(2) 獻給我的父母 To my parents.

(3) 致謝. Acknowledgment 中文部分 (Chinese-for my local friends) 博士班是一個漫長的過程，除了感謝在天上的父親與一直鼓勵我的母親之外， 這一路上有著許多幫助我的人，感謝我的指導教授劉家新老師讓我認識了這個將 群與環結合的領域，並在我研究遇到瓶頸時不斷鼓勵我，才能夠在研究上一直有 些許進展，這些累積出來的經驗以至於最後在滑鐵盧大學 (University of Waterloo) 的郭文堂老師參與之下有了一個漂亮的結果。在學習的過程中，也常常想起中央 研究院數學所王姿月老師的諄諄教誨，那段在中研院的日子使我後來的求學過程 獲益良多。感謝李華介老師與夏良忠老師除了給我學習與研究的建議之外，也提 供了我部分研究經費讓我可以專心做研究。也很感謝游森棚教官將我視為自己的 學生一樣照顧我，也讓我一同參與研究討論問題 一日衛兵，終身衛兵。論文口 試當天也特別感謝成功大學黃世昌老師與彰化師範大學劉承楷老師特地從南部上 來台北擔任口試委員。 感謝在這過程中一起打拚的夥伴朋友們：胡全燊、廖信傑、洪介興、陳冠洋、 陳柏宇、鍾明廷、施翔仁、崔庭緯、周鑫壯、徐佳安、蘇爽、石毓萱、高智強、 林書愷、簡旭麟、許辰頡、姚俊丞。 感謝在博士班第六年時獲教育部核定之國立臺灣師範大學「高等教育深耕計 畫」經費補助，前往布魯塞爾自由大學 (Vrije Universiteit Brussel) 訪問 Eric Jespers 教授六個月。最後由於「嚴重特殊傳染性肺炎」而不得不提前兩週結束訪問，於 此悼念在這特別的一年中受到災難的人們。. English-for my foreign friends. I would like to thank Professor Eric Jespers for giving me a chance to visit him and his research group at Vrije Universiteit Brussel in Belgium. Thanks to Ann Kiefer for helping me find accommodation and tell me how the cafeteria works. Thanks to Ilaria Colazzo and Leo Margolis for often drinking coffee and beer with me after lunch and work. (It was a wonderful time!) Thanks to Geoffrey Janssens and Łukasz Kubat for playing crazy table tennis together. Thanks to Andreas Bächle, Mauricio Caicedo, Timothy De Deyn, Theo Raedschelders, Doryan Temmerman, Arne Van Antwerpen and Charlotte Verwimp for having you in my life. Thanks to Tran The Dung and Nguyen Duy Dat for having you in the same office room.. i.

(4) ABSTRACT. We study the multiplicative Jordan decomposition property in integral group rings. The aim of this study is to find out which integral group rings have this property. This problem was proposed by A.W. Hales and I.B.S. Passi in 1991 and it is still open now. In this dissertation, we prove that this property holds when the group is the direct product of a quaternion group of order 8 and a cyclic group of certain prime order p. We also show negative statements for some different prime numbers p. These results give a great advance of this problem. Additionally, we study the nilpotent decomposition property in integral group rings where this concept comes from the multiplicative Jordan decomposition property. Moreover, this research leads us to another problem that when a rational group algebra of a finite group has only one Wedderburn component which is not a division ring. We classify these rational group algebras for finite SSN groups. Two related conjectures are presented in the content. Keywords: multiplicative Jordan decomposition, integral group ring, rational group algebra, Wedderburn component, Shoda pair, strong Shoda pair, nilpotent decomposition, SN group, SSN group.. ii.

(5) CONTENTS. Acknowledgment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. i. Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. ii. Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. iii. 0. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 1. I. Jordan Decomposition in Integral Group Rings . . . . . . . . . . . . . . . . I.1 Prerequisite . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I.2 Jordan Decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . I.2.1 Matrix Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . I.2.2 Group Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . I.3 Additive Jordan Decomposition in Integral Group Rings . . . . . . . . . I.4 Multiplicative Jordan Decomposition in Integral Group Rings . . . . . . . I.4.1 2-Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I.4.2 3-Groups and {2, 3}-Groups . . . . . . . . . . . . . . . . . . . . I.4.3 Groups C p oCnk . . . . . . . . . . . . . . . . . . . . . . . . . . I.5 Multiplicative Jordan Decomposition in Z[Q8 ×C p ] . . . . . . . . . . . . I.5.1 Nilpotent Elements in Q[Q8 ×C p ] . . . . . . . . . . . . . . . . . I.5.2 Semisimple and Nilpotent Parts . . . . . . . . . . . . . . . . . . I.5.3 The Main Theorem . . . . . . . . . . . . . . . . . . . . . . . . . I.5.4 A Proof for p = 5 . . . . . . . . . . . . . . . . . . . . . . . . . . I.5.5 The Density of Primes . . . . . . . . . . . . . . . . . . . . . . . I.6 Multiplicative Jordan Decomposition in Z[Q8 ×C p ], II . . . . . . . . . . I.7 Remaining Problems and Wedderburn Components . . . . . . . . . . . .. 6 7 10 13 18 23 25 27 29 29 31 31 33 37 41 44 46 52. II. Nilpotent Decomposition in Integral Group Rings . . . . . . . II.1 Prerequisite . . . . . . . . . . . . . . . . . . . . . . . . II.2 Nilpotent Decomposition Property, SN and SSN Groups II.2.1 Nilpotent Decomposition . . . . . . . . . . . . . II.2.2 SN Groups and SSN Groups . . . . . . . . . . . II.2.3 Examples . . . . . . . . . . . . . . . . . . . . .. 54 55 59 59 61 63. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..

(6) Contents. II.3 Nilpotent Decomposition for Nilpotent SSN Groups . . . II.3.1 p-Groups . . . . . . . . . . . . . . . . . . . . . II.3.2 Non-p-Groups . . . . . . . . . . . . . . . . . . II.4 Nilpotent Decomposition for Non-nilpotent SSN Groups II.4.1 Solvable SSN Groups with ND . . . . . . . . . . II.4.2 Solvable SSN Groups with ND, II . . . . . . . . II.4.3 Non-solvable SSN Groups . . . . . . . . . . . .. iv. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. . . . . . . .. 67 67 75 78 78 82 89. III. Concluding Remark and Future Work . . . . . . . . . . . . . . . . . . . . .. 91. Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 96. Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.

(7) 0. INTRODUCTION. Multiplicative Jordan Decomposition Let F be a perfect field and A be a finite dimensional F-algebra. It is well-known that every element α of A can be written uniquely in the form α = αs + αn where αs is semisimple (that is, it satisfies an F-polynomial with no multiple roots), αn is nilpotent and αs αn = αn αs . This is the Jordan decomposition for the algebra A. Now, assume F = Q and A = Q[G], a rational group algebra, for a finite group G. Consider the integral group ring Z[G] which is a subring of Q[G]. Then every element of Z[G] has the Jordan decomposition in Q[G]. The integral group ring Z[G] is said to have the additive Jordan decomposition property, AJD for short, if for every α ∈ Z[G], both αs , αn ∈ Z[G]. In other words, Z[G] has its own Jordan decomposition in itself. For instance, if G is abelian, then Z[G] has AJD since there is no nonzero nilpotent elements in Z[G] (see Theorem I.1.3). On the contrary, Z[D8 ] does not have AJD where D8 is the dihedral group of order 8. The integral group rings of finite groups with AJD have been classified by A.W. Hales, I.S. Luthar and I.B.S. Passi in [HLP90] and [HP91] (Theorem I.3.1). In the paper [HP91], Hales and Passi introduced another concept of classification of finite groups via focusing on the Jordan decomposition of units of integral group rings. This classification is one of main topics in this dissertation. We explain it below. Let G be a finite group and let U (Z[G]) be the unit group of Z[G]. For any unit α ∈ U (Z[G]), we still have the decomposition α = αs + αn in Q[G]. One can show that αs is also a unit in Q[G]. Thus, α = αs αu where αu = 1 + αs−1 αn is a unipotent unit (that is, αu is a unit and αu − 1 is nilpotent) in Q[G] and αs αu = αu αs . It follows that every unit can be decomposed into a product of a semisimple unit and a unipotent unit. This decomposition is unique in the following sense: if α = SU where S,U ∈ U (Q[G]), S is a semisimple unit, U is a unipotent unit and SU = US, then S = αs and U = αu . In general, αs and αu do not lie in U (Z[G]). The integral group ring Z[G] is said to have the multiplicative Jordan decomposition, MJD for short, if for every α ∈ U (Z[G]), both αs , αu ∈ U (Z[G]). Clearly, if Z[G] has AJD, then Z[G] have MJD. The converse is not true. For example, we have mentioned that Z[D8 ] does not have AJD but it has MJD (Theorem I.4.2). Thus, it is more interesting to characterize integral group rings of finite groups with MJD. In 2007, Hales, Passi and L.E. Wilson gave an important direction to this problem: Theorem ([HPW07, Theorem 29]). Let G be a finite group such that Z[G] has MJD. Then 1.

