Units in Integral Group Rings for Direct Products

Western Michigan University ScholarWorks at WMU

Dissertations Graduate College

6-1998

Richard M. Low Western Michigan University

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Recommended Citation Low, Richard M., "Units in Integral Group Rings for Direct Products" (1998). Dissertations. 1550. https://scholarworks.wmich.edu/dissertations/1550

This Dissertation-Open Access is brought to you for free and open access by the Graduate College at ScholarWorks at WMU. It has been accepted for inclusion in Dissertations by an authorized administrator of ScholarWorks at WMU. For more information, please contact [email protected]. UNITS IN INTEGRAL GROUP RINGS FOR DIRECT PRODUCTS

by Richard M. Low

A Dissertation Submitted to the Faculty of The Graduate College in partial fulfillment of the requirements for the Degree of Doctor of Philosophy Department of Mathematics and Statistics

Western Michigan University Kalamazoo, Michigan June 1998

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. UNITS IN INTEGRAL GROUP RINGS FOR DIRECT PRODUCTS

Richard M. Low. Ph.D.

Western Michigan University, 1998

Given a finite group G and the ring of integers, one can form the integral

group ring ZG. A natural problem to investigate is to find a description of the

group of units for this ring ZG. Since the unit problem for integral group rings

arises in the contexts of algebraic topology, number theory, and algebra, it is an

important question to try to answer. For this reason, it has drawn the attention

of researchers from diverse areas of mathematics.

Graham Higman (circa 1940) made substantial contributions to the solu

tion of this problem, in the case where G was a finite abelian group. From then

on. the main focus of the research in this area has been in the case where G was a

finite non-abelian group. Although there are some results for U{ZG) for various

groups G, very little is known with respect to explicit descriptions ofU(ZG).

In this dissertation, we address the following question: If we have a de

scription of U{ZG), what information can we obtain about U(ZG*), where G' =

G x Cp and p is prime? A general algebraic framework using short-exact se

quences is developed to study this problem. In addition, known and new results

are obtained in the case where p = 2.

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Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Copyright by Richard M. Low 1998

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. ACKNOWLEDGMENTS

So many people have encouraged, supported, and inspired me during the

last several years as I was pursuing my doctorate in mathematics. Indeed. I want

to thank the Department of Mathematics and Statistics for the education and

scholarship that it has offered me while I was a graduate student.

I extend my deep gratitude to my advisor Dr. Joseph Buckley for his

guidance and direction with regards to this dissertation. Without his insight,

intuition, and encouragement, this dissertation would not have become a reality.

My appreciation is also extended to the remaining members of my dissertation

committee, namely, Dr. John Martino, Dr. Thomas Richardson, Dr. Kung-

Wei Yang, and Dr. Michael Paxmenter of Memorial University of Newfoundland.

Each provided me with a great deal of support throughout the completion of the

dissertation. A sincere thank you is owed to Dr. Michael Slack, Dr. Arthur

Stoddart. and Jerome Helton Sensei. They have each contributed in significant

ways during my journey towards the doctoral degree.

Finally, I am deeply grateful to my family who have supported me in

countless ways throughout this entire time. To my mother Gwendolyn Low and

father Fred Low, thank you very much.

Richard M. Low

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. TABLE OF CONTENTS

ACKNOWLEDGMENTS...... ii

CHAPTER

I. INTRODUCTION...... 1

1.1 Basic Facts, Definitions, and Notation...... 1

1.2 The Unit Problem for the Integral Group R ing...... 2

II. U(Z[G x C2\)...... 5

2.1 The Split Extension...... 5

2.2 Applications...... 9

III. A GENERALIZATION...... 28

3.1 t/(ZG *), Where G* = G x Cp ...... 28

IV. PROBLEMS FOR FUTURE STU DY...... 34

4.1 An Example Leading to a Conjecture ...... 34

4.2 New Directions...... 38

REFERENCES...... 39

iii

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. CHAPTER I

INTRODUCTION

1.1 Basic Facts, Definitions, and Notation

Given a finite group G and a commutative ring R with identity, one can

form the group ring RG. This algebraic structure is in fact a ring with unity. Its

elements are all finite formal sums

rgg. where rg 6 R, g€G

along with the following operational rules:

= 5 Z s9g r9 = s9^9 € G: geG gee

Y lr99 + Y l s99 = + sg)9: g£G geG g€G

I r 9 9 ) I sa9 ) = 5 1 199, where tg = ^ rxsy. \geG / \geG / geG xy=g

There is a surjective ring homomorphism e: RG —> R, defined by

Y . r9 9 ^ Y l rr g€G g€G

e is called the augmentation map and its kernel A r(G) = (g — 1 : g E G) is the

augmentation ideal. The elements g — 1 generate A r(G ) as a free R-module. In

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. this dissertation, we will focus our attention on the integral group ring ZG\ that

is, the underlying ring R will be the integers. A few elementary remarks and

the standardization of notation should be made at this point. Clearly. ZG is a

commutative ring if and only if G is abelian. Also, note that there are natural

embeddings of G and R into RG, given by g —► 1/? • g and r —> r - lc respectively.

We will denote the group of units of ZG by U{ZG). U\(ZG) will denote the

units of augmentation one in U(ZG). Thus, Ui(ZG) is a subgroup of U(ZG) and

±Ui(ZG) = U(ZG). Observe that ±G < U(ZG). The elements ± G axe called

the trivial units of ZG.

1.2 The Unit Problem for the Integral Group Ring

Describing the units of the integral group ring is a classical and difficult

problem. Over the years, it has drawn the attention of those working in the areas

of algebra, number theory, and algebraic topology. For a compilation of results

in the unit problem, the reader is directed to the monograph by Sehgal [17]. We

will use the standard notation of that text. Most descriptions of U{ZG) in the

mathematical literature either give an explicit description of the units, the general

structure of U(ZG), or a subgroup of finite index of the unit group U(ZG). These

results were often obtained by using techniques from representation theory and

algebraic number theory. In 1940, substantial work on the unit problem was

done by Graham Higman [4-5]. He first showed that if U(ZG) = ±G, then

U(Z[G x C 2]) = ±(G x C 2). Using this, he showed that U(ZG) = ± G G

is abelian of exponent 2, 3, 4, or 6 or G = E x K$ where K$ is the quaternion

group of order 8 and E is an elementary abelian 2-group. Furthermore, Higman

gave a general structure theorem for U(ZA), where A is a finite abelian group.

Other results include: Aj and S4 by Allen-Hobby [1-2], D2p by Passman-Smith

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. 3

[14], G = Cp >i Cq, where q is a prime dividing p — 1 by Galovitch-Reiner-Ullom

[3], |G| = p3 by Ritter-Sehgal [15], and U{ZS3) by Hughes-Pearson [ 6]. More

recently, Jespers and Parmenter [9] gave a more explicit description ofU{ZS3).

In 1993, Jespers and Parmenter 10 [ ] completed the description ofU(ZG) for all

groups of order 16. Jespers [8], in 1995, gave a description of the U{ZG), for the

dihedral group of order 12 and for G = D8 x C 2. We state some of these results

for later reference.

Theorem A (Higman). Let A be a finite abelian group of order n. Then.

U(ZA) = ±A x F, where F is a free abelian group of rank |( n + 1 + n2 — 2/).

Here. n 2 denotes the the number of cyclic subgroups of order 2 and I is the number

of cyclic subgroups of A.

Theorem B (Higman). Let G* = G x (x).x2 = 1. Then U{ZG) = ±G = >

U{ZG*) = ± G \

Theorem C (Jespers—Parmenter). InU\{ZS3), S3 = (a.b : a 3 = b'2 = (ab ) 2 =

1) has a torsion-free normal complement K={u = 1 + q(1 — a):a€ A z(S 3),

u a unit}. V is a free group of rank 3 generated by the 3 distinct (up to inverses)

bicyclic units of ZS3.

Theorem D (Jespers). In f/1(Z D 12), D 12 has a torsion-free normal comple

ment V which is a semi-direct product of a free group of rank 5 by a free group of

rank 3. Furthermore, V is generated by the bicyclic units.

Theorem E (Jespers-Parmenter). In U\(Z[D 3 x C2]), D8 xC 2 has a torsion-

free normal complement V which is a semi-direct product of a free group of rank

9 by a free group of rank 3.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. 4 In the next chapter, we will obtain a description of U(Z[G x C2]) in terms

of U(ZG). Among the consequences, we will give new proofs of Theorems B, D,

and E as well as several new results.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. CHAPTER II

U(Z[G x C2])

2.1 The Split Extension

We wish to develop a general algebraic framework to study the unit problem

in integral group rings for a direct product with a cyclic group of order two. In

this section, we build the machinery which will allow us to answer the following

question: Assuming that we have a good description of U(ZG), can we obtain a

description for U(ZG*), where G* = G x C2?

