The Congruence Subgroup Problem for the Free Metabelian group on $n\geq4$ generators

TL;DR

This paper investigates the congruence subgroup problem for free metabelian groups, showing that for four or more generators, the kernel of the associated map is abelian, revealing a new structural property.

Contribution

It proves that the kernel $C(\

Findings

For $n \\geq 4$, $C(\\\Phi_{n})$ is abelian.

The dichotomy in the kernel's structure shifts at $n=4$ in the metabelian case.

Extends understanding of the congruence subgroup problem for free metabelian groups.

Abstract

The congruence subgroup problem for a finitely generated group $\Gamma$ asks whether the map $\hat{Aut\left(\Gamma\right)}\to Aut(\hat{\Gamma})$ is injective, or more generally, what is its kernel $C\left(\Gamma\right)$? Here $\hat{X}$ denotes the profinite completion of $X$. It is well known that for finitely generated free abelian groups $C\left(\mathbb{Z}^{n}\right)=\left\{ 1\right\}$ for every $n\geq3$, but $C\left(\mathbb{Z}^{2}\right)=\hat{F}_{\omega}$, 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 $C\left(\Phi_{2}\right)=\hat{F}_{\omega}$ and $C\left(\Phi_{3}\right)\supseteq\hat{F}_{\omega}$. In this paper we prove that $C\left(\Phi_{n}\right)$ for $n\geq4$ is abelian. So, while the dichotomy in the abelian case is between $n=2$ and $n\geq3$, in the metabelian case it is between $n=2,3$ and $n\geq4$.

Equations

$$G\left(M\right)$$

$$(\text{resp.}G\left(M\right)$$

$$\underleftarrow{\lim }G/G\left(M\right)$$

$$IA\left({\Phi }_n\right)=\ker \left(Aut\left({\Phi }_n\right)\to Aut\left({\Phi }_n/{\Phi }_n^{\prime }\right)=G{L}_n\left(\mathbb{Z}\right)\right).$$

$$\begin{array}[]{ccccccc}\widehat{IA\left(\Phi_{n}\right)}&\to&\widehat{Aut\left(\Phi_{n}\right)}&\to&\widehat{GL_{n}\left(\mathbb{Z}\right)}&\to&1\\ &\searrow&\downarrow&&\downarrow\\ &&Aut(\hat{\Phi}_{n})&\to&GL_{n}(\hat{\mathbb{Z}})\end{array}$$

$$IA\left({\Phi }_n\right)\twoheadrightarrow \ker (G{L}_{n-1}\left(\mathbb{Z}[x^{\pm 1}]\right)\stackrel{x\mapsto 1}{⟶}G{L}_{n-1}\left(\mathbb{Z}\right))$$

$$IG{L}_{n-1,i}:=\ker (G{L}_{n-1}\left(R_n\right)\stackrel{x_i\mapsto 1}{⟶}G{L}_{n-1}\left(R_n/(x_i-1)R_n\right)).$$

$$g=h_0{k}_1{h}_1{k}_2{h}_2\cdot \dots \cdot k_n{h}_n$$

$$⟨IA{\left({\Phi }_n\right)}^m⟩G\left(M_{n,m^4}\right)/⟨IA{\left({\Phi }_n\right)}^m⟩\text{(noticethe}{m}^4\text{)}$$

$$\Phi_{n}=\left\{\left(\begin{array}[]{cc}g&a_{1}t_{1}+\ldots+a_{n}t_{n}\\ 0&1\end{array}\right)\,|\,g\in\mathbb{Z}^{n},\,a_{i}\in R_{n},\,g-1=\sum_{i=1}^{n}a_{i}(x_{i}-1)\right\}$$

$$\left(\begin{array}[]{cc}x_{i}&t_{i}\\ 0&1\end{array}\right)\,\,\,\,\,1\leq i\leq n.$$

$$\alpha:\left(\begin{array}[]{cc}x_{i}&t_{i}\\ 0&1\end{array}\right)\mapsto\left(\begin{array}[]{cc}x_{i}&a_{i,1}t_{1}+\ldots+a_{i,n}t_{n}\\ 0&1\end{array}\right)$$

$$IA\left({\Phi }_n\right)$$

$$\text{defined by}\alpha$$

$$IA\left({\Phi }_n\right)$$

$$\begin{array}[]{c}\Phi_{n}=\left\{\left(\begin{array}[]{cc}g&a_{1}t_{1}+\ldots+a_{n}t_{n}\\ 0&1\end{array}\right)\,|\,g\in\mathbb{Z}^{n},\,a_{i}\in R_{n},\,g-1=\sum_{i=1}^{n}a_{i}(x_{i}-1)\right\}\\ \downarrow\\ \left\{\left(\begin{array}[]{cc}g&a_{1}t_{1}+\ldots+a_{n}t_{n}\\ 0&1\end{array}\right)\,|\,g\in\mathbb{Z}_{m}^{n},\,a_{i}\in\mathbb{Z}_{m}[\mathbb{Z}_{m}^{n}],\,g-1=\sum_{i=1}^{n}a_{i}(x_{i}-1)\right\}\end{array}$$

$$I{G}_{n,m}$$

$$IGL_{n-1,i}=\left\{I_{n}+A\in IA\left(\Phi_{n}\right)\,|\,\begin{array}[]{c}\textrm{The\,\,}i\textrm{-th\,\, row\,\, of\,\,}A\textrm{\,\, is\,\,0,}\\ I_{n-1}+A_{i,i}\in GL_{n-1}\left(R_{n},\sigma_{i}R_{n}\right)\end{array}\right\}$$

$$G{L}_{n-1}\left(R_n,{\sigma }_i{R}_n\right)=\ker (G{L}_{n-1}\left(R_n\right)⟶G{L}_{n-1}\left(R_n/{\sigma }_i{R}_n\right)).$$

$$IS{L}_{n-1,i}\left(H\right)$$

$$I{E}_{n-1,i}\left(H\right)$$

$$I{G}_{n,m^2}$$

$$I{G}_{n,m^4}$$

$$I{E}_{n-1,i}\left(H_{n,m^2}\right)\subseteq I{A}_n^m.$$

$$S{K}_1\left(R,H;d\right)=S{L}_d\left(R,H\right)/E{}_d\left(R,H\right)$$

$$S{L}_{n-1}\left(R_n,H_{n,m}\right)/E{}_{n-1}\left(R_n,H_{n,m}\right)=S{K}_1\left(R,H_{n,m};n-1\right)$$

$$IS{L}_{n-1,i}\left(H_{n,m}\right)/I{E}_{n-1,i}\left(H_{n,m}\right)\le S{L}_{n-1}\left(R_n,H_{n,m}\right)/E{}_{n-1}\left(R_n,H_{n,m}\right)$$

$$C\left(IA\left({\Phi }_n\right),{\Phi }_n\right)$$

$$I{A}_{n,m}=\cap \left\{N⊲IA\left({\Phi }_n\right)|[IA\left({\Phi }_n\right):N]|m\right\}$$

$$I{A}_{n,m}\cdot IS{L}_{n-1,i}\left({\sigma }_i{H}_{n,m^2}\right)/I{A}_{n,m}$$

$$I{A}_n^m\cdot IS{L}_{n-1,i}\left({\sigma }_i{H}_{n,m^2}\right)/I{A}_n^m$$

$$I{A}_n^m\cdot \prod _{i=1}^{n}IS{L}_{n-1,i}\left({\sigma }_i{H}_{n,m^2}\right)/I{A}_n^m$$

Full text

Considering $\Phi_{n}$, the free metabelian group on $n$ generators, it was also proven that $C\left(\Phi_{2}\right)=\hat{F}_{\omega}$ and $C\left(\Phi_{3}\right)\supseteq\hat{F}_{\omega}$. In this paper we prove that $C\left(\Phi_{n}\right)$ for $n\geq 4$ is abelian. So, while the dichotomy in the abelian case is between $n=2$ and $n\geq 3$, in the metabelian case it is between $n=2,3$ and $n\geq 4$.

Mathematics Subject Classification (2010): Primary: 19B37, 20H05, Secondary: 20E36, 20E18.

Key words and phrases: congruence subgroup problem, automorphism groups, profinite groups, free metabelian groups.

1. 1 Introduction

2. 2 Some properties of $IA\left(\Phi_{n}\right)$ and its subgroups

3. 3 The main theorem’s proof

4. 4 Some elementary elements of $\left\langle IA\left(\Phi_{n}\right)^{m}\right\rangle$

5. 4.1 Elementary elements of type 1

6. 4.2 Elementary elements of type 2

7. 5 A main lemma

8. 5.1 Reducing Lemma 3.1’s proof

9. 5.2 A technical lemma

10. 5.3 Finishing Lemma 3.1’s proof

11. 6 Remarks and problems for further research

12. 7 Index of notations

1 Introduction

The classical congruence subgroup problem (CSP) asks for, say, $G=SL_{n}\left(\mathbb{Z}\right)$ or $G=GL_{n}\left(\mathbb{Z}\right)$, whether every finite index subgroup of $G$ contains a principal congruence subgroup, i.e. a subgroup of the form $G\left(m\right)=\ker\left(G\to GL_{n}\left(\mathbb{Z}/m\mathbb{Z}\right)\right)$ for some $0\neq m\in\mathbb{Z}$. Equivalently, it asks whether the natural map $\hat{G}\to GL_{n}(\hat{\mathbb{Z}})$ is injective, where $\hat{G}$ and $\hat{\mathbb{Z}}$ are the profinite completions of the group $G$ and the ring $\mathbb{Z}$, respectively. More generally, the CSP asks what is the kernel of this map. It is a classical $19^{\underline{th}}$ century result that the answer is negative for $n=2$. Moreover (but not so classical, cf. [Mel], [L]), the kernel in this case is $\hat{F}_{\omega}$ - the free profinite group on a countable number of generators. On the other hand, it was proved in the sixties by Mennicke [Men] and Bass-Lazard-Serre [BLS] that for $n\geq 3$ the answer is affirmative, and the kernel is therefore trivial.

By the observation $GL_{n}\left(\mathbb{Z}\right)\cong Aut\left(\mathbb{Z}^{n}\right)=Out\left(\mathbb{Z}^{n}\right)$, the CSP can be generalized as follows: Let $\Gamma$ be a group and $G\leq Aut\left(\Gamma\right)$ (resp. $G\leq Out\left(\Gamma\right)$). For a finite index characteristic subgroup $M\leq\Gamma$ denote

[TABLE]

Such a $G\left(M\right)$ will be called a “principal congruence subgroup” and a finite index subgroup of $G$ which contains $G\left(M\right)$ for some $M$ will be called a “congruence subgroup”. The CSP for the pair $\left(G,\Gamma\right)$ asks whether every finite index subgroup of $G$ is a congruence subgroup. In some sense, the CSP tries to understand whether every finite quotient of $G$ comes from a finite quotient of $\Gamma$.

One can easily see that the CSP is equivalent to the question: Is the congruence map $\hat{G}=\underleftarrow{\lim}G/U\to\underleftarrow{\lim}G/G\left(M\right)$ injective? Here, $U$ ranges over all finite index normal subgroups of $G$, and $M$ ranges over all finite index characteristic subgroups of $\Gamma$. When $\Gamma$ is finitely generated, it has only finitely many subgroups of given index $m$, and thus, the charateristic subgroups $M_{m}=\cap\left\{\Delta\leq\Gamma\,|\,\left[\Gamma:\Delta\right]=m\right\}$ are of finite index in $\Gamma$. Hence, one can write $\hat{\Gamma}=\underleftarrow{\lim}_{m\in\mathbb{N}}\Gamma/M_{m}$ and have111By the celebrated theorem of Nikolov and Segal which asserts that every finite index subgroup of a finitely generated profinite group is open [NS], the second inequality is actually an equality. However, we do not need it.