(8) 0. Introduction. 2. one of the following holds. (i) G is either abelian or of the form Q8 × E × A where Q8 is the quaternion group of order 8, E is an elementary abelian 2-group and A is abelian of odd order such that 2 has odd multiplicative order modulo |A|; (ii) G has order 2a 3b for some nonnegative integers a, b; (iii) G = Q8 × C p for some prime p ≥ 5 where 2 has even multiplicative order modulo p; (iv) G is the split extension of C p for some prime p ≥ 5 by a cyclic group hgi of order nk for some n ∈ {2, 3}, k ≥ 1, and gn acts trivially on C p . The above theorem is a one-way result. For each class of groups, one has to check whether their integral group rings have MJD or not. First of all, the integral group rings of groups in the class (i) have MJD since their integral group rings have no nonzero nilpotent elements (Theorem I.1.3). Groups in the class (ii) have been classified by S.R. Arora, Hales, Passi, Wilson [AHP98, HPW07, HPW12], M.M. Parmenter [Par02], C.-H. Liu and D.S. Passman [LP09, LP10, Liu12, LP13, LP14]. Most of groups in the class (iv) are classified in [AHP98, LP13, HP17]. Some groups are still open for the MJD problem. For the groups in the class (iii), there is only one result for p = 5 by X.-L. Wang and Q.-X. Zhou [WZ17] that Z[Q8 ×C5 ] has MJD. In this dissertation, we give a new result for the groups Q8 ×C p . Specifically, we will show the following theorem which is a joint work with Wentang Kuo. Theorem I.5.14 ([KS19, Theorem 4.4]). If the multiplicative order of 2 modulo p is congruent to 2 modulo 4, then Z[Q8 ×C p ] has MJD. We have an immediate consequence. Corollary I.5.15 ([KS19, Corollary 1.2]). If the multiplicative order of 2 modulo p is even and p ≡ 3 (mod 4), then Z[Q8 ×C p ] has MJD. Moreover, we also prove that Z[Q8 ×C5 ] has MJD (Theorem I.5.17) where the proof is different from [WZ17]. We will illustrate that Z[Q8 × C p ] does not have MJD for several primes p (Corollary I.6.5). Thus, we think that the MJD property may be completely different for the remaining primes. In other words, Conjecture 1. If p > 5 and the multiplicative order of 2 modulo p is congruent to 0 modulo 4, then Z[Q8 ×C p ] does not have MJD..

(9) 0. Introduction. 3. Note that for a finite group G, Q[G] has the Wedderburn decomposition such that Q[G] is isomorphic to a direct sum of matrix rings over division rings. A Wedderburn component which is not a division ring is called a matrix component for convenience. We will list the Wedderburn decomposition of Q[G] for which Z[G] has MJD as we have known so far. One can observe that those Q[G] has at most one matrix component. Based on this observation, we conjecture that Conjecture 2. If Z[G] has MJD, then Q[G] has at most one matrix component. The converse is not true. For example, Q[A4 ] has only one matrix component but Z[A4 ] does not have MJD. Nilpotent Decomposition The above conjecture relates to a certain property in integral group rings. Specifically, an integral group Z[G] is said to have the nilpotent decomposition property, ND for short, if for every nilpotent α ∈ Z[G] and every central idempotent e ∈ Q[G], we have αe ∈ Z[G]. If Z[G] has ND and α ∈ Z[G] is nilpotent, then there are finitely many orthogonal nilpotent elements αi ∈ Z[G] such that α = ∑i αi . Those αi are taken from Wedderburn components. Clearly, if Q[G] has no matrix component or it has exactly one matrix component, then Z[G] has ND. The ND property is widely used to study the MJD problem by the following result. Theorem ([HPW07, Corollary 9]). If Z[G] has MJD, then Z[G] has ND. The ND property leads to study a special kind of groups. A group G is said to have SN if for every subgroup Y of G and every normal subgroup N of G, we have either Y ⊇ N or Y N E G. A group is said to have SSN if all its subgroups have SN. The following result comes from [LP09, Proposition 2.5] and [Liu12, Lemma 1.2]. Theorem. If Z[G] has ND, then G has SN. Thus, if Z[G] has MJD, then G has SSN. The second statement of the previous theorem holds because the MJD property is inherited by subgroups. Finite SSN p-groups have a nice group structure. This provides a more effective way given by Liu and Passman to classify 2-groups and 3-groups with the MJD property. Finite SN groups is a big class of finite groups. For example, finite simple groups are SN groups. In [LP16], Liu and Passman classify finite SSN groups with nice descriptions (see also Proposition II.2.15, II.2.16, Theorem II.2.17 and Theorem II.2.18). Using those descriptions, we classify SSN groups G such that Q[G] has only one matrix component. (Note that Q[G] has no matrix component if and only if it has no nonzero nilpotent elements. Such finite groups have been classified by Theorem I.1.3.) This is a joint work with Eric Jespers when the author visited him at Vrije Universiteit Brussel in Belgium. Specifically, we have the following results..

(10) 0. Introduction. 4. Theorem II.3.13 ([JS]). Let G be a finite nilpotent SSN group. Then Q[G] has only one matrix component if and only if G is one of the following cases. (a) G is a p-group and one of the following holds: m. n. m−1. (i) G = ha, b | a p = b p = 1, ab = a1+p n = 1, or p = m = n = 2;. i where p ≥ 2 is a prime, m ≥ 2 and. (ii) G = G0 ∗ Z, the central product of a nonabelian group G0 of order p3 with a cyclic group Z, where G0 ∩ Z = Z (G0 ) and if p = 2, then |Z| > 2; (iii) G = Q8 ×C4 ; (iv) G = Q16 ; (v) G = D8 ∗ Q8 , an extraspecial 2-group of order 25 . (b) G is not a p-group and G = Q8 ×C p for some odd prime p such that the multiplicative order of 2 modulo p is even. Theorem II.4.19 ([JS]). Let G be a finite non-nilpotent SSN group. Then Q[G] has only one matrix component if and only if G is one of the following: (i) G = C p oCn where Cn acts faithfully on C p ; (ii) G = Cnp oCq where n ≥ 2, pn − 1 = (p − 1)q and Cq acts nontrivially on Cnp ; (iii) G = C p o hyiqk where k ≥ 2, y acts nontrivially on C p , yq acts trivially on C p and one of the following holds: (a) q 6= 2, q | p − 1 and q2 - p − 1; (b) q = 2 and either k = 2, or k > 2 and p ≡ 5 (mod 8). Here p, q are distinct primes and n, k ∈ N. We also determine the structure of the matrix component for each group above. According to Corollary II.4.5 and Corollary II.3.14, we observe that for most finite SSN groups G, if Z[G] has ND, then Q[G] has at most one matrix component. Hence, it seems that these two concepts are equivalent: Conjecture 3. Z[G] has ND if and only if Q[G] has at most one matrix component. Clearly, Conjecture 3 implies Conjecture 2. Guideline The structure of this dissertation is as follows. Chapter I contains the topics of Jordan decomposition. We present some basic definitions, notations and well-known theorems.

(11) 0. Introduction. 5. in Section I.1. In Section I.2, we give the notions of the original Jordan decomposition, additive Jordan decomposition and multiplicative Jordan decomposition for general situations. Some well-known results for matrix ring and group rings are also presented. We survey the developments of additive Jordan decomposition and multiplicative Jordan decomposition in integral group rings in Section I.3 and I.4, respectively. The new results for integral group rings Z[Q8 ×C p ] are presented in Section I.5 and I.6. In Section I.7, we list the Wedderburn decomposition of Q[G] where Z[G] is already known to have MJD. This gives us the motivation of the next chapter. We study the ND property, SN groups and SSN groups in Chapter II. Some necessary knowledge are introduced in the first two sections II.1 and II.2. We classify finite SSN groups G such that Q[G] has only one matrix component in the last two sections. We separate the classification into Section II.3 for nilpotent SSN groups and Section II.4 for non-nilpotent SSN groups. In Chapter III, we list several related problems and a figure to illustrate the relations between MJD, ND, SN, SSN and Q[G] with at most one matrix component..

(12) I. JORDAN DECOMPOSITION IN INTEGRAL GROUP RINGS. In this chapter, we focus on the Jordan decomposition in integral group rings. In the first section, we give some basic definitions, notations and well-known results for group rings. We recall the classical Jordan decomposition and introduce the general concept of additive Jordan decomposition property and multiplicative Jordan decomposition property in Section I.2. In that section, we also focus on those two properties in matrix rings Mn (R) and group rings R[G] in two subsections, respectively, where R is an integral domain of characteristic 0 and G is a finite group. The additive Jordan decomposition problem in integral group rings Z[G] of finite groups G has been solved and we survey its development in Section I.3. The multiplicative Jordan decomposition problem in integral group rings of finite groups has not been completely solved yet and it is the main topic in this chapter. In Section I.4, we survey the development of this topic and present well-known results in three subsections. In Section I.5, we give a new result that the multiplicative Jordan decomposition property holds in Z[Q8 × C p ] for certain primes p. This is a joint work with Wentang Kuo. Moreover, we also give opposite results for several primes in Section I.6. However, this problem is still open. In the final section, we list the remaining finite groups for this problem. We observe the Wedderburn decomposition of Q[G] where Z[G] is already showed to have the multiplicative Jordan decomposition property. Furthermore, we give a conjecture from this observation and this conjecture relates to the next chapter.. 6.

