Computing large direct products of free groups in integral group rings

Computing large direct products of free groups in integral group rings



Angeldel Roand ManuelRuiz



Abstract

We construct explicitly a subgroup of nite minimalindex and min- imal rank in^ZGwhich is a direct products of free groups for each nite group^Gfor which this is possible.

1 Introduction and preliminaries

Let G be a nite group. If G is abelian then the structure of the group of units^ZGof the integral group ring^ZGis well known by the work of Higman, Bass and Milnor. IfG is non abelian then generators of subgroups of nite index of^ZG have been found for a large class of groups, (see e.g. [12] and [4]). However the structure of ^ZG(or of a subgroup of nite index) is not known except for some few groups. Some of the known cases appeared in [1], [2], [3], [7], [8] and [10].

In a series of three papers ([5], [9] and [6]) Leal, Jespers and the rst author characterized the groups G such that ^ZG contains a subgroup of nite index which is a direct product of free groups. In those papers the existence of such a large subgroup of ^ZG is proved theoretically. Let G be a group satisfying the mentioned property. The aim of these paper is to construct explicitly a subgroup of ^ZG which is a direct product of free groups and of minimal index in ^ZG.

We start with some notation:

The cyclic, dihedral and quaternion groups of order m are denoted by Cm,Dm and Qm respectively.

Let G be a group andR a ring. Then R denotes the group of units of Rand RGthe group ring of G with coecients in R. We refer to [12] for



Therstauthorhasb eenpartially supp orted bytheD.G.I.ofSpain andFundacion

SenecaofMurcia

notational matters concerning group rings. For any nite subsetH ofGlet Hb = ^j_H^{1j P}

h²Hh ^{2 Q}G. More generally, if : H ^{! Z}is a map, then we set Hc = ^j_H^{1j P}

h²H (h)h. If g²G, then set ^bg=^hcgⁱ. The augmentation map is denoted by! :^ZG^{! Z}. If H is a normal subgroup of G then⁴(G;H) = Ker!H is the augmentation ideal modulo H. We set ⁴(G) = ⁴(G;G).

The rank ofG, denoted byr(G), is the minimum of the cardinalities of the generating subsets ofG. Set G^ = Hom(G;^Z) and let inv :G^! Gdenote the map given byg^7!g^?1.

If K is another group and ':G ^!Aut(K) is a group homomorphism, thenK^o'Gdenotes the corresponding semidirect product.

The non abelian nite groups G such that ^ZGcontains a subgroup of nite index which is a direct product of free groups, are of the form G = H^Z where Z is an elementary abelian 2-group and H is of one of the following types (see [6]):

(a) ^hx;y^jx⁴ =y⁴= [x²;y] = [x;y²] = [x;[x;y]] = [y;[x;y]] = 1ⁱ,

(b) ^hx;y¹;::: ;yn^jx⁴ =y_i² = [yi;yj] = [x²;yi] = [[x;yi];yj] = [[x;yi];x] = 1ⁱ,

(c) ^hx;y¹;::: ;yn^jx⁴ =y_i⁴=y_i²[x;yi] = [yi;yj] = [x²;yi] = [y_i²;x] = 1ⁱ, (d) ^hx;y¹;::: ;yn^jx² =y_i²= [yi;yj] = [[x;yi];yj] = [x;yi]² = 1ⁱ,

(e) ^hx;y¹;::: ;yn^jx² =y_i⁴=y_i²[x;yi] = [yi;yj] = [[x;yi];x] = 1ⁱ, (f) ^hx;y¹;::: ;yn^jx⁴ =y_i⁴=x²y¹²=y_i²[x;yi] = [yi;yj] = [y_i²;x] = 1ⁱ, (g) ^hx;y¹;::: ;yn^jx⁴ =x²y⁴_i =y_i²[x;yi] = [yi;yj] = 1ⁱ,

(h) U^ohxⁱ where U is an elementary abelian 3-group, x has order 2 or 4 and(x) = inv.

(i) U ^oK where U is an elementary abelian 3-group, K =^hx;y^i= Q⁸ and(x) =(y) = inv.

All throughout this paperG=H^Z whereZ is an elementary abelian 2-group and H is of one of the types (a)-(i). The letter n is reserved to denote either the number ofy's for the groups of types (b)-(g) or the rank ofU for the groups of type (h) and (i). The rank ofZ is always denoted by k.