Let G' = G x (x), x2 = 1, with |G| = n. Decomposing G' into two

cosets, we have that G* = G U xG = {<71, g 2, ... , gn,xgi , . . . . xgn}. Thus, ZG' =

ZG ® xZG, a direct sum of abelian groups. Now, consider the surjective group

homomorphism 7~ : G' —* G defined by g *• g.x 1. Letting l denote the

inclusion map. this gives an exact sequence:

(x) G* - 2 — G.

which induces a ring homomorphism tt : ZG* —*• ZG\ where 7r(Pi+xP2) = P\ +P2

for all Pi, P2 € ZG. At the ring level, the kernel of 7r, say A*, is:

K' = {Pi + xP2 e ZG* : 7r(Pi + xP2) = 0; Pu P2 e ZG}

= [Pi + xP2 e ZG' : Pi + P2 = 0; P2 € ZG}

= {Pi - xPi e ZG' : Pi e ZG}

= (x - 1)ZG.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. 6 So, we have

K' = { 1 - x)ZG — ZG* —=-» ZG.

where i is the inclusion map. We also have the inclusion map r : ZG —*• ZG’ and

7r o r is the identity map. So, this short-exact sequence splits. Restricting 7r to

the group of units, we obtain the split-exact sequence of groups:

K — U{ZG*) — U(ZG).

where K = Ker(7r). Hence, U{ZG*) = K * U{ZG). Note that K = U(ZG') n

(1 + K*). Thus, a unit in K has the form 1 + (x — 1 )P. where P E ZG. and has

an inverse 1 + (x — 1 )Q. where Q E ZG.

Lemma 2.1. K < U\(ZG*).

Proof. Recall that K = U{ZG*) fl (1 + K *) and that K’ = (x — 1 )ZG. Observe

that 1 + K* Q 1 + Az(G*), where A^(G*) is the augmentation ideal in ZG*. □

Hence, we have the following commutative diagram:

K —!—> U(ZG’) — U{ZG)

= L L

K — UX{ZG*) Ui(ZG)

L L L

(x) — i—» G* G.

Note that UX(ZG*) = K x UX(ZG).

Lemma 2.2. Let G* = G x (x), where x has order 2 ; u = 1 + (x — 1 )P, v =

1 + {x — 1 )Q, where P, Q E ZG. Then u and v are multiplicative inverses of each

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. 7 other in K <=$■ 1 — 2P and 1 — 2Q are multiplicative inverses of each other in

U{ZG).

Proof. Let u, v € K; with uv = 1. Observe that

uv = [1 + (x — 1)P][1 -t- (x — 1)Q]

= 1 + (x — 1)P + (x - 1 )Q + (x — 1)2PQ

= 1 + (x - 1)P + (x - 1)Q + (x - 1)(—2PQ)

= l + { x -l){ P + Q - 2 PQ).

So, uv =- 1 <=> (x — 1)(P -(- Q — 2PQ) - 0

«=> (2PQ - P - Q ) + (P + Q - 2 PQ)x = 0

«<=» 2PQ - P - Q = 0

«=► 4PQ - 2P - 2Q = 0

1 - 2P - 2g + 4PQ = 1

<=► (1 - 2 P )( 1 - 2 Q) = 1.

□

Hence, A' = {1 + (x — 1)P : P 6 ZG and 1 — 2P G U(ZG)}.

Now, consider the surjective ring homomorphism p : ZG -» Z2G. where

p reduces the coefficients modulo 2. The kernel of p, say M* (as an ideal), is

M* = 2ZG. Thus, we have the following exact sequence of rings:

M* = 2ZG ——> ZG — Z 2 G.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. 8 Furthermore, p induces the following exact sequence of groups, which does not

necessarily split:

M —— > U{ZG) — 9-+ U(Z2 G),

where M is the kernel of the group homomorphism p. Observe that M = U{ZG)C\

(1 + M'). Thus, a unit in M has the form 1 + 2P. where P G ZG and has an

inverse 1 + 2Q, where Q G ZG. Notice that here at the group level, p is not

necessarily surjective.

Since G* = G x ( x) and x 2 = 1, we have the group homomorphism a : G* —*

U(ZG), where a(g) = g and a(x) = — 1. This extends to a ring homomorphism

a : ZG' —*■ ZG. So on the ring level, we have the following diagram:

K' —i—► ZG' - 2 — ZG

a a p

M' — ZG — Z 2 G.

Observe that p o ~ = p o a. Hence. cr(K') C M' and thus it easily follows that

cr{K) C M.

Theorem 2.3. The map a : K —* M is an isomorphism of groups.

Proof. It is immediate that cr is a group homomorphism. Note that we have

cr[ 1 + (x — 1)P] = 1 — 2P and hence by Lemma 2.2, a is surjective. Now. suppose

that 1 + (x — 1 )P G Ker(• —2P = 0 = > P = 0. So.

1 4- (x — l)P = 1 = » a is injective. □

Theorem 2.4. U(ZG') = K >i U{ZG) = M xj U{ZG).

Proof. The elements of the semi-direct product M xi U(ZG) should be viewed as

ordered pairs (u, w), where u e M and w e U{ZG). If k e K and w G U{ZG),

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. 9 then the isomorphism maps kw to (cr(k),w) with the action of U{ZG) on M

induced by conjugation in U(ZG). □

It is also of interest to note that M is a subgroup of finite index in U(ZG).

Summarizing, we have the following diagram:

K — U(ZG') U(ZG) 5? <7 P

M — !— * U(ZG) U(Z2 G).

The problem of describing U(ZG*) has been reduced to the problem of describing

M. In the next section, we do this for several interesting cases.

Although we will not use the following facts, they are remarked here for the

sake of completeness. With respect to the group mapa : U(ZG *) —>■ U(ZG). the

kernel is N = (1 + N*) fl U(ZG') where elements of N m are of the form (x + 1 )P.

P 6 ZG. Furthermore, w : N —► M and the map 7 : K —> N (where g >—* g.

x —x) are group isomorphisms. Lastly, k = a o 7.

2.2 Applications

In the previous section, we established that U{ZG') = K x U(ZG) =

M >1 U(ZG). The results obtained in Section 2.1 will be of use if we can get

enough information about M and U(ZG). We can then pull back to K via the

isomorphism a to get a description of U(ZG*). Here, we will apply this result to

a few cases where in U\(ZG), G has a normal complement V. First, we see that

Theorem B (Higman’s result) follows from Theorem 2.4.

Theorem B (Higman). Let G* = G x (x),x2 = 1. Then U(ZG) = ±G =>

U{ZG*) = ± G \

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. 10 Proof. We know that U(ZG*) = K x U(ZG). It suffices to show that K = (x)

and that this semi-direct product is actually a direct product. Recall the following

diagram:

K — U(ZG') — t/(ZG) p

M ——* Cf(ZG) —^ U{Z2 G).

Since U(ZG) = and p mapsg —>■ g and — g —> +<7, it is clear that Ker(p) =

M = ±1. The inverse image of M with respect to a is K = (x) = Co- Since x

is central in ZG*, this implies that U(ZG*) = (x) x U{ZG) = ±G*. Thus, the

theorem is established. □

As another application, we give a different proof of Theorem D. First, we

recall the definition of a bicyclic unit.

Definition. Let b G G and set:

o(b) 6 = ^ 6'. 1

Then for any a £ G. u^a = 1 + (1 — b)ab is a unit in U(ZG), with inverse

1 — (1 — b)ab. The units u^a where a,b € G are called the bicyclic units of ZG.

These units have played an important role in the study of U(ZG) in recent years

[cf. 8, 9, 12, 13]. Among other things, we develop criteria to determine when two

bicyclic units are equal to each other or whether they are inverses of each other.

This helps in calculating the number of distinct bicyclic units (up to inverses) for

U(ZG). Although these lemmas are not essential for what follows, we include them for completeness.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. 11 Lemma 2.6. The bicyclic unit ut,a is trivial <=> a normalizes (b).

Proof. (<£=) Let ba = abl and b have order k. Then

(1 — b)ab = (1 — 6)a(l + b + . •. -I- bk ')

= (a — 6a )(l + 6 + ... + bk~l)

= (a — a6‘)(l + 6 + bk~l)

= a(l — 6l) ( l + 6 - F ...+ bk~l)

= a(l -F b 4- ... + 61_1)(1 - b)( 1 + b + . . + bk~l)

= a(l + b -+- ... + 6* l)( 0)

= 0.

So, U6.a = 1 + (1 — b)ab = 1.