(13) I. Jordan Decomposition in Integral Group Rings. I.1. 7. Prerequisite. We introduce some definitions and notations. Let G be a group and R be a ring with the unity 1R . Unless otherwise stated, all the rings in this dissertation will be assumed to be rings with unity. The so-called group ring R[G] is the free left R-module with basis G equipped the “multiplication” defined by !. !. ∑ ag g. g∈G. ! =. ∑ bg g. g∈G. ∑ ∑ asbt. g∈G. g. st=g. where ag = 0 and bg = 0 for all but finitely many ag , bg ∈ R. It is easy to check that R[G] forms a unitary ring with the unity 1 = 1R[G] = 1R 1G where 1G is the identity of G. If R is a field, R[G] is also called a group algebra. Moreover, when R = Q (resp. R = Z), R[G] is called a(n) rational group algebra (resp. integral group ring). For any element α = ∑g∈G ag g of R[G], the support of α is a subset of G defined by supp(α) = {g ∈ G | ag 6= 0} and the augmentation map aug : R[G] → R is a ring homomorphism defined by aug(α) =. ∑ ag .. g∈G. The kernel of the augmentation map is called the augmentation ideal and denoted by ω(R[G]). If aug(α) = 1R , then α is said to have augmentation 1. If H is a finite subgroup of G, we denote b= ∑h H h∈H. and. e = 1 ∑ h. H |H| h∈H. Similarly, if g is a torsion element of G, then we denote c gb = hgi and f ge = hgi. Let U (R[G]) be the unit group of R[G] and U1 (R[G]) be the group of units in R[G] with augmentation 1. Recall that a left R-module M is called semisimple if every left submodule of M is a direct summand. Moreover, R is called semisimple if the regular left module R R is.

(14) I. Jordan Decomposition in Integral Group Rings. 8. semisimple. We present some well-known results below. Theorem I.1.1 (Wedderburn-Artin Theorem [PS02, Theorem 2.6.18 and 19]). A ring R is semisimple if and only if it is a direct sum of matrix algebras over division rings: R ' Mn1 (D1 ) ⊕ · · · ⊕ Mnr (Dr ) for some division rings D1 , . . . , Dr . Moreover, if R ' Mn1 (D1 ) ⊕ · · · ⊕ Mnr (Dr ) ' Mm1 (D01 ) ⊕ · · · ⊕ Mms (D0s ) where Di and D0j , 1 ≤ i ≤ r, 1 ≤ j ≤ s, are division rings. Then r = s and, after a suitable permutation of indices, we have that ni = mi and Di ' D0i . The above decomposition is called the Wedderburn decomposition of R and integers ni are called the reduced degrees of a semisimple ring. The following theorem states sufficient and necessary conditions that a group ring is a semisimple ring. Theorem I.1.2 (Maschke’s Theorem [PS02, Theorem 3.4.7]). Let G be a group and R be a ring. Then R[G] is semisimple if and only if the following conditions hold: (i) R is semisimple; (ii) G is finite; (iii) |G| is invertible in R. According to the previous theorem, every group algebra Q[G] is semisimple for every finite group G. Thus, Q[G] contains some matrix rings as subrings. It follows that Q[G] may contain some nonzero nilpotent elements. The following theorem gives a classification of finite groups G such that Q[G] does not have nonzero nilpotent elements. Theorem I.1.3 ([Seh78, Theorem VI.1.17] or [PS02, Theorem 7.4.11]). Let G be a finite group of order 2k m with odd m. Then Q[G] has no nonzero nilpotent elements if and only if G is either abelian or G = Q8 × E × A where Q8 is the quaternion group of order 8, E is an elementary abelian 2-group and A is an abelian group of order m such that the multiplicative order of 2 modulo m is odd. The above groups have no non-normal subgroups. Indeed, if G is a finite group and H is a non-normal subgroup, then g−1 hg 6∈ H for some h ∈ H and g ∈ G. It follows that b is a nonzero nilpotent element of Q[G]. A group (not necessarily the element (1 − h)gH finite) is called a Dedekind group if all its subgroups are normal. A nonabelian Dedekind group is also called a Hamiltonian group. Such groups have a nice description..

(15) I. Jordan Decomposition in Integral Group Rings. 9. Theorem I.1.4 ([Bae33]). A group G (not necessarily finite) is Dedekind if and only if G is either abelian or G = Q8 × E × A where E is an elementary abelian 2-group and A is an abelian group whose elements are of odd order..

(16) I. Jordan Decomposition in Integral Group Rings. I.2. 10. Jordan Decomposition. Let F be a field and V be a finite dimensional vector space over F. A linear operator φ of V is called semisimple if its minimal polynomial over F contains no repeated (or multiple) roots in F, the algebraic closure of F. For example, any central linear operator of EndF (V ) is semisimple since it is a scalar map. Note that a linear operator φ of V can induce a linear operator φ of V = F ⊗F V . Then φ is semisimple if and only if φ is “diagonalizable,” that is, V is spanned by eigenvectors of φ . We have the following useful property. Lemma I.2.1. Let φ , ψ be two semisimple linear operators of V . If φ ψ = ψφ , then both φ ψ and φ + ψ are semisimple. In particular, φ k and kφ are semisimple for every k ∈ N. Proof. If φ ψ = ψφ , then φ ψ = ψ φ . By the simultaneous diagonalization ([Gol12, Proposition 12.18]), V is spanned by common eigenvectors of φ and ψ . The result follows. The following theorem is a classical result and the reader can refer to one of the references [Bor91, p. 80], [Bou58, Chap. 7, Sec. 5 and Chap. 8, Sec. 9], [BC74, Proposition], [Gre75, Chap. XIII], [HK71, Chap. 7]and [SW86, p. 83] Theorem I.2.2 (Jordan Decomposition). Let φ be a linear operator of V into itself. If F is a perfect filed, then φ = φs + φn where φs and φn are unique linear operators such that φs is semisimple, φn is nilpotent and φs φn = φn φs . In particular, both φs and φn are F-polynomials in φ . The decomposition φ = φs + φn is called the Jordan decomposition of φ . We also call it the additive Jordan decomposition. We remark that if F is not perfect, the Jordan decomposition may fail. See the following example which was mentioned from [BC74]. Example I.2.3. Consider the field F = F2 (x) of rational functions over the field F2 of two elements. F is not perfect because x is transcendental over F2 and it can not be a square in F. Let V be an F-vector space of dimension 4 with basis {vi }4i=1 and define the linear operator φ of V by φ (v1 ) = v2 ,. φ (v2 ) = xv1 ,. φ (v3 ) = v4 ,. φ (v4 ) = v1 + xv3 .. We claim that φ does not have the Jordan decomposition. Note that φ 2 = xI + N where I is the identity map and N(v1 ) = N(v2 ) = 0, N(v3 ) = v1 and N(v4 ) = v2 . In particular, N 2 = 0. Suppose that φ has the Jordan decomposition, says φ = φs + φn . Then φ 2 = φs2 + φn2 since φs φn = φn φs and the characteristic of F2 is 2. Note also that φs2 is semisimple (Lemma I.2.1), φn2 is nilpotent and φs2 φn2 = φn2 φs2 ..

(17) I. Jordan Decomposition in Integral Group Rings. 11. By the uniqueness of Jordan decomposition, φs2 = xI and φn2 = N. Assume that φs = f0 + f1 φ + · · · + fm φ m for some fi ∈ F, and we have φs2 = f02 + f12 φ 2 + · · · + fm2 φ 2m = f02 + f12 (xI + N) + · · · + fm2 (xI + N)m . If i is even, then (xI + N)i = xi I; if i is odd, then (xI + N)i = (xI + N)i−1 (xI + N) = 2 i−1 N. Again, by the 2 i xi−1 I(xI + N) = xi I + xi−1 N. Thus, φs2 = ∑m i=0 f i x I + ∑odd i f i x uniqueness, we obtain that m. ∑ fi2xi = x. i=0. and. ∑. fi2 xi−1 = 0.. odd i. We then deduce that ∑odd i fi2 xi = 0 and so ∑even i fi2 xi = x. However, this means that x is a square of some element in F. As we mentioned in the beginning, x can not be a square, i.e., φ can not have the Jordan decomposition. The minimal polynomials of φ and φs have a strong connection by the following lemma which will be useful later. Lemma I.2.4. If minF (φ ) = ∏i ri (x)ni is the decomposition of distinct irreducible polynomials ri (x) of F[x] with ni ≥ 1, then minF (φs ) = ∏i ri (x). In particular, the eigenvalues of φ and φs (if exist) coincide. Proof. We first note that if g(x), f (x) ∈ F[x], then g(x − f (x)) = g(x) + f (x)h(x) for some h(x) ∈ F[x]. Let r(x) = ∏i ri (x) and φn = f (φ ) for some f (x) ∈ F[x]. Then it follows that r(φs ) = r(φ − φn ) = r(φ ) + f (φ )h(φ ) = r(φ ) + φn h(φ ) for some h(x) ∈ F[x]. Note that r(φ )m = 0 when m ≥ ni for all i, i.e., r(φs ) is both semisimple and nilpotent. Thus, r(φs ) = 0, i.e., minF (φs ) | r(x). Let m(x) = minF (φs ). Similarly, 0 = m(φs ) = m(φ −φn ) = m(φ ) + φnt(φ ) for some t(x) ∈ F[x]. Thus, m(φ ) = −φnt(φ ) is nilpotent and we have minF (φ ) | m(x)k for some k ∈ N. Since m(x) | r(x) = ∏i ri (x), it follows that m(x) must contain all factors ri (x) and then m(x) = r(x). According to Theorem I.2.2, φs and φn are F-polynomials in φ . Here we recall a method to find φs when char(F) = 0 and φ is given. This method is referred to [SW86, Proof of 3.1.6]. For arbitrary char(F), see [Bor91, 4.2 Proposition], [HK71, Theorem 13 in Chapter 7] or [Hum72, Proposition in Section 4.2]. Assume that char(F) = 0. Let f (X) ∈ F[X] be the minimal polynomial of φ and write f (X) = ∏ti=1 qi (X)ei be the factorization in F[X] into irreducible factors qi (X). Let g(X) = ∏ti=1 qi (X). Then gcd(g(X), g0 (X)) = 1 since char(F) = 0. There exist polynomials h(X) and k(X) in F[X] such that g(X)k(X) + g0 (X)h(X) = 1..

