Twisted group ring isomorphism problem

Transcription

1 Spa, Belgium Groups, Rings and the Yang-Baxter equation

2 Group ring isomorphism problem Denote by Ω the class of finite groups and by Ω n the groups of order n. For a commutative ring R denote by R an equivalence relation on Ω which is defined by G R H if and only if RG = RH.

3 Group ring isomorphism problem Denote by Ω the class of finite groups and by Ω n the groups of order n. For a commutative ring R denote by R an equivalence relation on Ω which is defined by G R H if and only if RG = RH. The group ring isomorphism problem [GRIP] For a given commutative ring R, determine the equivalence classes of Ω with respect to the relation R. Answer in particular, for which groups G R H implies G = H.

4 Facts and results If G, H are abelian groups of the same cardinality then G C H. Any finite abelian group is a Q singleton [Perlis and Walker 1950]. G Z H G R H, for any commutative ring R. Any p-group G is a Z singleton [Roggenkamp and Scott 1987]. There exist non-isomorphic groups X, Y such that X Z Y [Hertweck 2001].

5 Twisted group rings We wish to investigate a twisted version of the GRIP.

6 Twisted group rings We wish to investigate a twisted version of the GRIP. For a 2-cocycle α Z 2 (G, R ) the twisted group ring R α G is the free R-module with basis {u g } g G such that u g u h = α(g, h)u gh for all g, h G and any u g commutes with the elements of R.

7 Twisted group rings We wish to investigate a twisted version of the GRIP. For a 2-cocycle α Z 2 (G, R ) the twisted group ring R α G is the free R-module with basis {u g } g G such that u g u h = α(g, h)u gh for all g, h G and any u g commutes with the elements of R. The ring structure of R α G depends only on the cohomology class of α, [α] H 2 (G, R ) and not on the particular 2-cocycle.

8 refinement of We are now ready to introduce the relation of interest.

9 refinement of We are now ready to introduce the relation of interest. Definition Let R be a commutative ring. For G, H Ω, G R H if and only if there exists a group isomorphism ψ : H 2 (G, R ) H 2 (H, R ) such that for any [α] H 2 (G, R ), R α G = R ψ(α) H.

10 refinement of We are now ready to introduce the relation of interest. Definition Let R be a commutative ring. For G, H Ω, G R H if and only if there exists a group isomorphism ψ : H 2 (G, R ) H 2 (H, R ) such that for any [α] H 2 (G, R ), R α G = R ψ(α) H. Notice that R is a refinement of R.

11 Main Problem: The twisted group ring isomorphism problem [TGRIP] For a given commutative ring R, determine the equivalence classes of Ω with respect to the relation R. Answer in particular, for which groups G R H, implies G = H.

12 Main Problem: The twisted group ring isomorphism problem [TGRIP] For a given commutative ring R, determine the equivalence classes of Ω with respect to the relation R. Answer in particular, for which groups G R H, implies G = H. We will deal mostly with the case R = C. In this case the second cohomology group is called the Schur multiplier and it is denoted by M(G).

13 Main Problem: The twisted group ring isomorphism problem [TGRIP] For a given commutative ring R, determine the equivalence classes of Ω with respect to the relation R. Answer in particular, for which groups G R H, implies G = H. We will deal mostly with the case R = C. In this case the second cohomology group is called the Schur multiplier and it is denoted by M(G). The twisted group algebra C α G may be simple. In these cases α is called nondegenerate and G is called of central type.

14 The abelian case While all the abelian groups of the same cardinality are C equivalent, the following holds for C.

15 The abelian case While all the abelian groups of the same cardinality are C equivalent, the following holds for C. Lemma Any abelian group A is a C singleton.

16 The abelian case While all the abelian groups of the same cardinality are C equivalent, the following holds for C. Lemma Any abelian group A is a C singleton. This follows from the fact that abelian groups of the same cardinality admits non-isomorphic Schur multipliers.

17 Necessary conditions for G C H Lemma The following conditions are necessary conditions for G C H. None of them is sufficient and none of them implies the other. A) M(G) = M(H).

18 Necessary conditions for G C H Lemma The following conditions are necessary conditions for G C H. None of them is sufficient and none of them implies the other. A) M(G) = M(H). B) There exists a set bijection φ : M(G) M(H) such that C α G = C φ(α) H for any [α] M(G) and in particular CG = CH.

19 The case G = p 4 Lemma Let p be prime and let G and H be groups of order p 4 with the following properties 1 G, H are both not of central type. 2 M(G) = M(H). 3 CG = CH. Then, G C H.

20 The case G = p 4 Lemma Let p be prime and let G and H be groups of order p 4 with the following properties 1 G, H are both not of central type. 2 M(G) = M(H). 3 CG = CH. Then, G C H. Using the above and the classification of groups of central type of order p 4 we partition Ω p 4 to C -classes for any prime p. In particular we show that there exist G = H such that G C H.

21 The case G = p 2 q 2 Theorem Let G and H be groups of cardinality p 2 q 2 for primes p < q. If M(G) = M(H) and CG = CH, then G C H.

22 The case G = p 2 q 2 Theorem Let G and H be groups of cardinality p 2 q 2 for primes p < q. If M(G) = M(H) and CG = CH, then G C H. Using this theorem and a known description of groups of order p 2 q 2 we are able to partition Ω p 2 q 2 to C-classes for any primes p, q. Example In the partition of the 21 groups of order to C classes there are 11 singletons, 3 classes containing 2 groups and 1 class containing 4 groups.

23 It turns out that any group of central type of cardinality p 4, p 2 q 2 or 64 is a C -singleton.

24 It turns out that any group of central type of cardinality p 4, p 2 q 2 or 64 is a C -singleton. One might conjecture that any group of central type is a C -singleton.

25 Groups of Central Type Using the description of groups of central type of order n 2 for n square-free we prove. Theorem Let G be a group of central type of order n 2 where n is a square free number. Then in the following cases G is a C -singleton. 1 If G is divisible by at most two primes. 2 If G G Z(G) is square-free.

26 Groups of Central Type Using the description of groups of central type of order n 2 for n square-free we prove. Theorem Let G be a group of central type of order n 2 where n is a square free number. Then in the following cases G is a C -singleton. 1 If G is divisible by at most two primes. 2 If G G Z(G) is square-free. However there exist non-isomorphic groups of central type of order n 2 where n is a square-free number which are C equivalent.

27 Thanks for your attention.