اشترك بالحزمة الذهبية واحصل على وصول غير محدود شمرا أكاديميا
تسجيل مستخدم جديدThe Congruence Subgroup Problem for low rank Free and Free Metabelian groups
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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 $ngeq4$.
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In this paper we describe the profinite completion of the free solvable group on m generators of solvability length r>1. Then, we show that for m=r=2, the free metabelian group on two generators does not have the Congruence Subgroup Property.
The congruence subgroup problem for a finitely generated group $Gamma$ asks whether the map $hat{Autleft(Gammaright)}to Aut(hat{Gamma})$ is injective, or more generally, what is its kernel $Cleft(Gammaright)$? Here $hat{X}$ denotes the profinite comp
letion of $X$. It is well known that for finitely generated free abelian groups $Cleft(mathbb{Z}^{n}right)=left{ 1right}$ for every $ngeq3$, but $Cleft(mathbb{Z}^{2}right)=hat{F}_{omega}$, where $hat{F}_{omega}$ is the free profinite group on countably many generators. Considering $Phi_{n}$, the free metabelian group on $n$ generators, it was also proven that $Cleft(Phi_{2}right)=hat{F}_{omega}$ and $Cleft(Phi_{3}right)supseteqhat{F}_{omega}$. In this paper we prove that $Cleft(Phi_{n}right)$ for $ngeq4$ is abelian. So, while the dichotomy in the abelian case is between $n=2$ and $ngeq3$, in the metabelian case it is between $n=2,3$ and $ngeq4$.
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