1973
DOI: 10.1017/s1446788700015044
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The soluble subgroups of a one-relator group with torsion

Abstract: In a survey article [1] Baumslag posed the problem of determining the abelian subgroups of a one-relator group. The solution of this problem was stated but not proved in [5], and partly solved by Moldavanskii [4]. In this paper it will be proved that the centralizer of every non-trivial element in a one-relator group with torsion is cyclic, and that the soluble subgroups of a one-relator groups with torsion are cyclic groups or the infinite dihedral group. That both types of groups may occur as subgroups is ea… Show more

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Cited by 12 publications
(7 citation statements)
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“…The result for freely reduced words follows from Newman's Spelling Theorem (Theorem 3 of [7] Lemma 2. Let Z be a t-reduced word and let h be a t-free word.…”
Section: Aut(g) Of G Is Finitely Generated^)mentioning
confidence: 99%
“…The result for freely reduced words follows from Newman's Spelling Theorem (Theorem 3 of [7] Lemma 2. Let Z be a t-reduced word and let h be a t-free word.…”
Section: Aut(g) Of G Is Finitely Generated^)mentioning
confidence: 99%
“…[11]) If V ∈ N F (R)\{1}, then V contains a subword S n−1 S 0 ,where S ≡ S 0 S 1 ∈ W F (W ) and every generator appearing in W appears in S 0 . (2) ([17, Theorem]) The centralizer of every non-trivial element in G is cyclic.…”
mentioning
confidence: 99%
“…In particular, we shall show that a quasivariety generated by free groups cannot be defined in the class of torsion-free groups by any finite system of quasi-identities. ~ such that T(~,~,.. , n ) = ~ we define 6g(~)-6~ (7) . Since by assumption the sum of indices on ~.…”
mentioning
confidence: 99%