect product) on the setABcan be defined so that conjugation ofA f1gby elements off1g Bmirrors the given action.
(*) Indirizzo dell'A.: Department of Mathematics, Faculty of Science and Art, Balikesir University, Campus, 10145, Balikesir, Turkey.
E-mail: [email protected]
(**) Indirizzo dell'A.: Department of Mathematics, Faculty of Science, SelcÎuk University Campus, 42003 Konya, Turkey.
E-mail: [email protected]
2000Mathematics Subject Classification: 20E22, 20F05, 20F55, 20F32.
The next step along this path is theZappa-Szep product of any two groups, which requires neither of the factors to be normal in the product.
In other words, if the subgroupAis not assumed to be normal then a si- milar situation exists. We first look at groups to get the main parts. LetG be a group with identityf1g, subgroupsA,K satisfyingK\A f1gand GAK. Then eachg2Gis uniquely expressible asgakwithk2Kand a2A. We are now in a position to reserve certain products witha2Aand k2K by consideringak2G. We must have unique elements k02K and a02Asuch thatkaa0k0. This defines two functions
(k;a)7 !ka2K; (k;a)7 !k:a2A (1)
(in fact these were calledthe mutual actions defined by the multiplication by Brin in [2]) that are unique subject to the relation
ka(k:a)(ka);
(2)
for allk2Kanda2A. We remark that the details of the above material can be found in [2]. We also note that the terminology Zappa-Szep product was developed and suggested by G. Zappa in [20].
Moreover, in [14], it is proved that if a Lie algebra is the direct sum of two sub Lie algebras then one can write the bracket in a way that mimics semidirect products on both sides. This construction is called the knit product of graded Lie algebras. Additionally, in [14], the behaviour of homomorphisms with respect to knit products was investigated. The in- tegrated version of a knit product of Lie algebras will be called theknit product of groupswhich coincides with theZappa-Szep product(see [18]).
Throughout this paper the notationZn denotes the cyclic group of or- dern,Dn denotes the dihedral group of order 2nandSndenotes the sy- metric group of ordern!, wheren2N.
2. Preliminaries.
In this section we define the knit product of any two groups by using the action given (1) and then give a standard presentation for this product. We note that material similar to this section may also be found in [2] and [14].
LetAandKbe subgroups of a groupGas defined in the previous section.
LEMMA2.1. The knitted pair of actions(~a;~b)for(A;K)are mappings
~
a:KA !A; ~b:KA !K (k;a)7 !k:a; (k;a)7 !ka
such that
i) a:K !Aut(A) is a group homomorphism, soak1(ak2)ak1k2
anda1IdA, whereak(a):~a(k;a),
ii) b:A !Aut(K) is a group anti homomorphism, in other words,ba1(ba2)ba2a1 andb1IdK, whereba(k):b(k;~ a),
iii) ak(a1a2)ak(a1):aba1(k)(a2), iv) ba(k1k2)bak
2(a)(k1):ba(k2), for all a;a1;a22A, k;k1;k22K.
One may find the proof of the above lemma in [2, Lemma 3.2] or [14].
Moreover, by considering the actions given in (1), knit products of groups can be defined as follows.
DEFINITION2.2. Let A and K be groups, and leta,bbe homomorphisms defined by
b:A !Aut(K);a7 !ba anda:K !Aut(A);k7 !ak
respectively, for all a2A and k2K. Then the ``knit product'' GA(a;b)K of K by A is defined on the set AK by the following operation:
(a1;k1)(a2;k2)(a1ak1(a2);ba2(k1)k2):
(3)
The identity is(1;1)and the inverse of an element(a;k)is (a;k) 1(ak 1(a 1);ba 1(k 1)):
In fact A f1g andf1g K are subgroups of A(a;b) K which are iso- morphic to A and K, respectively.
Similar definitions of knit product can also be found in [2], [14] and [18].
(We note that ifAandKare topological groups or Lie groups and~a,~bare continuous or smooth, then A(a;b)K is also a topological group or Lie group, respectively, which will not be needed in this paper).
REMARK2.3. 1)In Definition2.2, ifaIdA(orbIdB) then the knit product A(a;b)K becomes the semidirect product.
2) We know that the semidirect product of any two groups is actu- ally equivalent to the split extension. Also, by the previous material, the knit product can be thought as a two-sided semidirect product by as- suming the factors not to be normal. Therefore knit products can also be regarded as a type of split extension. To show this fact, it is enough to generalize Propositions 2:1and2:3given in[3, Chapter IV].
One can find an earlier proof for the following result in [14].
