Central Units in Metacyclic Integral Group Rings

Communications in Algebra®, 36: 3708–3722, 2008Copyright © Taylor & Francis Group, LLCISSN: 0092-7872 print/1532-4125 onlineDOI: 10.1080/00927870802158028

CENTRAL UNITS IN METACYCLIC INTEGRAL GROUP RINGS

Raul Antonio Ferraz1 and Juan Jacobo Simón-Pınero21Institute of Mathematics and Statistics, University of São Paulo,São Paulo, Brazil2Department of Mathematics, University of Murcia, Murcia, Spain

In this article, we give a method to compute the rank of the subgroup of central unitsof �G, for a finite metacyclic group, G, by means of �-classes and �-classes. Thenwe construct a multiplicatively independent set � ⊂ ��U��Cp�q�� and by applying ourresults, we prove that � generates a subgroup of finite index.

Key Words: Central units; Finite groups; Group rings; Metacyclic groups.

2000 Mathematics Subject Classification: Primary 20C05; Secondary 16S34.

1. INTRODUCTION

A finite metacyclic group of type Cm � Cn is a group having a normal cyclicsubgroup of order m and index n or equivalently a group given by the followingpresentation

Gm�r�n�s = �a� b � am = 1� bn = as� bab−1 = ar�� (1)

where m� r� s, and n are positive integers satisfying the following conditions:

s �m� m � rn − 1� m � s�r − 1�� and r < m�

In this case it is known that the order of Gm�r�n�s is mn, and that its exponent isequal to lcm�mn/s�m� = mn/d where d = gcd�n� s�. See Hempel (2000, Lemma 2.1)for details.

Throughout this article, we denote by � the Euler totient function. If d and rare two positive integers, we shall denote by Od�r� the order of r in U��d�. We alsodenote by �X� the cardinality of a given set X.

As usual, the symbols �, �, and � denote the ring of rational integers, thefield of the rational numbers, and the field of real numbers, respectively. For a groupG and a ring R, we denote by RG the group ring of G over R and by ��U�RG�� thegroup of central units of RG. By the way, whenever H denotes a ring or a group,

Received April 24, 2007; Revised July 30, 2007. Communicated by S. K. Sehgal.Address correspondence to Juan Jacobo Simón-Pınero, Departamento de Matemáticas,

Universidad de Murcia, 30100 Murcia, Spain; Fax: +34-968364181; E-mail: [email protected]

3708

CENTRAL UNITS 3709

��H� denotes the center of H . It is known that ��U��G�� is equal to ±��G�×A, where A is a finitely generated free abelian subgroup of ��U��G�� (see PolcinoMilies and Sehgal, 2003, Theorem 7.1.14, and p. 281. See also Polcino Milies andSehgal, 1999).

It is known that for a finite group G, the number of simple components of�G (�G), is the number of �-classes (�-classes) of G (Berman, 1956; Curtis andReiner, 1988a, Theorem 42.8, 1988b, Theorem 21.25; Ferraz, 2004; Witt, 1952) (seedefinitions below). In Ferraz (2004) and Ritter and Sehgal (2005) it is proved thatthe rank of the subgroup A of ��U��G�� is the difference between the number of�-classes and the number of �-classes. In this article we shall give explicit formulasto compute the number of �-classes and �-classes of a finite metacyclic group Gby means of its parameters.

In the specific case when G is the non-abelian group of order pq, with pand q different odd primes, we shall construct explicitly a subgroup of ��U��G��and by applying our results, we will prove that such a subgroup is of finite indexin ��U��G��. Jespers et al. (1996) give a recipe to get central units for nilpotentgroups.

2. CONJUGACY CLASSES, ��-CLASSES, AND ��-CLASSES IN G

Let G be a group. If g ∈ G, we denote by Cg the conjugacy class of g in Gand by �g the class-sum of g in G. Two generalizations of the concept of conjugacyclasses are the following.

For a given element g in a group G of exponent e, the �-class of g is definedas the union Cg ∪ Cg−1 , and the �-class of g is defined as the union

⋃gcd�r�e�=1 Cgr .

Let G = Gm�r�n�s be the metacyclic group defined as in (1). For each pair ofintegers i and v, v ≥ 0, we have the following useful expression

bvai = airvbv� (2)

Let 0 ≤ v ≤ n− 1, and denote by Bv, the following set

Bv = �aibv � 0 ≤ i ≤ m�

From the expression (2) above, it follows that Bv is a normal subset of G. Thus,every conjugacy class is contained in some Bv, with 0 ≤ v ≤ n− 1.