[TABLE]

Therefore, when $\Gamma$ is finitely generated, the CSP is equivalent to the question: Is the congruence map $\hat{G}\to Aut(\hat{\Gamma})$ (resp. $\hat{G}\to Out(\hat{\Gamma})$) injective? More generally, the CSP asks what is the kernel $C\left(G,\Gamma\right)$ of this map. For $G=Aut\left(\Gamma\right)$ we will also use the simpler notation $C\left(\Gamma\right)=C\left(G,\Gamma\right)$.

The classical congruence subgroup results mentioned above can therefore be reformulated as $C\left(\mathbb{Z}^{2}\right)=\hat{F}_{\omega}$ while $C\left(\mathbb{Z}^{n}\right)=\left\{e\right\}$ for $n\geq 3$. So the finite quotients of $GL_{n}\left(\mathbb{Z}\right)$ are closely related to the finite quotients of $\mathbb{Z}^{n}$ when $n\geq 3$, but the finite quotients of $GL_{2}\left(\mathbb{Z}\right)$ are far from being understandable by the finite quotients of $\mathbb{Z}^{2}$.

Very few results are known when $\Gamma$ is non-abelian. Most of the results are related to $\Gamma=\pi_{g,n}$, the fundamental group of $S_{g,n}$, the closed surface of genus $g$ with $n$ punctures. In these cases one can take $G=PMod\left(S_{g,n}\right)$, the pure mapping class group of $S_{g,n}$, and can naturally view it as a subgroup of $Out\left(\pi_{g,n}\right)$ (cf. [FM], chapter 8). Considering these cases, it is known that:

Theorem 1.1**.**

For $g=0,1,2$ and every $n\geq 0,1,0$ respectively, we have $C\left(PMod\left(S_{g,n}\right),\pi_{g,n}\right)=\left\{1\right\}$.

Note that when $g=1$ and $n=0$, $\pi_{1,0}\cong\mathbb{Z}^{2}$ and $PMod\left(S_{1,0}\right)\cong SL_{2}\left(\mathbb{Z}\right)$, so $C\left(PMod\left(S_{1,0}\right),\pi_{1,0}\right)=C\left(SL_{2}\left(\mathbb{Z}\right),\mathbb{Z}^{2}\right)=\hat{F}_{\omega}$. The cases for $g=0$ were proved in [DDH] (see also [Mc]), the cases for $g=1$ were proved in [A] (see also [Bo1], [BER]), and the cases for $g=2$ were proved in [Bo1] (see also [Bo2] for the specific case where $g=2$ and $n=0$). In particular, as $PMod\left(S_{1,1}\right)$ is isomorphic to the special outer-automorphism group of $F_{2}$, we have an affirmative answer for the full outer-automorphism group of $F_{2}$, and by some standard arguments it shows that actually $C\left(F_{2}\right)$ is trivial (see [BER], [BL]). Note that for every $n>0$, $\pi_{g,n}\cong F_{2g+n-1}$ = the free group on $2g+n-1$ generators. Hence, the above solved cases give an affirmative answer for various subgroups of the outer-automorphism group of finitely generated free groups, while the CSP for the full $Aut\left(F_{d}\right)$ when $d\geq 3$ is still unsettled, and so is the situation with $PMod\left(S_{g,n}\right)$ when $g\geq 3$.

All the above settled cases have a common property which plays a crucial role in the proof of Theorem 1.1: There is an intrinsic description of $G$ by iterative extension process by virtually free groups (groups which have a finite index free subgroup). Actually, in these cases, in some sense, we do understand the finite quotients of $G$, and the CSP tells us that these quotients are closely related to the finite quotients of $\Gamma$. This situation changes when we pass to $G=Aut\left(F_{d}\right)$ for $d\geq 3$ or $PMod\left(S_{g,n}\right)$ for $g\geq 3$. In these cases we do not have a description of $G$ that can help to understand the finite quotients of $G$. So in some sense, all the known cases do not give us a new understanding of the finite quotients of $G$. Considering the abelian case, what makes the result of Mennicke and Bass-Lazard-Serre so special is that it not only shows that the finite quotients of $GL_{n}\left(\mathbb{Z}\right)$ are related to the finite quotients of $\mathbb{Z}^{n}$, but also gives us a description of the finite quotients of $GL_{n}\left(\mathbb{Z}\right)$, which we have not known without this result.

Denote now the free metabelian group on $n$ generators by $\Phi_{n}=F_{n}/F_{n}^{\prime\prime}$. Considering the metabelian case, it was shown in [BL] (see also [Be1]) that $C\left(\Phi_{2}\right)=\hat{F}_{\omega}$. In addition, it was proven there that $C\left(\Phi_{3}\right)\supseteq\hat{F}_{\omega}$. So, the finite quotients of $Aut\left(\Phi_{2}\right)$ and $Aut\left(\Phi_{3}\right)$ are far from being connected to the finite quotients of $\Phi_{2}$ and $\Phi_{3}$, respectively.

Here comes the main theorem of this paper:

Theorem 1.2**.**

For every $n\geq 4$, $C\left(IA\left(\Phi_{n}\right),\Phi_{n}\right)$ is central in $\widehat{IA\left(\Phi_{n}\right)}$, where

[TABLE]

Using the commutative exact diagram (see $\oint$6)

[TABLE]

and the fact that $\widehat{GL_{n}\left(\mathbb{Z}\right)}\to GL_{n}(\hat{\mathbb{Z}})$ is injective for $n\geq 3$, we obtain that $C\left(IA\left(\Phi_{n}\right),\Phi_{n}\right)$ is mapped onto $C\left(\Phi_{n}\right)$. Therefore we deduce that:

Theorem 1.3**.**

For every $n\geq 4$, $C\left(\Phi_{n}\right)$ is abelian.

This is dramatically different from the cases of $n=2,3$ described above. Theorem 1.3 tells us that when $n\geq 4$ the situation changes, and the finite quotients of $Aut\left(\Phi_{n}\right)$ are closely related to the finite quotients of $\Phi_{n}$ in the following manner:

Corollary 1.4**.**

Let $n\geq 4$. Then, for every finite index subgroup $H\leq G=Aut\left(\Phi_{n}\right)$, there exists a finite index characteristic subgroup $M\leq\Phi_{n}$ and $r\in\mathbb{N}$ such that $G\left(M\right)^{\prime}G\left(M\right)^{r}\subseteq H$.

Note that by a theorem of Bachmuth and Mochizuki [BM2], $Aut\left(F_{n}\right)\to Aut\left(\Phi_{n}\right)$ is surjective for every $n\geq 4$, and thus $G=Aut\left(\Phi_{n}\right)$ is finitely generated. Hence, the principal congruence subgroups of the form $G\left(M\right)$ are finitely generated, and thus, the subgroups of the form $G\left(M\right)^{\prime}G\left(M\right)^{r}$ are also of finite index in $Aut\left(\Phi_{n}\right)$. Therefore, the quotients of the form $Aut\left(\Phi_{n}\right)/G\left(M\right)^{\prime}G\left(M\right)^{r}$ describe all the finite quotients of $Aut\left(\Phi_{n}\right)$. In particular, our theorem gives us a description of the finite quotients of $Aut\left(\Phi_{n}\right)$ when $n\geq 4$ - just like the theorem of [Men] and [BLS] gives for $GL_{n}\left(\mathbb{Z}\right)$ when $n\geq 3$. Corollary 1.4 obviously does not hold for $n=2,3$. So, the picture is that while the dichotomy in the abelian case is between $n=2$ and $n\geq 3$, in the metabelian case we have a dichotomy between $n=2,3$ and $n\geq 4$.

In [KN], Kassabov and Nikolov showed that $\ker(\widehat{SL_{n}\left(\mathbb{Z}\left[x\right]\right)}\to SL_{n}(\widehat{\mathbb{Z}\left[x\right]}))$ is central and not finitely generated, when $n\geq 3$. In [Be2] we use their techniques and an interesting surjective representation

[TABLE]

to show also that:

Theorem 1.5**.**

For every $n\geq 4$, $C\left(IA\left(\Phi_{n}\right),\Phi_{n}\right)$ is not finitely generated.

We remark that despite the result of the latter theorem, we do not know whether $C\left(\Phi_{n}\right)$ is also not finitely generated. In fact we cannot even prove at this point that it is not trivial (for more, see $\oint$6).

We would like now to give a conceptual explanation for the dichotomy between $n=2,3$ and $n\geq 4$. Let us recall first the strategy of Bachmuth and Mochizuki [BM2], showing that the natural map $Aut\left(F_{n}\right)\to Aut\left(\Phi_{n}\right)$ is surjective for $n\geq 4$. They start with the observation that as $Aut\left(F_{n}\right)\to GL_{n}(\mathbb{Z})$ is surjective, it is enough to show that every element of $IA\left(\Phi_{n}\right)$ is induced by an element of $IA\left(F_{n}\right)=\ker\left(Aut\left(F_{n}\right)\to Aut\left(F_{n}/F^{\prime}_{n}\right)=GL_{n}\left(\mathbb{Z}\right)\right)$. From here, the basic background for their proof are the following facts:

• •

For every $n$ (not only for $n\geq 4$), $IA\left(\Phi_{n}\right)$ can naturally be viewed as a subgroup of $GL_{n}\left(R_{n}\right)$ where $R_{n}=\mathbb{Z}[x_{1}^{\pm 1},\ldots,x_{n}^{\pm 1}]$ is the free Laurent polynomial ring on $n$ commutative variables over $\mathbb{Z}$ (see $\oint$2 for the details). It can be easily shown that for every $1\leq i\leq n$, this embedding contains a copy of the group

[TABLE]

• •

By a classical result of Magnus ([MKS], Chapter 3, Theorem N4) the group $IA\left(F_{n}\right)$ is finitely generated by a well described generating set of its elements, say $S$ (also here, it is true for every $n$). Denote the image of $S$ in $IA\left(\Phi_{n}\right)$ by $\bar{S}$.

The technique of [BM2] was to show that when $n\geq 4$ the set $\bar{S}$ generates the whole of $IA\left(\Phi_{n}\right)$. In Section 3, [BM2] start with presenting a variety of types of elements that are contained in the subgroup of $IA\left(\Phi_{n}\right)$ generated by $\bar{S}$ - here, [BM2] already needed the assumption of $n\geq 4$. In section 5, [BM2] show that every element $g\in IA\left(\Phi_{n}\right)$ can be written as a product of elements

[TABLE]

where $k_{i}\in IGL_{n-1,i}$ and $h_{i}$ are elements generated by $\bar{S}$ (by Section 3) - so it remains to show that $IGL_{n-1,i}$ is generated by $\bar{S}$. Then, in the beginning of Section 4, by some classical results from Algebraic K-Theory, [BM2] manage to give a description for a generating set to $IGL_{n-1,i}$. From here, the rest of Section 4 is devoted to show that the generating set of $IGL_{n-1,i}$ can be built from the elements of $\bar{S}$.