(18) I. Jordan Decomposition in Integral Group Rings. 12. Choose an integer m such that 2m ≥ ei for i = 1, . . . ,t. Then φs = (φ − g(φ )h(φ ))m . This construction will be used in the proof of Theorem I.2.7. Next, we give the concept of “multiplicative” Jordan decomposition for invertible linear operators. Let φ be an invertible linear operator of V . Note that φ −1 φn = φn φ −1 because φn is an F-polynomial in φ . Thus, φ −1 φn is still nilpotent and 1 − φ −1 φn is invertible. It follows that φs = φ (1 − φ −1 φn ) is invertible. Let φu = 1 + φs−1 φn . Then φu = φs−1 φ is invertible and φu − 1 is nilpotent. The invertible operator φu is said to be unipotent. We call φ = φs φu the multiplicative Jordan decomposition of φ . This decomposition is unique in the following sense: if φ = SU for some invertible semisimple operator S and invertible unipotent operator U such that SU = US, then S = φs and U = φu . To see this, we can write φ = S + S(U − 1). Since S(U − 1) is nilpotent and it commutes with S, we have S = φs and S(U − 1) = φn by the uniqueness of additive Jordan decomposition. Thus, U = S−1 φn + 1 = φu . We can write the Jordan decomposition of φ −1 in φ , φs and φn . Note that φ −1 = φs−1 φu−1 = φu−1 φs−1 . Moreover, φs−1 is semisimple and φu−1 −1 = φu−1 (1−φu ) is nilpotent. Thus, by uniqueness, (φ −1 )s = φs−1. and (φ −1 )u = φu−1 .. In particular, (φ −1 )n = (φ −1 )s ((φ −1 )u − 1) = (φ −1 )s φu−1 (1 − φu ) = −φ −1 φs−1 φn .. Let F be a field and let A be a finite dimensional F-algebra. Then every element α of A can be regarded as a linear operator lα of A by the left multiplication. Here l : A → EndF (A) is an injective F-algebra homomorphism. Then lα = (lα )s + (lα )n where (lα )s and (lα )n are linear operators of A which are also polynomials in lα over F by Theorem I.2.2. Thus, both (lα )s and (lα )n have the form lβ and lγ , respectively, for some β , γ ∈ A. It follows that β satisfies an F-polynomial without multiple roots, γ is a nilpotent element and β γ = γβ since l is injective. Moreover, we have α = β + γ. Let r : A → EndF (A) be the right multiplication of A. If m(X) ∈ F[X] is the minimal polynomial of β , then m(rβ ) = rm(β ) = 0. It follows that rβ satisfies an F-polynomial without multiple roots. In other words, rβ is semisimple. Moreover, rγ is nilpotent and.

(19) I. Jordan Decomposition in Integral Group Rings. 13. rβ rγ = rγ rβ . Since rα = rβ + rγ , it follows that (rα )s = rβ and (rα )n = rγ by the uniqueness of Jordan decomposition. In this sense, β and γ are independent of the left multiplication or the right multiplication for every element α ∈ A. The element β is also said to be a semisimple element. Such β and γ are unique by the uniqueness of (lα )s and (lα )n . Hence, α = β + γ is the Jordan decomposition of the element α and we denote αs = β and αn = γ. These two elements αs and αn are called the semisimple part and the nilpotent part of α, respectively. When α is a unit, the unipotent unit (unipotent element) αu = 1 + αs−1 αn is called the unipotent part of α. Definition I.2.5. Let F be a field, A be a finite dimensional F-algebra and B be a subring of A. For every β ∈ B, we have β = βs + βn for some unique elements βs , βn ∈ A. We say that an element β ∈ B has the additive Jordan decomposition property (AJD) if both βs , βn ∈ B. The ring B is said to have the additive Jordan decomposition property (AJD) if every element of B has AJD. Similarly, we say that a unit β ∈ B has the multiplicative Jordan decomposition property (MJD) if both βs , βu ∈ B. The ring B is said to have the multiplicative Jordan decomposition property (MJD) if every unit of B has MJD. Lemma I.2.6. Let F, A and B be as in the previous definition. Let β ∈ U (B). Then the following are equivalent. (i) β has MJD. (ii) Both βs , βn ∈ B. (iii) Both βs , βu ∈ U (B). Proof. Assume that β has MJD. Then both βs , βu ∈ B. So βn = βs (βu −1) ∈ B. Moreover, we have βs−1 = βu β −1 ∈ B and βu−1 = β −1 βs ∈ B. Thus, both βs and βu are also units in B. In other words, (i) implies (ii) and (iii). Now assume that βs , βn ∈ B. Note that βs−1 is an F-polynomial in βs . Thus, βs−1 ∈ B. Then βu = 1+βs−1 βn ∈ B. So (ii) implies (i). Clearly, (iii) implies (i) since U (B) ⊆ B. The condition (iii) above means that the MJD property is a local property for the group of units. The AJD problem or the MJD problem for a subring B is to ask when B has the AJD property or the MJD property. It is easy to see that if B has AJD, then it also has MJD. However, the converse is not true in general. We will see some examples for matrix rings and group rings later. I.2.1. Matrix Rings. In this subsection, we will mention some well-known results for matrix rings having the AJD property or the MJD property. Let R be an integral domain of characteristic 0 which.

(20) I. Jordan Decomposition in Integral Group Rings. 14. is integrally closed in its quotient field F. Then every element in Mn (F), the ring of n × n matrices over F, has the Jordan decomposition by Theorem I.2.2. By Definition I.2.5, an element α of Mn (R) is said to have AJD if both semisimple and nilpotent parts of α are in Mn (R); and Mn (R) has AJD (resp. MJD) if every element (resp. every unit) has AJD. Clearly, AJD implies MJD. The following theorem will be used to the AJD problem and the MJD problem for group rings. For convenience to the reader, we present the proof. Theorem I.2.7 ([HLP90, p. 2333] and [AHP98, Proposition 3.1]). Assume R 6= F. Then (i) AJD holds in M1 (R) and M2 (R) and fails in Mn (R) for n ≥ 3. (ii) MJD holds in M1 (R) and M2 (R); holds in M3 (R) if and only if the units of R, together with 0, form a subfield, and fails in Mn (R) for n ≥ 4. Proof. When n = 1, M1 (R) has both AJD and MJD since there is no non-zero nilpotent element in M1 (F) = F. (Every element is semisimple.) Assume n = 2. Let α ∈ M2 (R) be non-semisimple. Then the minimal polynomial of α over F has a repeated root in the algebraic closure F, says (X − λ )2 for some λ ∈ F. This is also the characteristic polynomial of α over F so λ = tr(α)/2 ∈ F. Moreover, (αs − λ I)3 = ((α − λ I) − αn )3 = 0 since αn2 = 0 and ααn = αn α. It follows that the minimal polynomial of αs over F is X − λ by Lemma I.2.4. Thus, αs = λ I. Note that λ 2 = det(α) ∈ R or, equivalently, λ satisfies the polynomial X 2 − det(α) ∈ R[X] which implies that λ is integral over R. Since R is integrally closed in F, we have λ ∈ R. Hence, αs ∈ M2 (R). It follows that M2 (R) has AJD and so MJD. Now, assume n = 3. Choose r ∈ R such that r−1 6∈ R. Consider that . 0 1 0 α = 0 0 1 ∈ M3 (R). 0 0 r Then αs = r−1 α 2 and αn = α − r−1 α 2 . Indeed, α satisfies X 2 (X − r) which implies that r−1 α 2 satisfies X(X − r) and α − r−1 α 2 satisfies X 2 . Both are polynomials in α over F. However, 0 0 r−1 αs = r−1 α 2 = 0 0 1 6∈ M3 (R). 0 0. r. Thus, M3 (R) (and hence Mn (R) for n > 3) does not have AJD. We have completed (i) and parts of (ii). We now consider the MJD problem for n = 3 and n ≥ 4. Assume n = 3 and we are going to prove that M3 (R) has MJD if and only if, for every a, b ∈ U (R), a 6= b, a − b ∈ U (R). Let α ∈ M3 (R) and let f (X) be its characteristic polynomial. If f (X) has no repeated root, then α is semisimple. If f (X) = (X − a)3 for some a ∈ F, then the coefficient of X 2 is −3a ∈ F. Thus, a ∈ F. So the minimal.