For each 0 ≤ v ≤ n− 1, we set

tv = gcd�rv − 1�m�� (3)

and having in mind the division algorithm, it is clear that for any element ajbv, thereis a representative of Cajbv of the form aidbv, where i and tv/d are coprime numbers.

Using the fact that Bv is a normal subset of G, Vera-López and Arregi (1992)showed that each Bv contains exactly

∑d � tv ��d�/Od�r� distinct conjugacy classes,

and conclude that G has exactly∑n

v=1

∑d � tv ��d�/Od�r� conjugacy classes. Grove

and Mutio evaluate the number of conjugacy classes of G in a very different way(Grove and Mutio, 1973/74).

3710 FERRAZ AND SIMÓN-PINERO

In order to evaluate the number of �-classes, we will make use of the sets Bv,and we will compute the cardinality �Cg�, for each g ∈ G. To do this, we evaluatefirst the order of the centralizer of aibv in G.

Let axby, 0 ≤ x ≤ m− 1, 0 ≤ y ≤ n− 1 be an element of the centralizer ofaibv in G. Since axbyaibv = aibvaxby, then ax+iry = ai+xrv and then x�rv − 1� ≡ i�ry −1��modm�. Thus, tv divides i�ry − 1� which is equivalent to that y is a multiple ofOtv/gcd�tv�i�

�r�. There exist exactly n/Otv/gcd�tv�i��r� multiples of Otv/gcd�tv�i�

�r� between 0and n− 1.

Now, for a fixed yo as above we have exactly tv solutions for x�rv − 1� ≡i�ry0 − 1��modm�. Namely, the least non-negative solution x0 and the translationsx0 +m/tv, x0 + 2m/tv � � � x0 + �tv − 1�m/tv. Hence, the order of the centralizer of aibv

is tvn/Otv/gcd�tv�i��r�.

Since the order of G is mn, we conclude that the cardinality of its conjugacyclass is mOtv/gcd�tv�i�

�r�/tv.Consequently, the element aibv contributes with tv/mOtv/gcd�i�tv�

�r�� to thecomputation of the numbers of conjugacy classes in Bv. Suitable simplifications leadto the formula obtained by Vera-López and Arregi mentioned above.

We recall that the �-class of an element g ∈ G is Cg ∪ Cg−1 . Hence, the �-classof aibv has �m/tv��Otv/gcd�i�tv�

�r�� elements if Cg = Cg−1 and 2�m/tv��Otv/gcd�i�tv��r��

elements otherwise.This motivates the definition of the function �v:

�v�i� ={1 if �aibv�−1 is conjugate to aibv�

2 otherwise�

Denote by n� the number of �-classes. Then

n� =n−1∑v=0

m−1∑i=0

tvm�v�i�Otv/gcd�i�tv�

�r��

Since ai+tvbv is conjugate to aibv, we can simplify the last equation:

n� =n−1∑v=0

tv−1∑i=0

1�v�i�Otv/gcd�i�tv�

�r��

If v ≡ −v �mod n�, that is if 0 < v < n, and v = n/2, then �v�i� = 2 for everyi. Hence we have

n� =n−12∑

v=1

tv−1∑i=0

1Otv/gcd�i�tv�

�r�+

tn−1∑i=0

1�n�i�Otn/gcd�i�tn�

�r��

when n is odd and

n� =n2−1∑v=1

tv−1∑i=0

1Otv/gcd�i�tv�

�r�+

tn−1∑i=0

1�n�i�Otn/gcd�i�tn�

�r�+

t n2−1∑

i=0

1�n

2�i�Ot n

2/gcd�i�t n

2��r�

when n is even.

CENTRAL UNITS 3711

For the first summand, we can carry out the usual simplifications obtaining

n� =n−12∑

v=1

∑d � tv

��d�

Od�r�+

tn−1∑i=0

1�v�i�Otn/gcd�i�tn�

�r��

when n is odd and

n� =n2−1∑v=1

∑d � tv

��d�

Od�r�+

tn−1∑i=0

1�v�i�Otn/gcd�i�tn�

�r�+

t n2−1∑

i=0

1�n

2�i�Ot n

2/gcd�i�tv�

�r�

when n is even.The following lemma allows us to evaluate �v by using parameters of G.

Lemma 2.1. Let G = Gm�r�n�s be the metacyclic group defined as in (1).

(1) Suppose v = 0. Then ai is conjugate to a−i, if and only if m/gcd�i�m� divides rl + 1,for some l in �.

(2) Suppose n even and v = n/2.Then �aibv�−1 is conjugate to aibv, if and only if tv divides irl + i+ s for some lin �.

Proof. To prove (1), note that the equation axbyaib−ya−x = a−i has a solution ifand only if the equation axairyb−ybya−x = airy = a−i has a solution.