Our aim is to nd a concrete subgroupF =F^0Qn_i⁼¹Fiof nite index of

ZG, such thatF⁰ is free abelian andFi is free nonabelian and optimal in the sense that the index ofF in^ZGis minimal among all the possible subgroups of^ZGwhich are a direct product of free groups . There is a natural way to obtain such a group if we do not impose the optimal condition. For every primitive central idempotent eof^QGone of the following conditions holds:

(A) ^QGeis isomorphic to ^Q, an imaginary quadratic extension of^Q or a totally denite quaternion algebra over^Q.

(B) ^QGeis a totally denite quaternion algebra over a real quadratic ex- tension of^Q.

(C) ^QGeis isomorphic to M²(^Q).

LetA,B andC be the sets of primitive central idempotents of^QGof types (A), (B) and (C) respectively and I = A ^[B ^[C. For every e ² I, let Oe be an order in ^QGe. Then O_e is nite if e²A, virtually innite cyclic if e ² B and virtually free nonabelian if e ² C. Since O = ^Q_e²_IOe and

ZGare orders in ^QG, then O and ^ZGare commensurable, that is O^\ZG has nite index in both. Therefore ^ZG^\(1 +^QGe) contains a subgroup of nite index Fe which is trivial if e ²A, innite cyclic if e² B and free nonabelian ife²C. Then^Q_e²_IFe is a subgroup of nite index of^ZGwith the desired structure. However this does not give information on how big the constructed group is. Surprisingly what we are going to prove in this paper is that this naive approach provides the optimal subgroup in almost all the cases. Our rst theorem is:

Theorem 1.1

Let G= H^Z, where Z is an elementary 2-group and H is one of the types (a)-(i). Let B and C be the sets of primitive central idempotents of types (B) and (C) respectively. Then

1. F⁰ =^ZG^\(1 +^QGfB), wherefB = ^P

f²Bf, is free abelian of rank ^jB^j 2. If G^6=D⁶;D⁸, then for all e²C

Fe=^ZG^\(1 +^QGe)

is free non abelian and all the Fe's have the same rank.

3. F =F^{0 Q}

e²CFe has nite index in^ZG

So the groupF =F^{0 Q}

e²CFeof Theorem 1.1 has the desired properties. We prove Theorem 1.1 in Section 2. Besides we compute the ranks ofF⁰ and all theFe's for all the groups. Moreover every Fe is isomorphic to a subgroup of the modular group SL²(^Z) that we compute in Proposition 2.1. Using that it is theoretically easy to compute generators for all the Fe's. Also generators ofF⁰ are easy to compute by using Proposition 2.3. So there is a method to obtain generators forF.

In Section 3 we prove that the group F computed in Section 2 is the best possible. Explicity we prove the following Theorem:

Theorem 1.2

Let G = H^Z and F = F^{0  Q}

e²CFe as in Theorem 1.1.

IfE =E^0Q_j²_JEj is a subgroup of nite index of ^ZG, where E⁰ is free abelian and Ej is free non abelian for every j. Then

1. r(E⁰) =^jB^jand ^jC^j=^jJ^j.

Besides, if G^6=D⁸;D⁶;Q⁸ and Q¹⁶ then 2. [^ZG:F]^[^ZG:E].

3. r(Ej)^r(Fe) for every e²C and j²J.

Note that the exceptional groups D⁸, D⁶, Q¹² and Q¹⁶ belong to the list of groups we are considering with the following parameters: k = 0 and n= 1 for all of them; then D⁸ is of type (d) or (e),D⁶ andQ¹² are of type (h) withxof order 2 and 4 respectively and nally Q¹⁶is of type (g). These four groups have been studied separately in several papers and satisfactory results can be found in [3] forD⁸; [7] forD⁶; [10] forQ¹² and [8] forQ¹⁶.

2 Large subgroups: Proof of Theorem 1.1

The statement (3) of Theorem 1.1 follows by the arguments in Section 1.

The subsequent Proposition 2.1 implies (2) and Proposition 2.3 implies (1).

