Invariances in commutative and noncommutative group algebras
The paper under review sketches the proofs or announces some results on group algebras \(FG\) over a field \(F\) both in the commutative and the noncommutative cases. For Abelian group algebras the author shows that the exchange property of a \(p\)-group \(G\) can be recovered from its group algebra, in the sense that if \(G\) satisfies the exchange property and \(FG\cong FH\), then \(H\) also has the exchange property. He also states that if the direct factor problem has a positive solution and \(G\) is a torsion complete \(p\)-group, then \(FG\cong FH\) implies \(G\cong H\). Further the author studies invariant properties of subgroups of \(G\) and the factor group \(V(RG)/G\), where \(V(RG)\) is the group of units with augmentation 1, which properties hold for any group \(H\) such that \(FG\cong FH\).NEWLINENEWLINENEWLINEFor noncommutative group algebras the author announces that if the centre of a finite group \(G\) is of index \(p^2\) and \(FG\cong FH\) for all fields \(F\), then \(G\cong H\) which is in the spirit of results of Sandling and the reviewer from the 80's.NEWLINENEWLINENEWLINEFinally, the author claims (Theorem 5) that if \(G\) and \(H\) are the two nonisomorphic finite groups with isomorphic group rings \(\mathbb{Z} G\cong\mathbb{Z} H\), constructed recently by M. Hertweck, then there exists a ring \(R\) such that the group rings \(RG\) and \(RH\) are not isomorphic. Then he makes a general conjecture asking whether such a statement holds for any two finite groups. In the reviewer's opinion, this statement sounds very strange and is obviously wrong because it is well known that the isomorphism \(\mathbb{Z} G\cong\mathbb{Z} H\) implies automatically \(RG\cong RH\) for any ring \(R\) (as a consequence of the ring isomorphism \(RG\cong R\otimes_\mathbb{Z}\mathbb{Z} G\)).
scientific article; zbMATH DE number 1611749