The aforementioned results from Algebraic K-Theory are strongly leaning on the assumption $n\geq 4$, i.e. $n-1\geq 3$. Actually, when $n=3$, i.e. $n-1=2$, the situation is completely different, and leads to the fact that $Aut(\Phi_{3})$ is not finitely generated (See [BM1]). In some sense, what we show in this paper is that this difference between $Aut(\Phi_{3})$ and $Aut(\Phi_{n\geq 4})$, combined with the dichotomy between $n=2$ and $n\geq 3$ in the CSP for the abelian case, induces a dichotomy between $n=2,3$ and $n\geq 4$ in the CSP for the metabelian case.

The main line of the proof of Theorem 1.2, is as follows: For $G=IA\left(\Phi_{n}\right)$ we first take the principal congruence subgroups $G\left(M_{n,m}\right)$ where $M_{n,m}=\left(\Phi^{\prime}_{n}\Phi_{n}^{m}\right)^{\prime}\left(\Phi^{\prime}_{n}\Phi_{n}^{m}\right)^{m}$. By [Be1], $\hat{\Phi}_{n}=\underleftarrow{\lim}\left(\Phi_{n}/M_{n,m}\right)$, and thus we deduce that the subgroups of the form $G\left(M_{n,m}\right)$ are enough to represent the congruence subgroups of $IA(\Phi_{n})$ in the sense that every congruence subgroup contains one of these principal congruence subgroups. Then, we follow the steps of the theorem of Bachmuth and Mochizuki [BM2], showing that $Aut\left(F_{n}\right)\to Aut\left(\Phi_{n}\right)$ is surjective for $n\geq 4$, and we try to build $G\left(M_{n,m}\right)$ using elements of $\left\langle IA\left(\Phi_{n}\right)^{m}\right\rangle$.

Along this paper, mostly in $\oint$4 and Claim 5.12, we present a variety of types of elements that are contained in $\left\langle IA\left(\Phi_{n}\right)^{m}\right\rangle$. In $\oint$5 we prove a main lemma, which can be viewed as a counterpart of Section 5 in [BM2]. A counterpart of Section 4 in [BM2] is proven in Section 7 of [Be2] (see Lemma 3.2 in this paper). These parts are combined together in $\oint$3, and by some additional results from algebraic K-theory we get that for every $m$

[TABLE]

is finite and central in $IA\left(\Phi_{n}\right)/\left\langle IA\left(\Phi_{n}\right)^{m}\right\rangle$. Hence, $\left\langle IA\left(\Phi_{n}\right)^{m}\right\rangle$ is of finite index in $IA\left(\Phi_{n}\right)$. In particular, as every normal subgroup of index $m$ in $IA\left(\Phi_{n}\right)$ contains $\left\langle IA\left(\Phi_{n}\right)^{m}\right\rangle$, we deduce that the groups of the form $\left\langle IA\left(\Phi_{n}\right)^{m}\right\rangle$ are enough to represent the finite index subgroups of $IA\left(\Phi_{n}\right)$. From here, it follows easily that $C\left(IA\left(\Phi_{n}\right),\Phi_{n}\right)$ is central in $\widehat{IA\left(\Phi_{n}\right)}$ (see Corollary 3.4).

We hope that the solution of the free metabelian case will help to understand some new cases of non-abelian groups, such as the automorphism group of a free group and the mapping class group of a surface. The immediate next challenges are the automorphism groups of free solvable groups.

Let us point out that, as remarked in$\oint$5 in [BL], one can deduce from Theorem 1.3 that for every $n\geq 4$, $Aut\left(\Phi_{n}\right)$ is not large, i.e does not contain a finite index subgroup which can be mapped onto a free group. This is in contrast with $Aut\left(\Phi_{2}\right)$ and $Aut\left(\Phi_{3}\right)$ which are large.

The paper is organized as follows: In $\oint$2 we present some notations and discuss $IA\left(\Phi_{n}\right)$ and some of its subgroups. Then, up to a main lemma, in $\oint$3 we prove the main theorem of the paper, Theorem 1.2. In $\oint$4 we present some elements of $\left\langle IA\left(\Phi_{n}\right)^{m}\right\rangle$ which we use in the proof of the main lemma. In $\oint$5 we prove the main lemma. We end the paper with the proof of Theorem 1.3, and some remarks on the problem of computing $C\left(\Phi_{n}\right)$ and $C\left(IA\left(\Phi_{n}\right),\Phi_{n}\right)$.

Acknowledgements: I wish to offer my deepest thanks to my great supervisor Prof. Alexander Lubotzky for his sensitive and devoted guidance, and to the Rudin foundation trustees for their generous support during the period of the research.

2 Some properties of $IA\left(\Phi_{n}\right)$

and its subgroups

Let $G=IA\left(\Phi_{n}\right)=\ker\left(Aut\left(\Phi_{n}\right)\to Aut\left(\Phi_{n}/\Phi^{\prime}_{n}\right)=GL_{n}\left(\mathbb{Z}\right)\right)$. We start with recalling some of the properties of $G=IA\left(\Phi_{n}\right)$ and its subgroups, as presented in Section 3 in [Be2]. We also refer the reader to [Be2] for the proofs of the statements in this section. We start with the following notations:

• •

$\Phi_{n}=F_{n}/F^{\prime\prime}_{n}$= the free metabelian group on $n$ elements. Here $F^{\prime\prime}_{n}$ denotes the second derivative of $F_{n}$, the free group on $n$ elements.

• •

$\Phi_{n,m}=\Phi_{n}/M_{n,m}$, where $M_{n,m}=\left(\Phi^{\prime}_{n}\Phi_{n}^{m}\right)^{\prime}\left(\Phi^{\prime}_{n}\Phi_{n}^{m}\right)^{m}$.

• •

$IG_{n,m}=G(M_{n,m})=\ker\left(IA\left(\Phi_{n}\right)\to Aut\left(\Phi_{n,m}\right)\right).$

• •

$IA_{n}^{m}=\left\langle IA\left(\Phi_{n}\right)^{m}\right\rangle$.

• •

$R_{n}=\mathbb{Z}[\mathbb{Z}^{n}]=\mathbb{Z}[x_{1}^{\pm 1},\ldots,x_{n}^{\pm 1}]$ where $x_{1},\ldots,x_{n}$ are the generators of $\mathbb{Z}^{n}$.

• •

$\mathbb{Z}_{m}=\mathbb{Z}/m\mathbb{Z}$.

• •

$\sigma_{i}=x_{i}-1$ for $1\leq i\leq n$. We also denote by $\vec{\sigma}$ the column vector which has $\sigma_{i}$ in its $i$-th entry.

• •

$\mathfrak{A}_{n}=\sum_{i=1}^{n}\sigma_{i}R_{n}$ = the augmentation ideal of $R_{n}$.

• •

$H_{n,m}=\ker\left(R_{n}\to\mathbb{Z}_{m}[\mathbb{Z}_{m}^{n}]\right)=\sum_{i=1}^{n}\left(x_{i}^{m}-1\right)R_{n}+mR_{n}$.

By the well known Magnus embedding (see [Bi], [RS], [Ma]), one can identify $\Phi_{n}$ with the matrix group

[TABLE]

where $t_{i}$ is a free basis for $R_{n}$-module, under the identification of the generators of $\Phi_{n}$ with the matrices

[TABLE]

Moreover, for every $\alpha\in IA\left(\Phi_{n}\right)$, one can describe $\alpha$ by its action on the generators of $\Phi_{n}$, by

[TABLE]

and this description gives an injective homomorphism (see [Ba], [Bi])

[TABLE]

which gives an identification of $IA\left(\Phi_{n}\right)$ with the subgroup

[TABLE]

One can find the proof of the following proposition in [Be2] (Propositions 3.1 and 3.2):

Proposition 2.1**.**

Let $I_{n}+A\in IA\left(\Phi_{n}\right)$. Then:

• •

If one denotes the entries of $A$ by $a_{k,l}$ for $1\leq k,l\leq n$, then for every $1\leq k,l\leq n$, $a_{k,l}\in\sum_{l\neq i=1}^{n}\sigma_{i}R_{n}\subseteq\mathfrak{A}_{n}$.

• •

$\det\left(I_{n}+A\right)$* is of the form $\det\left(I_{n}+A\right)=\prod_{r=1}^{n}x_{r}^{s_{r}}$ for some $s_{r}\in\mathbb{Z}$.*

Consider now the map

[TABLE]

which induced by the projections $\mathbb{Z}^{n}\to\mathbb{Z}_{m}^{n}$, $R_{n}=\mathbb{Z}[\mathbb{Z}^{n}]\to\mathbb{Z}_{m}[\mathbb{Z}_{m}^{n}]$. Using result of Romanovskiĭ [Rom], it is shown in [Be1] that this map is surjective and that $\Phi_{n,m}$ is canonically isomorphic to its image. Therefore, we can identify the principal congruence subgroup of $IA\left(\Phi_{n}\right)$, $IG_{n,m}$, with

[TABLE]

Let us step forward with the following definitions:

Definition 2.2**.**

Let $A\in GL_{n}\left(R_{n}\right)$, and for $1\leq i\leq n$, denote by $A_{i,i}$ the minor which obtained from $A$ by erasing its $i$-th row and $i$-th column. Now, for every $1\leq i\leq n$, define the subgroup $IGL_{n-1,i}\leq IA\left(\Phi_{n}\right)$, by

[TABLE]

where:

[TABLE]

The following proposition is proven in [Be2] (Proposition 3.4):

Proposition 2.3**.**

For every $1\leq i\leq n$ we have $IGL_{n-1,i}\cong GL_{n-1}\left(R_{n},\sigma_{i}R_{n}\right)$.

We recall the following definitions from Algebraic K-Theory:

Definition 2.4**.**

Let $R$ be a commutative ring (with identity), $H\vartriangleleft R$ an ideal, and $d\in\mathbb{N}$. Then:

• •

$E_{d}\left(R\right)=\left\langle I_{d}+rE_{i,j}\,|\,r\in R,\,1\leq i\neq j\leq d\right\rangle\leq SL_{d}\left(R\right)$ where $E_{i,j}$ is the matrix which has $1$ in the $\left(i,j\right)$-th entry and [math] elsewhere.

• •

$SL_{d}\left(R,H\right)=\ker\left(SL_{d}\left(R\right)\to SL_{d}\left(R/H\right)\right)$.

• •

$GL_{d}\left(R,H\right)=\ker\left(GL_{d}\left(R\right)\to GL_{d}\left(R/H\right)\right).$

• •

$E_{d}\left(R,H\right)$ = the normal subgroup of $E_{d}\left(R\right)$, which is generated as a normal subgroup by the elementary matrices of the form $I_{d}+hE_{i,j}$ for $h\in H$.

Under the above identification of $IGL_{n-1,i}$ with $GL_{n-1}\left(R_{n},\sigma_{i}R_{n}\right)$, for every $1\leq i\leq n$ we define:

Definition 2.5**.**

Let $H\vartriangleleft R_{n}$. Then:

[TABLE]

3 The main theorem’s proof

Using the above notations we prove in $\oint$5 the following main lemma:

Lemma 3.1**.**

For every $n\geq 4$ and $m\in\mathbb{N}$ one has

[TABLE]

Observe that it follows that when $n\geq 4$, then for every $m\in\mathbb{N}$

[TABLE]

The following Lemma is proved in [Be2], using classical results from Algebraic K-theory (Lemma 7.1 in [Be2]):

Lemma 3.2**.**

For every $n\geq 4$, $1\leq i\leq n$ and $m\in\mathbb{N}$ one has

[TABLE]

Let us now quote the following proposition (see [Be2], Corollary 2.3):

Proposition 3.3**.**

Let $R$ be a commutative ring, $H\vartriangleleft R$ ideal of finite index and $d\geq 3$. Assume also that $E_{d}\left(R\right)=SL_{d}\left(R\right)$. Then:

[TABLE]

is a finite group which is central in $GL_{d}\left(R\right)/E{}_{d}\left(R,H\right)$.