The last equation has a solution if and only if m divides iry + 1; that is,m/�gcd�m� i�� divides ry − 1.

To prove (2), we consider elements of the form aibv. Since each aibv

is conjugate to bvai, it suffices to know if there exists x and y such thataxbybvaib−ya−x = a−ib−v�

Since axbybvaib−ya−x = ax+iry+v−xrvbv� then ax+iry+v−xrvbv = a−ib−v�Now having in mind that bn = b2v = as, multiplying both sides to bv yields

ax+iry+v−xrv+s = a−i�In terms of congruences, we have

x�1− rv�+ iry+v + s ≡ −i �modm��

As rv is congruent to 1 modulo tv, this congruence has a solution if and onlyif iry + s ≡ −i �mod tv� has a solution. �

Notice that when v = n, the equality �v�i� = �v�j� holds whenever gcd�i�m�equals gcd�j�m�. The same will occur when v = n/2 and tn/2 divides s. So we cansimplify the formula of R-classes and summarize it in the following result.

Theorem 2.2. Let G = Gm�r�n�s be the metacyclic group defined as in (1).

1. If n is odd, then the number of �-classes in G is

n−12∑

v=1

∑d � tv

��d�

Od�r�+ ∑

d �m

��d�

�n�m/d� · Od�r��


2. If n is even, then the number of �-classes in G is

n2−1∑v=1

∑d � tv

��d�

Od�r�+

t n2∑

i=1

1�n

2�i�Ot n

2/gcd�i�tn/2�

�r�+ ∑

d �m

��d�

�n�m/d� · Od�r��

3. If n is even, and tn/2 divides s, then the number of �-classes in G is

n2−1∑v=1

∑d � tv

��d�

Od�r�+ ∑

d � t n2

��d�

� n2�t n

2/d�Od�r�

+ ∑d �m

��d�

�n�m/d� · Od�r��

We do not need to test every l ∈ � to find eventual solutions for the equationsabove. As we will see below, there are natural candidates.

If v = n, we want l satisfying rl ≡ −1 �modm/d�, for a divisor d of m. So inparticular r2y ≡ 1 �modm/d�.

Except in the easy cases when m/d equals 1 or 2, there exists only onecandidate to satisfy the equation above. It is Om

d�r�/2 (when this number is an

integer).Now consider the case v = n/2. Suppose that l is a solution for the congruence

iry + s ≡ −i �mod tv�� Since m and consequently tv divides �r − 1�s, we have thatrs ≡ s �mod tv� and thus �rl − 1�s ≡ 0 �mod tv�. Hence, multiplying both sides of thecongruence above to �rl − 1�, we have

0 ≡ �r2l − rl�i+ �rl − 1�s + �rl − 1�i ≡ r2li− rli+ rli− i ≡ ir2l − i �mod tv��

That is ir2l ≡ i �mod tv� or equivalently r2l ≡ 1 �mod tv/�gcd�tv� i��So there are at most two candidates to be solution of iry + s ≡ −i �mod tv�;

one of them is l = 0 and the other one is

l = Otn/2/gcd�tn/2�i��r�

2�

When G = Gm�r�n�s has the property that mn is odd, the general formula has anice simplification, that arises from a basic theoretical argument, which and in factworks in general for finite groups.

Let H be a group of odd order. If g ∈ H is conjugate to g−1, there must existsh ∈ H such that g−1 = hgh−1, and hence that g = h2gh−2. Thus we conclude thath2 commutes with g. Having in mind that the order of H is odd, we may see thath commutes with g, and so that g = g−1. This implies that g = 1. So except wheng = 1, Cg = Cg−1 . Thus if G = Gm�r�n�s is the metacyclic group defined as in (1), andmn is odd, then the number of �-classes is

�nr. of conjugacy cl. of G�− 12

+ 1 =∑n

v=1

∑d � tv ��d�/Od�r�+ 1

2� (4)

Now we proceed to determine the number of �-classes of a metacyclic groupG = Gm�r�n�s.

We begin by showing some properties of �-conjugation in any finite group Gwith exponent e. We recall that two elements a and b of G are �-conjugate, if there

CENTRAL UNITS 3713

exist an h ∈ G and an integer t relatively prime with e such that at = hbh−1. It iseasy to see that it is an equivalence relation.

Lemma 2.3. Let G be a group of exponent e, and let a and b be elements of G. Thena and b are �-conjugate in G if and only if there exist an h ∈ G and a ∈ � relativelyprime with the order of a, such that a = hbh−1.

Proof. Suppose first that a and b are �-conjugate in G. Then there is an h ∈ Gand a t ∈ � with gcd�t� e� = 1 and then gcd�t� o�a�� = 1. By taking = t, we havefinished.

