Masaq Index
arXiv 2003-12-06 1 views

A proof of Higgins' conjecture

Braun, Gabor

Original · EN

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.

English translation

This paper has no Arabic translation yet. Be the first: it takes a few seconds, and the result is stored for every future reader.

Security check

Type the characters above

Up to 10 translations per person per day.