For every a;b^2Zset

?(b) = SL²(^Z)^\



1 +b



Z Z

Z Z

 

;

^a(b) = SL²(^Z)^\



1 +b^{ Z} a^Z

Z Z

 

and

a(b) = SL²(^Z)^\



1 +b

 a^{Z Z} a^Z a^Z

 

:

If ? is a subgroup of SL²(^Z), then ^b? denotes the image of ? in PSL²(^Z).

Proposition 2.1

Letebe a primitive central idempotent of^QGof type (C).

1. Assume that H is of type (a)-(g). Then Ge ^' D⁸ and there are a;b² Gesuch that ^fe;a;b;ab^g is an integral basis of ^ZGe. Furthermore if e = ⁰e+¹a+²b+³ab, with ⁰;¹;²;^{3 2 Z}and  ^{2 Z}G, then

!()^0+¹+²+³mod 2.

(i) If H is of type (a) then e=^hx^\2;y²ⁱ(1^?cG⁰)Z^c for some non trivial

^2hx²;y^2i and some ²Z^. Moreover Fe ^'²(2^k+3), so that Fe

is free of rank1 + 2^3k+6,

(ii) Assume thatHis of type (b)-(f). For everyi= 1;::: ;nletti = [x;yi].

Let:G^0!Gbe given by (tⁱ¹¹tⁱ²²


tⁱ_nⁿ) =y¹ⁱ¹y²ⁱ²


yⁱ_nⁿ. Then e=eS;'; =^cx²(S^b?cG⁰)^\(S)'Z^c

where S is a maximal subgroup of G⁰, ² Z^ and ':(S)^{! Z}is a map such that '^2S^. Moreover, if H is of type (f), then t¹²S. IfH is of type (b) or (c), then Fe ^'²(2^2n+k) which is free of rank 1 + 2^6n+3(k?1).

If H is of type (d)-(f) then Fe ^'²(2^2n+k?1). which is free of rank 1+ 2^6n+3(k?2), unlessH is of either type (d) or (e),n= 1 and k= 0, or equivalentlyG^'D⁸.

(iii) Assume thatHis of type (g). For everyi= 1;::: ;nletti be the image of [x;yi] in G=^hx²ⁱ. Let :G=^hx^{2i !}G be given by (tⁱ¹¹tⁱ²²


tⁱ_nⁿ) = yⁱ¹¹y²ⁱ²


y_nⁱⁿ. Then

e=eS;'; = (S^b?G^c0)^\(S)'Z^c

where  : G ^! G=^hx²ⁱis the projection, S is a maximal subgroup of G⁰ containing^hx²ⁱ, ²Z^ and ':^(S)^!Zis a map such that '^2(S=^hx²ⁱ)^. MoreoverFe^'²(2^2n+k), so that Fe is free of rank 1 + 2^6n+3(k?1)

2. Assume that H is of type (h) or (i). ThenGe^'D⁶ and there exist a;b²Ge such that

ZG(1^?a)e=^Ze(1^?a)^Za(1^?a)^Zb(1^?a)^Zab(1^?a):

(i) IfHis of type (h) withxof order 2 thene=eS = (S^b?U^b)Z^c whereS is a maximal subgroup of U and ²Z^. Moreover Fe ^'³(2^k3^n?1) which is free unless k = 0 and n = 1, that is to say, G = S³. So if G ⁶= S³ then the rank of Fe is 3^3n?4 if k = 0, and 2^3k?13^3n?4 otherwise.

(ii) IfHis of type (h) with xof order 4 thene=eS =^cx²(S^b?U^b)Z^c where S is a maximal subgroup of U and ²Z^. Then Fe ^'³(2^k+13^n?1) which is free of rank 1 + 2^3k+23^3(n?1).

(iii) IfH is of type (i), thene=eS; ;= (S^b?G^c0)Z^ch\xyⁱwhereSis a max- imal subgroup ofU, ²Z^ and ^2hxy^i. ThenFe^'³(2^k+23^n?1), which is free of rank 1 + 2^3k+53^3(n?1).

Proof.

The proof is done separately for each case. The rst step in each case is to identify the elements ofC. We leave this part to the reader. Then for eache²C one has to make a good selection of aand b. This selection induces a ring isomorphism:^QGe^!M²(^Q) and using this isomorphism one identies (Fe) which happen to be the free subgroup of the modular group in the proposition. Cases (a)-(g) are very similar. We only do case (c) and let the reader check that similar arguments work for the remaining cases. Cases (h) and (i) are also similar so we only do case (i).

