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Register a new userThe 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 completion 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 $ngeq4$.
The congruence subgroup problem for a finitely generated group $Gamma$ and $Gleq Aut(Gamma)$ asks whether the map $hat{G}to Aut(hat{Gamma})$ is injective, or more generally, what is its kernel $Cleft(G,Gammaright)$? Here $hat{X}$ denotes the profinite completion of $X$. In this paper we investigate $Cleft(IA(Phi_{n}),Phi_{n}right)$, where $Phi_{n}$ is a free metabelian group on $ngeq4$ generators, and $IA(Phi_{n})=ker(Aut(Phi_{n})to GL_{n}(mathbb{Z}))$. We show that in this case $C(IA(Phi_{n}),Phi_{n})$ is abelian, but not trivial, and not even finitely generated. This behavior is very different from what happens for free metabelian group on $n=2,3$ generators, or for finitely generated nilpotent groups.
The congruence subgroup problem for a finitely generated group $Gamma$ and $Gleq Aut(Gamma)$ asks whether the map $hat{G}to Aut(hat{Gamma})$ is injective, or more generally, what is its kernel $Cleft(G,Gammaright)$? Here $hat{X}$ denotes the profinite completion of $X$. In the case $G=Aut(Gamma)$ we denote $Cleft(Gammaright)=Cleft(Aut(Gamma),Gammaright)$. Let $Gamma$ be a finitely generated group, $bar{Gamma}=Gamma/[Gamma,Gamma]$, and $Gamma^{*}=bar{Gamma}/tor(bar{Gamma})congmathbb{Z}^{(d)}$. Denote $Aut^{*}(Gamma)=textrm{Im}(Aut(Gamma)to Aut(Gamma^{*}))leq GL_{d}(mathbb{Z})$. In this paper we show that when $Gamma$ is nilpotent, there is a canonical isomorphism $Cleft(Gammaright)simeq C(Aut^{*}(Gamma),Gamma^{*})$. In other words, $Cleft(Gammaright)$ is completely determined by the solution to the classical congruence subgroup problem for the arithmetic group $Aut^{*}(Gamma)$. In particular, in the case where $Gamma=Psi_{n,c}$ is a finitely generated free nilpotent group of class $c$ on $n$ elements, we get that $C(Psi_{n,c})=C(mathbb{Z}^{(n)})={e}$ whenever $ngeq3$, and $C(Psi_{2,c})=C(mathbb{Z}^{(2)})=hat{F}_{omega}$ = the free profinite group on countable number of generators.
We prove that the finitely presentable subgroups of residually free groups are separable and that the subgroups of type $mathrm{FP}_infty$ are virtual retracts. We describe a uniform solution to the membership problem for finitely presentable subgroups of residually free groups.
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