A Polycyclic Presentation for the q-Tensor Square of a Polycyclic Group

TL;DR

This paper develops a polycyclic presentation for the q-tensor square of polycyclic groups, extending existing methods for q=0, and provides tools for computing the q-exterior centre and related homology groups.

Contribution

It introduces a new polycyclic presentation for the q-tensor square of polycyclic groups and extends prior methods to all q ≥ 0.

Findings

Derived a polycyclic presentation for G ⊗^q G.

Presented presentations for G ∧^q G and H_2(G, Z_q).

Established a criterion for computing the q-exterior centre.

Abstract

Let $G$ be a group and $q$ a non-negative integer. We denote by $\nu^q(G)$ a certain extension of the $q$-tensor square $G \otimes^q G$ by $G \times G$. In this paper we derive a polycyclic presentation for $G \otimes^q G$, when $G$ is polycyclic, via its embedding into $\nu^q(G)$. Furthermore, we derive presentations for the $q$-exterior square $G \wedge^q G$ and for the second homology group $H_2(G, \mathbb{Z}_q).$ Additionally, we establish a criterion for computing the $q-$exterior centre $Z_q^\wedge (G)$ of a polycyclic group $G, $ which is helpful for deciding whether $G$ is capable modulo $q$. These results extend to all $q \geq 0$ existing methods due to Eick and Nickel for the case $q = 0$.

Equations

$$(g\otimes h{h}_1)=(g\otimes h)(g^{h_1}\otimes h^{h_1})$$

$$g{g}_1\otimes h=(g^{g_1}\otimes h^{g_1})(g_1\otimes h)$$

$$(g\otimes h{)}^{\hat{k}}=(g^{k^q}\otimes h^{k^q})$$

$$\widehat{k{k}_1}=\hat{k}\prod _{i=1}^{q-1}{\left(k\otimes (k_1^{-i})\right)}^{k^{q-1-i}}{\hat{k}}_1$$

$$\left[\hat{k},\widehat{k_1}\right]=k^q\otimes {k_1}^q$$

$$\widehat{\left[g,h\right]}=(g\otimes h{)}^q$$

$$G{\wedge }^{q}G=\frac{G{\otimes }^{q}G}{{\nabla }^q(G)}.$$

$$\varrho :G{\otimes }^{q}G\to G,g\otimes h\mapsto [g,h],\hat{k}\mapsto k^q,$$

$$\mathrm{Ker}\varrho /{\nabla }^q(G)\cong H_2(G,{\mathbb{Z}}_q),$$

$$1\to H_2(G,{\mathbb{Z}}_q)\to G{\wedge }^{q}G\to G^{\prime }{G}^q\to 1.$$

$$Z_q^{\wedge }(G)=\left\{g\in G|g\wedge x=1\in G{\wedge }^{q}G,\text{for all}x\in G\right\}.$$

$$\nu (G):=⟨G\cup G^{\phi }|[g,h^{\phi }{]}^k=[g^k,(h^k{)}^{\phi }]=[g,h^{\phi }{]}^{k^{\phi }},\forall g,h,k\in G⟩.$$

$$g^{-1}\hat{k}g{\widehat{(k^g)}}^{-1};$$

$$(g^{\phi }{)}^{-1}\hat{k}{g}^{\phi }{\widehat{(k^g)}}^{-1};$$

$$(\hat{k}{)}^{-1}[g,h^{\phi }]\hat{k}[g^{k^q},(h^{k^q}{)}^{\phi }{]}^{-1};$$

$$(\hat{k}{)}^{-1}\widehat{k{k}_1}(\widehat{k_1}{)}^{-1}(\prod _{i=1}^{q-1}[k,(k_1^{-i}{)}^{\phi }{]}^{k^{q-1-i}}{)}^{-1};$$

$$[\hat{k},\widehat{k_1}][k^q,(k_1^q{)}^{\phi }{]}^{-1};$$

$$\widehat{[g,h]}[g,h^{\phi }{]}^{-q}.$$

$${\nu }^q(G):=(\nu (G)\ast F(\hat{\mathcal{G}}))/J.$$

$${\nu }^q(G)=⟨G,G^{\phi },\hat{\mathcal{G}}|R,[g,h^{\phi }{]}^k[g^k,(h^k{)}^{\phi }{]}^{-1},[g,h^{\phi }{]}^{k^{\phi }}[g^k,(h^k{)}^{\phi }{]}^{-1},\forall g,h,k\in G⟩.$$