Conversely, suppose a = hbh−1 for some h ∈ G, and some ∈ � such thatgcd� � o�a�� = 1. Consider the usual epimorphism from U��e� onto U��o�a��. As ∈U��o�a��, then any preimage t ∈ U��e� will work. So we are done. �

Lemma 2.4. Let G be a group, and let a and b be elements of G. Then a and b are�-conjugate in G if and only if there exist an h ∈ G and a � ∈ � such that a� = hbh−1

and the elements a and b have the same order.

Proof. Sufficiency is obvious. To prove necessity, first note that a� and b havethe same order. Now it is clear that a and a� have the same order if and only ifgcd�� o�a�� = 1. Now apply Lemma 2.3, and we are done. �

There is yet another equivalence to �-conjugation.

Corollary 2.5. Let G be a group, and let a and b be elements of G. Then a and bare �-conjugate in G if and only if there exist h, k in G, and �1� �2 ∈ � such thata�1 = hbh−1 and b�2 = kak−1.

Proof. Sufficiency follows immediately from the symmetry of �-conjugation.To prove necessity, first note that for every integer � we have o�a� ≥ o�a��.Thus, o�a� ≥ o�a�1� = o�b� and o�b� ≥ o�b�2� = o�a�. Therefore o�a� = o�b�, and byLemma 2.4, the result follows. �

Now we go back to metacyclic groups. Recall the definition of tv in Eq. (3).

Proposition 2.6. Let G = Gm�r�n�s be the metacyclic group defined as in (1). Thengiven an element axby ∈ G, there exists a unique divisor v of n such that axby is �-conjugate to aibv for some i. Moreover, v = gcd�y� n�, and we can choose i such that0 ≤ i ≤ tv − 1.

Proof. If q is a positive integer, then �axby�q = azbyq for some integer z. Theonly divisor of n which is congruent to yq modulo n for some q is v = gcd�y� n�.Moreover, it is known that there exists q in U��n� such that yq ≡ v �mod n�. Sincen divides e, we may assume q is relatively prime to e, and this proves the first claim.To complete the proof of the proposition, observe that a��azbv�a−� = az+��1−rv�bv =az+��1−rv�+�mbv for all �, �. It follows that all elements of the form az+�tvbv areconjugate to azbv, given the result. �


As a direct consequence of the proposition above, we have the followingcorollary.

Corollary 2.7. Let G = Gm�r�n�s be the metacyclic group defined as in (1). Define

Dv = �aibv � 0 ≤ i ≤ tv − 1�

Then each element of G is �-conjugate to an element in⋃

v � n Dv, and no element in Dv1

may be conjugate to another in Dv2, if v1 = v2.

Thus to evaluate the number of �-classes of G, it suffices to determine thenumber of �-classes intersecting Dv, for each divisor v of n, and then take the sumof them.

We notice that the set of intersections of �-classes with a set Dv give apartition to Dv. We denote by Li�v the intersection of the �-class of aibv with Dv.

Proposition 2.8. In the setting of Proposition 2.6 and Corollary 2.7, let aibv be anelement of Dv. Then the cardinality of the set

Di�v = Dv ∩ �a��aibv�qa−� � � ∈ �� gcd�q� e� = 1

is ��tvn/li�vv�/��n/v�, where li�v = gcd�tv� in/v+ s�.

Proof. Set Ti�v = �a��aibv�qa−� � � ∈ �� gcd�q� e� = 1 and Hv = �a� bv�. Clearly,�atv� is the derived subgroup of Hv (see expression (2)) and thus the group Av =Hv/�atv� is an abelian group. We denote by ��h� the image of h under the naturalprojection of Hv onto Av.

We claim that the restriction ��Di�vis a monomorphism. Indeed, if ��akbv� =

��ajbv�, then ��akbvb−va−j� = ��ak−j� = 1 and thus ak−j = a�tv for some � ∈ �. As0 ≤ k� j ≤ tv − 1, then we have that k = j.

We first compute the cardinality of ��Ti�v�. To do this, we claim that

��Ti�v� = ��aibv�q � gcd�q� o��aibv�� = 1�

We will prove the claim. Since, for any integer q, such that gcd�q� e� =1, we have a−��bv�qa� = a�rvqa−�bvq = a��rvq−1�bvq and a��rvq−1� ∈ �atv�, then��a��aibv�qa−�� = ��aibv�q�. As o��aibv�� divides o�aibv�, and o�aibv� divides e,we have that gcd�q� o��aibv�� = 1. On the other hand, if an integer k is suchthat gcd�k� o��aibv�� = 1, then there must exist another integer q congruent to kmodulo o��aibv��, and relatively prime with e. Thus the equality of sets has beenestablished.

