Rational conjugacy of torsion units in integral group rings. group rings of non-solvable groups

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1 Rational conjugacy of torsion units in integral group rings of non-solvable groups Vrije Universiteit Brussels, University of Stuttgart RTRA Murcia, June 4th 2013

2 Notations and Setting G a finite group. RG group ring of G over the ring R. Q p the p-adic number field, Z p ring of integers of Q p. ε : RG R augmentation map, ε x : RG R partial augmentation on the conjugacy class x G for x G: ε( u g g) = u g and ε x ( u g g) = g G g G g G g x G u g U(RG) units in RG, V(RG) units of augmentation 1 aka normalized units (±V(ZG) = U(ZG)). exp(v(zg)) = exp(g). (Cohn-Livingstone 65)

3 Notations and Setting G a finite group. RG group ring of G over the ring R. Q p the p-adic number field, Z p ring of integers of Q p. ε : RG R augmentation map, ε x : RG R partial augmentation on the conjugacy class x G for x G: ε( u g g) = u g and ε x ( u g g) = g G g G g G g x G u g U(RG) units in RG, V(RG) units of augmentation 1 aka normalized units (±V(ZG) = U(ZG)). exp(v(zg)) = exp(g). (Cohn-Livingstone 65)

4 Notations and Setting G a finite group. RG group ring of G over the ring R. Q p the p-adic number field, Z p ring of integers of Q p. ε : RG R augmentation map, ε x : RG R partial augmentation on the conjugacy class x G for x G: ε( u g g) = u g and ε x ( u g g) = g G g G g G g x G u g U(RG) units in RG, V(RG) units of augmentation 1 aka normalized units (±V(ZG) = U(ZG)). exp(v(zg)) = exp(g). (Cohn-Livingstone 65)

5 Zassenhaus Conjecture and Prime Graph Question (First) Zassenhaus Conjecture (H.J. Zassenhaus, in the 60s) For u V(ZG) of finite order there exist x U(QG) and g G s.t. x 1 ux = g. A weaker version of this conjecture is: Prime Graph Question Let p and q be different primes s.t. V(ZG) has an element of order pq. Does this imply that G has an element of that order? I.e.: Do G and V(ZG) have the same prime graph?

6 Zassenhaus Conjecture and Prime Graph Question (First) Zassenhaus Conjecture (H.J. Zassenhaus, in the 60s) For u V(ZG) of finite order there exist x U(QG) and g G s.t. x 1 ux = g. A weaker version of this conjecture is: Prime Graph Question Let p and q be different primes s.t. V(ZG) has an element of order pq. Does this imply that G has an element of that order? I.e.: Do G and V(ZG) have the same prime graph?

7 Results on the Zassenhaus Conjecture The Zassenhaus Conjecture is known for some series of solvable groups (e.g. nilpotent groups (Weiss 91)) and for some non-solvable. E.g.: A 5 (Luthar-Passi 89), S 5 (Luthar-Trama 91), Central extensions of S 5 (Bovdi-Hertweck 08), PSL(2, p), p a prime, p 17. (Hertweck 07, Gildea 12, Kimmerle-Konovalov 12), PSL(2, 8) (Gildea 12, Kimmerle-Konovalov 12), A 6 = PSL(2, 9) (Hertweck 07).

8 Results on the Zassenhaus Conjecture The Zassenhaus Conjecture is known for some series of solvable groups (e.g. nilpotent groups (Weiss 91)) and for some non-solvable. E.g.: A 5 (Luthar-Passi 89), S 5 (Luthar-Trama 91), Central extensions of S 5 (Bovdi-Hertweck 08), PSL(2, p), p a prime, p 17. (Hertweck 07, Gildea 12, Kimmerle-Konovalov 12), PSL(2, 8) (Gildea 12, Kimmerle-Konovalov 12), A 6 = PSL(2, 9) (Hertweck 07).

9 Results on the Zassenhaus Conjecture The Zassenhaus Conjecture is known for some series of solvable groups (e.g. nilpotent groups (Weiss 91)) and for some non-solvable. E.g.: A 5 (Luthar-Passi 89), S 5 (Luthar-Trama 91), Central extensions of S 5 (Bovdi-Hertweck 08), PSL(2, p), p a prime, p 17. (Hertweck 07, Gildea 12, Kimmerle-Konovalov 12), PSL(2, 8) (Gildea 12, Kimmerle-Konovalov 12), A 6 = PSL(2, 9) (Hertweck 07).