$${\nu }^q(G)\cong G⋉(G⋉(G{\otimes }^{q}G));$$

$${\tau }^q(G)\cong ((G{\wedge }^{q}G)⋊G)⋊G.$$

$$\left\{\begin{array}[]{cccc}(g\otimes h)^{x}=g^{x}\wedge h^{x}\\ \;(\widehat{k})^{x}=\widehat{k^{x}}.\\ \end{array}\right.$$

$$(g{g}_1,h,k)\lambda =(g^{g_1},h^{g_1},1)\lambda (g_1,h,k)\lambda$$

$$(g,h{h}_1,k)\lambda =(g,h_1,1)\lambda (g^{h_1},h^{h_1},k)\lambda$$

$${\left((1,1,k)\lambda \right)}^{-1}(g,h,1)\lambda (1,1,k)\lambda =(g^{k^q},h^{k^q},1)\lambda$$

$$(1,1,k{k}_1)\lambda =(1,1,k)\lambda \prod _{i=1}^{q-1}\left\{(k,(k_1^{-i}{)}^{k^{q-1-i}},1)\lambda \right\}(1,1,k_1)\lambda$$

$$\left[(1,1,k)\lambda ,(1,1,k_1)\lambda \right]=(k^q,k_1^q,1)\lambda$$

$$(1,1,[g,h])\lambda ={\left((g,h,1)\lambda \right)}^q$$

$$(g\otimes h)\tilde{\lambda }=(g,h,1)\lambda ;$$

$$(\hat{k})\tilde{\lambda }=(1,1,k)\lambda .$$

$$G{\wedge }^{q}G\cong \frac{F^{\prime }{F}^q}{[R,F]R^q}\text{ and }{H}_2(G,{\mathbb{Z}}_q)\cong \frac{R\cap F^{\prime }{F}^q}{[R,F]R^q}.$$

$$M^{\prime }{M}^q\stackrel{\cong }{\to }{F}^{\prime }{F}^q/[R,F]R^q,$$

Full text

(*) The author acknowledges partial financial support from FAPDF, Brazil, during the preparation of this work.

1. Introduction

Let $G$ be a group and $q$ a non-negative integer. The $q$-tensor square $G\otimes^{q}G$ is a particular case of the $q$-tensor product $G\otimes^{q}H$ of groups $G$ and $H$ which act compatibly on each other; this construction was defined by Conduché and Rodrigues-Fernandez in [6], in the context of $q$-crossed modules (see also [12], [15] and [2]). It reduces to Brown and Loday’s non-abelian tensor product $G\otimes H$ when $q=0$ (cf. [3]).

For $x,y\in G$, we write the conjugate of $y$ by $x$ as $y^{x}=x^{-1}yx$; the commutator of $x$ and $y$ is then written as $[x,y]=x^{-1}y^{-1}xy$. Commutators are left normed: $[x,y,z]=[[x,y],z]$, and so on for commutators of higher weights.

For $q\geq 1$ let $\widehat{\mathcal{G}}:=\{\widehat{k}\mid k\in G\}$ be a set of symbols, one for each element of $G$. According to Ellis [12], the $q$-tensor square $G\otimes^{q}G$ is then defined to be the group generated by all symbols $g\otimes h$ and $\widehat{k}$, $g,h,k\in G$, subject to the following defining relations:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

for all $g,g_{1}$,$h,h_{1}$, $k,k_{1}\in G$. If $q=0$ then we set $\widehat{\mathcal{G}}=\emptyset$ to get the group generated by the symbols $g\otimes h,\;g,h\in G,$ subject to the relations (1) and (2) only; that is, $G\otimes^{0}G$ is the non-abelian tensor square $G\otimes G$. By the defining relations (1) – (6) we see that the diagonal $\bigtriangledown^{q}(G)=\langle g\otimes g\mid g\in G\rangle$ is a central subgroup of $G\otimes^{q}G$. The $q$-exterior square $G\wedge^{q}G$ is by definition the factor group (see [13]):

[TABLE]

We usually write $g\wedge h$ for the image of $g\otimes h$ in $G\wedge^{q}G$.