Hence, it is clear that ��Ti�v�� = ��o��aibv��. Set � = o��aibv��. Now wewant to compute �. Having in mind that for every q ∈ �, we have �aibv�q =af�i�q�bqv+kn = af�i�q�+ksbqv for some f�i� q�� k ∈ �, we see that n � � · v; that is, � mustbe of the form hn/v, for some integer h.

Then

��aibv��hn/v = ��aihn/vbnh� = ��aihn/vash� = ��ah�in/v+s��

CENTRAL UNITS 3715

It follows that, � must be the integer hn/v where h is the least positive integersuch that h�in/v+ s� is divisible by tv. That is, h�in/v+ s� = lcm�tv� in/v+ s�. Then

h = lcm��in/v�+ s� tv�

��in/v�+ s�= tv

gcd�tv� �in/v�+ s��

Thus ��Ti�v�� = ��ntv/�vgcd�tv� �in/v�+ s�� = ��ntv/vli�v�.Now, in order to determine ��Di�v��, we need to evaluate how many elements

of ��Ti�v� are images of elements of Dv

Since the elements of Dv are of the form ajbv, an element of the form��aibv�q = ��aibv�q� is an image of an element of Dv if and only if q is congruentto 1 modulo n/v. Thus we have to count those q ∈ U�� that are congruent to 1modulo n/v. These integers are precisely the kernel of the natural projection fromU�� onto U��n/v�, which has cardinality ��ntv/vli�v�/��n/v�.

It follows that

��ntv/vli�v�/��n/v� = ��Di�v�� = �Di�v�

because the restriction of � is a monomorphism. �

Theorem 2.9. Let G = Gm�r�n�s be the metacyclic group defined as in (1). Let v be adivisor of n, and set tv = gcd�m� rv − 1� and ji�v = li�v/gcd�li�v� i�. Then

�Li�v� =�(

ntvvli�v

)Oji�v

�r�

�(nv

)Proof. Set K = �0 < j ≤ tv − 1 � Di�v = bjDi�vb

−j, and let � be the least element ofK. Clearly, �Li�v� = ��Di�v�.

We recall from the proof of the proposition above that an element of the form��aibv�q = ��aibv�q� is an image of an element of Dv if and only if q = hn/v+ 1. Inthis case, ��aibv��hn/v+1 = ��ah�in/v+s�+i�. Thus, if � is the least positive integer suchthat ��air� � = ��ah�in/v+s�+i�, we must have

ir� ≡ k��in/v�+ s�+ i mod tv�

Equivalently,

i�r� − 1� ≡ k��in/v�+ s� mod tv�

Now, by the definition of li�v, we have

i�r� − 1� ≡ jli�v mod tv�

where j = k��in/v�+ s�/li�v. As li�v divides tv, we can consider the congruence i�r� −1� ≡ 0 �mod li�v�.

And the least positive solution for the congruence above is Oji�v�r�.

Then �Li�v� = Oji�v�r��Di�v� as claimed. �


Finally, we may give a formula to compute the number of �-classes of anymetacyclic group.

Corollary 2.10. Let G = Gm�r�n�s be the metacyclic group defined as in (1). For eachv � n, set tv = gcd�m� rv − 1� and ji�v = li�v/gcd�li�v� i�. Then the number of �-classes ofG is

∑v � n

tv−1∑i=0

�(nv

)�(

ntvvli�v

)Oji�v

�r��

An important family of metacyclic groups arises from semidirect products ofcyclic groups; that is, when Gm�r�n�s satisfies s = m. Here, for each k, 0 ≤ k ≤ tv − 1,we have that, lk�v = gcd�kn/v+m� tv��

As tv divides m, then lk�v = gcd�kn/v� tv��Thus, if i = gcd�k� tv�, we get

lk�v = gcd(kn

v� tv

)= i gcd

(k

i

n

v�tvi

)�

Since gcd(ki� tv

i

) = 1, then

gcd(k

i

n

v�tvi

)= gcd

(n

v�tvi

)�

and therefore

lk�v = igcd(n

v�tvi

)= gcd

(in

v� tv

)= li�v�

Moreover

jk�v =li�v

gcd�li�v� k�= li�v

gcd(tv� i

nv� k

)= li�v

igcd(tvi� n

v� k

i

) = li�vi

= gcd(n

v�tvi

)�

And

tvlk�v

n

v= 1

i

(tvlk�v

in

v

)= 1

i

(lcm

(tv�

in

v

)= lcm

(tvi�n

v

)�

Clearly, ��0 ≤ k ≤ tv − 1 � i = gcd�k� tv�� = ��tv/i�. Hence,

∑v � n

∑i � tv

�(nv

)�(tvi

)Oji�v

�r��(

tvn

li�vv

) = ∑v � n

∑i � tv

�(nv

)�(tvi

)Oji�v

�r��(lcm

(tvi� n

v

))= ∑

v � n

∑i � tv

�(gcd

(tvi� n

v

))Oji�v

�r�= ∑

v � n

∑i � tv

��ji�v�

Oji�v�r�

�

CENTRAL UNITS 3717

Therefore, when s = m, the number of �-classes will be given by the sum

∑v � n

∑i � tv

��ji�v�

Oji�v�r�

� (5)