(= > ) Let (1 — b)ab = 0. Then, (1 — 6)a(l + 6 + . . + bk~l) = 0 = a ab -F ... ~F

abk~l = ba + bab+ ...+ babk~l =>• a(b) = ba(b) ==> a~lba G (b) > a normalizes

(b). □

Lemma 2.7. Let o(r) = k, and suppose that r.s,a,b G G. where s does not

normalize (r), b does not normalize (a). Then,

a = rJ, (j, k) = 1:

ur.s = uab <—> b = sr*, some i;

s - 1r J_1s G (r).

Proof. (=>) Let = ua.b ^ L Then, 1 + (1— r)sf = 1 + (1—a)6a = > (l-r)sr =

(1 — a)ba [both ^ 0] =F sr — rsr = ba — aba. But the left cosets s(r) and rs(r)

are disjoint as are the left cosets 6(a) and ab{a). Hence, sr = ba and rsr = aba.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. 12

Thus, s(r) = 6(a). Therefore, 6 € s(r) and we have that b(r) = s(r) = b(a). This

implies that (r) = (a). Hence, a = rj, where (j,k) = 1. Also, since 6(r) = s(r).

we have that b = sr1, some i. Now to prove the third condition, observe that:

urs = utJ s where (j, k) = 1 (1 — r)sf = (1 — r; )sr-/

(1 — r)sf = (1 — rJ)sf

sf — rsr = sr — r3 sf

rsr = rJ sf

rs(r) = rJs(r)

(r) = s-lrJ_ 1s(r)

s _ 1rJ_1s € (r) = (r3).

(<=) VVe need to show that under the given hypothesis, ur 6 = urJ sr,. All of the

above steps are reversible, which gives (1 — r)sf = (1 — rJ )sr->. Since rJ = f =

r‘f = r'rJ. we conclude that (1 — r3 )sr3 = (1 — rJ)srlrJ. Thus. 1 + (1 — r)sf =

1 + (1 — rJ)srlri = > ur,s = u rJ_sr.. □

Lemma 2.8. Let o(r) = k, and suppose that r.s,a.b 6 G, where s does not

normalize (r), 6 does not normalize (a). Then,

a = r3. (j,k) = 1; -1 6 = rsr', some i;

s~lr3+ls 6 (r).

Proof. (= > ) Let = (ua-6) 1 ^ 1. Then, 1 + (1 — r)sr = 1 - (1 — a)ba =>

(1 — r)sf = — (1 — a)ba [both ^ 0] =*► sf — rsr = —ba + aba =>• sf = aba

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. and rsf = ba. Thus, s(r) = ab(a) and rs(r) = 6(a) This implies that 6 G rs(r)

and we have 6(r) = rs(r) = b{a). Hence, (r) = (a) and we get that a = r3.

where (j.k) = 1. Also, 6{r) = rs(r) => b € rs{r) ==> 6 = r s r1, some t. So.

«r,3 = (ua, 6)-1 = > ur,s = (urJ-rsr.)“\ (j. fc) = 1. But in general. uc,d — u ^ t.

Hence. urs = (urj rs)_1 where (j,k) = 1. Summarizing, ur-s = (ua_b)~l imphes:

uT'S = {urj%rs)~l =>■ 1 -F (1 — r)s f = 1 — (1 — r3 )rsr3

=> (1 — r)sf = — (1 — r3 )rsr3

=>• (1 — r)sr = — (1 — r3)rsf

=> sf — rsr = —rs r + r3+lsf

=> sf = r3+lsf

=> s(r) = r3 +ls(r)

=F (r) = s_ 1rJ+ls(r)

==> s~lr3+ls G (r) = (rJ).

(<=) We need to show that under the three conditions, ur.s = {ur] rsr,)~l. All of

the above steps are reversible, which gives us the implication (r) = s~lr3 +ls(r) =>

1 + (1 — r)sr = 1 — (1 — r3 )rsr3 . But, 1 — (1 — r3 )rsr3 = 1 — (1 — r3 )rsrlr3 . Hence.

Ur.s (^r-J.rsr*) * i-J

Now. we are ready to prove Jespers' result regarding U(ZDi2). Before we start,

two preliminary remarks should be made. First, it should be noted that his

techniques for D i2 = S3 x C2 also make use of the map 7r : D i 2 —» S3. However,

he combines this with properties of congruence subgroups of integral matrices.

In contrast, our proof involves the computation of the image of p : U(ZSz) —*

U{Z2 S$). Second, our proof only gives the structure of Ui(ZDl2). We do not

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. 14 show that the normal complement of UiiZDu) is generated by the bicyclic units.

Theorem D (Jespers). In U\(ZD\ 2 ), D\ 2 has a torsion-free normal comple

ment which is a semi-direct product of a free group of rank 5 by a free group of

rank 3. Furthermore, this complement is generated by the bicyclic units.

Proof. Let G = S3 = (a, b : a 3 = b2 = (ab ) 2 = 1) and note that Gm - Di2 =

S3 x (x),x 2 = 1. Theorem C tells us that U\{ZS3 ) = V x S3 , where V is a free

group of rank 3 generated by the three distinct bicyclic units (up to inverses) of

ZS3 , namely V = (u^a, Uba.a, Uba 2 ,a)- The fact that these bicyclic units are distinct

also follows from the preceding lemmas. Thus, we have the following diagram:

K U(ZDV2) — U U(ZS3)

M —^ U(ZS3) U(Z2 S3) L L

M + — ^ Ul(ZS3) p(£/i(ZS3))

m + nv' — v P(V).

Here. M + and M + D V are the appropriate kernels of the restriction ofp. Recall

that 7r maps g >-> g, x 1; a maps g >—» g. x 1—► — 1; and p is the reduction map,

modulo 2. We will proceed to show that M + is a free group of rank 5. First, we

begin by analyzing p(V). Observe that

pi'Ub.a) = p(l + a — a 2 — ba + ba2) = I + a + a 2 + ba + ba 2

piV'ba.a) = p(l + n — a 2 — ba 2 + 6) = l+cz + fl^ + 6 -|- ba 2

p(j^ba 2 ,a) = p(l a — a 2 — 6 -f ba) = 1 -t- a 4- a 2 + 6 -t- ba.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Now, for notations! purposes, set = u^a-, u 2 = Uba,a^v 3 = a in U(ZSz) and

set N = 1 + a + a 2 + b + ba + ba 2 in Z 2 S3 . Note the following: Vi = av 3 a~l ,v 2 =

avia~l, u 3 = av 2 a ~ 1 and (N - 3,)('V - g3) = 61 Y - gtN - Ng 0 + gtg 3 = g,g 3 in

Z 2 S3 . We have that p(ub,a) = N — b, p{uba .a) = N — ba , and p(ufe,2,a) = N — ba2.

In particular, [p{vi)][p{vj)\ = 1, a, or a2. Since V = (u^a^ba.a^^ a), we see that

p(I/) = {l,a, a2, N — b, N — ba, N — ba2} “ S3.

In the free group V with basis ui, v2, and v3, recall that a word w(vi) G V

is of even length if it can be written as a product of an even number of the V{

and v~l. Now. let W be the subgroup of V. consisting of the even words of V.

Note that W < V and p{W) = (a). It is easily shown that M + D V = {u G

V : p{v) = 1} < W. Suppose that P G U\{ZS3 ). Then P = vg, where v G V

and g G S3 . So, p(P) = p{v)g. Recall that v is a word, using the alphabet

{v1 .v2 .v3 }. If v is not an even word, then p(v)g = [N — g{]g = N — gig ^ 1.

So. M + = Ker(p) = {val : i = 0.1, or2;t’ G Wanda' = [p(w)]-1 }- Thus.

p[Ui{ZS3 )\ = {1. a. a2. N - 6, N - ba , N - ba2. b, 6a, ba2. N - 1. N - a. N - a2} =

S3(N — 1 ) = S3 x C2. Thus, we have the following exact sequence:

M+ Ui(ZS3) = VxS3 p[Ui{ZS3)\ = SZ{N - 1).

Now, the pre-image of (N — b) with respect to p is M +{vi). Every element

of M +(v 1) can be written in one of the following forms: we(vi), p[we(vi)\ = 1;

we(Vi) ■ a,p[we{vi)} = a2; we(vi) ■ a2. p[we{vi)] = a; w0(vi) • l,p[m0(wi)] = N - b:

w0(vi) ■ a, p[w0(vi)} = N — ba2; wQ{vi) ■ a 2, p[w0(vi)\ = N — ba. Here, we(vi) denotes

a word of even length and w0(vi) denotes a word of odd length. Let t\ = vx,

t2 = v2a2, and t3 = v3a. We will eventually show that A//+(u1) = (£ 1, 62, £3) and

that this group is free of rank 3.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. 16 Theorem 2.10. Suppose that w(Vi) [not nec. reducedJ is a word of length s.