Assume that H is of type (c) and let e=eS;'; ²C as in (1.ii). Then HS =^hx²;G⁰;(S);Zⁱis a normal subgroup of index 4 ofGand ift²G⁰ⁿS, then 1;x;y = (t);xy is a transversal of G modulo HS. Set a = xe and b=ye. Every element^2ZGcan be written as

=⁰+¹x+²y+³xy

where i ^{2 Z}Gand Supp(i) ^ HS. Moreover if h ² HS, then he = ^1 and therefore e = ⁰e+¹a+²b +³ab for some ⁰;¹;²;^{3 2 Z} and !() = ^P3_i⁼⁰!(i) ^{ 0} +¹ +² +³mod 2. It is easy to see that Ge ^'D⁸ and e;a;b;abis an integral basis of ^ZGe. Then there is an isomorphism:^QGe^!M²(^Q) given by

(e) =

 ^0?1+^2?3 2(^2?1) ¹+^{3 0}+^1?2+³



(see [3] or [12]). Since the support of e is HS, ie ^{2 Z}G for every i and considering the coecient of 1 in this element one deduce that 2^{2n+k j} i. Let i=i=2^2n+k. Then = 1 + 2^2n+k( ⁰+ ¹x+ ²y+ ³xy)e, so that

() = 1 + 2^2n+k

 ^{0? 1}+ ^{2? 3} 2( ^{2? 1}) ¹+ ^{3 0}+ ^{1? 2}+ ³



:

Therefore () = 1 + 2^2n+k

 a 2b c d



with a ^ dmod 2. Note that this implies that the determinant of () is 1. Thus (Fe) ^ ²(2^2n+k) and by solving a system of linear equation one can easily verify that the equal- ity holds. Moreover ²(2^2n+k) is a subgroup of index 2 of ?(2^2n+k) and

?(2^2n+k)^'?(2^\2n+k) is a subgroup of index 2^6n+3k?4of?(2) (see [11]). Since^d the last group is free of rank 2 then ²(2^2n+k) is free of rank 1+2^6n+3(k?1). Now assume that H is of type (i). Then e = eS; ; as in (2.iii). Set a = ue and b = xe where u ² U ⁿS. Then HS = ^hxy;S;Zⁱ is a normal subgroup ofGand 1;u;u²;x;xu;xu²is a right transversal of GmoduloHS. For the rst part we argue as in the previous case by noticing thathe=^e for everyh²HSand (1+u+u²)e= 0. On the other hand^ZGe=^ZG(1^?a)e so that (1^?a)e;a(1^?a)e;b(1^?a)e;ba(1^?a) is an integral basis of ^ZGe. Moreover, as in [12] or [7], there is an isomorphism :^QGe^! M²(^Q) that associates⁰(1^?a)e+¹(1^?a)e+²b(1^?a)e+³ba(1^?a)ewith

 3(^0?1?2) ^?0+ 2¹+ 2^2?3 3(^0?2^1?2?3) 3(¹+²)



Furthermore, if  ^{2 Z}G^\(1 +^QGe) = Fe then ^?1 ²
(G;^huⁱ)e=

ZG(1^?a)eand

= 1 + (^?1)e= 1 + (⁰(1^?a) +¹a(1^?a)e+²b(1^?a) +³ba(1^?a)e with i ^{2 Z}. Unfortunately in this case, unlike the previous one, the sup- ports of the base elements intersect. However this diculty can be overcome as follows. The coecients of 1, u, x and xu in ^?1 are ⁰ = ²k^+230n^?1, ¹ = ²k^+231n^?1, ² = ²k^+232n^?1 and ³ = ²k^+233n^?1, respectively. Thus

() = 1 + 2^k+23^n?1

 3( ^{0? 1? 2}) ^{? 0}+ 2 ¹+ 2 ^{2? 3} 3( ^0?2 ^{1? 2? 3}) 3( ¹+ ²)



so that by similar arguments as in the previous paragraph one shows that

(Fe) = ³(2^k+23^n?1). The ranks can be computed as in the previous case.