There is a map

[TABLE]

for all $g,h,k\in G$. Clearly, $\nabla^{q}(G)\leq\operatorname{Ker}\varrho$ and we have (see for instance [2, Proposition 18] or [5, Theorem 2.12]):

[TABLE]

the second homology group of $G$ with coefficients in the trivial $G$-module $\mathbb{Z}_{q}$. The image $\operatorname{Im}\varrho$ is the subgroup $G^{\prime}G^{q}\leq G$, where $G^{\prime}$ is the derived subgroup of $G,$ generated by all commutators $[g,h]$ with $g,h\in G$, and $G^{q}$ is the subgroup of $G$ generated by all $q-$th powers $g^{q},\,g\in G$. Thus, we get the exact sequence (cf. [2, Proposition 18]):

[TABLE]

A group $G$ is called $q$-perfect in case $G=G^{\prime}G^{q}$. If this is the case, then the above sequence shows that $G\wedge^{q}G$ is a $q$-central extension of $G$ and in addition it is the unique universal $q-$central extension of $G$ (see [2]). Notice that if $G$ is $q-$perfect then $G\otimes^{q}G\cong G\wedge^{q}G$.

The $q$-exterior square is also helpful in deciding whether a group $G$ is $q$-capable; recall that $G$ is $q$-capable if there exists a group $Q$ such that $Z(Q)=Z_{q}(Q)$ and $G\cong Q/Z(Q)$, where $Z(Q)$ is the center of $Q$ and $Z_{q}(Q)$ is the $q$-center, that is, the elements of the center $Z(Q)$ of order dividing $q$. The $q$-exterior center of $G$ is the subgroup of $G$ defined by

[TABLE]

In [12, Proposition 16] Ellis proved that the group $G$ is $q$-capable if, and only if, $Z_{q}^{\wedge}(G)=1$.

So getting a presentation for the $q$-tensor square of a group $G$ and for its subfactors is an interesting task.

It’s known that if $G$ is a polycyclic group then $G\otimes^{q}G$ is polycyclic, for all $q\geq 0$ (see for instance [5]). In [9] the authors describe algorithms to compute the non-abelian tensor square $G\otimes G$, the exterior square $G\wedge G$ and the Schur multiplier $M(G),$ among others, for a polycyclic group $G$ given by a consistent polycyclic presentation; the implementation of this algorithm is available in [10]. They manage to find such an algorithm to computing $G\otimes G$ by finding a presentation of the group $\nu(G)$, as introduced for instance in [20] (see also [11]), which turns out to be an extension of $G\otimes G$ by $G\times G$.

The present paper aims to extend that algorithms to all $q\geq 0$. Instead of group $\nu(G)$ we now consider the group $\nu^{q}(G)$, defined for instance in [5, Definition 2.1]. To ease reference we briefly describe this group early in the next section.

The paper is organized as follows.

In Sec. 2 we describe some basic constructs and preliminaries results. In Sec. 3 we give consistent polycyclic presentations for certain q-central extension of $G$, more specifically, the groups $E_{q}(G)$ and $\mathfrak{E}^{q}(G).$ In Sec. 4 we give polycyclic presentations for the second homology group $H_{2}(G,\mathbb{Z}_{q})$ and for the $q$- exterior square $G\wedge^{q}G$; use of the group $E_{q}(G)$ is made to exemplify the computation of the q-exterior centre of $G$. In Sec. 5 we provide a consistent polycyclic presentation for the group $\tau^{q}(G)$. Finally, in Sec. 6 we give an algorithm to compute a polycyclic presentation for $\nu^{q}(G)$ and the q-tensor square of a polycyclic group $G$.

Notation is fairly standard; for basic results on Group Theory, see for instance [18]. In this article all group actions are on the right. Basic notation and structural results concerning $\nu^{q}(G)$ can be found for instance in [5].

The main content of this article is part of the doctoral thesis [19] of the first named author, elaborated under the supervision of the second.