PROPOSITION 2.4. Let G be a group and A, K be subgroups of G.
Suppose that GAK and A\K f1g. Then GA(a;b) K.
PROOF. L etk:a~a(k;a):~b(k;a) be the unique decomposition ofk:ain GAK. Then
a1k1a2k2a1~a(k1;a2)~b(k1;a2)k2(a1ak1(a2)):(ba2(k1)k2):
(4)
Thus we just need to show that the knitted pair of actions (~a;~b) satisfies the four conditions of Lemma 2.1. It is clear that we have~a(1;a)a,~b(1;a)1,
~a(k;1)1, ~b(k;1)k. In fact comparing coefficients in the law of asso- ciativity ofGgives two equations. Then setting suitable elements in these equations to 1 gives all conditions of Lemma 2.1. p In [2], Brin defined a standard presentation for the knit product of groups (and monoids as well) by stating a similar version of the following lemma.
LEMMA 2.5 [2, Lemma 3.16]. Suppose that PAhX; Ri and PK
hY ; Siare presentations for the groups A and K, respectively under the maps y7 !ky(y2Y), x7 !ax(x2X)with X\Y ;. Then a presentation for the structure defined by(3)on AK is
P hX;Y;R;S;Ti
in which T consists of all pairs (yx;(y:x)(yx)), as given in (2), for (y;x)2KA.
3. The knit product of cyclic groups.
In this section, as an application of Lemma 2.5, we will define a pre- sentation (see Proposition 3.1 below) for the knit product of finite cyclic groups in terms of the generators and relators of these groups. After that, by taking A an infinite group and KZp (p a prime), we will give an example of the knit product for the infinite case.
LetAandKbe bothZnandZmgenerated byxandy, respectively, and let G be the knit product A(a;b) K, where a:K !Aut(A) and b:A !Aut(K). We then have
PROPOSITION3.1. LetPAhx; xni andPKhy; ymi be presenta- tions for the groups A and K, respectively. Suppose that xtm 11Aand
yln 11B (1 jtj5n,1 jlj5m). Then G has a presentation PGx;y; xn;ym;Tyx
; (5)
where Tyx consists of all pairs(yx;xtyl).
PROOF. L etdt(1 jtj5n) andcl(1 jlj5m) be automorphisms ofA andK, respectively. Assume thatxtmxandylny. Then we have map- pingsy !Aut(A) andx !Aut(K). By [7], these induce homomorphisms
a : K ! Aut(A) and b : A ! Aut(K)
y 7 ! dt; x 7 ! cl
if and only if
dmt idA and cnl idB:
By the assumption on the generatorx, the homomorphismsdmt andidAare equal if and only if [x]dmt [x]idA. Similarly, by the assumption ony,cnl and idB are equal if and only if [y]cnl [y]idB. These imply that yxxtyl. Hence, by Lemma 2.5, we obtain the presentationPGin (5) for the group
Zn(a;b)Zm. p
EXAMPLE3.2. For any prime p, let us determine Z2(a;b) Zp. So suppose that hx;x2i and hy;ypi are the presentations for Z2 and Zp, respectively. Then the homomorphisms b:Z2 !Aut(Zp)Zp 1
and a:Zp !Aut(Z2) are defined by x7 !bx:y7 !y 1 and y7 !ay:x7 !x 1, respectively. Hence, by Proposition 3.1, we have a presentation hx;y; x2;yp;yxx 1y 1i for the group Z2(a;b)Zp. In fact this presentation can be written as hx;y; x2;yp;(yx)2i, and so this implies that Z2(a;b) ZpDp. In particular, if we take p2 then we obtain Z2(a;b) Z3D3S3.
Actually if we choose any positive integer m instead of p in the above calculations then we obtainZ2(a;b) ZmDm, where m2.
The following example, obtained by considering the (extended) Hecke groups, is an application about infinite case for the presentationP, defined in Lemma 2.5. We note that the fundamental material about (extended) Hecke groups can be found, for instance, in [11, 17]. Recall that Hecke groupsH(lq) are presented byhx;y;x2;yqiwhile extended Hecke groups H(lq) are presented by hx;y;r;x2;yq;r2;(xr)2;(yr)2i. In fact H(lq)
D2Z2Dq.
EXAMPLE 3.3. Let A be the Hecke group H(lq), where q3 and q2Z. Also let K be the groupZp generated by r, where p is a prime.
Let us consider the homomorphisms
a : Zp ! H(lq) and b : H(lq) ! Zp
r 7 ! ar(x)x 1 x 7 ! bx(r)rt ar(y)yj y 7 ! by(r)rs where1t;s5p,1j5q such that x 1px, yjpy, rt2 r and rsqr.
