A proof of Higgins' conjecture

Details A proof of Higgins' conjecture A proof of Higgins' conjecture

2003-12-06

arxiv.org/abs/math/0312139v2

Bull. Austral. Math. Soc., Vol. 70 (2004) [207-212]

Mathematics

Group Theory

6 pages; corrected typos; added journal-ref, MSC-class 20L05; bibliography converted to amsrefs format

Scientific paper

Let 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.

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