2. Preliminary Results

We begin this section by defining the group $\nu^{q}(G)$ and giving a brief description of some of its properties.

To this end, let $G^{\varphi}$ be an isomorphic copy of $G,$ via an isomorphism $\varphi$ such that $\varphi:g\mapsto g^{\varphi}$, for all $g\in G.$ With these data we immediately get the group $\nu(G)$, as mentioned before, defined as follows:

[TABLE]

It’s well known (see [20], and also [11]) that the subgroup $[G,G^{\varphi}]$ of $\nu(G)$ is isomorphic to the non-abelian tensor square $G\otimes G$, so that the strategy of finding an appropriate representation of $\nu(G)$ can be useful to compute $G\otimes G$ and various of its relevant subfactors (see for instance [11], [20], [15], [1] and [9]).

Now for $q\geq 1$, let $\widehat{\mathcal{G}}=\{\widehat{k}\,|\,k\in G\}$ be a set of symbols, one for each element of $G$ (for $q=0$ we set $\widehat{\mathcal{G}}=\emptyset,$ the empty set) and let $F(\widehat{\mathcal{G}})$ be the free group over $\widehat{\mathcal{G}}.$ Write $\nu(G)\ast F(\widehat{\mathcal{G}})$ for the free product of $\nu(G)$ and $F(\widehat{\mathcal{G}}).$ As $G$ and $G^{\varphi}$ are embedded into $\nu(G)$ we shall identify the elements of $G$ (respectively of $G^{\varphi}$) with their respective images in $\nu(G)\ast F(\widehat{\mathcal{G}})$. Denote by $J$ the normal closure in $\nu(G)\ast F(\widehat{\mathcal{G}})$ of the following elements, for all $\widehat{k},\widehat{k_{1}}\in\widehat{\mathcal{G}}$ and $\ g,\,h\in G:$

[TABLE]

Definition 2.1**.**

The group $\nu^{q}(G)$ is defined to be the factor group

[TABLE]

Note that for $q=0$ the sets of relations $(\ref{RR1})$ to $(\ref{RR6})$ are empty; in this case we have $\nu^{0}(G)=\nu(G)\ast F(\widehat{\mathcal{G}}))/J\cong\nu(G)$. Let $R_{1},\ldots,R_{6}$ be the sets of relations corresponding to $(\ref{RR1}),\ldots,(\ref{RR6})$, respectively, and let $R$ be their union, $R=\bigcup^{6}_{i=1}R_{i}$. Therefore, $\nu^{q}(G)$ has the presentation:

[TABLE]

The above presentation of $\nu^{q}(G)$ is a variant of the one given by Ellis in [12].

There is an epimorphism $\rho:\nu^{q}(G)\twoheadrightarrow G,g\mapsto g,h^{\varphi}\mapsto h,\widehat{k}\mapsto k^{q}$. On the other hand the inclusion of $G$ into $\nu(G)$ induces a homomorphism $\imath:G\to\nu^{q}(G)$. We have $g^{\imath\rho}=g$ and thus $\imath$ is injective. Similarly the inclusion of $G^{\varphi}$ into $\nu(G)$ induces a monomorphism $\jmath:G^{\varphi}\to\nu^{q}(G)$. These embeddings allow us to identify the elements $g\in G$ and $h^{\varphi}\in G^{\varphi}$ with their respective images $g^{\imath}$ and $(h^{\varphi})^{\jmath}$ in $\nu^{q}(G)$.

Now let $\mathfrak{G}$ denote the subgroup of $\nu^{q}(G)$ generated by the images of $\widehat{\mathcal{G}}$. By relations $(\ref{RR3})$, $\mathfrak{G}$ normalizes the subgroup $[G,G^{\varphi}]$ in $\nu^{q}(G)$ and hence ${\Upsilon}^{q}(G):=[G,G^{\varphi}]\mathfrak{G}$ is a normal subgroup of ${\nu}^{q}(G)$. Hence we get $\nu^{q}(G)=G^{\varphi}\cdot(G\cdot\Upsilon^{q}(G)),$ where the dots mean internal semidirect products.