The classical quaternion algebra over an arbitrary ring Ris denoted by

H(R).

In order to prove statement (3) of Theorem 1.1 we need information about the central idempotents of ^QG of type (B). This is the role of next Lemma. The proof is straightforward.

Lemma 2.2

1. If G is not of type (g) or (i) thenB =^;. 2. Assume that G is of type (g) and let

K=^hy¹²y²²;::: ;y¹²y_n²;Zⁱ=^fy²¹ⁱ¹y²²ⁱ²:::y²_nⁱⁿ :^Xn

t⁼¹it^0 mod 2^gZ:

For every ²K^ and every2^i^n, let n=(y¹²y_i²), K() =^hy^12y²;::: ;y^1nyn;Zⁱ and T=^fy^{P1 nj=2jij}yⁱ²²:::y_nⁱⁿ : 0^ij ^1^g:

ThenT is a transversal of^hx²;K()ⁱmodulo K. Moreover each element of B is of the form

f =f;= (1^?cx²)K^[()_;

for some ²K^ and ²K()^ so that K^Ker and ;(kt) =(k)(t) for everyk²K and t²T. Furthermore

Gf;=^ha=xf;b=y¹f ^ja² =b⁴;b⁸= 1;aba^?1=b^?1i'Q¹⁶;

B = ^ff;b;b²;b³;a;ab;ab²;ab^3g is an integral basis of ^ZGf and the map

:^QGf ^!H(^Q[^p2]) given by

(^P_g^2Bgg) = f +^p22(b^?b³) + (a+^p22(ab^?ab³))i+ (_b² +^p22(b+_b³))j+ (_ab²+^p22(ab+_ab³))k is a ring isomorphism.

3. Assume that Gis of type (i). Then every element of B is of the form fS; = (1^?cx²)(S^b?U^b)Z^c

where S is a maximal subgroup of U and ² Z^ Moreover GfS; ^' Q¹² and if X = xf, Y = yf and W = wf, with w ² U ⁿ S, then ^B =

ff;X;Y;XY;W;WX;WY;WXY^g is an integral basis of^ZGf and the map

:^QGf ^!H(^Q[^p3]) given by

(^P_g^2Bgg) = ¹²[2^1?W +^p3XY W + (2XY ^?XY W ^?p3W)i (2X ^?XW ^?p3Y W)j+ (2Y ^?Y W +^p23XW)k] is a ring isomorphism.

Next proposition provides a proof of statement (3) of Theorem 1.1 and also computes the rank ofF⁰.

Proposition 2.3

The group F⁰ is embedded in the centre of ^QG. IfH is of type (g) then F⁰ is free abelian of rank 2^2n+k?2; if H is of type (i) then F⁰ is free abelian of rank 2^k?1(3^n?1). In the remaining cases F⁰ = 1.

Proof.

By Lemma 2.2, we may assume thatGis of type (g) or (i). By the comments prior to Theorem 1.1, it is enough to show that for everyf ²B, F⁰f is embedded in the centre of^QGf and realizing that the rank claimed forF⁰coincides with the cardinality ofB that can be easily computed using Lemma 2.2.

Assume that H is of type (g). Then fB = 1^?cx². If  ² F⁰ then ^?1 ²
(G;^hx²ⁱ) and hence (^?1)fB ^ 0 mod 2, in ^ZGfB. Therefore, if f ²B, then (^?1)f ^0 mod 2, in ^ZGf and, by Lemma 2.2, () is a unit of^H(^Z[^p2]), whereis the isomorphism of Lemma 2.2. Using that all the units of^H(^Z[^p2]) are of the form u;ui;ujoruk, where u^2Z[^p2], one easily deduces that()^2Z[^p2] and hence  is central.

Assume now thatGis of type (i). ThenfB= (1^?cx²)(1^?U^b). Therefore, F^{0 }
(G;^hX²ⁱ)^\
(G;U) and we argue as in the previous case.

3 Optimality: Proof of Theorem 1.2

By Theorem [12, Theorem 30.1] the following is a torsionfree normal com- plement of the trivial units of^ZG:

V =^ZG^\(1 +
(G)
(G;G⁰))

It is well known thatV is free nonabelian if G=D⁸ orG=D⁶. Using this fact and Proposition 2.1, it is easy to prove:

Lemma 3.1

For everye²C, V e is torsionfree.

