Tevfik Bilgin, Necat Gorentas, and I. Gokhan Kelebek

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1 J Korean Math Soc 47 (010), No 6, pp DOI /JKMS CHARACTERIZATION OF CENTRAL UNITS OF ZA n Tevfik Bilgin, Necat Gorentas, and I Gokhan Kelebek Abstract The structure of V (Z(ZA n)) is known when n 6 If n = 5 or 6, then a complete set of generators of V (Z(ZA n)) has been determined In this study, it was shown that V (Z(ZA n)) is trivial when n = 7, 8 or 9 and it is generated by a single unit u when n = 10 or 11 This unit u is characterized explicitly for n = 10 or 11 by using irreducible characters of A n 1 Introduction Let V = V (ZG) denote the group of normalized units of the integral group ring ZG of a finite group G Let Z(V ) denote the subgroup of the central units of V and N V (G) denote the normalizer of G in the normalized units V (ZG) In [4], the following is considered as the normalizer property: N V (G) = GZ(V )? Verification of the normalizer property reduces the problem to the construction of central units of normalized units of ZG [4, 5] On the other hand, in order to find a counter example one must find a unit u N V (G) \ Z(V ) In both cases, construction of Z(V ) is important Let V (Z(ZG)) denote the normalized units of the center of ZG Both of them are defined as follows []: Z(V ) = Z(V (ZG)) = V (ZG) Z(ZG) = V (Z(ZG)) The problem of finding the full structure of V (Z(ZG)), including a complete set of generators, has been determined for only a small number of special cases When G is finite, Patay [8] proved the following theorem satisfying necessary and sufficient conditions for U(Z(ZG)) to be trivial Proposition 11 Let G be a finite group All central units of ZG are trivial if and only if for every x G and for every j N relatively prime to G, x j is conjugate to x or x 1 Ritter and Sehgal constructed a finite set of generators for a subgroup of finite index in U(Z(ZG)) for a finite group G [9] On the other hand, Jespers, Received February 6, 009; Revised August 1, Mathematics Subject Classification 16S34, 16U60 Key words and phrases normalizer, centralizer, generators of central units 139 c 010 The Korean Mathematical Society

2 140 TEVFIK BILGIN, NECAT GORENTAS, AND I GOKHAN KELEBEK Parmenter, and Sehgal [6] found a different set of generators which works for finitely generated nilpotent groups and constructed these generators from Bass cyclic units in ZG The construction depended on the existence of a finite normal series in G If we choose A n (n > 4) as a finite group, it is impossible to construct generators for U(Z(ZG)) from Bass cyclic units since A n is a simple group However Aleev [1] constructed all central units of ZA 5 and ZA 6 Later, Li and Parmenter [7] independently constructed all central units of ZA 5, too In this study we have given a method to construct generators for V (Z(ZG)), which works not only for A n but also for any finite group G We have mentioned this method in the second part In the last part we have completed the construction of generators for V (Z(ZA n )), where n < 1 The main results are the following Theorem 1 V (Z(ZA n )) is generated by a single unit for n = 5, 6, 10, 11 as follows: i) V (Z(ZA 5 )) = 49υ 1a 16υ a + 6υ 5a 10υ 5b ii) V (Z(ZA 6 )) = 18433υ 1a 304υ 3a 304υ 3b + 378υ 5a 144υ 5b iii) V (Z(ZA 10 )) = γ 180, where γ = (5897υ 1a 7υ 3a 9υ 3c +1υ 5a +3υ 15a 3υ 5b 3υ 7a +7υ 1a 4υ 1b ) iv) V (Z(ZA 11 )) = γ 10, where γ = (8146υ 1a + 54υ a 54υ b 405υ 3a + 7υ 6a + 7υ 6b 36υ 4a 9υ 1a + 18υ 4c 9υ 1b + 81υ 5a + 9υ 10a + 9υ 0a 9υ 5b 9υ 7a 9υ 14a ) υ 1a υ 1b Let us first recall some basic concept about conjugacy classes and irreducible characters of A n (n < 1) prior to prove these theorems Motivation to construct generators for V (Z(ZA n )) 1 - Let us list all conjugacy classes of A n by regarding of appearance first in A n for n < 1: 1a = {(1)}, a = {g(1, )(3, 4)g 1 : g A n }, b = {g(1, )(3, 4)(5, 6)(7, 8)g 1 : g A n }, 3a = {g(1,, 3)g 1 : g A n }, 3b = {g(1,, 3)(4, 5, 6)g 1 : g A n }, 3c = {g(1,, 3)(4, 5, 6)(7, 8, 9)g 1 : g A n }, 4a = {g(1, )(3, 4, 5, 6)g 1 : g A n },