3. CENTRAL UNITS OF ��Cp�q

Let p and q be two odd primes with q > p, and let us suppose that p dividesq − 1. Under these conditions, there exists only one non-abelian group of order pqusually denoted by Cp�q. This group has presentation

G = �a� b � aq = bp = 1� bab−1 = ar��

That is, G is the metacyclic group G = Gq�p�r�q.In this section we shall construct a group of multiplicatively independent set

of central units of �Cp�q, which generates a subgroup of finite index in ��U��Cp�q��.These units are of two kinds. The first kind arises from the subgroup generatedby b and the second is simply the suitable product of some units of ��a�. Theconstruction of both kinds are based in the Bass cyclic units.

In order to guarantee that the group we shall construct is indeed of finiteindex, we proceed to evaluate the rank of ��U��Cp�q��, using the methods developedin the previous section.

As the order of G is odd, the number of �-classes are given by the Eq. (4). Adirect computation yields that the number of conjugacy classes of G is

�p− 1�+ 1+ q − 1p

�

thus the number of �-classes is

�p− 1�+ 2+ q−1p

2= p− 1

2+ q − 1

2p+ 1�

Since there exists only one q-Sylow subgroup in G, and all the p-Sylowsubgroups of G are conjugate, it follows easily that the number of �-classes isalways 3.

Finally, the desired rank is

rank of ��U��Cp�q�� =p− 12

+ q − 12p

− 2� (6)

Now we proceed to construct the announced units. Let Cn be the cyclic groupof order n with n = 1� 2� 3� 4, and 6, and let i be an integer with 1 < i < n− 1 and�i� n� = 1. It is well known that the elements

ui = �1+ g + g2 + · · · + gi−1��n� − kg�

where g = 1+ g + · · · + gn−1 and k = �i��n� − 1�/n are units in �Cn. (See PolcinoMilies and Sehgal, 2003, pp. 237–238). If n is a prime number, the units


u2� u3� � � � � u�n−1�/2 are multiplicatively independent, and they generate a subgroupof finite index in U��Cn� (see Karpilovsky, 1989, Theorem 8.9.36). Note that thenumber of units is �n− 3�/2.

3.1. Central Units of the First Kind

Suppose that p, the order of b, is greater than 3, and we have constructedin the group of units of ��b� the �p− 3�/2 Bass cyclic units mentioned above, sayub�2� � � � � ub��p−1�/2�

Clearly, these units are not central. Having in mind that ba is a central elementin �Cp�q, it is easy to see that the products ub�ja, with 2 ≤ j ≤ �p− 1�/2 are centralelements; but now they are not units. So, in order to get central units, we shall makesome modifications. We begin with the following lemma.

Lemma 3.1. Let G be a finite group, let u be a unit of �G, and n a positive integer.Then there exists a positive integer m such that all coefficients of the element um − 1 aremultiples of n.

Proof. Let �n be the ring of integers modulo n, and let � be the natural projectionfrom �G onto �nG; that is, if � denotes the projection from � to �n, then��

∑g∈G �gg� = �

∑g∈G ��g�g�.

Clearly, if u is a unit, then ��u� is a unit too. Since G and �n are finite groups,then the ring �nG is also finite and then, for some positive integer m, we have that��u�m = 1; that is, um − 1 belongs to the kernel of �. But this is equivalent to saythat all coefficients of um − 1 are multiples of n. �

Now we are ready to construct central units. Let m be such that q divides allthe coefficients of um

b�j − 1, with 2 ≤ j ≤ �p− 1�/2, and set Ub�j = 1+ �umb�j − 1�a/q.