We then get the relators
ryar(y)by(r)yjrs and rxar(x)bx(r)x 1rt: Hence, by Lemma2.5,
P x;y;r;x2;yq; rp;ryyjrs;rxx 1rt (6)
is a presentation for the group H(lq)(a;b)Zp.
In presentation(6), we can choose p2, s1and jt 1since we have x 12x, y 12y, r1qr and r 12r. Therefore
H(lq)(a;b)Z2H(lq)D2Z2Dq:
3.1 ±The Schur multiplier.
In this part of the section we define the Schur multiplier (or, equiva- lently, the second homology group) of the knit product of two finite cyclic groups.
LetKbe a cyclic group of ordermwith a presentationPK hy; ymi, and let Abe cyclic group of order p (p is a prime) with a presentation PA hx; xpi. Then, by Proposition 3.1, a presentation forGA(a;b)K is given by
P hx;y; xp;ym;yxxtyli;
(7)
wherextm 11 andylp 11 such that 1 jtj5nand 1 jlj5m. Suppose that
(l 1;md)dwithd(l 1;m) and lp1(modmd).
We then have the following result.
THEOREM3.4. Let G be the knit product with presentationPas in(7).
Then
H2(G) f1g if d1;
Zp if dp:
(
Before giving a proof of this result, we should note that the equivalence class containing a factor set g ofG will be denoted byfgg. L et M be an algebraically closed field of characteristic zero with its multiplicative group MM f0g. The set H2(G;M) (which we denote it by H2(G)) of all equivalence classes of factor sets ofGoverMforms an abelian group under the multiplication defined by fg1gfg2g fg1g2g. This group is called the Schur multiplierofG, and it is actually the second homology group where M is a trivialG-module. We refer the reader to [12] for the details and applications of the Schur multiplier.
PROOF. Suppose thatd1. Thenpdoes not dividemand hence every Sylow subgroups ofGis cyclic, that isH2(G) f1g.
Now assume thatdp. In that case we will show thatH2(G) is a cyclic group of orderpwith
G1 hy1;x1; ymp1 ;xp1;(xt1) 1y1x1yl1i
is a representation group ofG. SinceGis generated by two elements and has three relators, by [15],H2(G) must be cyclic. On the other hand,Ghas a subgrouphxiof indexp. HencejH2(G)j p. Moreover, by the assumption, sincepjmandylp 11 then (l;m)1. It follows that (l;mp)1. Then we have lp1(modmp). Hence G1 is a well-defined group. By [15, Pro- position 2.2], we have
[G1;G1] hyl1 1i hyp1i and Z(G1) hym1i:
Therefore hym1i [G1;G1]\Z(G1) and, by the isomorphism theorem, we obtainG1=hym1i G. Hence it follows thatjH2(G)j p. Therefore,H2(G) is cyclic of orderp, andG1is a representation group ofG. p
3.2 ±Efficiency.
In this part, let us recall the definition of efficiency on groups and then give a result about efficiency for the knit product of finite cyclic groups by considering Theorem 3.4.
Let G be a finitely presented group, and let P hx; ri be a finite presentation for G. Then the Euler characteristic of P is defined by x(P)1 jxj jrj, wherej:jdenotes the number of elements in the set.
Letd(G)1 rkZ(H1(G))d(H2(G)) whererkZ(:) denotes theZ-rank of the torsion-free part andd(:) means the minimal number of generators.
Then, by [9], for the presentationP, it is always true thatx(P)d(G). We then define x(G)minfx(P):P is a finite presentation forGg: In view of these facts, for a groupG, we say thata presentationP0for G is called minimal ifx(P0)x(P), for all presentationsP of Ganda presentation P0is called efficient ifx(P0)d(G).
A brief survey of known results on efficiency can be found in [6]. In fact there is interest not only in finding efficient presentations, but also finding presentations which are efficient on the minimal number of generators (see [19]).
Letdpin Theorem 3.4. Then we have the following result.
THEOREM3.5. Let GZp(a;b)Zm with a presentationP as in(7).
Suppose that(l 1;mp)p with p(l 1;m)and lp1(modmp). Then Pis an efficient presentation on2-generators for the group G.
PROOF. First of all let us consider the lower boundd(G) ofG. Since the order ofGismp(by considering Remark 2.3-2)), we get rkZ(H1(G))0.
Thus we haved(G)1d(H2(G)). Actually, by Theorem 3.4,H2(G)Zp and so d(H2(G))1. Hence d(G)2. In the second part of the proof, a simple calculation shows that the Euler characteristic ofP is equal to 2.