10 Results on the Prime Graph Question The Prime Graph Question is known for: Solvable Groups (Kimmerle 06), PSL(2, p), p a prime (Hertweck 07), Many sporadic simple groups (Bovdi, Konovalov et al ), Groups, whose order is divisible by at most three primes, if there are no units of order 6 in V(Z PGL(2, 9)) and in V(ZM 10 ). (Where M 10 denotes the Mathieu group of degree 10.) (Kimmerle-Konovalov 12)

11 Our results Theorem The Zassenhaus Conjecture holds for PSL(2, 19) and PSL(2, 23). Theorem There are no units of order 6 in V(Z PGL(2, 9)) and in V(ZM 10 ). Corollary If the order of G is divisible by at most three primes, then the Prime Graph question has a positive answer for G.

12 Our results Theorem The Zassenhaus Conjecture holds for PSL(2, 19) and PSL(2, 23). Theorem There are no units of order 6 in V(Z PGL(2, 9)) and in V(ZM 10 ). Corollary If the order of G is divisible by at most three primes, then the Prime Graph question has a positive answer for G.

13 Partial augmentations The connection between rational conjugacy and partial augmentations is: Lemma u V(ZG) is conjugate in QG to an element of G if and only if ε g (v) 0 for every v u and every g G. Genral knowledge on partial augmenations of a torsion unit u V(ZG): ε 1 (u) = 0, if u 1. (Berman-Higman 39/ 51) ε x (u) 0, then the order of x divides the order of u. (Marciniak-Ritter-Sehgal-Weiss 87, Hertweck 07)

14 Partial augmentations The connection between rational conjugacy and partial augmentations is: Lemma u V(ZG) is conjugate in QG to an element of G if and only if ε g (v) 0 for every v u and every g G. Genral knowledge on partial augmenations of a torsion unit u V(ZG): ε 1 (u) = 0, if u 1. (Berman-Higman 39/ 51) ε x (u) 0, then the order of x divides the order of u. (Marciniak-Ritter-Sehgal-Weiss 87, Hertweck 07)

15 The HeLP-method The HeLP-method (Hertweck 07, Luthar-Passi 89) establishes a connection between partial augmentations and eigenvalues under representations: Let u V(ZG) be a torsion unit of order n, K an alg. closed field of characteristic not dividing n and D a K -representation of G with character χ. Let ζ be a prim. n-th root of unity and ξ some n-th root of unity. Then the number of times ξ appears as an eigenvalue of D(u) is 1 n d n, d 1 Tr Q(ζ d )/Q (χ(ud )ξ 1 )+ 1 n x G, p o(x) ε x (u)tr Q(ζ)/Q (χ(x)ζ 1 ).

16 The HeLP-method The HeLP-method (Hertweck 07, Luthar-Passi 89) establishes a connection between partial augmentations and eigenvalues under representations: Let u V(ZG) be a torsion unit of order n, K an alg. closed field of characteristic not dividing n and D a K -representation of G with character χ. Let ζ be a prim. n-th root of unity and ξ some n-th root of unity. Then the number of times ξ appears as an eigenvalue of D(u) is 1 n d n, d 1 Tr Q(ζ d )/Q (χ(ud )ξ 1 )+ 1 n x G, p o(x) ε x (u)tr Q(ζ)/Q (χ(x)ζ 1 ).

17 Example computation Part of the ordinary character table of A 6 : 1a 2a 3a 3b χ Let D be a representaion affording χ and ζ a prim. 3rd root of unity. Assume u V(ZA 6 ) is of order 6, s.t. ε 2a (u) = 2, ε 3a (u) = 2, ε 3b (u) = 1 and u 4 is rationally conjugate to a element in 3b. Then χ(u 3 ) = 1 and χ(u 4 ) = 1, so D(u 3 ) diag(1, 1, 1, 1, 1), D(u 4 ) diag(1,ζ,ζ 2,ζ,ζ 2 ). The eigenvalues of D(u) are products of the eigenvalues of D(u 3 ) and D(u 4 ) and χ(u) = ε 2a (u)χ(2a)+ε 3a (u)χ(3a)+ε 3b (u)χ(3b) = 1. This gives D(u) diag(1,ζ,ζ 2, ζ, ζ 2 ).