3 CHARACTERIZATION OF CENTRAL UNITS OF ZA n 141 4b = {g(1,, 3, 4)(5, 6, 7, 8)g 1 : g A n }, 4c = {g(1, )(3, 4)(5, 6)(7, 8, 9, 10)g 1 : g A n }, 5a = {g(1,, 3, 4, 5)g 1 : g A n }, 5b = {g(1,, 3, 4, 5)(6, 7, 8, 9, 10)g 1 : g A n }, 6a = {g(1, )(3, 4)(5, 6, 7)g 1 : g A n }, 6b = {g(1, )(3, 4, 5, 6, 7, 8)g 1 : g A n }, 6c = {g(1, )(3, 4)(5, 6, 7)(8, 9, 10)g 1 : g A n }, 6d = {g(1, )(3, 4)(5, 6)(7, 8)(9, 10, 11)g 1 : g A n }, 6e = {g(1, )(3, 4, 5)(6, 7, 8, 9, 10, 11)g 1 : g A n }, 7a = {g(1,, 3, 4, 5, 6, 7)g 1 : g A n }, 8a = {g(1, )(3, 4, 5, 6, 7, 8, 9, 10)g 1 : g A n }, 9a = {g(1,, 3, 4, 5, 6, 7, 8, 9)g 1 : g A n }, 10a = {g(1, )(3, 4)(5, 6, 7, 8, 9)g 1 : g A n }, 11a = {g(1,, 3, 4, 5, 6, 7, 8, 9, 10, 11)g 1 : g A n }, 1a = {g(1, )(3, 4, 5)(6, 7, 8, 9)g 1 : g A n }, 1b = {g(1,, 3, 4)(5, 6, 7, 8, 9, 10)g 1 : g A n }, 1c = {g(1,, 3)(4, 5, 6, 7)(8, 9, 10, 11)g 1 : g A n }, 14a = {g(1, )(3, 4)(5, 6, 7, 8, 9, 10, 11)g 1 : g A n }, 15a = {g(1,, 3)(4, 5, 6, 7, 8)g 1 : g A n }, 15b = {g(1,, 3)(4, 5, 6)(7, 8, 9, 10, 11)g 1 : g A n }, 0a = {g(1, )(3, 4, 5, 6)(7, 8, 9, 10, 11)g 1 : g A n }, 1a = {g(1,, 3)(4, 5, 6, 7, 8, 9, 10)g 1 : g A n } Using Gap 44 [10] it is easy to determine the size of conjugacy classes, the order of centralizers and the irreducible characters of A n (n < 1) - If we denote the class sum by υ kx, then we write υ kx = g g kx All class sums of A n form a basis for the center of the rational group algebra QA n That is, Z(QA n ) = υ kx : kx A n Q Since Z(QA n ) = i F i and F i = Z(D i ); the center of the division ring D i, Z(ZA n ) is an order of Z(QA n ) and also we say that Z(ZA n ) is a suborder