Then, w(vi) = w'{ti)al, where w'{ti) [not nec. reduced] has length s, and I = 0.1,

or 2. Conversely, if w{ti) [not nec. reduced] has length s, then w(ti) = w'(vi)al.

where w'(vt) [not nec. reduced] has length s, I = 0.1, or 2.

Proof. {=>) We will use induction on the length of w(vi) : Let s = 1. Then,

iq = t\. V2 = t2 a, v 3 = t3 a2, u f l = t f l, v f 1 = t^la, and v f 1 = t f la2. Hence, the

claim holds when s = 1. Now, assume that the claim holds for all words w(vi) of

length less than or equal to k. Now, consider a word w(vi) of length k + 1:

w(vi) = ■ ■ ■ v\kv-]]', where e, = ± 1.

= wl(ti)anv ^ 1i, length of w'(ti) = k.n = 0.1 or 2.

= w'(ti)anv£ ;a -nan

= w'(ti)vjk+lan. where j = ik+i + n (Mod 3)

= w'(ti)veJk+la 4 ' Jan+:’~4,j = ik+i + n (Mod 3) / w'{ti)t]an+3~4. if ek+l = 1;

w,(ti)tfla n + 3 ~ 4 if ek+l = -1 ,j = 1; = <

w,(ti)t^la n + J - 4 if Cfc+i = —1, j = 3;

w ' i t ^ a ^ - 4 if efc+1 = —1. j = 2.

Thus, we have a word of length k + 1 (in the alphabet of tt). followed by a power

of a.

(<=) Let the length of w(ti) be equal to s. For s = 1 : fL = iq, t2 = v2a2,

t3 = v3a , t[{ = v f 1, t f 1 = and t f 1 = v fla2. Hence, the claim holds when

s = 1. Now, assume that the claim holds for all words w(t,) of length less than or

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. 17 equal to k. Let us consider a word w(fj) of length s = k + 1:

w (ti) = % ■ ■ ■ tgtg+l, where e, = ± 1 .

= w'(vi)ant*j‘^. length of w'(v,)= k, n = 0,1 or 2.

= w \Vi)ant ?kl\a 'nan

•in (vi)vja 4 ~lk+lan, if efc+i = 1 \j = ik+l + n (Mod 3)]:

in {Vi)v2 lan, if efc+1 = -1 , n = 1, ik+l = 1;

w (vi)vl laan, if efc+i = - 1, n = 1, ik+1 = 2;

w (Vi)v^la 2an . if ek+i = -1, n = 1, ik+i - 3;

w (t/i)v3 ■lan. if efc+1 = - 1, n = 2, ik+1 = 1;

w (vi )vflaan. if ek+i = - 1. n = 2, zfc+1 = 2;

w {vi)v± la 2an. if efc+i = -1 . n = 2. z'fc+1 = 3:

w if ek+i = - L n = 0, z'fc+i = 1:

w {vi)vfla. if ek+l = - 1, 7i = 0. zfc+1 = 2:

w {vi)v2la2. if e/t+i = - 1 . 7i = 0. ik+l = 3.

Thus, we have a word of length k + 1 (in the alphabet of Uj), followed by a power

of a. □

Corollary 2.11. If ws(vi) is a reduced word of length s and w 3 (Vi) = w's(ti)al.

then w's(ti) is a reduced word of length s.

Proof. Assume that ws(v{) = w's(tt)al = wk(ti)am, where wk(ti) is reduced and

k < s. Thus, ws(vi) = wk(vi)ajam, where k < s. Since every element in V x (a)

is written uniquely, we get the desired contradiction. □

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. 18

Now for every s > 1, let R3 (vi) and Rs{ti) denote the sets of reduced words in

V{ of length s and reduced words in £, of length s respectively. Consider the map

0 : Rs(vi) —► Rs(ti) defined by 0 [uA,(ut)] = w'3 (ti). where ws(vi) = w's(ti)ak, for

some ak.

Theorem 2.12. 0 : Rs(vi) —► RS(U) is a bisection.

Proof. First, we show that 0 is injective. Suppose that we have two reduced words

in Vi. ws(i'i) = ws(ti)ak and = uis(ti)al. Then, ws(vi)a~k = w's{vi)a~l. But

these elements are in V xi (a). Hence, w s (v i) = w's {vi) and we have shown that 0

is one-to-one. Since 0 is injective and Rs{vi) and Rs{ti) are finite sets of the same

cardinality, 0 is onto. □

Corollary 2.13. Let ws(ti) be a reduced word of length s. Then there exists a

unique reduced word w's(vi) of length s and ak such that ws(ti) = w'^v^a1^.

Proof. Since 0 is a bijection. there exists a reduced word w's(vi) and ak such that

w's{vi) = ws(ti)ak. So, ws(ti) = w's(vi)a~k and we have existence. Uniqueness

follows from the fact that ws(ti) is an element of V >i (a) and is written uniquely

in the form ius(vi)am. □

Corollary 2.14. T = (£i,£2^ 3) is a free group of rank 3.

Proof. Assume that T is not free. Then, there exists a reduced word ws(ti) of

length s > 1, where ws(ti) = 1. By Corollary 2.13, this gives a reduced word

w's(vi) of length s, where w's(vi)ak = 1. Since w's(vi)ak 6 V x (a), this implies that

ws(vi) = 1- a reduced word of length s > 1. This contradicts the fact that V is free. □

Theorem 2.15. M + = TE := {even words using the alphabet £j}.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. 19 Proof. (D)

titi = V\V\ G M +

t2 t2 = (v2 a 2 )(v2a2) = v2v\a G M +

h h = (v3 a)(u 3 a) = v 3 vxa 2 G M +

t\t2 = V\V2a 2 G M +

t\t3 = v\v3a G M +

t3 t]_ = v3avi = G iV /+

t2 t\ = v2arv\ac? = v 2v3a 2 G M +

t2 t3 = v2a 2v3a = v 2v2 G A / +

M 2 = v3av 2a 2 = v3v3 G M +

M 2 1 = tqz/^a G A/+

M 3 1 = Uit’2 la2 G M +

M 3 1 = {v2 a2)(vf 1a 2 ) = v2a 2v flaa 2 a 2 = v2v^la G A / +

t2 lti = (v f la)vi = v flav\a2a = v flu2a G A / +

= (v f la 2 )v\ = v^cPviaa 1 = v flv3 a 2 G M +

^3 lt2 = (vf la 2 )(v2 a2) = v f la 2 v2aa = uflvia G M +

Since M + is a group, all the inverses of these elements are in M +. Thus, any even

word involving tu t2, t3, t f l, and f j 1 is in M +. Hence, TE < M +.

Therefore, we(vi)a 1 = vue{ti)ak+l G M +. Since TE < M + and the length of we(vi)

equals the length of we(U), this implies that we{t{) G M +. Thus, ak+i G M +. But

M + n (a) = (1). Therefore, ak+l = 1. Hence, Weiv^a' = we(ti). We conclude that

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. 2 0 M + < T E. □

So. we have that M + < T = (ti, t2 ,t3), with [T : M +] = [T : TE] = 2. Using

Schreier’s Theorem and the Reidemeister-Schreier method, we get that M + is

a free group of rank 5, with M + = (t\. tit2, t\t$, t 2 t f l.^ f 1)- summary: for

G — S3. G* = S3 x C2 = Di2 - we have

K — U^ZG*) — — Ui(ZG), with

Ui(ZDl2) = K x (U x S3)

“ M x (U x S3)

= [Af+ x C2] x (V x S3)

= [M+ x V] x (C 2 x S3)

= [M+ x U] x D 12.

This finishes the proof of Jespers? result. Of course, just as M — M + x ( — 1). it

is easily shown that K =K + x (x). where K + = [1 + (x —l)A^(G)] D U(ZG").

Then. Ui(ZDi2) = [K+ x V] x (S 3 x (x)) = [K+ x V]x £>i2 and K + is the free

group of rank 5 generated by the inverse images under a of the five generators of A/+. □

As another application of Theorem 2.4, let us give an alternative proof of

the following theorem.