(21) I. Jordan Decomposition in Integral Group Rings. 15. polynomial of αs is X − a by Lemma I.2.4. Thus, αs = aI. Moreover, a = tr(α)/3 ∈ F, a3 = − det(α) ∈ R and R is integrally closed in F. It follows that a ∈ R and αs ∈ M3 (R). Consequently, for M3 (R) having MJD or not, it only depends on elements with exactly two distinct eigenvalues. Assume further that α is invertible with f (X) = (X − a)2 (X − b) for some a, b ∈ F with a 6= b. We claim that a, b ∈ U (R). Since f (X) has a repeated root and char(F) = 0, it follows that f (X) is not irreducible in F[X]. Thus, either a or b is in F. Note that tr(α) = 2a + b ∈ R. We obtain that both a, b are in F. Moreover, a, b satisfy the monic polynomial f (X) in R[X]. We can conclude that a, b ∈ R since R is integrally closed in F. Now, α is invertible in M3 (R) which implies that a, b ∈ U (R) since det(α) = a2 b and det(α −1 ) ∈ R. Let c = (a − b)−1 in F. One can check that α − c(α − aI)(α − bI) satisfies the polynomial (X − a)(X − b) where c(α − aI)(α − bI) is nilpotent. Thus, αs = α − c(α − aI)(α − bI) and αn = c(α − aI)(α − bI). Hence, if a − b ∈ U (R) for every distinct a, b ∈ U (R), then M3 (R) has MJD. Conversely, if M3 (R) has MJD, then the invertible matrix . a 1 0 α = 0 a 1 0 0 b for distinct a, b ∈ U (R) has the nilpotent part 0 1 (a − b)−1 αn = 0 0 0 ∈ M3 (R). 0 0 0 . Hence, a − b ∈ U (R). Finally, we assume n = 4. Choose c ∈ R such that c2 − 4 6= 0 and c − 2 6∈ U (R). Such element c exists. Otherwise, if c − 2 ∈ U (R) for every c 6= ±2, then we obtain r = (r + 2) − 2 ∈ U (R) for each r ∈ R \ {0, −4}. Moreover, 4 = 6 − 2 ∈ U (R) so −4 ∈ U (R). Thus, R = F, a contradiction. Now, c − 2 6∈ U (R) implies c2 − 4 = (c − 2)(c + 2) 6∈ U (R). Consider the matrix 0 1 1 1 −1 −c 0 1 ∈ M4 (R) α = 0 0 0 1 0 0 −1 −c with the characteristic polynomial (X 2 + cX + 1)2 . Note that it is also the minimal poly-.

(22) I. Jordan Decomposition in Integral Group Rings. 16. nomial of α since α 2 + cα + I 6= 0. We can compute that 2 αs = α − d(α + cα + I)(2α + cI) 0 1 cd (c2 − 2)d −1 −c −2d −cd 6∈ M4 (R) = 0 0 0 1 0 0 −1 −c. where d = (c2 − 4)−1 . Therefore, M4 (R) (and hence Mn (R) for n > 4) does not have MJD. We remark from the above proof that AJD fails in M3 (R) and MJD fails in M4 (R) even if R is not integrally closed in F. Example I.2.8. If R = OE is the ring of integers of an algebraic number field E, then U (R) ∪ {0} can not form a field. Otherwise, U (R) ∪ {0} would contain Q but R ∩ Q = Z. Thus, M3 (OE ) does not have MJD. If R = K[X] for some field K, then R is integrally closed in K(X) and U (R) ∪ {0} = K. Thus, M3 (K[X]) has MJD. Remark I.2.9. (i) Recall that if f (X) = ∏i qi (X)ei is the minimal polynomial of α and g(X) = ∏i qi (X) where qi (X) are distinct irreducible polynomials, then the semisimple part αs = (α − g(α)h(α))m where 2m ≥ ei for each i and g(X)k(X) + g0 (X)h(X) = 1 for some k(X), h(X) ∈ F[X]. However, it is complicated to compute k(X) and h(X) in general. We consider a special case that we can construct αs without finding h(X). Assume that q1 (X)e1 = (x−a)2 for some a ∈ F and ei = 1 for i > 1. In this case, m = 1. Then we have (X − a)g(X) = f (X) or, equivalently, Xg(X) ≡ ag(X) (mod f (X)). Thus, g(X)h(X) ≡ g(X)h(a) (mod f (X)). Note that g(X)k(X) + g0 (X)h(X) = 1 and it follows that h(a) = g0 (a)−1 since g(a) = 0. We obtain g(X)h(X) ≡ g(X)g0 (a)−1 (mod f (X)). Consequently, we have αs = α − g(α)g0 (a)−1 because f (α) = 0. As an application, we consider another matrix to illustrate that M4 (R) does not have MJD as R 6= F. Let 0 6= r ∈ R with r 6∈ U (R) and let . 1 0 α = 0 0. 1 0 1 1 0 1+r 0 r. 0 0 ∈ M4 (R). 1 1. This matrix is referred to [HP99, p. 80]. Then α is invertible and one can check that its minimal polynomial is (X − 1)g(X) where g(X) = (X − 1)(X 2 − (r + 2)X + 1). Then.

(23) I. Jordan Decomposition in Integral Group Rings. 17. g0 (1) = −r and 0 −1 0 r−1 0 0 0 0 −1 ∈ αs = α − g(α)(−r) = α − 6 M4 (R). 0 0 0 0 0 0 0 0 We mention here that one can find k(X) = and h(X) =. 1 r3 + 4r2. 1 r3 + 4r2. (−(6r + 18)X + (4r2 + 21r + 18)). ((2r + 6)X 2 − (2r2 + 11r + 12)X + (r2 + 5r + 6)). by using the website [Wol]1 (ii) We use the idea of the previous remark (i) to find semisimple parts of another two matrices below. These two matrices will be used in the proof of Theorem I.2.11. First, consider the matrix 0 a 0 α = 0 0 a ∈ M3 (R) 0 0 ra2. for r 6= 0. Its minimal polynomial is X 2 (X − ra2 ) so g(X) = X(X − ra2 ) and g0 (0) = −ra2 . Then we have 0 0 r−1 αs = α − α(α − ra2 I)(−ra2 )−1 = (ra2 )−1 α 2 = 0 0 a ∈ M3 (F). 0 0 ra2 For another matrix, we consider a generalization of the 4 × 4 matrix in the previous remark. Let c a 0 0 0 c a 0 α = 0 0 c + ra3 a ∈ M4 (R) 2 0 0 ra c for r, a 6= 0 and ra3 6= −4. To compute αs , we write β = α − cI because it is easier to compute βs and αs = cI + βs by Lemma I.2.1. The minimal polynomial of β is Xg(X) where g(X) = X(X 2 − ra3 X − ra3 ). Note that X 2 − ra3 X − ra3 has no repeated root in F 1 In [Wol], typing the code “PolynomialExtendedGCD[m(x), n(x), x]” will present the result “{d(x), {k(x), h(x)}}” where d(x) = gcd(m(x), n(x)) = m(x)k(x) + n(x)h(x)..

(24) I. Jordan Decomposition in Integral Group Rings. 18. since ra3 6= 0, −4. Thus g0 (0) = −ra3 and, by the previous remark, . 0 0 βs = β − g(β )(−ra3 )−1 = 0 0. 0 0 r−1 0 a 0 . 3 0 ra a 2 0 ra 0. Hence, . c 0 αs = 0 0. 0 0 r−1 c a 0 ∈ M4 (F). 0 c + ra3 a 2 0 ra c. The AJD and MJD problems are difficult for group rings. One reason is that it is not easy to find the minimal polynomial of every element in group rings. So it is not easy to decompose elements into their Jordan decomposition. I.2.2. Group Rings. For any finite group G and any field F of characteristic 0, the group ring F[G] is a semisimple ring by Theorem I.1.2. Then we can ask whether R[G] has AJD or MJD if R is a subring of F. In this subsection, we focus on the reduced degrees (see the below of Theorem I.1.1) of group rings where AJD or MJD holds. We give a lemma for the next theorem. Lemma I.2.10. Let R be an integral domain with its quotient field F and R 6= F. Let D be a vector space over F and F ⊆ D. If M is a finitely generated R-submodule of D, then M can not contain F. Proof. Let M = ∑ni=1 Rxi and let B = {yλ | λ ∈ I} be a basis for D over F with y1 = 1F ∈ B where I is an index set. For each i, write xi = ∑λ ci,λ yλ for ci,λ ∈ F where ci,λ = 0 for all but finitely many λ . Suppose that F ⊆ M. Then each nonzero element a ∈ F can be written as a = ∑i ri xi = ∑λ ∑i ri ci,λ yλ for some ri ∈ R. It follows that only yλ = y1 = 1 term can be nonzero and we have a = ∑i ri ci,1 . Thus, F = ∑i Rci,1 , a finitely generated R-module. Note that there is 0 6= r ∈ R such that rci,1 ∈ R for all i. Hence, F = rF = ∑i Rrci,1 ⊆ R, a contradiction. Now the techniques in the proof of Theorem I.2.7 can be used to prove the following result. Theorem I.2.11 ([HLP90, 3.1 Theorem] and [AHP98, Theorem 4.1]). Let R be an integral domain of characteristic 0 with its quotient field F and R 6= F. Let G be a finite group.

(25) I. Jordan Decomposition in Integral Group Rings. and write F[G] '. m M. 19. Mni (Di ). i=1. as the Wedderburn decomposition where Di are division rings with center containing F. (i) If R[G] has AJD, then ni ≤ 2 for all i. (ii) If R[G] has MJD, then ni ≤ 3 for all i. Proof. (i) Assume that R[G] has AJD. Suppose that n = n1 ≥ 3 and write D = D1 . Let π : F[G] → Mn (D) be the natural projection. Note that there are some elements α, β in F[G] such that . 0 1 0 π(α) = 0 0 1 0 0 0. and. 0 0 0 π(β ) = 0 0 0 . 0 0 1. Moreover, we have aα, bβ ∈ R[G] for some nonzero a, b ∈ R since F is the quotient field of R. Let γ = ab(α + rabβ ) ∈ R[G] for r ∈ R and r 6= 0. Then . 0 ab 0 π(γ) = 0 0 ab . 0 0 ra2 b2 Note that π preserves semisimplicity, nilpotence and commutativity. Thus, (π(γ))s = π(γs ) ∈ π(R[G]) by assumption. According to Remark I.2.9(ii), we have 0 0 r−1 (π(γ))s = 0 0 ab ∈ π(R[G]). 0 0 ra2 b2 . This implies that 0 0 r−1 r0 (π(γ))s π(r0 ) = 0 0 ∗ ∈ π(R[G]) 0 0 ∗ for any r0 ∈ R. Let M = π(R[G])13 , the R-submodule of D consisting of the (1, 3)-entries of the matrices in π(R[G]). Then we have r−1 r0 ∈ M for every r, r0 ∈ R with r 6= 0. In other words, F ⊆ M. Since G is finite, it follows that M is finitely generated over R. Now, applying Lemma I.2.10, we have a contradiction. Thus, ni ≤ 2 for all i. (ii) Assume that R[G] has MJD. Suppose that n = n1 ≥ 4 and write D = D1 . As in the previous argument, we also assume that π : F[G] → Mn (D) is the natural projection. The.