Set

Fe⁰ =^fu^2ZG:ue=efor everye²C^g and for everye²C, let

Fee =^fu^2ZG:uf =f for allf ²C^nfe^gg: Plainly F^e0 =F^ee^1\F^ee² for every two dierente¹ and e² in C.

The group of type (h) with n= 1,k= 0 and x of order 4 is denoted by C^3oC⁴.

The crux of our argument relies on the following technical lemma.

Lemma 3.2

IfG⁶=C^3oC⁴ andG⁶=Q¹⁶, then for every e²C, F^ee^\V ^ F^0Fe and F^e0\V ^F⁰.

Proof.

Lete²C.

Claim 1. If f is a primitive central idempotent, such that^QGf is com- mutative or isomorphic to ^H(^Q), then V f =f.

If ^QGf is commutative, then f(1^?cG⁰) = 0 and the claim follows. If

QGf ^{' H}(^Q) then Gf ^' Q⁸ = ^ha;bⁱ and T = ^ff;a;b;ab^g is a rational basis of ^QGf. Moreover for every g ² G⁰, gf = ^f. This implies that every element of
(G)
(G;G⁰)f is of the form 2f for some  ²
(G).

Moreover f = ^P_t²_Ttt, where t ^{2 Z}W, W being the kernel of the canonical map G ^! Gf ^! Gf=^ha²ⁱ. Then tf = tf where t ^{2 Z}and t ^ !(t) mod 2. Therefore ^P_t²_Tt ^ !() = 0 mod 2. Thus, if u ² V, thenuf = 1 + 2^P_t²_Ttt, with^P_tt even. Since the unique units of^H(^Z) are^1,^i,^j and ^ij, we conclude that u=f. This proves Claim 1.

Claim 2. If G⁶=Q¹² andf ²A, then (F^ee^\V)f =f.

By Claim 1, we may assume that ^QGf is not commutative and^QGf ⁶=

H(^Q). This implies that H is of type (h) with x of order 4 and ^QGf is isomorphic to the generalized quaternion algebraA=^Q[i;j:i² =^?1;j²=

?3;ij=k=^?ji] (see [9]). By Proposition 2.1 and [9]

e=eS; =^cx²(S^b?U^b)Z^c and f =fS¹; ¹ = (1^?cx²)(^cS^1?U^b)Z^c;

where S and S¹ are two maximal subgroups of U and ; ^{1 2} Z^. Fix y ²U ⁿS¹. Then B¹ =^ff;a=xf;b=yf;ab^g is an integral basis of ^ZGf and there is an isomorphism:^QGf ^!Aso that(a) =iand(b) = ^1+2j. Therefore(^ZGf) is the subring ofAgenerated by iand ^1+2j. Furthermore the only units of this ring are^1,^i,^1?2j,^1+2j,^i^1?2j and ^i^1+2j [10].

Letu²V ^\F^ee. We have to prove that uf =f.

Assume rst that (S; ) ⁶= (S¹; ¹). Let e⁰ = eS¹; ¹. Then B² =

fe⁰;xe⁰;ye⁰;xye^0g is an integral basis of ^ZGe⁰. Moreover for an  ^{2 Z}G the coecients off ande⁰in the basisB¹ and B² are pairwise congruent modulo 2. Since (u^?1)e⁰= 0, the coecients of (u^?1)f in the basis B¹ are even. Therefore uf = 1 + 2(^P_t²_B¹tt) is a unit in ^ZGf. By the previous

paragraphuf =^f. Ifuf =^?f, then (u^?1)f =^?2f. Sinceu²V, we can write (u^?1)f = (⁰+¹x)^1?2x2cS¹Z^d1 where⁰;^12ZGand its support is embedded in a xed transversal ofGmodulo ^hx²;S¹;Zⁱcontaining 1. Then the coecient of 1 in (u^?1)f is ²k^+13m^?1, where  is the coecient of 1 in ⁰. However the coecient of^?2f is^?2k¹³m, which yields to a contradiction.

Thusuf =f as desired.