Clearly, for every j, Ub�j ∈ �Cp�q and in fact is a central element. Setting Mb�j =1+ �u−m

b�j − 1�a/q, we have that

Ub�jMb�j = 1+ �umb�j − 1�a

q+ �u−m

b�j − 1�a

q+ �um

b�j − 1��u−mb�j − 1�a2

q2

= 1+ umb�ja

q+ u−m

b�j a

q− 2

a

q+ a

q− um

b�ja

q− u−m

b�j a

q+ a

q= 1�

Thus Ub�j ∈ ��U��Cp�q��, for every 2 ≤ j ≤ �p− 1�/2.It remains to prove that they are multiplicatively independent. To do this, first

note that, in a similar way as above, we have that Ub�jUb�k = 1+ �umb�ju

mb�k − 1�a/q

and Uib�j = 1+ �umi

b�j − 1�a/q, for any 2 ≤ j� k ≤ �p− 1�/2 and any i ∈ �.Suppose now that we have

Um2b�2 · · ·Um�p−1�/2

b��p−1�/2 = 1

CENTRAL UNITS 3719

for some integers m2� � � � � m�p−1�/2. From Lemma 3.1 and the equalities above, wehave that there exists a positive integer m, such that

�um·m2b�2 · · · um·m�p−1�/2

b��p−1�/2 − 1�a

q= 0�

Analyzing the supports, we have

um·m2b�2 · · · um·m�p−1�/2

b��p−1�/2 − 1 = 0�

and so

um·m2b�2 · · · um·m�p−1�/2

b��p−1�/2 = 1�

Since the units ub�2 · · · ub��p−1�/2 are multiplicatively independent, we mayconclude that all mi are equal to 0 and consequently, Ub�2 · · ·Ub��p−1�/2 aremultiplicatively independent units.

3.2. Central Units of the Second Kind

We will build these central units by taking products of certain Bass cyclic unitsof ��a�.

Let q > 3 be a prime number and consider the ring ��q� where �q denotesa qth primitive root of unity. Set ui = 1+ �q + · · · + �iq, for 1 ≤ i ≤ �q − 3�/2. Itis known that �u1� � � � � u�q−3�/2� is a subgroup of finite index in U��q�� and that�u1� � � � � u�q−3�/2 is a multiplicatively independent set in U��q�� (see Washington,1980, Theorem 8.3). We shall determine another multiplicatively independent set ofunits which generates a subgroup of finite index in U��q��.

Let �q be the ring of integers modulo q. By definition of Cp�q, there exists agenerator � of the group U��q� such that r ≡ ��q−1�/p �mod q�. Set vi = 1+ ��

i

q +· · · + ��

i��−1�q .

Theorem 3.2. The set �v0� � � � � v�q−5�/2 is multiplicatively independent (and generatesa subgroup of finite index in U��q��).

Proof. Since �u1� � � � � u�q−3�/2, is a multiplicatively independent set which generatesa subgroup of finite index in U��q��, it is enough to prove that the subgroup�u1� � � � � u�q−3�/2� is contained in H = �−1� �q� v0� � � � � v�q−5�/2�.

Note that v0 = u�−1� Furthermore,

v0 · v1 = �1+ �q + · · · + ��−1q ��1+ ��q + · · · + ��−1�

q �

= 1+ �q + · · · + ��−1+��−1�q = 1+ �q + · · · + ��

2−1q �

Extending products, we have for each k

v0v1 · · · vk = 1+ �q + · · · + ��r+1−1

q �


Now for each t with 1 < t ≤ �q − 3�/2, there exists a j ∈ U��q� for which t =�j+1 − 1 in U��q�. We consider two cases. The first case is j ≤ �q − 5�/2. Then wecan write

ut = v0v1 · · · vj

and then ut ∈ H . In the second case (that is, when j > �q − 5�/2�, we notice thatsince ��q−3�/2+1 = −1, then j = �q − 3�/2. The same argument shows that j = q − 2,too. Hence, �q − 1�/2 ≤ j ≤ q − 3 under the assumption j > �q − 5�/2.

So that, as j ≥ �q − 1�/2, we have

uj = 1+ ��j

q + · · · + ��j��−1�

q

= ��j��−1�

q �1+ �−�j + · · · + �−�j��−1�q �

= ��j��−1�

q �1+ ��q−1�/2�j + · · · + ��

�q−1�/2�j��−1�q �

= ��j��−1�

q �1+ ��j−�q−1�/2 + · · · + ��

j−�q−1�/2��−1�q �

= ��j��−1�

q vj−�q−1�/2��

Viewing the inequality �q − 1�/2 ≤ j ≤ q − 3 as 0 ≤ j − �q − 1�/2 ≤ �q − 5�/2,we have that

ut = v0 · · · vj = �lqv0v1 · · · v�q−3�/2v0 · · · vj−�q−1�/2�

for a suitable l. Having in mind that v0v1 · · · v�q−3�/2 = −��q−1�/2, we may see that

ut = −�2l+q−1

2q v0 · · · vj−�q−1�/2�

Hence ut is in H . �

Now we shall make use of the units constructed above to yield amultiplicatively independent set in ��a� of Bass cyclic units. Thus we associate toeach vi = 1+ ��

i

q + · · · + ��i��−1�

q the following Bass cyclic unit

ua�i = �1+ a�i + · · · + a�i��−1��q� − ka�

where a is a generator of Cq (recall that Cp�q = Gq�p�r�q described in (1)). Set, as usual,a = 1+ a+ · · · + aq−1 and k = ��q� − 1�/q.