Theorem E (Jespers-Parmenter). In Ui{Z[D 8 xC 2\), D8 xC 2 has a torsion-

free normal complement which is a semi-direct product of a free group of rank 9

by a free group of rank 3.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Proof. Let G' = Ds x C2: with D$ = (a, b : aA = b 2 = 1 = abab ), the dihedral

group of order 8. In 1990, Parmenter [12] showed that in Ui(ZDs). Ds has a free

torsion-free normal complement V which is generated by any three of the four

distinct bicyclic units (up to inverses). Now, consider the following commutative

diagram:

K — ^ U^ZGT) Ui(ZDs)

M — U(ZDs) — U(ZM

M + — UdZD8) — p(Ui(ZD8))

M + n v —v p(V).

with V = (v\,v2, u3), where ui = Ub.a = 1 + (1 — 6)a6, v2 = u^a = 1 + (1 — ba)aba.

and v3 = Uba2 a = 1 + (1 — ba 2 )aba2. Now in C/(Z2Z?8), let N = l + a + a2 + a3-F6 +

ba + ba 2 + baz. Straight-forward computations show that p(vi) = N+a 2 -hb+ba2 =

p(v3). p(v2) = N + a 2 + ba + ba 3, and p{v\)p(v2) = p(v2 )p(v\) = N + a + a 2 -Fa3.

Furthermore, p(v 1), p{v2), and p(vi)p(v2) are all of order 2. Thus. p(V) = C2 x C2.

Observe that p(V) Pi D8 = 1. Thus, p[Ui(ZDs)] = [C2 x C2\ x D8. Note that

p[w(vi) ■ g] = 1 if and only if g = 1, p[w(vi)] = 1. Thus, we deduce that Ker(p) =

M + is contained in V. Hence, M + = V D M +. Since [V : V fl M +] = 4 and V

is a free group of rank 3, it follows by the Schreier Index Formula and Schreier's

Theorem that M + is a free group of rank 9. Note that generators for M + could

easily be obtained as before. In summary,

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. 22

Ui(Z[D6 x Co}) = K x (V x Da)

= M x (V' x £>8)

= [M+ x C2] x (1/ x D8)

= [M+ x V] x (D$ x C2).

As before. Ui(Z[D8 x C2]) = [F+ x 1/] x (£>8 x C2), where K + = a~l{M+). □

Let us use the main structure result in Section 2.1 to prove some new

theorems. Let F„ denote a free group of rank n.

Theorem 2.17. Let P be the group of order 16 with the presentation (a, b: a 4 =

1 = b4 ,bab~la - 1 = a2). Then, UV{Z[P x C2]) = [F65 x F9] x (F x C2).

Proof. Let P* = P x Co, with P = (a, b : a 4 = I = b4, bab~la~l = a2). Parmenter

[12] showed that U\{ZP) = V x P; with V a free group of rank 9 generated by:

Vi = 1 + (1 - a 2) 1 + b2) a + b)

Vo = 1 + (1 - a2) 1 + b2) —a + ab)

v3 = 1 + (1 - a 2) i + h2) 13a + 56 - 12a6)

^4 = 1 + (1 - a 2) 1 + b2) 17a + 156 — 8a6)

vs = 1 + (1 - a 2) 1 + b2) -125a - 446+ 117a6)

Vs = 1 + (1 - a 2) 1 +b2) 149a + 516 — 140a6)

V7 = 1 + (1 - a 2) 1 + b2) — 2 + a — 2a6)

vs = 1 + (1 - a 2) 1 + b2) -8 - 19a - 146 + 15a6) to

Vg = 1 + (1 - a 2) + —2 — 7a — 46 + 6a6).

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Now, consider the following diagram:

K — —> U ^ZP') —=-> U,{ZP)

2; P

M — tf(ZP) —^ t/(Z 2P) t[

M+ —i-, tft(ZP) -£-> p(Ui(ZP))

1 1 L

M +n V V —! p(V).

In I/(Z 2P), we have the following: p(vi) = p(v3) = p(^) = p(^6) = 1 + (1 4-

a2)(l 4- b2)(a + 6); p(u2) = p(v5) = p(v8) = 1 -I- (1 4- a2)(l 4- b2)(a 4- ab ); and

p(ur) = p(u9) = 1 + (1 + a2)(l 4- b2 )(a). So, p(V) = (p(nt ), p(v2). p{v7)).

Observe that the center of P is C(P) = {1, a2, b2, a 2 b2}. Now, let Nc =

1 + a 2 + b2 + a 2b2. Note that in U(Z2 P)- [p{vi)]2 = [1 4- Nc(a 4- b)]2 = 14-

Nc(a 4- b)Nc(a 4- 6) = 1 4- (a + b)(4Nc)(a + b) = 1, [p(v2 )]2 = [1 4- Nc(a 4- ab )}2 =

l + Nc(a+ab)Nc(a+ab) = l + (a+ab)(4Nc)(a+ab) = 1, and [p(v?)]2 = [l4-iVca]2 =

1 4- NcaNca = 1 4- a(4;Vc)a = 1.

We will now show that p(v 1), p(v2). and p(u7) commute with each other

pair-wise: p{v\)p(v2) = [l + vVc(a4-&)][l4-Arc(a4-a6)] = 1 +Nca +Ncb+NcaNcab =

1 4- Ncb + Ncab = [l 4- Nc(a 4- a6)][l 4- Nc(a + 6)] = p{v2 )p(v 1); p(v{)p{v7) =

[l 4- Nc(a 4- &)][l + Nca] — 1 4- Nc(a 4- b) 4- Nca = 1 4" Ncb = 1 4- *Vca 4- iVc(a 4- b) =

[1 4- JVca][l 4- Nc(a + 6)] = p(v7 )p(vt); p(v2 )p(v7) = [1 4- Nc(a 4- a6)][l 4- Nc(a)\ =

1 4- Nc(a 4- ab) 4- Nca = 1 4- Ncab = [1 4- iVc(a)][l 4- Nc(a 4- a6)] = p(v7 )p(v2).

Also, note that p(vi)p(v2 )p(v7) = (1 4- Nca)( 1 4- Ncb 4- Ncab) = 14- Nca 4-

Ncb+Ncab. It is clear that 1, p(v 1), p{v2), p(v7), p{vi)p(v2), p(vi)p(v7), p{v 2 )p(v7).

and p(vi)p(v2 )p(v7) axe all distinct since they are all different sums of cosets of

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. Nc. Hence, we deduce that p(V) = C2 x C2 x C2. Since p(V) n P = 1 and

p[w(vi) ■ g] = 1 if and only if g = 1, p[w{vi)\ = 1, we see that Ker(p) = M + < V .

Thus, M + = V fl M +. Since [V : V n M +) = 8 and V is a free group of rank 9. it

follows by the Schreier Index Formula and Schreier’s Theorem that M + is a free

group of rank 65. To summarize, we have that U\{Z[P x C2]) = K xi (V' x P) =

M xi (K xi P) = [M+ x C2] x (V x P) = [M+ x V] x (P x C2). □

The following two theorems verify Theorem A in specific cases. However, they

also give additional information.

Theorem 2.18. Let Cl = Cs x Co = (c) x (x). Then. Ui(ZCl) = (1 + (x —

1 )P) x (v) x Cl = Z x Z x Q ; where P = —4 — 3c 4- 3c3 + 4c4 + 3c° — 3c7,

v = 2 + c — c3 — c4 — c5 + c '.

Proof. Karpilovsky [11] showed the following result: If Cs = (c), then Ij\{ZCg) =

C8 x (v) = Cg x Z . where u = 2 + c — c3 — c4 — c° + c7. Now. let V = {v). Then,

we have the following commutative diagram:

K U^ZCl) — U^ZCs) S'

A/ — U (Z C s) U {Z 2C S)

L L L

M + I/i(ZC8) -2— p(Ui(ZCs))

L l L

M + fl y — v V" — ^ p{V).

Now, set iV = l+ c + c2+c3-fc4+c5 + c6 + c7. Then, p{v) = c + c3 + c4 +c5 4-c7 =

N 4-1 4-c2 4-c6. Observe that in t/l(Z 2C8), \p{v) ] 2 = [iV 4- 1 4-c2 4-c6]2 = (1 4-c2 4-

c6)2 = 1 4-c44-c12 = 1. Thus, p{y) has order 2 in U\(Z2 C%) and hence, p(V) = C2;

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. 25 where p(V) = {N + 1 + c2 + c6, 1}. It is clear that M + = {vl • 1, where i is even}.

So, M = (v2) x (—1), where v = 2 + c — c3 — c4 — c5 + c7.

Recall that a : K —* M defined by g g and n -> -1 ; [so, 1 + (x — 1 )P >—►

1 — 2P] is an isomorphism of groups. A straight-forward calculation shows that

v2 = ( 2 + c — c3 — c4 — c° + c1) 2

= 9 + 6c — 6c3 — 8c4 — 6c° + 6c'

= 1 + 2(4 + 3c - 3c3 - 4c4 - 3c5 + 3c7)

= I - 2(—4 - 3c + 3c3 + 4c4 + 3c5 - 3c7).

