2000
DOI: 10.1070/sm2000v191n11abeh000524
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On primitive representations of soluble groups of finite rank

Abstract: Within the framework of the thermally activated process of the flux line or flux line bundles, and by time integration of the 1D equation of motion of the circulating current density J (ρ, t), which is suitable for thin superconducting films (R d, λ), we present numerical calculations of the current profiles, magnetization hysteresis loops and ac susceptibility χ n = χ n + iχ n for n = 1, 3 and 5 of a thin disc immersed in an axial time-dependent external magnetic field B a (t) = B dc + B ac cos(2πνt). Our cal… Show more

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Cited by 8 publications
(9 citation statements)
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“…But then (17) implies that P / ∈ µ kA (akL/akL) and the obtained contradiction shows that µ kA (akL/akL) ⊆ µ kA (bkL/bkLI) If P ∈ µ kA (bkL/bkLI) and P / ∈ µ kA (akL/akL) then it follows from (17) that there is Q ∈ µ kA (akL/akLI) such that Q < P and (19) implies that Q ∈ µ kA (ckL/ckLI). But then (18) implies that P / ∈ µ kA (bkL/bkL) and the obtained contradiction shows that µ kA (akL/akL) ⊇ µ kA (bkL/bkLI). Thus, we can conclude that µ kA (akL/akL) = µ kA (bkL/bkLI).…”
Section: A Finite Set Of Commutative Invariants For Modules Over Grou...mentioning
confidence: 92%
See 3 more Smart Citations
“…But then (17) implies that P / ∈ µ kA (akL/akL) and the obtained contradiction shows that µ kA (akL/akL) ⊆ µ kA (bkL/bkLI) If P ∈ µ kA (bkL/bkLI) and P / ∈ µ kA (akL/akL) then it follows from (17) that there is Q ∈ µ kA (akL/akLI) such that Q < P and (19) implies that Q ∈ µ kA (ckL/ckLI). But then (18) implies that P / ∈ µ kA (bkL/bkL) and the obtained contradiction shows that µ kA (akL/akL) ⊇ µ kA (bkL/bkLI). Thus, we can conclude that µ kA (akL/akL) = µ kA (bkL/bkLI).…”
Section: A Finite Set Of Commutative Invariants For Modules Over Grou...mentioning
confidence: 92%
“…The subgroup H ≤ G which provides (1) is said to be a control subgroup for the RG-module M. Various types of control subgroups for modules over group rings of groups of finite rank were considered in [15][16][17][18].…”
Section: Introductionmentioning
confidence: 99%
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“…It is well known (see [14]) that any polycyclic group is finitely generated soluble of finite rank and meets the maximal condition for subgroups (in particular, for normal subgroups). In [10] we showed that in the class of soluble groups of finite rank with the maximal condition for normal subgroups only polycyclic groups may have faithful irreducible primitive representations over a field of characteristic zero. In [7][8][9] we studied irreducible primitive representations of metabelian groups of finite rank over a field of characteristic zero.…”
mentioning
confidence: 99%