Clearly, �ua�0� ua�1� � � � � ua��q−5�/2 is a multiplicatively independent set whichgenerates a subgroup of finite index in U��Cq�. Now we shall make the units ua�i

symmetric in regard to the involution ∗ in KG; that is,

(∑g∈G

�gg

)∗= ∑

g∈G�gg

−1�

CENTRAL UNITS 3721

By definition of ua�i, we have that ua�i = a�i��−1�u∗a�i. Furthermore, u∗

a�i = �1+a−�i + · · · + a−�i��−1��q� − ka, and hence, as ��q−1�/2 = −1, we have that u∗

a�i =a−�i��−1�ua�i+�q−1�/2.

Let k ∈ U��q� such that 2k = �i�� − 1�. If we set wa�i = a−kua�i, then w∗a�i =

wa�i, and �wa�0� wa�1� � � � wa��q−5�/2 is again a multiplicatively independent set.However, neither the units ua�i nor the units wa�i are central in Cp�q. We need

one more step. Recall that � is the generator of U��q� which satisfies ��q−1�/p = r.Then

bua�ib−1 = �1+ �bab−1��

i + · · · + �bab−1��i��−1��q� − ka

= �1+ ar�i + · · · + ar�i��−1��q� − ka

= �1+ a��q−1�/p�i + · · · + a��q−1�/p�i��−1��q� − ka = ua�i+�q−1�/p�

Similarly,

bwa�ib−1 = wa�i+�q−1�/p�

Thus, if t = �q − 1�/p then, for any j, Ua�j = wa�jwa�j+t · · ·wa��p−1�t is a centralunit of Cp�q. Now since p and q are positive integers, then �p+ 1�/2�t > �q − 1�/2.Moreover, t�p+ 1�/2�− �q − 1�/2 = �q − 1�/2p and so wa�t�p+1�/2� = wa��q−1�/2p.Thus we can rewrite Ua�j as

Ua�j = wa�j · wa�j+t/2 · · ·wa�jt�p−1�/2�

The last step in our construction is to determine the greatestindex j, for which �Ua�1� Ua�2� � � � � Ua�j is a multiplicatively independentset. Since �wa�0� wa�1� � � � � wa��q−5�/2 is multiplicatively independent, and�wa�0� wa�1� � � � � wa��q−5�/2� wa��q−3�/2 is not, then the maximum j must be the onewhich satisfies the following equation:

q − 52

= j + t

2�p− 1� = j + q − 1

2p�p− 1��

Hence j = �q − 1�/2p− 2, and then �Ua�1� Ua�2� � � � � Ua� q−12p has �q − 1�/2p− 1

elements. This ends our construction.The proof of next theorem is quite simple, having in mind Eq. (6).

Theorem 3.3. The set � = �Ub�2� � � � � Ub��p−1�/2� Ua�0� � � � � Ua��q−1�/2p−2 is multiplica-tively independent and generates a subgroup of finite index in ��U��Cp�q��.

Proof. Since the rank of ��U��Cp�q�� coincides with ��, it suffices to prove that� is a multiplicatively independent set.

Suppose that m2� � � � � m�p−1�/2� n0� � � � � n�q−1�/2p−2 are integers such that

Um2b�2 · · ·Um�p−1�/2

b��p−1�/2Un0a�0 · · ·Un�q−1�/2p−2

a��q−1�/2p−2 = 1�


A simple support analysis will lead us to the following equations:

Um2b�2 · · ·Um�p−1�/2

b��p−1�/2 = 1�

Un0a�0 · · ·Un�q−1�/2p−2

a��q−1�/2p−2 = 1�

From the multiplicative independence of the two sets of units above, weconclude that all mi and nj are 0, and consequently that the union of two sets isitself multiplicatively independent. �

ACKNOWLEDGMENTS

The authors thank the referee for several helpful remarks, which contributedto improve a first version of this article.

The first author thanks the Department of Mathematics of the Universityof Murcia, and the second one that of University of São Paulo for their warmhospitality.

Research supported by FAPESP, Procs. 02/02933-0, 00/07291-0 and CNPqProc. 300243/79-0 (RN), CAPES of Brazil, D.G.I. of Spain, and Fundación Sénecaof Murcia.

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Central Units in Metacyclic Integral Group Rings