Now. set P = —4 — 3c + 3c3 + 4c4 + 3c5 — 3c7. So. the inverse image with respect

to cr of M + = (v2) is (1 + (x — 1 )P). Hence, (1 + (x — 1)P) = Z. Also, the

inverse image with respect to a of {—1) is x. Since C| is an abelian group. ZC$

is a commutative ring. Hence. Ui(ZC%) is an abelian group. Thus.

Ux{ZCl) — I< x U^ZCs)

= (x) x (1 + (x — 1)P) x Ui{ZC$)

= {x) x (1 + (x - 1)P) x C8 x (v)

= (1 + (x - 1)P) x (v) X Cg

= Z x Z x C g*.

□

Theorem 2.19. Let C% = C5 x C2 = (c) x (x). Then, Ui(ZC^) = (1 + (x -

1)P) x (v) x Cl = Z x Z x Cl; where P = —3 — c + 3c2 -t-3c3 — c4, v = (c+ l)2 — c.

Proof. It has been shown [17] that if C5 = (c), then Ui(ZC5) = C5x (v) = C5 xZ ,

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. 26 where v = (c+1)2 —c. Now, let V = (v). Then, we have the following commutative

diagram:

K — Ui(ZCZ) U,{ZC5)

M UiZCs) —^ U(Z2 C5)

L L

M+ Ui{ZC5) Pm Z C s))

L L L M+nv v —?—> P(v).

Now. set N = 1 + c + c2 -+- c3 + c4. Observe that v = (c + l)2 — c = (c + l) 2 —

(1 + c + c2 + c3 + c4) = c2 + 2c + 1 — 1 — c — c2 — c3 — c4 = c — c3 — c4. Thus.

p(v) = c + c3 +c4 = N + l+ c 2. In U\(Z2 C=,), we see that:

\p{v 2 = (c2 + l)iV + (c2 + l)N + 5 N + (c2 + l)2 = N + c4 + 1 = c + c2 + c3

[p{v 3 = (.V + c4 + 1)(N + c2 + 1) = 5iV + 2N + 2N+ c6+ c 4 +c2+ 1 = c3

[p(v 4 = (N + c4 + 1 )(N + c4 + 1) = 5Ar + 2N + 2N + c8 + 2c4 + 1 = N + c3 + 1

[p(v 5 = c3(N + c4 + 1) = N + c7 + c3 = N + c2 + c3

[p{v « = c3c3 = c6 = c

\p(v 7 = c(N + c2 + 1) = N + c3 + c

\p(v 8 = (N + c3 + l)2 = 5 iV + c6 + 1 = N + c + 1

[p(v 9 = (N + c2 + 1)(N + c + 1) = N + c3 + c2 + c + 1 = c4

\p[v 10 = c4(N + c2 + l) = N + c6 + c4 = N + c4 + c

Io{v 11 = c3(jV + c + l ) = N + c4 + c3

[p(v 12 = ( C3 )4 = c 12 = c 2

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. 27

[p(u)]13 = c2(N + c2 -t- 1) = N + c4 + c2

[p(u)]14 = c2(N + c4 + I) = N + c6 + c2 = N + c2 + c

[p(c)]15 = (iV + c2 + c){N + c2 + 1) = N + c4 + c3 + c2 + c = 1.

So. p(V) = Ci5. Observe that M + = (w(t;3) • g : g = p[iu(t/3)]-1} = (v3 • c2) = Z.

Notice that c and v commute with each other. Thus, M = (v3 • c2) x (—1). Now,

v2 = (c—c3 — c4)2 = c2 —c4 —1 —c4+ c + c 2 —l+ c^ + c3 = — 2 + c + 3 c2 + c 3 — 2 c4. This

implies that v3 = (—2 + c + 3c2 + c 3 — 2c4)(c — c3 — c4) = —6 — 6c + 2c2 + 7c3 + 2c4.

Hence, v3 • c2 = —6c2 — 6c3 + 2c4 + 7 + 2c = 1 — 2(—3 — c + 3c2 + 3c3 — c4). Now,

set P = —3 — c + 3c2 -I- 3c3 — c4. The inverse image with respect to a of M + is

(1 + (x — l)P) = Z. Also, the inverse image with respect to a of — 1 is x. Since C$

is an abelian group, ZC~ is a commutative ring. Hence, Ui(ZC?) is an abelian

group. Thus,

Ul{ZCl) = KxUy{ZC 5)

= (x) x ( l + ( x - 1 )P) x U\{ZCs)

= (x) x (1 + ( x — 1 )P) X Cs X (v)

= (l + (x -l )P )x (v )x C ;

= Z x Z x C j.

□

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. CHAPTER III

A GENERALIZATION

3.1 U{ZGm), Where G' = G x Cp

It is now natural for one to ask if the main result in Section 2.1 can be

generalized to the case where G* = G x Cn. In this section, we answer this

question for n, a prime number. Throughout this chapter, ® denotes a direct sum

of abelian groups.

Let p be an odd prime, G* = C x C p = G x(i), and G be a complex primitive

p-th root of unity. We have the following surjective ring maps: p : Z -» Zp,

sending z z\ cr : ZCP -» Z[Q\, sending i h (, j m and a : Z[Q] -» Zp,

sending G 1—1> L - ► z. Note that a is well-defined for prime p. For the sake of

completeness, we give the following lemmas.

Lemma 3.1. Ker(a) = (N). the ideal generated by N = 1 + x + • • • + xp~l.

Proof. (D) It is clear that N G Ker(cr). Hence, (N) C Ker(cr).

Qo + aiG + • • • 4- ap_iGp 1 = 0- But, Gp_1 = — 1 — G *' * — Gp_2- So, ao + aiG + • • • +

ap_2Gp_2 - aP- i - ap_!G - ap_tG2 ap_1Gp_2 = 0. Thus, (a0 - ap_t) + (ax -

ap~i)C H 1- (aP-2 - aP-i)Gp~2 = 0. Therefore, a0 - ap_L = 0, a! - ap_i = 0, etc. This implies that a0 = ax = a2 = ■ ■ ■ = ap- X. Thus, P = aoN. □

Lemma 3.2. Ker(a) = (Q — 1), the ideal generated by G — 1.

Proof. (D) It is clear that G - 1 € Ker(a). Hence, (G - 1) C Ker(a).

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. So in Z, a0 + ai + • • • + ap_2 = kp. Thus, U — kp = — 1) + a2(C2 — 1) +

1- ap_2((7-2 — 1) = (£ —1 )Q G (C — 1)- Now, observe that p = 1 + 1 -I-----t-1

[p terms] = 1 + [C + (1 - 0 ] + [C2 + (1 - C2)] H- + [Cp_1 + (1 - C*"1)] =

1 + C + C2 + • • • + CP'1 + (1 ~ OS = (1 - QS. Thus, p G {C - 1). Since

U — kp € (C — 1) and p G ( ( - 1), we conclude that U G (C — 1)- □

The projection mapn : G* -» G induces the surjective ring map it : ZG" -*• ZG.

In addition, we have that the surjective ring maps p : Z -» Zp, cr ZCP: -*» Z[Q\,

and q : Z[C] —► Zp induce maps at the group ring level. Note that as group rings.

ZCP(G) = ZG*. Hence at the ring level, we have the following diagram, where

K * = Ker(7r) and M* = Ker(o:):

K* ZG * —?-> ZG

cr a P

A/* —i— Z{Q\G —2— ZPG.

Note that a : ZG * —► Z[£]G is defined by g ■—► g, x i—► 0 Clearly, (p o tt)(g) =

(a o cr)(g): for all g G G. Also, (p o tt)(x) = p(l) = 1 and (a o a)(x) = a(£) = I.

Thus, p o 7r = a o a. Note that cr maps K* into A/* because the right-hand part

of the diagram commutes. Let us decompose G* in the following way: G' =

G U xG U • • • U xp~lG. So, ZG* = ZG 0 xZG ®---@xp~lZG.

Proof. Let U = A0 + xA{ h xp~lAp-t G K*,Aj G ZG. Then. U G K* <=»

Ao + A\ + • • • + Ap_i = 0 (in ZG) <=> U = (Ao + xA\ + • • • + xp~l Ap_!) — (Ao +

Ai + ■ • • + Ap_l) <=>• U = (x — l)A i + (x2 — 1)A2 H -I- (xp~l — 1)AP_!. Now,

we need to show that this sum is direct. Let (x — l)Ai -j + (xp~l — l)Ap_i =

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. xp lAp_i — — + • ■ • -+- Bp— i) + xB\ + x* Bo + • • • + xp ^ Bp— i [*]. But in ZG*,

this is a direct sum. So [*] holds •£=>• Ai — Bi, for all i. □

Q) corresponding to Q is q(x) = xp~x+xp~2-\----- hr+1. So. Cp-1-FCp-2H K + l =

0. But, 1, £, • ■ • -Cp_2 are linearly independent over Q. Thus, {1,C, ••• •Cp-2}

forms a Z-basis for Z[Q\. Suppose that G = {9 1,9 2, ■ ■■ ■ 9 n] and U E Z[QG.