Henced(G)x(P) and soPis an efficient presentation. Moreover, sinceG is a 2-generator group, we cannot apply any reduction on the generating set ofP, which means this efficient presentationPmust have 2 generators, as
required. p
As an application of Theorem 3.5, let us choose t1, l 1 and p2. Then we get the dihedral group Dm of order 2m (m2) pre- sented by
P hx;y; x2;ym;(yx)2i:
(8)
Thus a straightforward computation shows that
COROLLARY3.6. P, as in(8), is an efficient presentation for the group Dm ifmis even integer greater than or equal to4.
REMARK3.7. In Theorem3.5, if d1then H2(G)is trivial (by The- orem 3.4), and so d(H2(G))0. Thus this gives the inefficiency of the presentationP. Therefore it might be useful to study whether or not the presentationPin(7)is minimal when d1.
REMARK 3.8. In Example 3.3, up to isomorphism, we obtained a presentation for the extended Hecke group H(lq). In fact the efficiency of H(lq)has been examined under certain conditions in[8].
4. Subgroup separability ofF(a;b)Z.
In [10, Problem 5], Scott asked a question which was whether all semidirect productsFaZwitha2Aut(F) are subgroup separable. We should note that the first negative answer was given by Burns, Karras and Solitar in [4]. After that it has been positively answered by Metaftsis and Raptis [13, Corollary 3].
Now our aim is to lift the problem above to knit products. Thus, by considering the knit product of a free group of rank n by infinite cyclic group, we give a partial answer to it.
By using the following lemma, we will give necessary and sufficient conditions for the knit product F(a;b)Z to be subgroup separable. We recall that a groupGis said to besubgroup separableif, for every finitely generated subgroupHofG,His the intersection of finite index subgroup ofG.
LEMMA4.1 ([13)]. Let K be a finitely generated abelian group and let A, B be subgroups of K such that G is the HNN-extension
G ht;K; t 1AtBi:
Then G is subgroup separable if and only if A\B is a subgroup of finite index in both A and B, and there is a finitely generated normal subgroup of G, say H, such that H has finite index in A\B.
Before giving our theorem, let us give the definition of right layered basis which will be needed for our result. Suppose that a is an auto- morphism of the free groupFhaving ranknandX fx1;x2; ;xngis a basis forF. ThenXis a right layered basis foraif
a(x1)x1 and a(xi)xiwi; (9)
wherewi2F(x1; ;xi 1) for 2in. The existence of a right layered basis for an automorphism of free groups with ranknhas been shown in [5].
Let us considerGF(a;b) Z. We then have the following result.
THEOREM4.2. G is subgroup separable if and only ifais the identity automorphism.
PROOF. Let us suppose the set Z is generated by y. Also let us assume 16a2Aut(F) such that there exists a right layered basis X fx1;x2; ;xng for a as in (9). Therefore, by Lemma 2.5, one can easily show thatGhas a presentation
G hx1;x2; ;xn;y; yx1x1yj1;yx2x2w2yj2; ;yxnxnwnyjni;
wherewi's are words as in (9) and eachji(1in) is an integer. Since a61, at least one element of thewi's is a non-trivial word. Letws be the first such a word, where ws2F(x1;x2; ;xs 1). Then we can choose wsxm1xm2 xms 1such thatxm1;xm2; ;xms 12 fx1;x2; ;xs 1g. Also let H be a subgroup of G generated by fxs;ws;yg. Then H has a pre- sentation of the form
H hxs;ws;y;ywsy ji1 xm1yji1 1xm2yji1 1 xms 1yji1 1; yxsy ji2 xsws (1ji1;ji2 n)i
since yxmky ji1 xmk for all 1mk s 1. In this step, let us take ji1 1ji2. Then the above presentation can be rewritten as follows:
H hxs;ws;y; ywsy 1ws;xs1yxsywsi:
This meansHis an HNN-extension with base groupKthat is a free abelian group of rank two generated byfws;ygand stable letterxs. Let the iso- morphic subgroups ofKbeCandDgenerated byfygandfywsg, respec- tively. It is obvious thatC\D f1gand, by Lemma 4.1,His not subgroup separable. ThereforeGcannot be subgroup separable since it containsH which gives a contradiction by the assumptiona61.
On the other hand, let a1. Then GZbF which is subgroup se- parable since it has index two subgroup isomorphic to ZF which is subgroup separable. ThusGis subgroup separable (by the results given
in [1, 16]), as required. p
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Manoscritto pervenuto in redazione il 18 aprile 2007.
