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The monomorphism problem in free groups

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 Publication date 2009
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and research's language is English




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Let $F$ be a free group of finite rank. We say that the monomorphism problem in $F$ is decidable if for any two elements $u$ and $v$ in $F$, there is an algorithm that determines whether there exists a monomorphism of $F$ that sends $u$ to $v$. In this paper we show that the monomorphism problem is decidable and we provide an effective algorithm that solves the problem.



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118 - Tara Brough 2011
We consider the class of groups whose word problem is poly-context-free; that is, an intersection of finitely many context-free languages. We show that any group which is virtually a finitely generated subgroup of a direct product of free groups has poly-context-free word problem, and conjecture that the converse also holds. We prove our conjecture for several classes of soluble groups, including metabelian groups and torsion-free soluble groups, and present progress towards resolving the conjecture for soluble groups in general. Some of the techniques introduced for proving languages not to be poly-context-free may be of independent interest.
The congruence subgroup problem for a finitely generated group $Gamma$ asks whether $widehat{Autleft(Gammaright)}to Aut(hat{Gamma})$ is injective, or more generally, what is its kernel $Cleft(Gammaright)$? Here $hat{X}$ denotes the profinite completion of $X$. In this paper we first give two new short proofs of two known results (for $Gamma=F_{2}$ and $Phi_{2}$) and a new result for $Gamma=Phi_{3}$: 1. $Cleft(F_{2}right)=left{ eright}$ when $F_{2}$ is the free group on two generators. 2. $Cleft(Phi_{2}right)=hat{F}_{omega}$ when $Phi_{n}$ is the free metabelian group on $n$ generators, and $hat{F}_{omega}$ is the free profinite group on $aleph_{0}$ generators. 3. $Cleft(Phi_{3}right)$ contains $hat{F}_{omega}$. Results 2. and 3. should be contrasted with an upcoming result of the first author showing that $Cleft(Phi_{n}right)$ is abelian for $ngeq4$.
225 - Derek Holt , Sarah Rees 2020
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