18 Basic idea Natural question: What can we do with the eigenvalues? Let: p a prime dividing the order of u, D an ordinary representation of G, K the p-adic completion of a number field affording D with minimal ramification index over Q p, R the ring of integers of K with maximal ideal P, s.t. p P, L an RG-lattice affording D, k = R/P the quotient field of R, the reduction mod P. Eigenvalues of D(u) Poss. iso types of L as k ū -module. K u -composition factors of L Poss. iso types of L as R u -lattice

19 Basic idea Natural question: What can we do with the eigenvalues? Let: p a prime dividing the order of u, D an ordinary representation of G, K the p-adic completion of a number field affording D with minimal ramification index over Q p, R the ring of integers of K with maximal ideal P, s.t. p P, L an RG-lattice affording D, k = R/P the quotient field of R, the reduction mod P. Eigenvalues of D(u) Poss. iso types of L as k ū -module. K u -composition factors of L Poss. iso types of L as R u -lattice

20 Basic idea Natural question: What can we do with the eigenvalues? Let: p a prime dividing the order of u, D an ordinary representation of G, K the p-adic completion of a number field affording D with minimal ramification index over Q p, R the ring of integers of K with maximal ideal P, s.t. p P, L an RG-lattice affording D, k = R/P the quotient field of R, the reduction mod P. Eigenvalues of D(u) Poss. iso types of L as k ū -module. K u -composition factors of L Poss. iso types of L as R u -lattice

21 Basic idea Natural question: What can we do with the eigenvalues? Let: p a prime dividing the order of u, D an ordinary representation of G, K the p-adic completion of a number field affording D with minimal ramification index over Q p, R the ring of integers of K with maximal ideal P, s.t. p P, L an RG-lattice affording D, k = R/P the quotient field of R, the reduction mod P. Eigenvalues of D(u) Poss. iso types of L as k ū -module. K u -composition factors of L Poss. iso types of L as R u -lattice

22 Basic Proposition Proposition Notation as above, let (u) = p a m with p m, let ζ a prim. m-th root of unity s.t. ζ R. Let A i be tuples of p a -th roots of unity s.t. the eigenvalues of D(u) are ζa 1 ζ 2 A 2... ζ m A m. Then as R u -lattice L = M 1... M m s.t. rang R (M i ) = A i = dim k ( M i ) and M i has only one composition factor up to isomorphism. Easiest case: K unramified over Q p, (u) = p. There are three indecomposable R u -lattices R, I(RC p ), RC p of rank 1, p 1, p resp. with corresponding eigenvalues{1}, {ξ p,...,ξp p 1 }, {1,ξ p,...,ξp p 1 }, where ξ p is a primitive p-th root of unity. The reduction of any such lattice stays indecomposable.

23 Basic Proposition Proposition Notation as above, let (u) = p a m with p m, let ζ a prim. m-th root of unity s.t. ζ R. Let A i be tuples of p a -th roots of unity s.t. the eigenvalues of D(u) are ζa 1 ζ 2 A 2... ζ m A m. Then as R u -lattice L = M 1... M m s.t. rang R (M i ) = A i = dim k ( M i ) and M i has only one composition factor up to isomorphism. Easiest case: K unramified over Q p, (u) = p. There are three indecomposable R u -lattices R, I(RC p ), RC p of rank 1, p 1, p resp. with corresponding eigenvalues{1}, {ξ p,...,ξp p 1 }, {1,ξ p,...,ξp p 1 }, where ξ p is a primitive p-th root of unity. The reduction of any such lattice stays indecomposable.

24 Zassenhaus Conjecture for PSL(2, 19) Let G = PSL(2, 19). After applying HeLP the only critical case left is: (u) = 10, (ε 5a (u),ε 5b (u),ε 10a (u)) = (1, 1, 1). Let the setting be as above with p = 5 and let ζ be a primitive 5th root of unity. There are ordinary representations D 18 and D 19 s.t. D 18 is a Z 5 [ζ +ζ 1 ]-representation and D 19 a Z 5 -representation. If L 18 and L 19 are RG-lattices (R is different) then L 18 L 19 and L 19 / L 18 = trivial kg-module.

25 Zassenhaus Conjecture for PSL(2, 19) Let G = PSL(2, 19). After applying HeLP the only critical case left is: (u) = 10, (ε 5a (u),ε 5b (u),ε 10a (u)) = (1, 1, 1). Let the setting be as above with p = 5 and let ζ be a primitive 5th root of unity. There are ordinary representations D 18 and D 19 s.t. D 18 is a Z 5 [ζ +ζ 1 ]-representation and D 19 a Z 5 -representation. If L 18 and L 19 are RG-lattices (R is different) then L 18 L 19 and L 19 / L 18 = trivial kg-module.