4 14 TEVFIK BILGIN, NECAT GORENTAS, AND I GOKHAN KELEBEK of maximal order (1) M = i Oi, where Oi denotes the algebraic integers of F i [3] If we denote the central units of ZA n by U(Z(ZA n )) and the normalized central units of ZA n by V (Z(ZA n )), then we say that V (Z(ZA n )) is trivial if and only if V (Z(ZA n ))=1 For n < 1 we can classify the central units of ZA n Proposition 1 Let n < 1 V (Z(ZA n )) = 1 if and only if n 5, 6, 10, 11 Proof By considering the equation (1) and the character tables for A n one by one: For n =, ZA n = Z so V (Z(ZAn )) is trivial For n = 3 and 4, O i = Z or Z[ω], where ω = 1+i 3 For n = 5 and 6, O i = Z or Z[α], where α = 1+ 5 For n = 7 and 8, O i = Z or Z[ω], where ω = 1+i 7 For n = 9, O i = Z or Z[ω], where ω = 1+i 15 For n = 10, O i = Z or Z[α], where α = 1+ 1 For n = 11, O i = Z, Z[ω] or Z[α], where ω = 1+i 11 and α = 1+ 1 So by Proposition 11 for n =, 3, 4, 7, 8, 9 we get V (Z(ZA n )) = 1, and for n = 5, 6, 10, 11 we get V (Z(ZA n )) 1 Proposition If χ 1 is a trivial character and γ V (Z(ZA n )), then χ 1 (γ) = 1 Proof Let us consider the augmentation map: ε : ZG Z γg g γ g Let T n be transversal for the alternating group A n, l g be the class size of g and g be the conjugacy class of g Then we have A n = g, υ g = g g T n g g If χ 1 is a trivial character, then we write g G, χ 1 (g) = 1 For any γ V (Z(ZA n )) we have χ 1 (γ) = χ 1 ( γ g υ g ) = γ g χ 1 (υ g ) g T n g T n = γ g χ 1 ( g) = γ g χ 1 (g) g T n g g g T n g g = γ g l g = γ g g T n g g g T n

5 CHARACTERIZATION OF CENTRAL UNITS OF ZA n 143 = γ g = ε(γ) = 1 g A n 3 Construction of generators of central units of ZA n Proof of Theorem 1 i) If γ V (Z(ZA 5 )), then we can write γ = γ 1a υ 1a + γ a υ a + γ 3a υ 3a + γ 5a υ 5a + γ 5b υ 5b = g T 5 γ g υ g By using Gap 44 [10] we can get the character table of A 5 Therefore, the values of γ at the irreducible characters are χ j (γ) = g T 5 γ g χ j (υ g ) = g T 5 γ g l g χ j (g) Let us consider these values one by one By Proposition, we have χ 1 (γ) = 1 (31) γ 1a + 15γ a + 0γ 3a + 1γ 5a + 1γ 5b = 1, χ (γ) = 3γ 1a 15γ a + 1αγ 5a + 1αγ 5b = 3λ, λ U(Z[ 5 ]), (3) γ 1a 5γ a + 4αγ 5a + 4αγ 5b = λ, χ 3 (γ) = 3γ 1a 15γ a + 1αγ 5a + 1αγ 5b = 3λ 3, λ 3 U(Z[ 5 ]), (33) γ 1a 5γ a + 4αγ 5a + 4αγ 5b = λ 3, χ 4 (γ) = 4γ 1a + 0γ 3a 1γ 5a 1γ 5b = 4λ 4, λ 4 U(Z), (34) γ 1a + 5γ 3a 3γ 5a 3γ 5b = ±1, χ 5 (γ) = 5γ 1a + 15γ a 0γ 3a = 5λ 5, λ 5 U(Z), (35) γ 1a + 3γ a 4γ 3a = ±1 Since d (χ 4 (γ) χ 1 (γ)) and d (χ 5 (γ) χ 1 (γ)) for d > we can write χ 4 (γ) = χ 5 (γ) = 1 So we can get the following system of linear equations γ 1a α 4α γ a α 4α γ 3a γ 5a = λ λ γ 5b 1 If we consider the 1st, 4th and 5th equations and the first three variables we get γ 1a γ a = 1 1(γ 5a + γ 5b ) 1 + 3(γ 5a + γ 5b ) γ 3a 1 So the parametric solution is γ 1a = 1 + 3(γ 5a + γ 5b ), γ a = (γ 5a + γ 5b ), γ 3a = 0