(26) I. Jordan Decomposition in Integral Group Rings. 20. following argument is similar to (i). There are α, β ∈ F[G] such that . 0 0 π(α) = 0 0. 1 0 0 0. 0 1 0 0. 0 0 1 0. and. 0 0 π(β ) = 0 0. 0 0 0 0. 0 0 0 1. 0 0 , 0 0. and the images of α, β in Mni (Di ), i > 1, are 0 via the Wedderburn decomposition. Thus, α, β are nilpotent. Moreover, aα, bβ ∈ R[G] for some nonzero elements a, b ∈ R. Let γ = (1 + abα)(1 + ra2 b2 β ) where r ∈ R such that ra3 b3 6= 0, −4 and we have . 1 ab 0 0 0 1 ab 0 . π(γ) = 3 b3 ab 0 0 1 + ra 2 2 1 0 0 ra b Then γ is a unit in R[G] since 1 + abα and 1 + ra2 b2 β are unipotents. According to Remark I.2.9(ii), we have . 1 0 (π(γ))s = 0 0. 0 0 r−1 1 ab 0 . 0 1 + ra3 b3 ab 2 2 0 ra b 1. Since R[G] has MJD, we have (π(γ))s = π(γs ) ∈ π(R[G]). Let M = π(R[G])14 , the Rsubmodule of D consisting of the (1, 4)-entries of the matrices in π(R[G]). Multiplying (π(γ))s by π(r0 ) for any r0 ∈ R , we can conclude that r−1 r0 ∈ M for every r, r0 ∈ R with r 6= 0 and ra3 b4 6= −4. For the remaining r0 ∈ R such that r0 a3 b4 = −4, we observe that −r0 6= 0 and (−r0 )a3 b4 6= −4. It follows that (−r0 )−1 r0 ∈ M for all r0 ∈ R. Thus, r0−1 r0 ∈ M. Consequently, F ⊆ M. This is a contradiction by Lemma I.2.10. Therefore, we must have ni ≤ 3 for all i. Remark I.2.12. In the original result [HLP90, 3.1 Theorem], R is further assumed to be Noetherian, R ∩ Q = Z and integrally closed in F. These additional conditions can be dropped via the note below the proof of [AHP98, Theorem 4.1]. The above proof of Theorem I.2.11(i) just follows that note. See also the proof of [HP99, 5.1 Proposition]. In general, reduced degrees are divisors of the order of a given group. Lemma I.2.13. Let G be a finite group and F be a field of characteristic 0. Assume that F[G] '. m M i=1. Mni (Di ).

(27) I. Jordan Decomposition in Integral Group Rings. 21. is a Wedderburn decomposition where each Di is a division ring with center containing F. Then ni divides |G| for each i. Proof. By [CR62, (69.11) Theorem], there is a finite extension splitting field L of F for G. (A field L is said to be a splitting field for G if for every irreducible (left) L[G]-module M, the (left) E[G]-module E ⊗L M is irreducible for every field E ⊇ L.) Then L[G] ' L ⊗F F[G] ' L ⊗F. m M. ! Mni (Di ). '. i=1. m M. (L ⊗F Mni (Di )). i=1. as L-algebras. Note that Mn (A) ' A ⊗F Mn (F) as F-algebras for any F-algebra A. It follows that Mni (Di ) ' Di ⊗F Mni (F) and (L ⊗F Di ) ⊗F Mni (F) ' Mni (L ⊗F Di ). Thus, L ⊗F Mni (Di ) ' Mni (L ⊗F Di ) as L-algebras. Note that L is a finite separable extension of F since char(F) = 0. Thus, L ⊗F Di is a semisimple L-algebra by [CR62, (27.16) and Theorem 69.4]2 . We have Li L ⊗F Di ' tj=1 Ms j (D0i j ) for some division ring D0i j and thus L[G] '. ti m M M. Mni s j (D0i j ).. i=1 j=1. Since L is a splitting field of G, it follows that (D0i j ' L and) ni s j divides |G| for each i, j by [CR62, Theorem 33.7]. The result follows. Now we have the following conclusion. Theorem I.2.14 ([AHP98, Corollary 4.2]). Let R be an integral domain of characteristic 0 which is not a field and let G be a finite group. (i) If 2 - |G|, then R[G] has AJD if and only if G is abelian. (ii) If 2, 3 - |G|, then R[G] has MJD if and only if G is abelian. Proof. Let F be the quotient field of R. By Theorem I.2.11 and Lemma I.2.13, if R[G] has AJD in (i) or R[G] has MJD in (ii), then we can conclude that each Wedderburn component of F[G] has reduced degree 1. In other words, F[G] is a direct sum of division rings. Thus, F[G] has no nonzero nilpotent elements. It follows that Q[G] has no nonzero nilpotent elements. By Theorem I.1.3, G is either abelian or Hamiltonian. Since G is of odd order, G must be abelian. The converse statements of (i) and (ii) are trivial when G is abelian. 2. The semisimplicity in [CR62] is defined by having zero Wedderburn radical. For the connection between Wedderburn radical and Jacobson radical, one can see [Lam01b, Chapter 2]..

(28) I. Jordan Decomposition in Integral Group Rings. 22. Therefore, to classify nonabelian groups G for which R[G] has AJD (resp. MJD), we only need to consider that |G| is divided by 2 (resp. 2 or 3)..

(29) I. Jordan Decomposition in Integral Group Rings. I.3. 23. Additive Jordan Decomposition in Integral Group Rings. Now we focus on the AJD problem in integral group rings Z[G]. This problem has been solved completely by Hales and Passi [HP91, Theorem 4.1]. We briefly introduce the ideas of their proof. First of all, assume that Z[G] has AJD. Then the Wedderburn components of Q[G] have reduced degrees ≤ 2 by Theorem I.2.11(i). If p is an odd prime, then G has an abelian normal Sylow p-subgroup including the trivial subgroup by [GH87, 2.2 Theorem]. As a consequence, G has an abelian normal subgroup N of odd order which is a 2-complement (including N = 1). Then G = NS for some Sylow 2-subgroup S. If S is abelian, then one can show that either G is abelian or G ' D2p , the dihedral group of order 2p for some odd prime p. If S is not abelian, then one can show that Z[G] does not have nonzero nilpotent elements. Thus, by Theorem I.1.3, G ' Q8 × E × A where E is an elementary abelian 2-group and A is an abelian group of odd order so that the multiplicative order of 2 modulo |A| is odd. For both situations of S, the proofs use the fact that the AJD property can be inherited by subgroups and quotient groups. Consequently, if Z[G] has AJD, then G is either abelian, Q8 × E × A or D2p . Conversely, when G is one of the first two types, Z[G] has only semisimple elements and AJD holds. When G ' D2p , then Z[G] has AJD by [HLP90, 3.4 Theorem]. Hence, we have the following. Theorem I.3.1 ([HP91, Theorem 4.1]). Let G be a finite group. Then Z[G] has AJD if and only if G is one of the following: (i) abelian; (ii) Q8 × E × A where E is an elementary abelian 2-group and A is an abelian group of odd order so that the multiplicative order of 2 modulo |A| is odd; (iii) D2p for odd prime p. From Theorem I.3.1, we can observe that the dihedral group D8 of order 8 is the smallest group for which AJD fails in integral group rings. For a specific example, we consider α = x + xt ∈ Z[D8 ] where D8 = hx,t | x4 = t 2 = 1, txt −1 = x−1 i. Note that {x, x3 } and {xt, x3t} are two conjugacy classes of D8 . Thus, the linear combinations of x + x3 and xt + x3t over Q are central in Q[D8 ]. Note also that each central element is semisimple since it maps onto a tuple of diagonal matrices from the Wedderburn decomposition. Moreover, one can compute that the square of (x − x3 ) + (xt − x3t) is zero. Thus, 1 1 αs = [(x + x3 ) + (xt + x3t)] and αn = [(x − x3 ) + (xt − x3t)] 2 2 which are obviously not in Z[D8 ]..

(30) I. Jordan Decomposition in Integral Group Rings. 24. Remark I.3.2. Note that Q[D8 ] ' Q ⊕ Q ⊕ Q ⊕ Q ⊕ M2 (Q). Specifically, the isomorphism can be generated by the following mappings x 7→. 1, 1, −1, −1,. 0 −1 1 0. !! and t 7→. 0 1 1 0. !!. −1 −1 1 1. !!. 1, −1, 1, −1,. .. That is, Z[D8 ] ,→ Z ⊕ Z ⊕ Z ⊕ Z ⊕ M2 (Z). For α = x + xt, we have αs 7→. 2, 0, −2, 0,. 0 0 0 0. !! and αn 7→. 0, 0, 0, 0,. which lie in the image of Z[D8 ]. As a consequence, from the image of an integral group ring under the Wedderburn decomposition of its rational group algebra, we still can not directly ensure whether the semisimple and nilpotent parts of an integral element are integral. The idea of the element α = x + xt ∈ Z[D8 ] which we used above refers to the proof of the following lemma. Lemma I.3.3 ([HP91, Lemma 2.1]). Let G be a group in which every conjugacy class has size at most two. Then AJD holds in Z[G] if and only if Z[G] does not have nonzero nilpotent elements. We remark here that the proof of Theorem I.3.1 uses the fact that AJD is inherited by subgroups and by quotient groups. In other words, if Z[G] has AJD, then so does Z[H] and Z[G/N] for H ≤ G and N E G since Q[H] ∩ Z[G] = Z[H] and the natural map Z[G] → Z[G/N] preserves semisimplicity, nilpotence and commutativity. However, MJD is only inherited by subgroups because units need not lift by the natural map. This makes MJD a much less tractable property..

