2018
DOI: 10.1016/j.jalgebra.2017.11.016
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Groups in which each subgroup is commensurable with a normal subgroup

Abstract: A group G is a cn-group if for each subgroup H of G there exists a normal subgroup N of G such that the index |HN : (H ∩ N)| is finite. The class of cn-groups contains properly the classes of core-finite groups and that of groups in which each subgroup has finite index in a normal subgroup.\ud In the present paper it is shown that a cn-group whose periodic images are locally finite is finite-by-abelian-by- finite. Such groups are then described into some details by considering automorphisms of abelian groups. … Show more

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Cited by 6 publications
(3 citation statements)
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“…Clearly all the above results rely on corresponding previous results on groups in which all subgroups are nn, cf , cn resp. (see [13,1,3] resp.). A similar approach was adopted in [7] where finitely generated groups in which subnormal subgroups are inert have been considered, where the term inert refers to a different generalization of both nn, cf (namely, an inert subgroup is a subgroup which is commensurable with each of its conjugates).…”
Section: Introduction and Statement Of Resultsmentioning
confidence: 99%
See 1 more Smart Citation
“…Clearly all the above results rely on corresponding previous results on groups in which all subgroups are nn, cf , cn resp. (see [13,1,3] resp.). A similar approach was adopted in [7] where finitely generated groups in which subnormal subgroups are inert have been considered, where the term inert refers to a different generalization of both nn, cf (namely, an inert subgroup is a subgroup which is commensurable with each of its conjugates).…”
Section: Introduction and Statement Of Resultsmentioning
confidence: 99%
“…Since G is periodic and A has infinite rank, it follows that the socle S of A is a normal subgroup of G which is the direct product of infinitely many non-trivial cyclic subgroups. Application of Lemma 1 yields that all subgroups of S are cn-subgroups of G and hence by Lemma 2.8 of [3], there exist G-invariant subgroups S 0 ≤ S 1 of S such that S 0 and S/S 1 are finite and all subgroups of S lying between S 0 ad S 1 are normal in G Let X be any subnormal subgroup of finite rank of G. Then X ∩ S 1 is finite, hence S 2 = S 0 (X ∩ S 1 ) is likewise finite. Since S 2 X is commensurable with X, we may assume S 2 = {1}.…”
Section: Lemma 1 Let G Be a T [+]-Group And Let A Be A Subnormal Subg...mentioning
confidence: 94%
“…Moreover, there exist also soluble nitely generated groups with subgroups which are inert but not uniformly inert (see Remark 2.15 below). Groups in which all subgroup are uniformly inert have been studied in [4].…”
Section: Theorem 14 [3 Theorem 3] a Subgroup Of A Group Is Uniformmentioning
confidence: 99%