26 Zassenhaus Conjecture for PSL(2, 19) Using the given partial augmentations we compute: D 18 (u) diag(a, 1, ζ, ζ 2, ζ 3, ζ 4, 1, ζ, ζ 4, ζ, ζ 4 ), D 19 (u) diag(a, 1, ζ, ζ 2, ζ 3, ζ 4, 1, ζ, ζ 2, ζ 3, ζ 4 ), where A and A are some 5th roots of unity. As k ū -modules we have L 19 = M 1 19 M 1 19 and L 18 = M 1 18 M 1 18, where all compositions factors of M1 are trivial and all composition factors of M 1 are non-trivial. Then M 1 19 {2(k) 2I(kC 5 ),(k) I(kC 5 ) (kc 5 ), 2(kC 5 ) }. As L 19 / L 18 is a trivial module, we have M 1 18 = M 1 19, but this is impossible by the composition factors of M 1 18 (by results of Jacobinski 67 and Gudivok 65). The Zassenhaus Conjecture holds for PSL(2, 19).

27 Zassenhaus Conjecture for PSL(2, 19) Using the given partial augmentations we compute: D 18 (u) diag(a, 1, ζ, ζ 2, ζ 3, ζ 4, 1, ζ, ζ 4, ζ, ζ 4 ), D 19 (u) diag(a, 1, ζ, ζ 2, ζ 3, ζ 4, 1, ζ, ζ 2, ζ 3, ζ 4 ), where A and A are some 5th roots of unity. As k ū -modules we have L 19 = M 1 19 M 1 19 and L 18 = M 1 18 M 1 18, where all compositions factors of M1 are trivial and all composition factors of M 1 are non-trivial. Then M 1 19 {2(k) 2I(kC 5 ),(k) I(kC 5 ) (kc 5 ), 2(kC 5 ) }. As L 19 / L 18 is a trivial module, we have M 1 18 = M 1 19, but this is impossible by the composition factors of M 1 18 (by results of Jacobinski 67 and Gudivok 65). The Zassenhaus Conjecture holds for PSL(2, 19).

28 Zassenhaus Conjecture for PSL(2, 19) Using the given partial augmentations we compute: D 18 (u) diag(a, 1, ζ, ζ 2, ζ 3, ζ 4, 1, ζ, ζ 4, ζ, ζ 4 ), D 19 (u) diag(a, 1, ζ, ζ 2, ζ 3, ζ 4, 1, ζ, ζ 2, ζ 3, ζ 4 ), where A and A are some 5th roots of unity. As k ū -modules we have L 19 = M 1 19 M 1 19 and L 18 = M 1 18 M 1 18, where all compositions factors of M1 are trivial and all composition factors of M 1 are non-trivial. Then M 1 19 {2(k) 2I(kC 5 ),(k) I(kC 5 ) (kc 5 ), 2(kC 5 ) }. As L 19 / L 18 is a trivial module, we have M 1 18 = M 1 19, but this is impossible by the composition factors of M 1 18 (by results of Jacobinski 67 and Gudivok 65). The Zassenhaus Conjecture holds for PSL(2, 19).

29 Zassenhaus Conjecture for PSL(2, 19) Using the given partial augmentations we compute: D 18 (u) diag(a, 1, ζ, ζ 2, ζ 3, ζ 4, 1, ζ, ζ 4, ζ, ζ 4 ), D 19 (u) diag(a, 1, ζ, ζ 2, ζ 3, ζ 4, 1, ζ, ζ 2, ζ 3, ζ 4 ), where A and A are some 5th roots of unity. As k ū -modules we have L 19 = M 1 19 M 1 19 and L 18 = M 1 18 M 1 18, where all compositions factors of M1 are trivial and all composition factors of M 1 are non-trivial. Then M 1 19 {2(k) 2I(kC 5 ),(k) I(kC 5 ) (kc 5 ), 2(kC 5 ) }. As L 19 / L 18 is a trivial module, we have M 1 18 = M 1 19, but this is impossible by the composition factors of M 1 18 (by results of Jacobinski 67 and Gudivok 65). The Zassenhaus Conjecture holds for PSL(2, 19).

30 Thank You! Thank you for your attention!