6 144 TEVFIK BILGIN, NECAT GORENTAS, AND I GOKHAN KELEBEK By substituting this solution into (3), we get (1 + 10(γ 5a + γ 5b )) + (γ 5a γ 5b ) 5 U(Z[ 5 ]) or equivalently, (1 + 10(γ 5a + γ 5b )) + (γ 5a γ 5b ) 5 = ( + 5) k for some k N We get the smallest solution when k = 4 γ 5a = 6, γ 5b = 10, γ 3a = 16, γ 1a = 49, γ a = 0 Thus the generator is γ = 49υ 1a 16υ a + 6υ 5a 10γ 5b On the other hand the solution of the following equation (1 + 10(γ 5a + γ 5b )) + (γ 5a γ 5b ) 5 = ( 5) k for the same k N gives us its inverse Hence for k = 4 γ 5a = 10, γ 5b = 6, γ 3a = 16, γ 1a = 49, γ a = 0 So the inverse of γ is γ 1 = 49υ 1a 16υ a 10υ 5a + 6υ 5b ii) If γ V (Z(ZA 6 )), then we can write γ = γ 1a υ 1a + γ a υ a + γ 3a υ 3a + γ 3b υ 3b + γ 4a υ 4a + γ 5a υ 5a + γ 5b υ 5b = γ g υ g g T 6 By using Gap 44 [10], we can get the character table of A 6 Thus, the values of γ at the irreducible characters are χ j (γ) = g T 6 γ g χ j (υ g ) = g T 6 γ g l g χ j (g) Let us consider these values one by one By Proposition, we have χ 1 (γ) = 1 (36) γ 1a + 45γ a + 40γ 3a + 40γ 3b + 90γ 4a + 7γ 5a + 7γ 5b = 1, χ (γ) = 5γ 1a + 45γ a + 80γ 3a 40γ 3b 90γ 4a = 5λ, λ U(Z), (37) γ 1a + 9γ a + 16γ 3a 8γ 3b 18γ 4a = ±1, χ 3 (γ) = 5γ 1a + 45γ a 40γ 3a + 80γ 3b 90γ 4a = 5λ 3, λ 3 U(Z), (38) γ 1a + 9γ a 8γ 3a + 16γ 3b 18γ 4a = ±1, χ 4 (γ) = 8γ 1a 40γ 3a 40γ 3b + 7αγ 5a + 7αγ 5b = 8λ 4, λ 4 U(Z[α]), (39) γ 1a 5γ 3a 5γ 3b + 9αγ 5a + 9αγ 5b = λ 4 U(Z[α]), χ 5 (γ) = 8γ 1a 40γ 3a 40γ 3b + 7αγ 5a + 7αγ 5b = 8λ 5, λ 5 U(Z[α]), (310) γ 1a 5γ 3a 5γ 3b + 9αγ 5a + 9αγ 5b = λ 5 U(Z[α]),

7 CHARACTERIZATION OF CENTRAL UNITS OF ZA n 145 χ 6 (γ) = 9γ 1a + 45γ a + 90γ 4a 7γ 5a 7γ 5b = 9λ 6, λ 6 U(Z), (311) γ 1a + 5γ a + 10γ 4a 8γ 5a 8γ 5b = ±1, χ 7 (γ) = 10γ 1a 90γ a + 40γ 3a + 40γ 3b = 10λ 7, λ 7 U(Z), (31) γ 1a 9γ a + 4γ 3a + 4γ 3b = ±1 Since d (χ j (γ) χ 1 (γ)) for d > and j =, 3, 6, 7 we can write χ j (γ) = 1 (j =, 3, 6, 7) If we consider the equations (36)-(31) we obtain the following system of linear equations So the solution is γ 1a γ a γ 3a γ 3b γ 4a = 1 7(γ 5a + γ 5b ) (γ 5a + γ 5b ) 1 γ 1a = 1+8(γ 5a +γ 5b ), γ a = 0, γ 3a = (γ 5a +γ 5b ), γ 3b = (γ 5a +γ 5b ), γ 4a = 0 Therefore, by substituting these values in to the equation (39) we get (γ 5a + γ 5b ) + 9( )γ 5a + 9( 1 5 )γ 5b U(Z [α]), or simply we have (γ 5a + γ 5b ) + 9 5(γ5a γ 5b ) = ( ) k for some k N We can get the smallest solution when k = 4 as follows: γ 1a =18433, γ a =0, γ 3a = 304, γ 3b = 304, γ 4a =0, γ 5a =378, γ 5b = 144 Thus, the generator is γ = 18433υ 1a 304υ 3a 304υ 3b + 378υ 5a 144υ 5b By considering the conjugate of the equation we get the inverse of γ So, we have (γ 5a + γ 5b ) + 9 5(γ5a γ 5b ) = ( 1 5 ) k for the same k γ 1a =18433, γ a =0, γ 3a = 304, γ 3b = 304, γ 4a =0, γ 5a = 144, γ 5b =378 Hence, the inverse of γ is γ 1 = 18433υ 1a 304υ 3a 304υ 3b 144υ 5a + 378υ 5b iii) Let γ V (Z(ZA 10 )) Then we can write γ as follows according to the appearance of the conjugacy classes in the software Gap 44 [10]: γ = γ 1a υ 1a + γ a υ a + γ b υ b + γ 3a υ 3a + γ 6a υ 6a + γ 3b υ 3b + γ 6b υ 6b +γ 3c υ 3c + γ 4a + γ 4b υ 4b + γ 4b υ 4b + γ 4b υ 4b + γ 1a υ 1a + γ 4c υ 4a