Then.

U = (a i,i + a i,2C + • • • + Qi.p-2CP 2)Pi +

(a2.1 + U2.2C H + a2.p-2CP 2)92 +

(an,i + o.noC, H + an.p_2Cp ~)gn-

Recombining terms, we get that U = A 0 + C,Ai ^ hCp_2Ap_2, where A i E ZG.

Clearly, this sum is unique. □

Proof. Let U = >4.o d-C 1 d- * • -+CP 2Ap_2. Ai E ZG. Then U £ Ker(a:) = M* < ■>

&(U) = p{Ao)+p{A\)+- ■ -+p(Ap- 2 ) = 0 = p{Aa+- ■ -+Ap- 2 ) <=> A0-f- ■ -+AP_2 =

pC <- - > A0 = ~{A\ + • • • + Ap-f) + pC <=•> U = — (Ai + • • • + Ap-2 ) + pC + £Ai +

• • • + Qp~2 Ap—2 <=► U = (C - lM i + (C2 - 1)A2 + • • • + (C '2 - 1)V2 +pC. Now, we will show that this is a direct sum. Suppose that (£ — 1) A\ + (C2 — 1)A2 H F

(Cp"2-l)Ap_2+pC = (C-l)5i+(C2- l) 5 2+--- + (Cp-2-l)5p_2+pD. By Lemma 3.4. we have that A,- = Bl, for all i. Furthermore. —(Ai + • • • + Ap_2) + pC =

—{B\ -\------1- Bp-2 ) + pD. This implies that pC = pD, and hence, C = D. □

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. 31 Theorem 3.6. The map cr : K ' —> M" is an isomorphism of rings.

Proof. Let U = (x — l).4i -b • • • + (xp_1 — 1)j4p_i G K*. N o w .

1)^! -b • - • + (CP-1 — 1)j4p_i. Observe that £p_l = —(1 -b C + • • * + O’-2)- This

implies that Cp_l ~ 1 = —1 — £ — ••• —Q v ~ 2 — 1 + (Ylp 1) — P- Recombining

terms, we have that C?-1 — 1 = (1 — 0 + (1 — C2) + •' • + U — Cp~2) ~ P• So,

a{U) = (C — l)(^4i — Ap_i) H------b (CP-2 — l)(-^p-2 — Ap- 1) -bp(—Ap-i). Now, let us show that a is injective.

G(U) ~ (C — l)(^i — "4p-i) H + (Cp 2 ~ l)(^4p-2 — Ap^i) + p{— >lp-i) = 0

•F=> {A\ — >lp_i) = 0,... , [Ap_2 — ^4.p-i) = 0, — Ap-i = 0

Ai = 0; for i = 1 .2 ,... . p — 1.

Hence, o is injective. We conclude by showing that a is surjective. Let V G M*.

where V = (£ — l)/li H b (Cp_2 ~ 1MP_2 +pC. Then its pre-image with respect

to a is U — (x — l)(^4i — C) H b ( xp~2 — l)(.4p_2 — C) + (xp~l — 1)(—C). Thus.

a is surjective. □

We have the following commutative diagrams:

K ' — !—> ZG* — ► ZG

M*Z[Q)G - 2 — Z f p

K — > U(ZG *) U{ZG)

M — U{Z[Q]G) — 2 - f U{ZjjG).

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. 3 2 where K and M are the kernels of ir : U(ZG') —* U(ZG) and a : U{Z[C]G) —►

U(ZpG) respectively. Observe that cr(K) C M, since the right-hand portion of

the diagram commutes. Also, note that K = (1 + K *) fl U(ZGm) and M =

(1 -F M*) fl U{Z[Q\G). Since a : K* —* M * is bijective, and

that cr : (1 + K *) — ► (1 -I- M*) is bijective and preserves products. Note that this

can be viewed as an isomorphism of monoids.

Theorem 3.7. a : K —► M is an isomorphism of groups.

Proof. The multiplicative monoids 1 -I- K * and 1 + M m are isomorphic under the

map a. K and M are the respective groups of units of these monoids. □

Theorem 3.8. For any odd prime p. U(Z[Gx.C p]) = K >iU(ZG) = M ~xU(ZG).

Proof. This is immediate from the diagram since if j : U(ZG) —► U(ZG') is the

inclusion map, then tt o j is the identity map. The elements of the semi-direct

product M >c U{ZG) should be viewed as ordered pairs ( u. w). where u £ M

and w € U(ZG). U k £ K and w £ U{ZG). then the isomorphism mapskw to

(a(k), w) with the action of U(ZG) on M induced by conjugation in U{ZG). □

Observe that M is a subgroup of finite index in U(Z[Q]G). Summarizing,

we have the following diagram:

K — !— U(ZGm) —^ U{ZG) p M U{Z[C]G) U{ZPG).

It should be noted that using this result to calculate is more difficult than in the

case where p = 2. Here, in order to describe M, we must work in U(Z[C}G).

At the time of this writing, the author is looking for applications for this general

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. result. It seems that knowledge of U(Z[(]G), where £ is a primitive p-th root of

unity, would be required in order to use the result. As far as the author is aware

of, not much is known about group rings, where the coefficients are from complex

integral domains [cf. 16].

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. CHAPTER IV

PROBLEMS FOR FUTURE STUDY

4.1 An Example Leading to a Conjecture

We return to an application in the case where G* = G x C2. Three of our

results (Thms. D. E, and 2.17) deal with G = S 3 , D$, and P. It was shown by

Jespers [7] that there are only four finite groups G with the property thatG has

a non-abelian free normal complement in Ui(ZG). namely G = S 3 , D&, P and

T = (a, b : a 6 = 1, a3 = b2,ba = a 5 b).

Here, we present our partial results for £/i(ZT*), where T* = T x Co

in 1993, Parmenter [13] showed that U\(ZT) = V x T. where V =

(vi, vo- 0 3 . V4 . U5) is a free group of rank five. He also gave the generators of V

to be:

Vi = 1 + (1 + a3)(—a2 + ba2)( 1 — a2)

v2 = 1 + (1 + a3) (—a2 + 6a) (1 — a2)

U3 = 1 + (1 + a3) (—a2 + 6)(1 — a2)

v4 = 1 + [—1 + (1 + a 3 )a 2 (a 2 + ba2)](l — a2)

v5 = 1 + [-1 - a2 + (1 + a3)a(l - a - 26a2)](l - a2).

Let us determine p(V). It is straight-forward to verify the following facts. First,

P(vt)p(vj) = p{vj)p(vi), where 1 < i,j < 3. Also, p(v 1)2 = p(v2 ) 2 = p(v3 ) 2 = 1

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. 35

and it is easily shown that E = (p(t^), p(v2 ).p(v3)) = C2 x C2 x C2. Now. calcula

tions show that a2p(ui)a4 = p(v2 ),a 2p(u2 )ai = p{v3), a 2p{v3 )aA = p(wt), a2p(t/i) =

p(u4), and [p(n4)]3a4 = p(v5). Thus, (a2,p(vi)) = (a2,p(u2)) = (a2,p(t;3)) = p(V).

Lemma 4.1. p(V) = E x (a2), a group of order 24.

Proof. Since i? = (p(vi). p(v2), p(v3)) is normalized by p(vi), p{v2), p(v3), and a2,

we have that E < p(V). So, E ■ (a2) < p(V). In fact, E ■ (a2) = p{V) and

E fl (a2) = 1. Thus, p(V) = E *i (a2) = [C2 x C2 x C2] x C3, a group of order

24. □

Lemma 4.2. p[Ui(ZT)\ = £ x T .

Proof. Clearly, p[Ui{ZT)\ = p(V x T) = p(V) • T. Since E < p{V), we have

that p[(7i(Zr)] = E -T. Since ap{v\)aa = p(v3), ap(v 3 )a° = p(v2 ).ap(u 2 )a° =

p(u1).6p(t'1)63 = p(vi),bp(v2 )b3 = p(v3 ).bp{v3 )b3 = p{v2), we see that is is nor

malized by T. Note that E fl T = 1. Hence, the lemma is established. □

A remark should be made at this point. Since p [U \(Z T)] has order 96. \p{V)\ = 24.

and |T| = 12, this implies that p(V) | n T\ = 3. But (a2) < p(V) fl T, where the

order of a2 is 3. Hence, p(V) PI T = (a2). Now, we have the diagram:

K —i—► U(ZTm) U(ZT) p

M — U{ZT) U(Z2 T)

M + U\{ZT) onto E x T

M+nv vonto E x (a2).