Now, according to Proposition 3.3 and the fact that $E_{d}\left(R_{n}\right)=SL_{d}\left(R_{n}\right)$ for every $d\geq 3$ [Su], we obtain that for every $n\geq 4$

[TABLE]

is a finite group. Thus

[TABLE]

is also a finite group. Hence, the conclusion from Lemmas 3.1 and 3.2 is that for every $m\in\mathbb{N}$, one can cover $IG_{n,m^{4}}$ with finite number of cosets of $IA_{n}^{m}$. As $IG_{n,m^{4}}$ is obviously a finite index subgroup of $IA\left(\Phi_{n}\right)$ we deduce that $IA_{n}^{m}$ is also a finite index subgroup of $IA\left(\Phi_{n}\right)$. Therefore, as every normal subgroup of $IA\left(\Phi_{n}\right)$ of index $m$ cotains $IA_{n}^{m}$ we deduce that one can write explicitely $\widehat{IA\left(\Phi_{n}\right)}=\underleftarrow{\lim}\left(IA\left(\Phi_{n}\right)/IA_{n}^{m}\right)$. On the other hand, it is proven in [Be1] that $\hat{\Phi}_{n}=\underleftarrow{\lim}\Phi_{n,m}$, and thus:

Corollary 3.4**.**

For every $n\geq 4$

[TABLE]

Now, Proposition 3.3 gives us also that for every $m\in\mathbb{N}$ and $n\geq 4$, the subgroup $SK_{1}\left(R_{n},H_{n,m};n-1\right)$ is central in $GL_{n-1}\left(R_{n}\right)/E{}_{n-1}\left(R_{n},H_{n,m}\right)$. This fact is used in Section 5 of [Be2] to prove that if we define

[TABLE]

then for every $n\geq 4$, $m\in\mathbb{N}$ and $1\leq i\leq n$ the subgroup

[TABLE]

is central in $IA\left(\Phi_{n}\right)/IA_{n,m}$. Completely similar arguments yield the following result222The only property of $IA_{n,m}$ used in Chapter 5 of [Be2] is that $IA_{n}^{m}\subseteq IA_{n,m}$.:

Proposition 3.5**.**

For every $n\geq 4$, $m\in\mathbb{N}$ and $1\leq i\leq n$ the subgroup

[TABLE]

is central in $IA\left(\Phi_{n}\right)/IA_{n}^{m}$.

Corollary 3.6**.**

For every $n\geq 4$ and $m\in\mathbb{N}$ the elements of the set

[TABLE]

belong to the center of $IA\left(\Phi_{n}\right)/IA_{n}^{m}$.

The conclusion from the latter corollary is that for every $n\geq 4$ and $m\in\mathbb{N}$, the set

[TABLE]

is an $abelian$ $group$ which contained in the center of $IA\left(\Phi_{n}\right)/IA_{n}^{m}$. In particular, $IA_{n}^{m}\cdot IG_{n,m^{4}}/IA_{n}^{m}$ is contained in the center of $IA\left(\Phi_{n}\right)/IA_{n}^{m}$, and thus, by Corollary 3.4, $C\left(IA\left(\Phi_{n}\right),\Phi_{n}\right)$ is in the center of $\widehat{IA\left(\Phi_{n}\right)}$. This finishes, up to the proof of Lemma 3.1, the proof of Theorem 1.2.

So it remains to prove Lemma 3.1. But before we start to prove this lemma, we need to present some elements of $IA_{n}^{m}$. We will do this in the following section.

4 Some elementary elements of $\left\langle IA\left(\Phi_{n}\right)^{m}\right\rangle$

In this section we introduce some elements of $IA_{n}^{m}=\left\langle IA\left(\Phi_{n}\right)^{m}\right\rangle$ which are needed through the proof of Lemma 3.1. As one can see below, we separate the elementary elements to two types. In addition, we separate the treatment of the elements of type 1, to two parts. We hope this separation will make the process clearer.

Additionally to the previous notations, on the section, and also later on, we will use the notation

[TABLE]

4.1 Elementary elements of type 1

Proposition 4.1**.**

Let $n\geq 3$, $1\leq u\leq n$ and $m\in\mathbb{N}$. Denote by $\vec{e}_{i}$ the $i$-th row standard vector. Then, the elements of $IA\left(\Phi_{n}\right)$ of the form (the following notation means that the matrix is similar to the identity matrix, except the entries in the $u$-th row)

[TABLE]

when $\left(a_{u,1},\ldots,a_{u,u-1},0,a_{u,u+1},\ldots,a_{u,n}\right)$ is a linear combination of the vectors

[TABLE]

with coefficients in $R_{n}$, belong to $IA_{n}^{m}$.

Before proving this proposition, we present some more elements of this type. Note that for the following proposition we assume $n\geq 4$:

Proposition 4.2**.**

Let $n\geq 4$, $1\leq u\leq n$ and $m\in\mathbb{N}$. Then, the elements of $IA\left(\Phi_{n}\right)$ of the form

[TABLE]

when $\left(a_{u,1},\ldots,a_{u,u-1},0,a_{u,u+1},\ldots,a_{u,n}\right)$ is a linear combination of the vectors

[TABLE]

with coefficients in $R_{n}$, belong to $IA_{n}^{m}$.

Proof.

(of Proposition 4.1) Without loss of generality, we assume that $u=1$. Observe now that for every $a_{i},b_{i}\in R_{n}$ for $2\leq i\leq n$ one has

[TABLE]

Hence, it is enough to prove that the elements of the following forms belong to $IA_{n}^{m}$ (when we write $a\vec{e}_{i}$ we mean that the entry of the $i$-th column in the first row is $a$):

[TABLE]

We start with the elements of form 1. Here we have

[TABLE]

We pass to the elements of form 2. In this case we have

[TABLE]

We finish with the elements of form 3. If $k=i$, it is a special case of the previous case, so we assume $k\neq i$. So we assume that $i,j,k$ are all different from each other and $i,j,k\neq 1$ - observe that this case is interesting only when $n\geq 4$. The computation here is more complicated than in the previous cases, so we will demonstrate it for the special case: $n=4$, $i=2$, $j=3$, $k=4$. It is clear that symmetrically, with similar argument, the same holds in general when $n\geq 4$ for every $i,j,k\neq 1$ which different from each other. So

[TABLE]

∎

We pass now to the proof of Proposition 4.2.

Proof.

(of Proposition 4.2) Also here, without loss of generality, we assume that $u=1$. Thus, all we need to show is that also the elements of the following forms belong to $IA_{n}^{m}$:

[TABLE]

Also here, to simplify the notations, we will demonstrate the proof in the special case: $n=4$, $i=2$, $j=3$. We start with the first form. From Proposition 4.1 we have (an element of form 2 in Proposition 4.1)

[TABLE]

Therefore, we also have

[TABLE]

We pass to the elements of form 2. From Proposition 4.1 we have (an element of form 3 in Proposition 4.1)

[TABLE]

and therefore, we have

[TABLE]

∎

4.2 Elementary elements of type 2

Proposition 4.3**.**

Let $n\geq 4$, $1\leq u<v\leq n$ and $m\in\mathbb{N}$. Then, the elements of $IA\left(\Phi_{n}\right)$ of the form

[TABLE]

for $f\in H_{n,m}$, belong to $IA_{n}^{m}$.

Proof.

As before, to simplify the notations we will demonstrate the proof in the case: $n=4$, $u=1$ and $v=2$, and it will be clear from the computation that the same holds in the general case, provided $n\geq 4$.

First observe that for every $f,g\in R_{n}$ we have

[TABLE]

[TABLE]

so it is enough to consider the cases $f\in mR_{4}$ and $f\in\sigma_{r}\mu_{r,m}R_{4}$ for $1\leq r\leq 4$, separately. Consider now the following computation. For an arbitrary $f\in R_{n}$ we have

[TABLE]

Therefore, we conclude that if

[TABLE]

then also

[TABLE]

Thus, the cases $f\in mR_{4}$ and $f\in\sigma_{r}\mu_{r,m}R_{4}$ for $r\neq 4$, are obtained immediately from Proposition 4.1. Hence, it remains to deal with the case $f\in\sigma_{r}\mu_{r,m}R_{4}$ for $r=4$. However, it is easy to see that by switching the roles of $3$ and $4$, the remained case is also obtained by similar arguments. ∎

5 A main lemma

In this section we prove Lemma 3.1 which states that for every $n\geq 4$ and $m\in\mathbb{N}$ we have

[TABLE]

The proof will be presented in a few stages - each of which will be covered in a separate subsection. In this sections $n\geq 4$ will be constant, so we will make notations simpler and write

[TABLE]

We will also use the following notations:

[TABLE]

Notice that it follows from the definitions, that $H_{m}=\sum_{r=1}^{n}\sigma_{r}U_{r,m}+O_{m}$ (we note that in [Be2] the notation $U_{r,m}$ is used for $\sigma_{r}\mu_{r,m}R$).

Before we get deeply into the details, let us give an outline of the proof of the above main lemma. Given $0\leq u\leq n$, denote the ideal

[TABLE]

The lemma is proven by induction on $1\leq u\leq n$. Note that by Proposition 2.1 $IA(\Phi_{n})\subseteq GL_{n}(R,\mathfrak{A})=GL_{n}(R,\mathfrak{\tilde{A}}_{0})$. Now, let $g\in IG_{n,m^{2}}\cap GL_{n}(R,\mathfrak{\tilde{A}}_{u-1})$. If one could show that by multiplying it by elements of $IA^{m}$ and an element of $ISL_{n-1,u}\left(\sigma_{u}H_{n,m}\right)$ we can “push” $g$ to an element of $IG_{n,m^{2}}\cap GL_{n}(R,\mathfrak{\tilde{A}}_{u})$, then as $GL_{n}(R,\mathfrak{\tilde{A}}_{n})=\{I_{n}\}$, it will certainly be sufficient for proving the lemma. The issue is that the elements of $IA^{m}$ take us out from $IG_{n,m^{2}}$. Hence, we extend $IG_{n,m^{2}}\cap GL_{n}(R,\mathfrak{\tilde{A}}_{u-1})$ to a larger subgroup, denoted by $\mathbb{\mathbb{\tilde{J}}}_{m,u-1}$. In general, these subgroups do not satisfy $\mathbb{\mathbb{\tilde{J}}}_{m,u}\subseteq\mathbb{\mathbb{\tilde{J}}}_{m,u-1}$. However, in the delicate process described below we show that we can “push” $g\in\mathbb{\mathbb{\tilde{J}}}_{m,u-1}$ to an element of $\mathbb{\mathbb{\tilde{J}}}_{m,u}$ by elements of $IA^{m}$ and an element of $ISL_{n-1,u}\left(\sigma_{u}H_{n,m}\right)$. We go out from $\mathbb{\mathbb{\tilde{J}}}_{m,u-1}$ and get into $\mathbb{\mathbb{\tilde{J}}}_{m,u}$. The process ends when we get into $\mathbb{\mathbb{\tilde{J}}}_{m,n}=\left\{I_{n}\right\}$. The definition of $\mathbb{\mathbb{\tilde{J}}}_{m,u}$ is quite delicate, and so is the process.