8 146 TEVFIK BILGIN, NECAT GORENTAS, AND I GOKHAN KELEBEK +υ 4c + γ 5a υ 5a + γ 10a υ 10a + γ 15a υ 15a + γ 5b υ 5b + γ 6c υ 6c + γ 1b υ 1b +γ 7a υ 7a + γ 1a υ 1a + γ 1b υ 1b + γ 8a υ 8a + γ 9a υ 9a + γ 9b υ 9b By using Gap 44 [10], we can get all irreducible characters Only few values of the characters are irrational, the others are integers We can find all values of irreducible characters of γ χ j (γ) = γ g χ j (υ g ) = γ g l g χ j (g) g T 10 g T 10 By Proposition, we have χ 1 (γ) = 1 That is, 1 = υ 1a + 630υ a + 475υ b + 40υ 3a + 500υ 6a υ 3b + 500υ 6b + 400υ 3c υ 4a υ 4b (313) υ 1a υ 4c υ 5a υ 10a υ 15a υ 15b υ 6c υ 1b υ 7a υ 1a υ 1b υ 8a υ 9a υ 9b For j =,, 19,, 3, 4 we get χ j (γ) = γ g l g χ j (g) = χ j (1)λ j, λ j U(Z), g T 10 χ j(γ) χ j (1) = ±1 We can easily observe that d (χ j (γ) χ 1 (γ)) for d > and j =,,19,, 3, 4 So (314) χ j (γ) = 1 for j =,, 19,, 3, 4 χ j (1) For j = 0, 1 we get (315) χ j (γ) = γ g l g χ j (g) = χ j (1)λ j, λ j U(Z[α]), g T 10 (316) χ j(γ) χ j (1) U(Z[α]) By transfering the 0th and 1st terms to the right hand side in the equations (313) and (314) we get the system of linear equations AX T = Y T, where A is the matrix of coefficients (See Appendix I), X = [γ 1a,, γ 7a, γ 8a, γ 9a, γ 9b ], Y = [ (γ 1a + γ 1b ), (γ 1a + γ 1b ), 1, 1 400(γ 1a + γ 1b ), 1, 1 115(γ 1a + γ 1b ), 1, (γ 1a + γ 1b ), 1, (γ 1a + γ 1b ),

