A proof of Higgins' conjecture

dc.creatorBraun, Gabor
dc.date2003-12-06
dc.date2004-09-05
dc.date.accessioned2026-07-07T05:03:37Z
dc.date.available2026-07-07T05:03:37Z
dc.descriptionLet f: G=* G(i) -> B=* B(i) be a group homomorphism between free products of groups. Suppose that G(i)f=B(i) of all i. Let H be a subgroup of G such that Hf=B. Then H decomposes into a free product H=*H(i) with H(i)f=B(i). Furthermore, H(i) decomposes into a free product of a free group and the intersection of H(i) with some conjugate of G(i). Higgins conjectured this in 1971 and now we prove it.
dc.description6 pages; corrected typos; added journal-ref, MSC-class 20L05; bibliography converted to amsrefs format
dc.identifierhttps://arxiv.org/abs/math/0312139
dc.identifierhttp://arxiv.org/abs/math/0312139
dc.identifierBull. Austral. Math. Soc., Vol. 70 (2004) [207-212]
dc.identifier.urihttp://salesiana.dossiersoluciones.com/handle/123456789/69493
dc.subjectGroup Theory
dc.subject20E06; 20L05
dc.titleA proof of Higgins' conjecture
dc.typetext

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