In Subsection 5.1 we describe the above definitions and process is details. Then, in Subsection 5.2 we show that given an element of $\mathbb{\mathbb{\tilde{J}}}_{m,u-1}$, before “pushing” it into $\mathbb{\mathbb{\tilde{J}}}_{m,u}$, one can fix it a bit with elements of $\mathbb{\mathbb{\tilde{J}}}_{m,u-1}\cap IA^{m}$ to a more convenient form. Then, in Subsection 5.3 we define the “pushing elements” from $IA^{m}$ and $ISL_{n-1,u}\left(\sigma_{u}H_{n,m}\right)$.

5.1 Reducing Lemma 3.1’s proof

We start this subsection with introducing the following objects:

Definition 5.1**.**

Let $m\in\mathbb{{N}}$. Define

[TABLE]

Proposition 5.2**.**

For every $m\in\mathbb{N}$ we have

[TABLE]

Proof.

Let $x\in R$. Notice that $\sum_{i=0}^{m-1}x^{i}\in\left(x-1\right)R+mR$. In addition, by replacing $x$ by $x^{m}$ we obtain $\sum_{i=0}^{m-1}x^{mi}\in\left(x^{m}-1\right)R+mR$. Hence

[TABLE]

Thus, we obtain that $H_{m^{2}}=\sum_{r=1}^{n}(x_{r}^{m^{2}}-1)R+m^{2}R\subseteq J_{m}+O_{m}^{2}$. Now, let $I_{n}+A\in IG_{m^{2}}=IA\left(\Phi_{n}\right)\cap GL_{n}\left(R,H_{m^{2}}\right)$. From the above observation and from Proposition 2.1, it follows that every entry of $A$ belongs to $\left(J_{m}+O_{m}^{2}\right)\cap\mathfrak{A}=J_{m}$. In addition, by Proposition 2.1, the determinant of $I_{n}+A$ is of the form $\prod_{r=1}^{n}x_{r}^{s_{r}}$. On the other hand, we know that under the projection $R_{n}\to\mathbb{Z}_{m^{2}}[\mathbb{Z}_{m^{2}}^{n}]$ we have $I_{n}+A\mapsto I_{n}$ and thus also $\prod_{r=1}^{n}x_{r}^{s_{r}}=\det\left(I_{n}+A\right)\mapsto 1$. Therefore, $\det\left(I_{n}+A\right)$ is of the form $\prod_{r=1}^{n}x_{r}^{m^{2}s_{r}}$, as required. ∎

Corollary 5.3**.**

Let $n\geq 4$ and $m\in\mathbb{N}$. Then, for proving Lemma 3.1 it suffices to prove that

[TABLE]

We continue with defining the following objects:

Definition 5.4**.**

For $0\leq u\leq n$ and $1\leq v\leq n$, define the following ideals of $R=R_{n}=\mathbb{Z}[x_{1}^{\pm 1},\ldots,x_{n}^{\pm 1}]$:

[TABLE]

and for $0\leq u\leq n$ define the groups $\tilde{\mathbb{A}}_{u}=IA\left(\Phi_{n}\right)\cap GL_{n}(R,\mathfrak{\tilde{A}}_{u})$, and

[TABLE]

Remark 5.5*.*

If $I_{n}+A\in\mathbb{\mathbb{\tilde{J}}}_{m,u}$, the entries of the columns of $A$ may belong to different ideals in $R$, so it is not obvious that $\mathbb{\tilde{\mathbb{J}}}_{m,u}$ is indeed a group, i.e. closed under matrix multiplication and the inverse operation. However, showing that $\mathbb{\tilde{\mathbb{J}}}_{m,u}$ is a group is not difficult and we leave it to the reader.

Notice now the extreme cases:

1. For $u=0$ we have (for every $v$ and $m$) $\mathfrak{\tilde{A}}_{0}=\mathfrak{A}$, and $J_{m}\subseteq\tilde{J}_{m,0,v}$. Hence, we have $\mathbb{\mathbb{J}}_{m}\subseteq\mathbb{\mathbb{\tilde{J}}}_{m,0}$.

2. For $u=n$ we have (for every $v$ and $m$) $\mathfrak{\tilde{A}}_{n}=\tilde{J}_{m,n,v}=0$. Hence, we also have $\mathbb{\mathbb{\tilde{J}}}_{m,n}=\left\{I_{n}\right\}$.

Corollary 5.6**.**

For proving Lemma 3.1, it is enough to prove that for every $1\leq u\leq n$

[TABLE]

Proof.

Using that $IA^{m}$ is normal in $IA\left(\Phi_{n}\right)$ and the latter observations, under the above assumption, one obtains that

[TABLE]

which is the requirement of Corollary 5.3. ∎

We continue with defining the following objects:

Definition 5.7**.**

For $0\leq u\leq n$ and $1\leq v\leq n$, define the following ideals of $R$:

[TABLE]

and for $0\leq u\leq n$ define the group

[TABLE]

It follows from the definitions that for every $1\leq u\leq n$ we have:

$J_{m,u-1,v}\subseteq J_{m,u,v}$, but $\mathfrak{\tilde{A}}_{u-1}\supseteq\mathfrak{\tilde{A}}_{u}$. Thus, we have also 2. 2.

$\mathbb{J}_{m,u-1}\subseteq\mathbb{J}_{m,u}$, but $\tilde{\mathbb{A}}_{u-1}\supseteq\tilde{\mathbb{A}}_{u}$.

Here comes the connection between the latter objects to the objects defined in Definition 5.4.

Proposition 5.8**.**

For every $0\leq u\leq n$ and $1\leq v\leq n$ we have $J_{m,u,v}\cap\mathfrak{\tilde{A}}_{u}=\tilde{J}_{m,u,v}$, and hence $\mathbb{J}_{m,u}\cap\tilde{\mathbb{A}}_{u}=\mathbb{\mathbb{\tilde{\mathbb{J}}}}_{m,u}$.

Proof.

It is clear from the definitions that we have $\tilde{J}_{m,u,v}\subseteq J_{m,u,v}\cap\mathfrak{\tilde{A}}_{u}$, so we have to show an opposite inclusion. Let $a\in J_{m,u,v}\cap\mathfrak{\tilde{A}}_{u}$. As

[TABLE]

we can assume that $a\in\mathfrak{A}\left(\sum_{r=1}^{u}\mathfrak{A}\sigma_{r}U_{r,m}+\mathfrak{A}O_{m}+O_{m}^{2}\right)\cap\mathfrak{\tilde{A}}_{u}$.

Observe now that by dividing an element $b\in R$ by $\sigma_{u+1},\ldots,\sigma_{n}$ (with residue), one can present $b$ as a summand of an element of $\mathfrak{\tilde{A}}_{u}$ with an element of $R_{u}=\mathbb{Z}[x_{1}^{\pm 1},\ldots,x_{u}^{\pm 1}]$. Hence, $R=\mathfrak{\tilde{A}}_{u}+R_{u}$ and $\mathfrak{A}=\mathfrak{\tilde{A}}_{u}+\mathfrak{A}_{u}$, where $\mathfrak{A}_{u}$ is the augmentation ideal of $R_{u}$. Hence

[TABLE]

Hence, we can assume that $a\in\left(\mathfrak{A}_{u}^{2}\sum_{r=1}^{u}\sigma_{r}\mu_{r,m}R_{u}+\mathfrak{A}_{u}^{2}mR_{u}+\mathfrak{A}_{u}m^{2}R_{u}\right)\cap\mathfrak{\tilde{A}}_{u}=\left\{0\right\}$, i.e. $a=0\in\tilde{J}_{m,u,v}$, as required. ∎

Due to the above, we can now reduce Lemma 3.1’s proof as follows.

Corollary 5.9**.**

For proving Lemma 3.1 it suffices to show that given $1\leq u\leq n$, for every $\alpha\in\mathbb{\tilde{\mathbb{J}}}_{m,u-1}$ there exist $\beta\in IA^{m}\cap\mathbb{J}_{m,u}$ and $\gamma\in ISL_{n-1,u}\left(\sigma_{u}H_{m}\right)\cap\mathbb{\mathbb{J}}_{m,u}$ such that $\gamma\alpha\beta\in\tilde{\mathbb{A}}{}_{u}$.

Proof.

As clearly $\mathbb{J}_{m,u}\supseteq\mathbb{\mathbb{J}}_{m,u-1}\supseteq\mathbb{\mathbb{\tilde{J}}}_{m,u-1}$, we obtain from Proposition 5.8 that $\gamma\alpha\beta\in\tilde{\mathbb{A}}{}_{u}\cap\mathbb{J}_{m,u}=\mathbb{\mathbb{\tilde{J}}}_{m,u}$. Thus

[TABLE]

This yields that $\mathbb{\mathbb{\tilde{J}}}_{m,u-1}\subseteq IA^{m}\cdot ISL_{n-1,u}\left(\sigma_{u}H_{m}\right)\cdot\tilde{\mathbb{\mathbb{J}}}_{m,u}$ which is the requirement of Corollary 5.6. ∎

5.2 A technical lemma

In this section we will prove a technical lemma, which will help us in subsection 5.3 to prove Lemma 3.1. In the following subsections $1\leq u\leq n$ will be constant. We will use the following notations:

• •

For $a\in R$ we denote its image in $R_{u}$ under the projection $x_{u+1},\ldots,x_{n}\mapsto 1$ by $\bar{a}$. In addition, we denote its image in $R_{u-1}$ under the projection $x_{u},\ldots,x_{n}\mapsto 1$ by $\bar{\bar{a}}$.

• •

For $\alpha\in GL_{n}\left(R\right)$ we denote its image in $GL_{n}\left(R_{u}\right)$ under the projection $x_{u+1},\ldots,x_{n}\mapsto 1$ by $\bar{\alpha}$.

• •

Similarly, we will use the following notations for every $m\in\mathbb{N}$:

• –

$\mathfrak{\bar{A}}=\mathfrak{A}_{u}=\sum_{i=1}^{u}\sigma_{i}R_{u}$, $\bar{U}_{r,m}=\mu_{r,m}R_{u}$ for $1\leq r\leq u$, $\bar{O}_{m}=mR_{u}$ and $\bar{H}_{m}=H_{u,m}=\sum_{r=1}^{u}\sigma_{r}\bar{U}_{r,m}+\bar{O}_{m}$.

• –

$\bar{\bar{\mathfrak{A}}}=\mathfrak{A}_{u-1}=\sum_{i=1}^{u-1}\sigma_{i}R_{u-1}$, $\bar{\bar{U}}_{r,m}=\mu_{r,m}R_{u-1}$ for $1\leq r\leq u-1$ and $\bar{\bar{O}}_{m}=mR_{u-1}$.

Now, let $\alpha=I_{n}+A\in\mathbb{\mathbb{\tilde{J}}}_{m,u-1}$, and denote the entries of $A$ by $a_{i,j}$. Consider the $u$-th row of $A$. Under the above assumption, for every $v$ we have

[TABLE]

Hence we have

[TABLE]

We can state now the technical lemma:

Lemma 5.10**.**

Let $\alpha=I_{n}+A\in\mathbb{\tilde{\mathbb{J}}}_{m,u-1}$. Then, there exists $\delta\in IA^{m}\cap\mathbb{\mathbb{\tilde{J}}}_{m,u-1}$ such that for every $v\neq u$, the $\left(u,v\right)$-th entry of $\overline{\alpha\delta^{-1}}$ belongs to $\sigma_{u}^{2}\bar{H}_{m}$.