9 CHARACTERIZATION OF CENTRAL UNITS OF ZA n 147 1, 1, 1, 1 384(γ 1a + γ 1b ), 1, 1 300(γ 1a + γ 1b ), (γ 1a + γ 1b ), 1, 1, (γ 1a + γ 1b ), 1, 1] So the solution of the system is γ 1a = (γ 1a + γ 1b ) γ 4a = 0 γ 6c = 0 γ a = 0 γ 4b = 0 γ 1b = 0 γ b = 0 γ 1a = 0 γ 7a = (γ 1a + γ 1b ) γ 3a = 4(γ 1a + γ 1b ) γ 4c = 0 γ 8a = 0 γ 6a = 0 γ 5a = 4(γ 1a + γ 1b ) γ 9a = 0 γ 3b = 0 γ 10a = 0 γ 9b = 0 γ 6b = 0 γ 15a = (γ 1a + γ 1b ) γ 3c = 3(γ 1a + γ 1b ) γ 5b = (γ 1a + γ 1b ) Thus, by substituting these values in to the equation (315) we get the following Pell-equation (317) (γ 1a + γ 1b ) + 5 (γ 1a γ 1b ) ( 5 + ) k 1 1 = This can be obtained when k = 180 by using Maple 70 [11] But it is not useful to express the other coefficients in terms of γ 1a and γ 1b Therefore, we will solve this for 5+ 1 instead of ( 5+ 1 ) 180 Hence, the solution gives us the 180th root of the solution γ = (5897γ 1a 7γ 3a 9γ 3c + 1γ 5a + 3γ 15a 3γ 5b 3γ 7a + 7γ 1a 4γ 1b ) By regarding rational conjugate of the right hand side of the equation (316) we get the inverse of γ as follows: γ 1 = (5897γ 1a 7γ 3a 9γ 3c +1γ 5a +3γ 15a 3γ 5b 3γ 7a +7γ 1a 4γ 1b ) Hence, V (Z(ZA 10 )) = γ 180 iv) For γ V (Z(ZA 11 )), we can write γ as follows according to the appearance of conjugacy classes in the software Gap 44 [10]: γ = γ 1a υ 1a + γ a υ a + γ b υ b + γ 3a υ 3a + γ 6a υ 6a + γ 6b υ 6b + γ 3b υ 3b γ 6c υ 6c + γ 3c υ 3c + γ 4a υ 4a + γ 4b υ 4b + γ 1a υ 1a + γ 4c υ 4c + γ 1b υ 1b +γ 5a υ 5a + γ 10a υ 10a + γ 15a υ 15a + γ 15b υ 15b + γ 0a υ 0a + γ 5b υ 5b +γ 6d υ 6d + γ 6e υ 6e + γ 1c υ 1c + γ 7a υ 7a + γ 14a υ 14a + γ 1a υ 1a +γ 1b υ 1b + γ 8a υ 8a + γ 9a υ 9a + γ 11a υ 11a + γ 11b υ 11b We can determine all irreducible characters by the same software There are 4 irreducible characters without integer values, two of which have irrational

10 148 TEVFIK BILGIN, NECAT GORENTAS, AND I GOKHAN KELEBEK values while the other two have complex values We can find all values of γ at irreducible characters χ j (γ) = γ g χ j (υ g ) = γ g l g χ j (g) g T 11 g T 11 Let us consider these values one by one By Proposition, we have (318) χ 1 (γ) = 1 On the other hand, for j =,, 17, 0,, 31 we get χ j (γ) = γ g l g χ j (g) = χ j (1)λ j, λ j U(Z), g T 11 χ j(γ) χ j (1) = ±1 But we can easily guess that d (χ j (γ) χ 1 (γ)) for d > and j =,, 17, 0,, 31 Thus we write (319) χ j (γ) = 1 for j =,, 17, 0,, 31 χ j (1) For j = 18, 19 we get (30) (31) χ j (γ) = γ g l g χ j (g) = χ j (1)λ j, λ j U(Z[α]), g T 11 χ j(γ) χ j (1) U(Z[α]) By transfering the 18th and 19th terms to the right hand side in the equations (317) and (318) we get the system of linear equations AX T = Y T, where A is the matrix of coefficients (See Appendix II), X = [γ 1a,, γ 14a, γ 8a,, γ 11b ], Y =[ (γ 1a + γ 1b ), 1, (γ 1a + γ 1b ), 1, (γ 1a + γ 1b ), 1 790(γ 1a +γ 1b ), 1, 1, 1+700(γ 1a +γ 1b ), 1, 1, 1, 1 880(γ 1a +γ 1b ), 1, 1, 1, (γ 1a +γ 1b ), (γ 1a +γ 1b ), 1, 1+115(γ 1a +γ 1b ), 1, 1, 1, 1 864(γ 1a + γ 1b ), 1, 1, 1, 1, 1]