By [5, Proposition 2.9] there is an isomorphism $\mu:{\Upsilon}^{q}(G)\to G\otimes^{q}G$ such that $[g,h^{\varphi}]\mapsto g\otimes h,\;\widehat{k}\mapsto\widehat{k}$, for all $g,h,k\in G$ and for all $q\geq 0$. We then get (see [5, Corollary 2.11])

[TABLE]

This decomposition of $\nu^{q}(G)$ is analogous to one due to Ellis in [12]; it generalizes a similar result found in [20] for $q=0.$

In view of the above isomorphism, unless otherwise stated from now on we will identify $G\otimes^{q}G$ with the subgroup $\Upsilon^{q}(G)=[G,G^{\varphi}]\mathfrak{G}\leq\nu^{q}(G)$ and write $[g,h^{\varphi}]$ in place of $g\otimes h$, for all $g,h\in G$. Following [5] we write $\Delta^{q}(G)$ for the subgroup $\langle[g,g^{\varphi}]|g\in G\rangle\leq\Upsilon^{q}(G),$ which by Remark 2.2 below is a central subgroup of $\nu^{q}(G).$ The isomorphism $\mu$ restricts to an isomorphism $\Delta^{q}(G)\stackrel{{\scriptstyle\mu}}{{\cong}}\nabla^{q}(G)$ and, consequently, the factor group $\Upsilon^{q}(G)/\Delta^{q}(G)$ is isomorphic to the q-exterior square $G\wedge^{q}G$. In this case, as usual we simply write $g\wedge h$ to denote the coset $[g,h^{\varphi}]\Delta^{q}(G)$ in $G\wedge^{q}G.$ We shall eventually write $\mathcal{T}$ to denote the subgroup $[G,G^{\varphi}]$ of ${\nu^{q}(G)}$ in order to distinguish it from the non-abelian tensor square $G\otimes G\equiv[G,G^{\varphi}]\leq\nu(G)$ in the case $q=0$. We also write $\tau^{q}(G)$ for the factor group $\nu^{q}(G)/\Delta^{q}(G);$ thus we get

[TABLE]

Remark 2.2**.**

It should be noted that the actions of $G$ and $G^{\varphi}$ on $\Upsilon^{q}(G)$ are those induced by the defining relations of $\nu^{q}(G)$: for any elements $g,x\in G$, $h^{\varphi},y^{\varphi}\in G^{\varphi}$ and $\widehat{k}\in\widehat{\mathcal{G}}$, we have $[g,h^{\varphi}]^{x}=[g^{x},(h^{x})^{\varphi}]$ and $(\widehat{k})^{x}=\widehat{(k^{x})}$. In view of the isomorphism $\Upsilon^{q}(G)\equiv G\otimes^{q}G$, these correspond to the action of $G$ on $G\otimes^{q}G$ as given for instance in [12]:

[TABLE]

Similarly, $[g,h^{\varphi}]^{y^{\varphi}}=[g^{y},(h^{y})^{\varphi}]$ and $(\widehat{k})^{y^{\varphi}}=\widehat{(k^{y})}.$ In addition, for any $\tau\in\Upsilon^{q}(G)$, $(g\tau)^{y^{\varphi}}=g[g,y^{\varphi}]\tau^{y^{\varphi}}\in G\Upsilon^{q}(G)$. Similar actions are naturally induced on the q-exterior square $G\wedge^{q}G$.

It is known that if $G$ is polycyclic, then $\nu^{q}(G)$ is polycyclic for all $q\geq 0$ and thus, as mentioned before, $G\otimes^{q}G$ is polycyclic. In [5] the authors proved that, for a polycyclic group $G$ given by a consistent polycyclic presentation, the defining relations of $\nu^{q}(G)$ can be reduced to relations among the polycyclic generators, with the only exception of relations (16) which have a more complicated handling characteristic. Even so, they were able to use the GAP System [7] to compute $\nu^{q}(G),$ $G\otimes^{q}G$ and $G\wedge^{q}G$, for some small groups $G$ and particular values of $q.$ In addition, in [21] it is given a description of the q-tensor square of a n-generator nilpotent group of class 2, $n>1$, for all $q>1$ and $q$ odd.