We will prove the lemma in two steps. Here is the first step:

Proposition 5.11**.**

Let $\alpha=I_{n}+A\in\mathbb{\tilde{\mathbb{J}}}_{m,u-1}$. Then, there exists $\delta\in IA^{m}\cap\mathbb{\mathbb{\tilde{J}}}_{m,u-1}$ such that for every $v<u$, the $\left(u,v\right)$-th entry of $\overline{\alpha\delta^{-1}}$ belongs to $\sigma_{u}^{2}\bar{H}_{m}$.

Proof.

So let $\alpha=I_{n}+A\in\mathbb{\mathbb{\tilde{J}}}_{m,u-1}$, and observe that for every $1\leq v\leq u-1$ one can write $\bar{a}_{u,v}=\sigma_{u}\bar{b}_{u,v}$ for some $\bar{b}_{u,v}\in\sum_{r=1}^{u-1}\mathfrak{\bar{A}}\sigma_{r}\bar{U}_{r,m}+\sigma_{u}^{2}\bar{U}_{u,m}+\mathfrak{\bar{A}}\bar{O}_{m}+\bar{O}_{m}^{2}$. In addition, as it is easy to see that

[TABLE]

[TABLE]

one can write $\bar{b}_{u,v}=\sigma_{u}\bar{c}_{u,v}+\bar{\bar{b}}_{u,v}$ for every $1\leq v\leq u-1$, for some

[TABLE]

Notice, that as $A$ satisfies the condition $A\vec{\sigma}=\vec{0}$ we have the equality $\sigma_{1}a_{u,1}+\ldots+\sigma_{n}a_{u,n}=0$, which yields the following equalities as well:

[TABLE]

Observe now that for every $1\leq v\leq u-1$ we have

[TABLE]

and thus, if we define

[TABLE]

then $\delta\in\tilde{\mathbb{\mathbb{J}}}_{m,u-1}$. We claim now that we also have $\delta\in IA^{m}$. We will prove this claim soon, but assuming this claim, we can now multiply $\alpha$ from the right by $\delta^{-1}\in\tilde{\mathbb{\mathbb{J}}}_{m,u-1}\cap IA^{m}$ and obtain an element in $\mathbb{\mathbb{\tilde{J}}}_{m,u-1}$ such that the image of its $\left(u,v\right)$-th entry for $1\leq v\leq u-1$, under the projection $x_{u+1},\ldots,x_{n}\mapsto 1$, is

[TABLE]

as required. ∎

So it remains to prove the following claim:

Claim 5.12*.*

Let $n\geq 4$, $1\leq u\leq n$, and $\bar{\bar{b}}_{u,v}\in\sum_{r=1}^{u-1}\bar{\bar{\mathfrak{A}}}\sigma_{r}\bar{\bar{U}}_{r,m}+\bar{\bar{\mathfrak{A}}}\bar{\bar{O}}_{m}+\bar{\bar{O}}_{m}^{2}$ for $1\leq v\leq u-1$ which satisfy the condition

[TABLE]

Then

[TABLE]

Proof.

It will be easier to prove a bit more - we will prove that if for every $1\leq v\leq u-1$

[TABLE]

then the vector $\vec{b}=(\bar{\bar{b}}_{u,1},\ldots,\bar{\bar{b}}_{u,u-1},0,\ldots,0)$ is a linear combination of the vectors

[TABLE]

with coefficients in $R_{u-1}.$ This will show that $\sigma_{u}(\bar{\bar{b}}_{u,1},\ldots,\bar{\bar{b}}_{u,u-1},0,\ldots,0)$ is a linear combination of the vectors in Propositions 4.1 and 4.2, so the claim will follow.

We start with expressing $\bar{\bar{b}}_{u,1}$ explicitly by writing

[TABLE]

for some $p_{i,r},\,q_{i,j},\,r\in R_{u-1}$. Now, Equation 5.3 gives that under the projection $\sigma_{2},\ldots,\sigma_{u-1}\mapsto 0$, $\bar{\bar{b}}_{u,1}\mapsto 0$. It follows that $\bar{\bar{b}}_{u,1}\in\sum_{i=2}^{u-1}\sigma_{i}R_{u-1}\subseteq\bar{\bar{\mathfrak{A}}}$. In particular, as obviously

[TABLE]

we also have $mr\in\bar{\bar{\mathfrak{A}}}$ and hence $r\in\bar{\bar{\mathfrak{A}}}$. Hence, we can write

[TABLE]

for some $p_{i,r},\,q_{i,j},\,r_{i}\in R_{u-1}$.

Observe now that by dividing $r_{1}$ by $\sigma_{2},\ldots,\sigma_{u-1}$ (with residue) we can write $r_{1}=r^{\prime}_{1}+\sum_{i=2}^{u-1}\sigma_{i}r^{\prime}_{i}$ where $r^{\prime}_{1}$ depends only on $x_{1}$. Therefore, by replacing $r_{1}$ by $r^{\prime}_{1}$ and $r_{i}$ by $r_{i}+\sigma_{1}r^{\prime}_{i}$ for $2\leq i\leq n$, we can assume that $r_{1}$ depends only on $x_{1}$. Similarly, by dividing $q_{1,1}$ by $\sigma_{2},\ldots,\sigma_{u-1}$, we can assume that $q_{1,1}$ depends only on $x_{1}$. Now, by replacing $\vec{b}$ with

[TABLE]

we can assume that $\bar{\bar{b}}_{u,1}$ is a polynomial which depends only on $x_{1}$. On the other hand, we already saw that Equation 5.3 yields that $\bar{\bar{b}}_{u,1}\in\sum_{i=2}^{u-1}\sigma_{i}R_{u-1}$, so we can actually assume that $\bar{\bar{b}}_{u,1}=0$.

We continue in this manner by induction. In the $1\leq v\leq u-1$ stage we assume that $\bar{\bar{b}}_{u,1}=\ldots\bar{\bar{b}}_{u,v-1}=0$. Then we write

[TABLE]

for some $p_{i,r},\,q_{i,j},\,r\in R_{u-1}$. The condition $\bar{\bar{b}}_{u,1}=\ldots=\bar{\bar{b}}_{u,v-1}=0$ and Equation 5.3 give that $\sigma_{v}\bar{\bar{b}}_{u,v}+\sigma_{v+1}\bar{\bar{b}}_{u,v+1}+\ldots+\sigma_{u-1}\bar{\bar{b}}_{u,u-1}=0$ and thus, under the projection $\sigma_{v+1},\ldots,\sigma_{u-1}\mapsto 0$, $\bar{\bar{b}}_{u,v}\mapsto 0$, so $\bar{\bar{b}}_{u,v}\in\sum_{i=v+1}^{u-1}\sigma_{i}R_{u-1}\subseteq\bar{\bar{\mathfrak{A}}}$. In particular, $r\in\bar{\bar{\mathfrak{A}}}$, so we can write

[TABLE]

for some $p_{i,r},\,q_{i,j},\,r_{i}\in R_{u-1}$.

Now, as we explained previously, by dividing $p_{i,r},\,q_{i,j},\,r_{i}$ for $1\leq i,j,r\leq v$ by $\sigma_{v+1},\ldots,\sigma_{u-1}$, we can assume that these polynomials depend only on $x_{1},\ldots,x_{v}$. Thus, by replacing $\vec{b}$ with

[TABLE]

we can assume that $\bar{\bar{b}}_{u,v}$ is a polynomial which depends only on $x_{1},\ldots,x_{v}$, without changing the assumption that $\bar{\bar{b}}_{u,w}=0$ for $w<v$. But we saw that in this situation Equation 5.3 yields that $\bar{\bar{b}}_{u,v}\in\sum_{i=v+1}^{u-1}\sigma_{i}R_{u-1}$, so we can actually assume that $\bar{\bar{b}}_{u,v}=0$, as required. ∎

This finishes the proof of Proposition 5.11. Here is the second step of the technical lemma’s proof:

Proposition 5.13**.**

Let $\alpha=I_{n}+A\in\mathbb{\tilde{\mathbb{J}}}_{m,u-1}$ such that for every $v<u$, $\bar{a}_{u,v}\in\sigma_{u}^{2}\bar{H}_{m}$. Then, there exists $\delta\in IA^{m}\cap\mathbb{\mathbb{\tilde{J}}}_{m,u-1}$ such that for every $v\neq u$, the $\left(u,v\right)$-th entry of $\overline{\alpha\delta^{-1}}$ belongs to $\sigma_{u}^{2}\bar{H}_{m}$.

Proof.

So let $\alpha=I_{n}+A\in\mathbb{\tilde{\mathbb{J}}}_{m,u-1}$ such that for every $v<u$, , $\bar{a}_{u,v}\in\sigma_{u}^{2}\bar{H}_{m}$. We remined that by Equation 5.2, for every $v>u$ we have $\bar{a}_{u,v}\in\sigma_{u}\left(\sum_{r=1}^{u-1}\mathfrak{\bar{A}}\sigma_{r}\bar{U}_{r,m}+\sigma_{u}^{2}\bar{U}_{u,m}+\mathfrak{\bar{A}}\bar{O}_{m}+\bar{O}_{m}^{2}\right)$. Hence, we can write explicitly

[TABLE]

for some $p_{r,i},\,q,\,r_{i},\,s\in R_{u}$. Clearly, as $\mathfrak{\bar{A}}\bar{O}_{m}\supseteq\mathfrak{\bar{A}}\bar{O}_{m}^{2}$, by dividing $s$ by $\sigma_{i}$ for $1\leq i\leq u$ (with residue), we can assume that $s\in\mathbb{Z}$. Consider now the following element:

[TABLE]

By the computation in the proof of Proposition 5.2, we obtain that

[TABLE]

and thus (we remind that $v>u$)

[TABLE]

In addition, the determinant of $\delta^{\prime}$ is $x_{v}^{m^{2}}$. Therefore, $\delta^{\prime}\in\mathbb{\mathbb{\tilde{J}}}_{m,u-1}$. Observe now that as $v>u$, under the projection $\sigma_{u+1},\ldots,\sigma_{n}\mapsto 0$, $x_{v}\mapsto 1$, and $\delta$ is therefore maped to

[TABLE]

Thus, if we multiply $\alpha$ from the right by $\delta^{\prime s}$ we obtain that the value of the entries in the $u$-th row under the projection $\sigma_{u+1},\ldots,\sigma_{n}\mapsto 0$ does not change, besides the value of the entry in the $v$-th colmun, which changes to (see Equation 5.2 for the ideal which contains $\bar{a}_{u,u}$)

[TABLE]

Hence, we can assume that $\bar{a}_{u,v}\in\sigma_{u}\sum_{i=1}^{u-1}\sigma_{i}f_{i}+\sigma_{u}^{2}\left(\sum_{r=1}^{u}\sigma_{r}\bar{U}_{r,m}+\bar{O}_{m}\right)=\sigma_{u}\sum_{i=1}^{u-1}\sigma_{i}f_{i}+\sigma_{u}^{2}\bar{H}_{m}$, for some $f_{i}\in\sum_{r=1}^{u-1}\sigma_{r}\bar{U}_{r,m}+\bar{O}_{m}$. Define now (the coefficient of $\vec{e}_{v}$ is the value of the $\left(u,v\right)$-th entry)

[TABLE]

By proposition 4.1, we obviously have $\delta_{v}\in IA^{m}$. In addition, as $v>u$, under the projection $\sigma_{u+1},\ldots,\sigma_{n}\mapsto 0$ we have

[TABLE]

Thus, by multiplying $\alpha$ from the right by $\bar{\delta}_{v}^{-1}$ we obtain that the value of the entries in the $u$-th row under the projection $\sigma_{u+1},\ldots,\sigma_{n}\mapsto 0$ does not change, besides the value of the entry in the $v$-th colmun, which changes to

[TABLE]

Thus, defininig $\delta=\prod_{v=u+1}^{n}\delta_{v}$ finishes the proof of the proposition, and hence, also the proof of the technical lemma. ∎

5.3 Finishing Lemma 3.1’s proof

We remind that we fixed a constant $1\leq u\leq n$. We remind also that by Corollary 5.9, it suffices to show that given $\alpha\in\mathbb{\tilde{\mathbb{J}}}_{m,u-1}$ there exist $\beta\in IA^{m}\cap\mathbb{J}_{m,u}$ and $\gamma\in ISL_{n-1,u}\left(\sigma_{u}H_{m}\right)\cap\mathbb{\mathbb{J}}_{m,u}$ such that $\gamma\alpha\beta\in\tilde{\mathbb{A}}{}_{u}$.