11 CHARACTERIZATION OF CENTRAL UNITS OF ZA n 149 So the solution is γ 1a = (γ 1a + γ 1b ) γ 4b = 0 γ 6d = 0 γ a = 6(γ 1a + γ 1b ) γ 1a = (γ 1a + γ 1b ) γ 6e = 0 γ b = 6(γ 1a + γ 1b ) γ 4c = (γ 1a + γ 1b ) γ 1c = 0 γ 3a = 45(γ 1a + γ 1b ) γ 1b = (γ 1a + γ 1b ) γ 7a = (γ 1a + γ 1b ) γ 6a = 3(γ 1a + γ 1b ) γ 5a = 9(γ 1a + γ 1b ) γ 14a = (γ 1a + γ 1b ) γ 6b = 3(γ 1a + γ 1b ) γ 10a = (γ 1a + γ 1b ) γ 8a = 0 γ 3b = 0 γ 15a = 0 γ 9a = 0 γ 6c = 0 γ 15b = 0 γ 11a = 0 γ 3c = 0 γ 0a = (γ 1a + γ 1b ) γ 11b = 0 γ 4a = 4(γ 1a + γ 1b ) γ 5b = (γ 1a + γ 1b ) Therefore, by substituting these values in to the equation (319) we get the following Pell-equation (3) (γ 1a + γ 1b ) + 800(γ 1a γ 1b ) 1 = ( ) k Using Maple 70 [11] we get the solution when k = 10 But it is not useful to express the other coefficients in terms of γ 1a and γ 1b Therefore, we will solve this for instead of ( ) 10 Hence, we obtain the 10th root of the solution as follows: γ = (8146υ 1a + 54υ a 54υ b 405υ 3a + 7υ 6a + 7υ 6b 36υ 4a 9υ 1a + 18υ 4c 9υ 1b + 81υ 5a + 9υ 10a + 9υ 0a 9υ 5b 9υ 7a 9υ 14a ) υ 1a υ 1b If we consider the rational conjugate of the right hand side of the equation (30) we obtain the inverse of γ as follows: γ 1 = (8146υ 1a + 54υ a 54υ b 405υ 3a + 7υ 6a + 7υ 6b 36υ 4a 9υ 1a + 18υ 4c 9υ 1b + 81υ 5a + 9υ 10a + 9υ 0a 9υ 5b 9υ 7a 9υ 14a ) As a result υ 1a υ 1b V (Z(ZA 11 )) = γ 10 Aknowledgement This work is supported by the Head of Scientific Research Projects and Sciences, University of Yuzuncu Yıl, Van, Turkey

12 150 TEVFIK BILGIN, NECAT GORENTAS, AND I GOKHAN KELEBEK Appendix I The coefficient matrix for the characterization of V (Z(ZA 10 ))

13 CHARACTERIZATION OF CENTRAL UNITS OF ZA n 151 Appendix II The coefficient matrix for the characterization of V (Z(ZA 11 ))

14 15 TEVFIK BILGIN, NECAT GORENTAS, AND I GOKHAN KELEBEK References [1] R Z Aleev, Higman s central unit theory, units of integral group rings of finite cyclic groups and Fibonacci numbers, Internat J Algebra Comput 4 (1994), no 3, [], Central elements of integral group rings, Algebra Log 39 (000), no 5, , 630; translation in Algebra and Logic 39 (000), no 5, [3] C W Curtis and I Reiner, Representation Theory of Finite Groups and Associative Algebras, Interscience Publishers, a division of John Wiley & Sons, New York-London 196 [4] M Hertweck and E Jespers, Class-preserving automorphisms and the normalizer property for Blackburn groups, J Group Theory 1 (009), no 1, [5] M Hertweck and W Kimmerle, Coleman automorphisms of finite groups, Math Z 4 (00), no, [6] E Jespers, M M Parmenter, and S K Sehgal, Central units of integral group rings of nilpotent groups, Proc Amer Math Soc 14 (1996), no 4, [7] Y Li and M M Parmenter, Central units of the integral group ring ZA 5, Proc Amer Math Soc 15 (1997), no 1, [8] Z F Patay, Structure of the center of the multiplicative group of a group ring, Visnik Kiiv Univ Ser Mat Mekh No 7 (1985), 90 91, 17 [9] J Ritter and S K Sehgal, Integral group rings with trivial central units, Proc Amer Math Soc 108 (1990), no, [10] The GAP Group, GAP Groups, Algorithms, and Programming, Version 441, 008 ( [11] Waterloo Maple Inc, Maple, Version 700, 008 ( Tevfik Bilgin Department of Mathematics Faculty of Arts and Sciences Fatih University Istanbul, Turkey address: tbilgin@fatihedutr Necat Gorentas Department of Mathematics Faculty of Arts and Sciences Yuzuncu Yil University Van, Turkey address: negorentas@yyuedutr I Gokhan Kelebek Department of Mathematics Faculty of Arts and Sciences Fatih University Istanbul, Turkey address: gkelebek@fatihedutr