(31) I. Jordan Decomposition in Integral Group Rings. I.4. 25. Multiplicative Jordan Decomposition in Integral Group Rings. A group ring Z[G] has AJD then it also has MJD. However, the converse is false. In the previous subsection, we know that Z[D8 ] does not have AJD. In fact, Z[D8 ] has MJD. We present a proof below to illustrate that it is complicated to show that an integral group ring Z[G] has MJD even if the order of G is small. Proposition I.4.1 ([AHP93, 3.1 Proposition]). Z[D8 ] has MJD. Proof. Write D8 = hx,t | x4 = t 2 = 1, txt −1 = x−1 i. Let R1 , . . . , R5 be the natural projections of Q[D8 ] projecting to each component of Wedderburn decomposition as in Remark I.3.2. Specifically, given α = α0 + α1 x + α2 x2 + α3 x3 + α4t + α5 xt + α6 x2t + α7 x3t ∈ Q[D8 ] with αi ∈ Q. If we write a = α0 + α2 , b = α1 + α3 , c = α4 + α6 and d = α5 + α7 , then we have R1 (α) = a + b + c + d, R2 (α) = a + b − c − d, R3 (α) = a − b + c − d, R4 (α) = a − b − c + d and R5 (α) =. α0 − α2 − α5 + α7 −α1 + α3 + α4 − α6 α1 − α3 + α4 − α6 α0 − α2 + α5 − α7. ! .. Assume that α ∈ U (Z[D8 ]) with augmentation 1. Then αi ∈ Z for all i. Furthermore, we assume that α is not semisimple. Then R5 (α) is not semisimple since Ri (α) is semisimple for i = 1, 2, 3, 4. Thus the minimal polynomial of R5 (α) over Q has multiple roots in Q, says (X − λ )2 for some λ ∈ Q. Note that this is also the characteristic polynomial of R5 (α). Thus, 2λ , λ 2 ∈ Z because αi ∈ Z for all i. In particular, λ ∈ Q. By Lemma I.2.4, the minimal polynomial of the semisimple part of R5 (α) is X − λ . In particular, R5 (αs ) = (R5 (α))s = λ I is central in M2 (Q). Since Ri (αs ) is central in Q for each i = 1, 2, 3, 4, we can conclude that αs is central in Q[D8 ]. In other words, αs is a linear combination of class sums of conjugacy classes. Let β = β0 + β2 x2 + b0 (x + x3 ) + c0 (t + x2t) + d 0 (xt + x3t) ∈ Q[D8 ] be a central element where βi , b0 , c0 , d 0 ∈ Q. Let δ = α − β . Observe that if we set β0 = α0 , β2 = α2 , b0 = b/2, c0 = c/2 and d 0 = d/2, then Ri (δ ) = 0 for i = 1, 2, 3, 4. Moreover, tr(R5 (δ )) = 0. If det(R5 (δ )) 6= 0, then R5 (δ ) is semisimple and so is δ . This is a contradiction since α = β + δ is semisimple by Lemma I.2.1. Thus, det(R5 (δ )) = 0 and R5 (δ )2 = 0. We obtain δ 2 = 0. By the uniqueness of Jordan decomposition, we have β = αs and δ = αn where β0 = α0 , β2 = α2 , b0 = b/2, c0 = c/2 and d 0 = d/2. We claim that αs ∈ Z[D8 ]. Since α −1 ∈ Z[D8 ], it follows that Ri (α) = ±1 for i = 1, 2, 3, 4 and λ 2 = det(R5 (α)) = ±1. Note that R1 (α) = aug(α) = 1. We can conclude that a + b, a + c, a + d ∈ {0, 1}. Since λ ∈ Q, we have λ = ±1. Moreover, tr(R5 (α)) =.

(32) I. Jordan Decomposition in Integral Group Rings. 26. 2(α0 − α2 ) = 2λ = ±2. Thus, a ≡ α0 − α2 ≡ 1 (mod 2). So b + c + d is even. Suppose two of b, c, d are odd, says b, c. We have a + b = a + c = 0 and a + d = 1. Thus, 1 = a + b + c + d = 1 − 2a. Then a = 0, a contradiction. By symmetry, neither two of b, c, d are odd. As a consequence, b, c and d are even. Therefore, b0 , c0 , d 0 , ∈ Z and αs ∈ Z[D8 ]. Furthermore, αn = α − αs ∈ Z[D8 ]. Hence, Z[D8 ] has MJD. In particular, we have a + b = a + c = a + d = 1. So a = 1 and b = c = d = 0. This means that αs = α0 + α2 x2 .. More generally, the MJD property for dihedral groups is clear by the following result. Theorem I.4.2 ([AHP98, Theorem 5.2]). MJD holds in Z[D2n ] if and only if n is either 2, 4, or an odd prime. Recalling Theorem I.2.11(ii), if Z[G] has MJD, then every reduced degree of Q[G] is less than or equal to 3. This result is best possible. For instance, let G = C7 oC3 = hx,t | x7 = t 3 = 1,txt −1 = x2 i and let ω be a primitive cube root of unity in C. Then √ Q[G] ' Q ⊕ Q(ω) ⊕ M3 (Q( −7)) and Z[G] has MJD, see [Aro94, Proposition II.5.1]. In contrast, Q[A4 ] ' Q ⊕ Q(ω) ⊕ M3 (Q) has a reduced degree 3 and Z[A4 ] does not have MJD, see [Aro94, Proposition II.4.3]. It seems that reduced degrees can not give us further information to deal with the MJD problem. The MJD problem is first proposed by Hales and Passi [HP91, Concluding remarks] in 1991 after they complete the classification of the AJD problem (Theorem I.3.1). In 2007, they together with Wilson give a direction to deal with the MJD problem by only focusing on some specific finite groups. More precisely, they prove the following result. Theorem I.4.3 ([HPW07, Theorem 29]). Let G be a finite group such that Z[G] has MJD. Then one of the following holds. (i) G is either abelian or of the form Q8 × E × A where E is an elementary abelian 2group and A is abelian of odd order such that 2 has odd multiplicative order modulo |A|;.

(33) I. Jordan Decomposition in Integral Group Rings. 27. (ii) G has order 2a 3b for some nonnegative integers a, b; (iii) G = Q8 ×C p for some prime p ≥ 5 so that 2 has even multiplicative order modulo p; (iv) G is the split extension of C p for some prime p ≥ 5 by a cyclic group hgi of order 2k or 3k for some k ≥ 1, and g2 or g3 acts trivially on C p . However, this problem is still open in the following sense: characterize finite groups G in the above list such that Z[G] has MJD. Note that for groups G in the type (i), Z[G] has AJD and so MJD by Theorem I.3.1. In the following subsections, we will mention results for groups of type (ii) and (iv) in Theorem I.4.3. For groups of type (iii), we will present some new results in the next section. For convenience, we call that a group G has MJD (resp. AJD) if Z[G] has MJD (resp. AJD). I.4.1 2-Groups All finite 2-groups with MJD have been classified. For groups of order 8, Q8 has MJD since it is in the type (ii) of Theorem I.3.1 and D8 has MJD by Proposition I.4.1. For groups of order 16, there are some results in [AHP98]. In [Par02], M.M. Parmenter completed the classification of groups of order 16 with MJD by using his earlier work [JP93] with E. Jespers on the units in the group rings of groups of order 16. Before we state his result, we give the following definition, see also [Gor80, p. 29]. Definition I.4.4. A group G is called an internal central product of its subgroups H and K if G = HK and every element of H commutes with every element of K, and we denote G = H ∗ K. If A and B are two groups such that their centers are isomorphic via an isomorphism ϕ, then the quotient group A × B/{(a, ϕ(a)−1 ) | a ∈ Z (A)} is called an external central product of A and B via ϕ and it is denoted by A ∗ϕ B. Here, Z (A) is the center of A. In particular, when Z (A) = Z (B) = Z, we denote A ∗Z B the external central product of A and B via the identity map. Note that an internal central product can be regarded as an external central product and vice versa. So we simply call it a central product. Recall that for a positive integer n, Q4n = ha, b | a2n = 1, b2 = an , bab−1 = a−1 i is the generalized quaternion group3 (also called the dicyclic group or the binary dihedral group) of order 4n. We now present the classification below for all nine non-isomorphic nonabelian groups of order 16. 3. In some books, a generalized quaternion group is restricted to the case where the order is a power of 2..