So let $\alpha=I_{n}+A\in\mathbb{\mathbb{\tilde{J}}}_{m,u-1}$. By the above technical lemma, there exists $\delta\in IA^{m}\cap\mathbb{\mathbb{\tilde{J}}}_{m,u-1}\subseteq IA^{m}\cap\mathbb{J}_{m,u}$ such that for every $v\neq u$, the $\left(u,v\right)$-th entry of $\overline{\alpha\delta^{-1}}$ belongs to $\sigma_{u}^{2}\bar{H}_{m}$. Thus, by replacing $\alpha$ with $\alpha\delta^{-1}$, with out loss of generality, we can assume that we have $\bar{a}_{u,v}\in\sigma_{u}^{2}\bar{H}_{m}$ for every $v\neq u$. I.e. for every $v\neq u$ one can write $\bar{a}_{u,v}=\sigma_{u}^{2}\bar{b}_{u,v}$ for some $\bar{b}_{u,v}\in\bar{H}_{m}$.

Now, for every $v\neq u$ define the matrix

[TABLE]

which is equals, by direct computation, to the product of the matrices

[TABLE]

for $k\neq u,v$ and the matrix (the following is an example for $v>u$)

[TABLE]

i.e. $\delta_{v}=\eta_{v}\cdot\prod_{u,v\neq k=1}^{n}\varepsilon_{v,k}$ (observe that the matrices $\varepsilon_{v,k}$ commute, so the product is well defined). One can see that by Propositions 4.1 and 4.2, $\varepsilon_{v,k}\in IA^{m}$ for every $k\neq u,v$. Moreover, by Proposition 4.3, $\eta_{v}\in IA^{m}$. Hence, $\delta_{v}\in IA^{m}\cap\mathbb{J}_{m,u}$. Now, as for every $1\leq i\leq n$ we have $\sum_{j=1}^{n}a_{i,j}\sigma_{j}=0$ (by the condition $A\vec{\sigma}=\vec{0}$), $\alpha\cdot\prod_{u\neq v=1}^{n}\delta_{v}$ is equals to

[TABLE]

[TABLE]

It is easy to see now that if we denote $\alpha\cdot\prod_{u\neq v=1}^{n}\delta_{v}=I_{n}+C$, then for every $v\neq u$, $\bar{c}_{u,v}=0$, when $c_{i,j}$ is the $\left(i,j\right)$-th entry of $C$. Hence, we also have

[TABLE]

Thus, we can write $\overline{\alpha\cdot\prod_{u\neq v=1}^{n}\delta_{v}}=I_{n}+\bar{C}$ when the matrix $\bar{C}$ has the following properties:

• •

The entries of the $u$-th row of $\bar{C}$ are all [math].

• •

As $a_{i,v}\in\tilde{J}_{m,u-1,v}$ for every $i,v$, by the computation for Equation 5.2 we have $\bar{a}_{i,v}\in\sigma_{u}\left(\sum_{r=1}^{u-1}\mathfrak{\bar{A}}\sigma_{r}\bar{U}_{r,m}+\sigma_{u}^{2}\bar{U}_{u,m}+\mathfrak{\bar{A}}\bar{O}_{m}+\bar{O}_{m}^{2}\right)$ for every $i,v\neq u$. Hence, for every $i,v\neq u$ we have

[TABLE]

Now, as $\det(\delta_{v})=1$ for every $v\neq u$, $\det(\overline{\alpha\cdot\prod_{u\neq v=1}^{n}\delta_{v}})=\det(\overline{\alpha})=\prod_{i=1}^{u}x_{i}^{s_{i}m^{2}}$. However, as the entries of $\bar{C}$ have the above properties, this determinant is mapped to $1$ under the projection $\sigma_{u}\mapsto 0$. Thus, $\det(\overline{\alpha\cdot\prod_{u\neq v=1}^{n}\delta_{v}})$ is of the form $x_{u}^{s_{u}m^{2}}$. Now, set $i_{0}\neq u$, and denote

[TABLE]

By the computation in the proof of Proposition 5.2, we obtain that

[TABLE]

and thus

[TABLE]

so $\zeta\in IA^{m}\cap\mathbb{J}_{m,u}$. In addition $\det\left(\zeta\right)=x_{u}^{m^{2}}$. Therefore, $\overline{\alpha\cdot\prod_{v\neq u}\delta_{v}\zeta^{-s_{u}}}$, writen as $I_{n}+\bar{C}$, has the following properties:

• •

The entries of the $u$-th row of $\bar{C}$ are all [math].

• •

For every $i,v\neq u$ we have $\bar{c}_{i,v}\in\sigma_{u}\left(\mathfrak{\bar{A}}\bar{H}_{m}+\bar{O}_{m}^{2}\right)$, so we can write $\bar{c}_{i,v}=\sigma_{u}d_{i,v}$ for some $d_{i,v}\in\mathfrak{\bar{A}}\bar{H}_{m}+\bar{O}_{m}^{2}$.

• •

For every $1\leq i\leq n$ we have $\sum_{k=1}^{u}\sigma_{k}\bar{c}_{i,k}=0$, so $\bar{c}_{i,u}=-\sum_{k=1}^{u-1}\sigma_{k}d_{i,k}$ .

• •

$\det\left(I_{n}+\bar{C}\right)=1$.

In other words

[TABLE]

for some $d_{i,j}\in\mathfrak{\bar{A}}\bar{H}_{m}+\bar{O}_{m}^{2}$, and $\det\left(I_{n}+\bar{C}\right)=1$.

Define now $\beta=\prod_{v\neq u}\delta_{v}\zeta^{-s_{u}}$, so $\beta\in IA^{m}\cap\mathbb{J}_{m,u}$. In addition, define $\gamma$ to be the inverse of $\gamma^{-1}=I_{n}+\tilde{C}$ where

[TABLE]

is the $\left(i,j\right)$-th entry of $\tilde{C}$. Notice that $\gamma^{-1}\in IA\left(\Phi_{n}\right)$, and that $\overline{\gamma^{-1}}=I_{n}+\bar{C}=\overline{\alpha\beta}$. In addition

[TABLE]

Moreover, as $d_{i,j}\in\mathfrak{\bar{A}}\bar{H}_{m}+\bar{O}_{m}^{2}\subseteq H_{m}$, $\gamma\in ISL_{n-1,u}\left(\sigma_{u}H_{m}\right)$. Additionally, $\gamma\in\mathbb{J}_{m,u}$. Hence, we obtained $\beta\in IA^{m}\cap\mathbb{J}_{m,u}$ and $\gamma\in ISL_{n-1,u}\left(\sigma_{u}H_{m}\right)\cap\mathbb{J}_{m,u}$ such that $\overline{\gamma\alpha\beta}=I_{n}$, i.e. $\gamma\alpha\beta\in\tilde{\mathbb{A}}{}_{u}$, as required.

6 Remarks and problems for further research

We will prove now Theorem 1.3, which asserts that $C\left(\Phi_{n}\right)$ is abelian for every $n\geq 4$. But before, let us state the following proposition, which is slightly more general than Lemma 2.1. in [BER], but proven by similar arguments:

Proposition 6.1**.**

Let $1\to G_{1}\overset{\alpha}{\to}G_{2}\overset{\beta}{\to}G_{3}\to 1$ be a short exact sequence of groups. Assume also that $G_{1}$ is finitely generated. Then:

1. The sequence $\hat{G}_{1}\overset{\hat{\alpha}}{\to}\hat{G}_{2}\overset{\hat{\beta}}{\to}\hat{G}_{3}\to 1$ is also exact.

2. The kernel $\ker(\hat{G}_{1}\overset{\hat{\alpha}}{\to}\hat{G}_{2})$ is central in $\hat{G}_{1}$.

Proof.

(of Theorem 1.3) By Proposition 6.1, the commutative exact diagram

[TABLE]

gives rise to the commutative exact diagram

[TABLE]

Now, as $n\geq 4$, by the CSP for $GL_{n}\left(\mathbb{Z}\right)$, the map $\widehat{GL_{n}\left(\mathbb{Z}\right)}\to GL_{n}(\hat{\mathbb{Z}})$ is injective, so one obtains by diagram chasing, that $C\left(IA\left(\Phi_{n}\right),\Phi_{n}\right)=\ker(\widehat{IA\left(\Phi_{n}\right)}\to Aut(\hat{\Phi}_{n}))$ is mapped onto $C\left(\Phi_{n}\right)=\ker(\widehat{Aut\left(\Phi_{n}\right)}\to Aut(\hat{\Phi}_{n}))$ through the map $\widehat{IA\left(\Phi_{n}\right)}\to\widehat{Aut\left(\Phi_{n}\right)}$. In particular, as by Theorem 1.2 $C\left(IA\left(\Phi_{n}\right),\Phi_{n}\right)$ is central in $\widehat{IA\left(\Phi_{n}\right)}$ for every $n\geq 4$, it is also abelian, and thus $C\left(\Phi_{n}\right)$ is an image of an abelian group, and therfore abelian, as required. ∎

Problem 6.2**.**

Is $C\left(\Phi_{n}\right)$ not finitely generated? trivial?

We proved in [Be2] that $C\left(IA\left(\Phi_{n}\right),\Phi_{n}\right)$ is not finitely generated for every $n\geq 4$. This may suggest that also $C\left(\Phi_{n}\right)$ is not finitely generated, or at least, not trivial. Moreover, if $C\left(IA\left(\Phi_{n}\right),\Phi_{n}\right)$ were not central in $\widehat{IA\left(\Phi_{n}\right)}$, we could use the fact that $IA\left(\Phi_{n}\right)$ is finitely generated for every $n\geq 4$ [BM2], and by the second part of Proposition 6.1 we could derive that the image of $C\left(IA\left(\Phi_{n}\right),\Phi_{n}\right)$ in $\widehat{Aut\left(\Phi_{n}\right)}$ is not trivial. However, we showed that $C\left(IA\left(\Phi_{n}\right),\Phi_{n}\right)$ is central in $\widehat{IA\left(\Phi_{n}\right)}$, so it is possible that $C\left(IA\left(\Phi_{n}\right),\Phi_{n}\right)\subseteq\ker(\widehat{IA\left(\Phi_{n}\right)}\to\widehat{Aut\left(\Phi_{n}\right)})$ and thus $C\left(\Phi_{n}\right)$ is trivial.

