脱毛 发表于 2025-3-30 10:43:05
Auf dem Weg zur Zweiten Moderne which the ring will act. We solve here a similar problem: given a ℚ-matrix . in the suitable class, build from it the Butler ..-groups for which . is a base change. The building procedure we give is an algorithm, easy to perform by hand in one-digit ranks.Commission 发表于 2025-3-30 14:41:46
http://reply.papertrans.cn/15/1432/143129/143129_52.pngESO 发表于 2025-3-30 18:54:21
Von der zweiten Gründerzeit zur Postmoderne a constructive fashion the question when an almost completely decomposable group has non-zero completely decomposable direct summands. A new invariant, the rank-width difference of . at τ, given by . is the exact rank of a maximal τ-homogeneous (completely decomposable) direct summand of .. We showPituitary-Gland 发表于 2025-3-30 21:13:44
http://reply.papertrans.cn/15/1432/143129/143129_54.pngNOT 发表于 2025-3-31 01:00:00
http://reply.papertrans.cn/15/1432/143129/143129_55.pngprobate 发表于 2025-3-31 08:05:24
https://doi.org/10.1007/978-3-322-86659-2(.)-module isomorphism from . to its endomorphism ring .(.). Groups which are isomorphic to the additive group of their endomorphism rings are called weak .-groups. The purpose of this article is to consider the apparently yet weaker condition that there be a homorphism from . onto the additive grouFsh238 发表于 2025-3-31 11:23:43
https://doi.org/10.1007/978-3-531-90027-8We consider the connection and interplay between some well-known problems in modular group algebras and the structure of simply presented .-groups. New results in both areas are obtained.变化无常 发表于 2025-3-31 17:01:39
http://reply.papertrans.cn/15/1432/143129/143129_58.pngTAIN 发表于 2025-3-31 18:36:30
http://reply.papertrans.cn/15/1432/143129/143129_59.pngFibrinogen 发表于 2025-3-31 22:13:44
Auf dem Weg zur Zweiten ModerneWe will find normal forms of matrices which describe almost completely decomposable groups in the class of uniform groups defined by Dugas and Oxford up to near-isomorphism. This problem is equivalent to the diagonal equivalence of matrices over ℤ.. ℤ.