Now assume that S = S¹. By Claim 1 and the previous paragraph we have

u^?1 = (u^?1)(fS; +eS; ) = (u^?1)(S^b?U^b)Z^c =^X3

i⁼⁰ixⁱ(S^b?U^b)Z^c withi=^P2_j⁼⁰ijy^{j 2
h}yⁱ. The coecient ofxⁱy^j in u^?1 is

3ij ^?i^0?i^1?i²

2^k3ⁿ = ij

2^k3^{n?1 2Z}:

Therefore 2^k3^{n?1 j} ij and hence uf ² f + 2^k3^n?1ZGf is a unit of ^ZGf. If n > 1, then uf = f. Assume now that n = 1. Since we are assuming that G⁶=C^3oC⁴, then k ^1 and we argue as at the end of the previous paragraph to prove thatuf =f. This nishes the proof of Claim 2.

Now we prove thatF^e0\V ^F⁰. By Claim 1, the result is obvious unless H is of type (h) and the order of x is 4. In this case if G⁶=Q¹², then the cardinality ofC is greater than 2 and the result is a consequence of Claim 2. Now we prove F^ee ^\V ^ F^{0 }Fe for every e ² C. If H is neither of type (g) or (i), thenB =^;and hence the Lemma is a direct consequence of Claim 2.

Assume now that H is of type (g) and G ⁶= Q¹⁶, so that either n > 1 or k > 0. The sum of the elements of B is fB = 1^?cx². Without lost of generality one may assume that

e=R^b(S^b?G^c0)Z^c

where R=^hy²;::: ;ynⁱ,S =^hy²²;::: ;y_n²ⁱ and ²Z^. The support of e is L=^hy¹²;y²;::: ;yn^iZandTe=^f1;x;y¹;xy^1gis a transversal ofGmodulo L.

Let u ² V ^\F^ee and  = u^?1. If l ² L, then le = ^e. Therefore, e= ⁰e with ^{0 2 Z}G, Supp(⁰) ^ Te and !(⁰) ^ !() = 0 mod 2. By Claim 2,u= 1+⁰e+fB. It is enough to prove that g^?gx² is even for

everyg²G. Indeed, in this case fB ^2ZGand sou= (1 +⁰e)(1 +fB), 1 +⁰e²Fe and 1 +fB²F⁰.

Let g² Gand t ²Te so thatg ^tmodL. Then the coecient of g in u^?1 is



⁰_t

2^2n+k +g^?gx²

2 :

Therefore⁰_t= 2^2n+k?1t for some t^2Zand g =g^?_x²_g ^tmod 2.

So it is enough to show thatt is even for everyt²Te.

To obtain our goal we are going to consider the image of uf under the isomorphism=f :^QGf ^'H(^Q(^p2)), given in Lemma 2.2, for everyf ² B. Recall that an element ofB is of the formf =f;= (1^?cx²)K^\()_; as in Lemma 2.2. We use all the notation of that lemma. Thenf =^P_b^2B bb where

b= ^X

k²K;(k)(bk^?bkx²):

Note thatK()^\Lhas index 2 inK() and hence the cardinality ofK()^\L is 2^2(n?1)+k. Since either n > 1 or k > 0, these cardinality is even. Since _g¹_x^{2 ?}_g¹_x^{2 }_g²_x^{2 ?}_g²_x² mod 2 if g^{1 }g² modL,

t= ^X

k²K⁽^)\L;(k)(th^?thx²) + ^X

k²K⁽⁾ⁿL;(k)(th^?thx²) is even. By Lemma 2.2,(^ZGf)^H(^Z[^p2]). Since every unit of^H(^Z[^p2]) is of the formu,ui,uj orukwithu a unit of^Z[^p2] then by Lemma 2.2 we deduce that

a= _b² = _ab² = ab= _ab³ = b+ _b³ = 0:

Writing these equations in terms of the 's we obtain a system of linear

equations ^X

h²K⁽⁾;(h)(th^?_thx²) = 0;

for everyt=x;y¹²;xy²¹;xy¹;xy³¹. For a xed ²K^andk²Kthe previous system of linear equations becomes

X

t¹²T_(t¹)^X

k²K(k)(tt¹k^?_tt¹_kx²) = 0:

But the matrix ((t¹))t¹²T_;²⁽K⁽⁾=K^) is a Hadamard matrix, that is a matrix of 1's and ^?1's with orthogonal rows. In particular its determinant