(34) I. Jordan Decomposition in Integral Group Rings. 28. Theorem I.4.5 ([AHP98, Par02]). Let G be a nonabelian group of order 16. When G is one of the followings, Z[G] has MJD: • Q8 ×C2 (AJD holds), • P = ha, b | a4 = b4 = 1, bab−1 = a−1 i ([AHP98, Theorem 6.1]), • Q8 ∗C4 ([Par02, Theorem 2(i)]), 8 2 −1 = a5 i ([Par02, Theorem 2(ii)]), • D+ 16 = ha, b | a = b = 1 bab. • Q16 ([Par02, Theorem 6]). For otherwise: • D16 ([AHP98, Proposition 5.1(v)]), • H = hg, a, b | g2 = a4 = b2 , ab = ba, gb = bg, gag−1 = a3 bi ([Par02, Theorem 7(i)]), • D8 ×C2 ([Par02, Theorem 7(ii)]), 8 2 −1 = a3 i ([Par02, Theorem 7(iii)]). • D− 16 = ha, b | a = b = 1 bab. In [HPW07], they classified groups of order 32 with MJD. Based on this classification, results for groups of order 64 with MJD were solved. Using induction, they completed the classification of groups of order higher than 64 with MJD. In summary, they proved Theorem I.4.6 ([HPW07, Theorem 19, 20 and 28]). Out of the forty-four nonabelian groups of order 32, the integral group rings of exactly the following four groups have the MJD property: • Q8 ×C2 ×C2 (AJD holds), • Q8 ×C4 , • Q8 ∗C8 , • Q8 ∗ D8 . If G is of order 2n for n ≥ 6, then Z[G] has MJD if and only if G is a Hamiltonian group. There are 256 non-isomorphic nonabelian groups of order 64. In [HPW07], they essentially checked all 300 groups, each of which has MJD or not. This was a huge work. Moreover, they used induction to check groups of order great than 64. We remark that there was an error in [HPW07] and they had corrected it in [HPW12]. In [Liu12], C.-H. Liu reproved the classification of 2-groups having MJD by checking only few classes of groups without the classification of groups of order 32 and 64 and without induction. The reason that Liu could reduce the number of groups is via the following observation..

(35) I. Jordan Decomposition in Integral Group Rings. 29. Theorem I.4.7 ([LP09, Proposition 2.5], [Liu12, Proposition 2.2 and 3.3]). (i) Let G have MJD and N E G. If Y ≤ G, then either Y ⊆ N or Y N E G. Such group G is said to have the SN property. (ii) If G has MJD, then every subgroup of G has SN. Such group G is said to have the SSN property. (iii) If G is a finite p-group and G has SSN, then every non-cyclic subgroup of G is normal. Such group G is said to have the NCN property. The above result (ii) follows from the fact that MJD is inherited by subgroups. Finite p-groups with NCN have been studied by F.N. Liman [Lim68], D.S. Passman [Pas70] and Z. Boˇzikov and Z. Janko [BJ09]. The above theorem is also used to classified 3-groups and {2, 3}-groups. The SN groups, SSN groups and NCN p-groups will be mentioned again in Chapter II. I.4.2 3-Groups and {2, 3}-Groups Using Theorem I.4.7(i), (ii), Liu and Passman prove the following results. Theorem I.4.8 ([LP09, Theorem 2.1, 3.5], [LP10, Lemma 2.1, 2.2] and [LP14, Theorem 2.4]). Let G be a nonabelian 3-group. Then G has MJD if and only if |G| = 27. Theorem I.4.9 ([LP10, Theorem 3.6]). Let G be a nonabelian {2, 3}-group with order divisible by 6. Then G has MJD if and only if G is in precisely the following cases: (i) G ' S3 , (ii) G ' Q12 , (iii) G ' Q8 ×C3 . They also use Theorem I.4.7(iii) to simplify the proof of Theorem I.4.8 in [LP13, Theorem 3.9]. I.4.3. Groups C p oCnk. Let G(p, n, k) = C p oCnk be nonabelian where p is an odd prime ≥ 5, n ∈ {2, 3}, Cnk = hgi and gn centralizes C p . When n = 2, we note that G(p, 2, 1) = D2p which has AJD by Theorem I.3.1 and so MJD. For k = 2, one has G(p, 2, 2) = Q4p . We remark that if m is an odd integer, then Q4m has another presentation Q4m = hx, y | xm = y4 = 1, yxy−1 = x−1 i by setting x = ab2 and y = b from the definition above Theorem I.4.5. Using this presentation, one has the following result..

(36) I. Jordan Decomposition in Integral Group Rings. 30. Theorem I.4.10 ([AHP98, Theorem 6.1]). Q4p has MJD. For k > 2, one has Theorem I.4.11 ([HP17, Theorem 4.1]). If p ≡ 3 (mod 4) and k > 2, then G(p, 2, k) does not have MJD. For n = 3, we observe that p ≥ 7. To see this, we let C p = hti. Since G(p, 3, k) is 3 nonabelian, we have t g 6= t. Moreover, g forms an automorphism of C p and t g = t. It follows that g forms an automorphism of C p of order 3. Since Aut(C p ) ' C p−1 , we have 3 | p − 1. Thus, p ≡ 1 (mod 3) and so p 6= 5. Liu and Passman prove the following result. Theorem I.4.12 ([LP13, Theorem 4.2]). If p > 7, then G(p, 3, k) does not have MJD for every k ≥ 1. They prove this result by finding a specific unit b in Z[C p ] to construct a central unit in Z[G(p, 3, k)] of infinite order whose semisimple part has non-integer coefficients. However, such unit b does not exist in Z[C7 ]. When p = 7, there is only one result. Proposition I.4.13 ([Aro94, Proposition 5.1]). G(7, 3, 1) has MJD..

(37) I. Jordan Decomposition in Integral Group Rings. I.5. 31. Multiplicative Jordan Decomposition in Z[Q8 ×C p ]. In this section, we will prove that Z[Q8 ×C p ] has MJD for certain primes p. This result is a joint work with Wentang Kuo and the content follows from the author’s paper [KS19]. Nilpotent Elements in Q[Q8 ×C p ]. I.5.1. We recall the following notations. For a group G and α = ∑g∈G αg g ∈ Q[G] with αg ∈ Q, we have aug(α) = ∑g∈G αg and α is said to have augmentation 1 if aug(α) = 1. Moreover, U (Z[G]) is the unit group of Z[G] and U1 (Z[G]) is the group of units in U (Z[G]) with augmentation 1. Let p be an odd prime such that the multiplicative order of 2 modulo p is even. Let G = Q8 ×C p where Q8 = ha, b | a4 = 1, a2 = b2 , bab−1 = a−1 i and C p = ht | t p = 1i. Denote c = ab and z = a2 = b2 = c2 . In this subsection, we will study nilpotent elements in Q[G]. It will be used when we try to find the Jordan decomposition of a non-semisimple unit in the next subsection. Since the multiplicative order of 2 modulo p is even, there exist r, s ∈ Z[ε] such that r2 + s2 = −1 where ε is a primitive p-th root of 1 in C (see [GS95, p. 153] or Section I.6). Then we have a ring homomorphism ρ : Q[G] → M2 (Q(ε)) defined by a 7→. 0 1 −1 0. ! ,. b 7→. r s s −r. ! ,. t 7→. ε 0 0 ε. ! .. Clearly, ρ(Z[G]) ⊆ M2 (Z[ε]). We also consider the following two ring homomorphisms φ : Q[G] → Q[G/G0 ] = Q[G/hzi] defined by a 7→ a, b 7→ b and t 7→ t with ker(φ ) = (1 − z)Q[G] and θ : Q[G] → Q[G/C p ] = Q[Q8 ].

(38) I. Jordan Decomposition in Integral Group Rings. 32. defined by a 7→ a, b 7→ b and t 7→ 1 with ker(θ ) = (1 − t)Q[G]. Note that the center of G is hzi×hti and we have the following immediate consequence which will be used in Subsection I.5.2. Lemma I.5.1. Let g ∈ G. (i) If g is not central, then tr(ρ(g)) = 0. (ii) If g = zit k , then tr(ρ(g)) = (−1)i 2ε k . (iii) tr(ρ(Z[G])) ⊆ 2Z[ε]. Now, consider the following ring isomorphism ∆ : Q[C p ] → Q ⊕ Q(ε) by sending t to (1, ε). For an element α(t) ∈ Q[C p ], we denote ∆(α(t)) = (α(1), α(ε)) for convenience. Note that α(1) = aug(α). Under the isomorphism ∆, if α(1) = 0 and α(ε) = 0, we can conclude that α(t) = 0. Next, we consider a certain situation under ρ. Lemma I.5.2. Let α = ∑g∈Q8 αg (t)g ∈ Q[G] for some αg (t) ∈ Q[C p ] and we set ξg = αg (ε) − αgz (ε) for g ∈ {1, a, b, c}. Then ρ(α) is nilpotent if and only if ξ1 = 0 and ξa2 + ξb2 + ξc2 = 0. Proof. Observe that ρ(α) =. ∑. αg (ε)ρ(g) =. g∈Q8. ξ1 + rξb + sξc ξa + sξb − rξc −ξa + sξb − rξc ξ1 − rξb − sξc. ! .. Then tr(ρ(α)) = 2ξ1 and det(ρ(α)) = ξ12 + ξa2 + ξb2 + ξc2 since r2 + s2 = −1. The result holds since X 2 − tr(ρ(α))X + det(ρ(α)) is the characteristic polynomial of ρ(α). We recall Theorem I.1.3 below for convenience. Theorem I.5.3 ([Seh78, p. 172] or [PS02, Theorem 7.4.11]). Let H be a finite group of order 2k m with odd m. Then Q[H] has no nonzero nilpotent elements if and only if H is either abelian or H = Q8 × E × A where E is an elementary abelian 2-group and A is an abelian group of order m such that the multiplicative order of 2 modulo m is odd..