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Register a new userThe IA-congruence kernel of high rank free Metabelian groups
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David El-Chai Ben-Ezra
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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.
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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 $ngeq4$.
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 $ngeq4$.
In the high rank limit, the fraction of non-zero character table entries of finite simple groups of Lie type goes to zero.
For each odd prime p>=5, there exist finite p-groups G with derived quotient G/D(G)=C(p)xC(p) and nearly constant transfer kernel type k(G)=(1,2,...,2) having two fixed points. It is proved that, for p=7, this type k(G) with the simplest possible case of logarithmic abelian quotient invariants t(G)=(11111,111,21,21,21,21,21,21) of the eight maximal subgroups is realized by exactly 98 non-metabelian Schur sigma-groups S of order 7^11 with fixed derived length dl(S)=3 and metabelianizations S/D(D(S)) of order 7^7. For p=5, the type k(G) with t(G)=(2111,111,21,21,21,21) leads to infinitely many non-metabelian Schur sigma-groups S of order at least 5^14 with unbounded derived length dl(S)>=3 and metabelianizations S/D(D(S)) of fixed order 5^7. These results admit the conclusion that d=-159592 is the first known discriminant of an imaginary quadratic field with 7-class field tower of precise length L=3, and d=-90868 is a discriminant of an imaginary quadratic field with 5-class field tower of length L>=3, whose exact length remains unknown.
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