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. 36

Lemma 4.3. p{V) = {p{vx), p{v2), p{v3), p{v4)) = {[p(ui)]il • [p(n2)]i2 • [p(ua)]i3 •

[p(t?4)]14 : 0 < * 1, 221*3 < LO < *4 < 2}. Furthermore, this canonical representa

tion is unique.

Proof. Note that p(V) = (a 2 ,p(vx)) = (p(vx), p(v2), p(v3), p(v4)). Also, routine

calculations show the following:

p(v4 )p(v i) = p(v2 )a 2p(vl ) = p{v2 )2 a 2 = a 2 = p(v2 )p{v4)

p(v4 )p(v2) = p(v2 )a 2p(v2) = p{v2 )p(u3 )a 2 = p(v2 )p(v3 )p(v2 )p(v4) = p(v3 )p(v4)

p(v4)p(v3) = p{v2)a 2 p(v3) = p(v2)p(vi)a 2 = p(v2)p(vi)p(vz)p(v4) = p(vi)p(v4)

p{v4 )4 p(vi) = p(vi)p(v2 )p(v3 )p(v4 )p(vi) = p(v2 )p(v3 )p(v 4)

p(v4 )4 p(v2) = p(vx)p(v2 )p(v3 )p(v4 )p{v2) = p(vx)p{v3 )p{v4)

p{v4 )Ap{v3) = p{v\)p(v2 )p{v3 )p{v4 )p(v3) = p(vi)p(v2 )p(v4)

p{v4 fp{vx) = p{v\)p{v2 )p(v3 )p{v4 )2 p(vi) = p{v\)p{v2 )p{v4 ) 2

p{v4 fp{v2) = p{vx)p{v2 )p{v3 )p{v4 )2p(v2) = p(v2 )p{v3 )p{v4 ) 2

p(u4 )°p(v3) = p(vi)p(v2 )p(v3 )p{v4 )2p(v3) = p(yx)p(v3 )p{v 4)2.

Also, note that p(v4 )2 p(vx) = p(v3 )p(v4)2, p(v4 )2p{v2) = p{vx)p(v4)2, p(v4 )2p{v3) =

p{v2 )p{v4)2, p{v4f p{vx) = p(v2 )p{v3 ),p(v4 )3p(v2) = p(vx)p(v3), p(v4 fp(v3) = p(ux)-

p{v2). Thus, every word in p(V) can be put into the canonical form p(t*i )]11 •

[p(v2 )]12 ■ [p(n3)]13 - [p(u4)]l-*;0 < *l,* 2,*3 < 1;0 < i4 < 2. The fact that this

representation is unique follows from the fact that Ip ^ )! = 24. □

Lemma 4.4. Let w[p(v\). p(v2), p{v3)] € E.t 6 T, with w[p(vx), p(v2), p(v3)\ ■t =

1. Then, t = l^.

Proof. Suppose that w[p(vx), p(v2), p{v3)\ ■ t = 1 = E D T. Then, we have

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. w[p(v\), p(v2), p{v3)\ = t~l G T and w[p{v\), p(v2), p(v3)\ G E. This implies that

w[p(vi), p(v2), p(v3)\ G E n T = lr - Thus. w[p(vi), p(v2), p(v3)\ = 1, which implies

that t~l = 1-r- Hence, t = It- □

Theorem 4.5. M + < V xi (a2).

Proof. Suppose that w(vi, u2, u3, u4, u5) - 1 G M +, where t e T . This implies that

p[w(vi, v2, v3, V4 , v3) • f] = 1. By Lemma 4.3, we have that ([p(vi)]11 • [p(u2)]12 •

[p(u3)]‘3 • [p(u 4)]t4) ■ t = 1, where 0 < ii,i2,i3 < 1;0 < z4 < 2:t £ T. Now,

[p(xr!)]11 • [p(u2)]‘2 • [p(u3)]13 • [p(u4)]u has three possible forms:

p{vx)il ■ p(v2 ) 12 ■ p(v3 ) 13 ■ [p{v2 )a2\,if u = 1;

< p(u1)il • p(u2)'2 • p(y3)t3 • \p{v2 )p(v3 )a% if z4 = 2:

p(vi)il ■ p(v2 ) 12 - p(u3)13, if z4 = 0.

Using Lemma 4.4. ([p(ui)]‘‘ • [p(t’2)]‘2 • [p(u3)]13 • [p(u4)]‘4) • t = 1 implies that

t G (a2). □

Note that M + < p~l[E(a2)). Since p : V —* E xi (a2) is surjective. this induces an

isomorphism p : M±nV E x (a2) and hence we have that [V : M + n V] = 24.

As V is a free group of rank 5, this implies that M + D V is a free group of rank

(24)(5) — 24+1 = 97. Let F97 = M+ fl V. As a set, M + = F97 U F97xa 2 U F^ya4.

where x G V D p~l(a4),y G V fl p_1(a2). Note that [M + : M + fl V] = 3 . The

author has reason to believe that M+ is a free group, but has not yet proved this.

If this can be established, we can use the Schreier Index Formula to obtain that

M + is a free group of rank 33. This leads to the following conjecture:

Conjecture. Let T* = T x C2, where T = (a, b : a 6 = 1, a 3 = b2,ba = a 5 b).

Then, Ui(ZT') = [F33 xi F 5 ] xi T", where Fi is a free group of rank i.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. 3 8 4.2 New Directions

There axe a few natural directions in which the research can proceed. The

most obvious one is to continue to describe U(Z[G x Co]), for various G. In the

case where p is an odd prime, research into U(Z\Q\G), where £ is a primitive p-th

root of unity may prove fruitful as it may provide natural applications to Theorem

3.8. Another avenue to explore would be the case where G’ = G x Cn, where n

is composite. In a slightly different direction, one could try to develop analogous

algebraic results for G* = G xi Cp. For example, it would be interesting to see if

one can obtain Theorem C using this type of approach. It is clear that there are

many open and interesting questions in this rich field.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. REFERENCES

[1] P.J. Allen and C. Hobby, A characterization of units of Z[Aa\. J. Algebra 66 (1980) 534 - 543.

[2] P.J. Allen and C. Hobby, A characterization of units in Z[SA]. Comm. Algebra 16 (1988) 1479 — 1505.

[3] S. Galovich, I. Reiner, and S. Ullom, Class groups for integral represen tations of metacyclic groups. Mathematika 19 (1972) 105 — 111.

[4] G. Higman, Units in Group Rings. D. Phil. Thesis. University of Oxford, Oxford (1940).

[5] G. Higman, The units of group rings. Proc. London Math Soc. 46 (1940) 231 - 248.

[6] I. Hughes and K.R. Pearson, The group of units of the integral group ring ZSz■ Canad. Math. Bull. 15 (1972) 529 — 534.

[7] E. Jespers, Free normal complements and the unit group of integral group rings. Proc. Amer. Math. Soc. 122 (1994) 59 — 66.

[8] E. Jespers, Bicyclic units in some integral group rings. Canad. Math. Bull. 38 (1995) 80-86.

[9] E. Jespers and M.M. Parmenter, Bicyclic units in ZSz- Bull. Belg. Math. Soc. 44 (1992) 141 - 146.

[10] E. Jespers and M.M. Parmenter, Units of group rings of groups of order 16. Glasgow Math. J. 35 (1993) 367 - 379.

[11] G. Karpilovsky, Commutative Group Algebras. Marcel Dekker, New York (1983).

[12] M.M. Parmenter, Torsion-free normal complements in unit groups of integral group rings. C.R. Math. Rep. Acad. Sci. Canada 12 (1990) 113 - 118.

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. 4 0 [13] M.M. Parmenter, Free torsion-free normal complements in integral group rings. Commun. in Algebra 21 (1993) 3611 — 3617.

[14] D.S. Passman and P.F. Smith, Units in integral group rings. J. Algebra 69 (1981) 2 1 3 -2 3 9 .

[15] J. Ritter and S.K Sehgal, Integral group rings of some p-groups. Canad. J. Math. 34 (1982) 233 - 246.

[16] J. Ritter and S.K Sehgal, Units of group rings of solvable and Frobenius groups over large rings of cyclotomic integers. J. Algebra 158 (1993) 116 - 129.

[17] S.K Sehgal, Units in Integral Group Rings. Longman Scientific & Technical, Essex (1993).

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission. IMAGE EVALUATION TEST TARGET (Q A -3 ) /

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