We saw in [Be2] that for every $i$ there is a natural surjective map

[TABLE]

These maps enabled us to show in [Be2] that for every $n\geq 4$, $C\left(IA\left(\Phi_{n}\right),\Phi_{n}\right)$ can be written as

[TABLE]

where

[TABLE]

are central in $\widehat{IA\left(\Phi_{n}\right)}$. Here we showed that also $C\left(IA\left(\Phi_{n}\right),\Phi_{n}\right)\cap_{i=1}^{n}\ker\hat{\rho}_{i}$ lie in the center of $\widehat{IA\left(\Phi_{n}\right)}$ but we still do not know to determine whether:

Problem 6.3**.**

Is $C\left(IA\left(\Phi_{n}\right),\Phi_{n}\right)=\prod_{i=1}^{n}C_{i}$ or does it contain more elements?

It seems that having the answer to Problem 6.3 will help to solve Problem 6.2.

7 Index of notations

• •

$F_{n}$ = the free group on $n$ elements, Section 2.

• •

$\Phi_{n}=F_{n}/F^{\prime\prime}_{n}$= the free metabelian group on $n$ elements, Section 2.

• •

$\Phi_{n,m}=\Phi_{n}/M_{n,m}$, where $M_{n,m}=\left(\Phi^{\prime}_{n}\Phi_{n}^{m}\right)^{\prime}\left(\Phi^{\prime}_{n}\Phi_{n}^{m}\right)^{m}$, Section 2.

• •

$IA(\Phi_{n})=\ker\left(Aut\left(\Phi_{n}\right)\to Aut\left(\Phi_{n}/\Phi_{n}^{\prime}\right)\right)$, Section 2.

• •

$IG_{m}=IG_{n,m}=G(M_{n,m})=\ker\left(IA\left(\Phi_{n}\right)\to Aut\left(\Phi_{n,m}\right)\right)$, Section 2.

• •

$IA^{m}=IA_{n}^{m}=\left\langle IA\left(\Phi_{n}\right)^{m}\right\rangle$, Section 2.

• •

$IA_{n,m}=\cap\left\{N\vartriangleleft IA\left(\Phi_{n}\right)\,|\,[IA\left(\Phi_{n}\right):N]\,|\,m\right\}$, Section 3.

• •

$R=R_{n}=\mathbb{Z}[x_{1}^{\pm 1},\ldots,x_{n}^{\pm 1}]$ where $x_{1},\ldots,x_{n}$ are free commutative variables, Section 2.

• •

$\mathbb{Z}_{m}=\mathbb{Z}/m\mathbb{Z}$, Section 2.

• •

$\sigma_{i}=x_{i}-1$ for $1\leq i\leq n$, Section 2.

• •

$\vec{\sigma}$ = the column vector which has $\sigma_{i}$ in its $i$-th entry, Section 2.

• •

$\mu_{r,m}=\sum_{i=0}^{m-1}x_{r}^{i}$, Section 4.

• •

$\mathfrak{A}=\mathfrak{A}_{n}=\sum_{i=1}^{n}\sigma_{i}R_{n}\vartriangleleft R_{n}$ = the augmentation ideal of $R_{n}$, Section 2.

• •

$\mathfrak{\bar{A}}=\mathfrak{A}_{u}=\sum_{i=1}^{u}\sigma_{i}R_{u}\vartriangleleft R_{u}$, where $1\leq u\leq n$, Subsection 5.2.

• •

$\bar{\bar{\mathfrak{A}}}=\mathfrak{A}_{u-1}=\sum_{i=1}^{u-1}\sigma_{i}R_{u-1}\vartriangleleft R_{u-1}$, where $1\leq u\leq n$, Subsection 5.2.

• •

$\mathfrak{\tilde{A}}_{u}=\sum_{r=u+1}^{n}\sigma_{r}R_{n}\vartriangleleft R_{n}$, where $0\leq u\leq n$, Subsection 5.1.

• •

$\tilde{\mathbb{A}}_{u}=IA\left(\Phi_{n}\right)\cap GL_{n}(R,\mathfrak{\tilde{A}}_{u})$, where $0\leq u\leq n$, Subsection 5.1.

• •

$O_{m}=mR_{n}\vartriangleleft R_{n}$, Section 5.

• •

$\bar{O}_{m}=mR_{u}\vartriangleleft R_{u}$, where $1\leq u\leq n$, Subsection 5.2.

• •

$\bar{\bar{O}}_{m}=mR_{u-1}\vartriangleleft R_{u-1}$, where $1\leq u\leq n$, Subsection 5.2.

• •

$U_{r,m}=\mu_{r,m}R_{n}\vartriangleleft R_{n}$, Section 5.

• •

$\bar{U}_{r,m}=\mu_{r,m}R_{u}\vartriangleleft R_{u}$, where $1\leq u\leq n$, Subsection 5.2.

• •

$\bar{\bar{U}}_{r,m}=\mu_{r,m}R_{u-1}\vartriangleleft R_{u-1}$, where $1\leq u\leq n$, Subsection 5.2.

• •

$H_{m}=H_{n,m}=\sum_{i=1}^{n}\left(x_{i}^{m}-1\right)R_{n}+mR_{n}\vartriangleleft R_{n}$, Section 2.

• •

$\bar{H}_{m}=H_{u,m}=\sum_{r=1}^{u}\sigma_{r}\bar{U}_{r,m}+\bar{O}_{m}\vartriangleleft R_{u}$, where $1\leq u\leq n$, Subsection 5.2.

• •

$J_{m}=\sum_{r=1}^{n}\sigma_{r}^{3}U_{r,m}+\mathfrak{A}^{2}O_{m}+\mathfrak{A}O_{m}^{2}\vartriangleleft R_{n}$, Subsection 5.1.

• •

\mathbb{\mathbb{J}}_{m}=\left\{I_{n}+A\,|\,\begin{array}[]{c}I_{n}+A\in IA\left(\Phi_{n}\right)\cap GL_{n}\left(R,J_{m}\right)\\ \det\left(I_{n}+A\right)=\prod_{r=1}^{n}x_{r}^{s_{r}m^{2}},\,\,s_{r}\in\mathbb{Z}\end{array}\right\}, Subsection 5.1.

• •

$\tilde{J}_{m,u,v}=\begin{cases}\mathfrak{\tilde{A}}_{u}\left(\sum_{r=1}^{u}\mathfrak{A}\sigma_{r}U_{r,m}+\mathfrak{A}O_{m}+O_{m}^{2}\right)+\\ \sum_{r=u+1}^{n}\sigma_{r}^{3}U_{r,m}&v\leq u\\ \mathfrak{\tilde{A}}_{u}\left(\sum_{r=1}^{u}\mathfrak{A}\sigma_{r}U_{r,m}+\mathfrak{A}O_{m}+O_{m}^{2}\right)+\\ \sum_{v\neq r=u+1}^{n}\sigma_{r}^{3}U_{r,m}+\mathfrak{A}\sigma_{v}^{2}U_{v,m}&v>u\end{cases}$,

where $0\leq u\leq n$ and $1\leq v\leq n$, Subsection 5.1.

• •

\mathbb{\mathbb{\tilde{J}}}_{m,u}=\left\{I_{n}+A\in IA\left(\Phi_{n}\right)\,|\,\begin{array}[]{c}\det\left(I_{n}+A\right)=\prod_{i=1}^{n}x_{i}^{s_{i}m^{2}}\textrm{,\,\,every\,\,entry\,\,in}\\ \textrm{the\,\,}v\textrm{-th\,\,colmun\,\,of\,\,}A\textrm{\,\,belongs\,\,to\,\,}\tilde{J}_{m,u,v}\end{array}\right\},

where $0\leq u\leq n$, Subsection 5.1.

• •

$J_{m,u,v}=\begin{cases}\mathfrak{A}\left(\sum_{r=1}^{u}\mathfrak{A}\sigma_{r}U_{r,m}+\mathfrak{A}O_{m}+O_{m}^{2}\right)+\\ \sum_{r=u+1}^{n}\sigma_{r}^{3}U_{r,m}&v\leq u\\ \mathfrak{A}\left(\sum_{r=1}^{u}\mathfrak{A}\sigma_{r}U_{r,m}+\mathfrak{A}O_{m}+O_{m}^{2}\right)+\\ \sum_{v\neq r=u+1}^{n}\sigma_{r}^{3}U_{r,m}+\mathfrak{A}\sigma_{v}^{2}U_{v,m}&v>u\end{cases}$,

where $0\leq u\leq n$ and $1\leq v\leq n$, Subsection 5.1.

• •

\mathbb{J}_{m,u}=\left\{I_{n}+A\in IA\left(\Phi_{n}\right)\,|\,\begin{array}[]{c}\det\left(I_{n}+A\right)=\prod_{i=1}^{n}x_{i}^{s_{i}m^{2}}\textrm{,\,\,every\,\,entry\,\,in}\\ \textrm{the\,\,}v\textrm{-th\,\,colmun\,\,of\,\,}A\textrm{\,\,belongs\,\,to\,\,}J_{m,u,v}\end{array}\right\},

where $0\leq u\leq n$, Subsection 5.1.

• •

$E_{d}\left(R\right)=\left\langle I_{d}+rE_{i,j}\,|\,r\in R,\,1\leq i\neq j\leq d\right\rangle\leq SL_{d}\left(R\right)$, where $R$ is a ring and $E_{i,j}$ is the matrix that has $1$ in its $\left(i,j\right)$-th entry and [math] elsewhere, Section 2.

• •

$SL_{d}\left(R,H\right)=\ker\left(SL_{d}\left(R\right)\to SL_{d}\left(R/H\right)\right)$, where $R$ is a ring and $H\vartriangleleft R$, Section 2.

• •

$GL_{d}\left(R,H\right)=\ker\left(GL_{d}\left(R\right)\to GL_{d}\left(R/H\right)\right)$, where $R$ is a ring and $H\vartriangleleft R$, Section 2.

• •

$E_{d}\left(R,H\right)$ = the normal subgroup of $E_{d}\left(R\right)$, generated as a normal subgroup by the matrices of the form $I_{d}+hE_{i,j}$ for $h\in H$, Section 2.

• •

IGL_{n-1,i}=\left\{I_{n}+A\in IA\left(\Phi_{n}\right)\,|\,\begin{array}[]{c}\textrm{The\,\,}i\textrm{-th\,\, row\,\, of\,\,}A\textrm{\,\, is\,\,0,}\\ I_{n-1}+A_{i,i}\in GL_{n-1}\left(R_{n},\sigma_{i}R_{n}\right)\end{array}\right\},

for $1\leq i\leq n$, Section 2.

• •

$ISL_{n-1,i}\left(H\right)=IGL_{n-1,i}\cap SL_{n-1}\left(R_{n},H\right)$, under the identification of $IGL_{n-1,i}$ with $GL_{n-1}\left(R_{n},\sigma_{i}R_{n}\right)$, Section 2.

• •

$IE_{n-1,i}\left(H\right)=IGL_{n-1,i}\cap E{}_{n-1}\left(R_{n},H\right)$, under the identification of the group $IGL_{n-1,i}$ with $GL_{n-1}\left(R_{n},\sigma_{i}R_{n}\right)$, Section 2.