Our purpose in this article is to overcome in some way the difficulty of dealing with relations (16) and give a polycyclic presentation for the groups $\nu^{q}(G)$, $G\otimes^{q}G$, $G\wedge^{q}G$ and $H_{2}(G,\mathbb{Z}_{q})$, for all $q\geq 0$, when $G$ is polycyclic given by a consistent polycyclic presentation. Our approach is based on ideas of Eick & Nickel [9] for the case $q=0$.

The concept of a crossed pairing (biderivation) has been used in order to determine homomorphic images of the non-abelian tensor square $G\otimes G$ (see [4, Remark 3]). We need to extend this concept in order to the context of the q-tensor square.

Definition 2.3**.**

Let $G$ and $L$ be arbitrary groups and $q$ a non-negative integer. A function $\lambda:G\times G\times G\to L$ is called a $q$-biderivation if the following properties hold:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

for all $g,g_{1},h,h_{1},k_{1}\in G$.

By the defining relations 2.1 of $\nu^{q}(G)$, it is easy to see that a $q$-biderivation provides a universal property of the $q$-tensor square of a group $G$; we record this property as

Proposition 2.4**.**

Let $G$ and $L$ be arbitrary groups and $\lambda:G\times G\times G\to L$ a $q$-biderivation. Then, there exists a unique homomorphism $\widetilde{\lambda}:G\otimes^{q}G\to L$ such that the following hold, for all $g,h,k\in G$:

[TABLE]

[TABLE]

To ease reference we include the next Lemma, which relates the $q$-exterior square of $G$ and the second homology group $H_{2}(G,\mathbb{Z}_{q})$ with an arbitrary free presentation $F/R$ of $G$ (see [15] and also [13]).

Lemma 2.5**.**

Let $F/R$ be a free presentation for the group $G$. Then,

[TABLE]

Thus we have $H_{2}(G,\mathbb{Z}_{q})\cong(G\wedge^{q}G)\cap M^{q}(G),$ where $M^{q}(G)=R/[R,F]R^{q}$ is the $q$-multiplier.

Notice that when $F$ is a free group, then we find that $F\wedge^{q}F\cong F^{\prime}F^{q}.$ A similar result is also valid for projective q-crossed $G$-modules:

Proposition 2.6**.**

[15, Proposition 1.3.11]** Let $\delta:M\rightarrow G$ be a projective $q$- crossed $G$-module and let $F/R$ be a free presentation of $G$, with $\pi:F\rightarrow G$ being the natural epimorphism. Then there exists an isomorphism

[TABLE]

such that $[m,m^{\prime}]^{\delta}=[f,f^{\prime}][R,F]R^{q}$ and $(m^{q})^{\delta}=f^{q}[F,R]R^{q}$, where $(m)^{\delta}=(f)^{\pi}$.

3. Consistent polycyclic presentations for the groups

$E_{q}(G)$ and $\mathfrak{E}^{q}(G)$, $q$-central extensions of $G$

In this section we describe a method for computing consistent polycyclic presentations for certain $q$-central extensions of a polycyclic group $G$ given by a consistent polycyclic presentation. Our method is a generalization of the one given by Eick and Nickel in [9] for the case $q=0$.

Let $G$ be a polycyclic group defined by a consistent polycyclic presentation $F_{n}/R$, where $F_{n}$ is the free group in the generators $g_{1},...,g_{n}$ and let $H$ be a finitely presented group defined by a finite presentation $F_{m}/S$, where $F_{m}$ is the free group on the generators $f_{1},...,f_{m}$. For our purposes we shall assume that $m\leq n$. Suppose that $\xi:H\rightarrow G$ is an epimorphism, such that $(f_{i})\xi=w_{i},\;1\leqslant i\leqslant m,$ where $w_{i}$ is a word in the generators $g_{1},\ldots,g_{n}$. Denote by $K/S$ the kernel $\operatorname{Ker}\xi.$ Thus, $G\cong F_{m}/K$. Define the groups

[TABLE]

and

[TABLE]

which, by construction, are $q$-central extensions of $G$.

The following result in the context of crossed modules will be helpful.

Proposition 3.1**.**

[15, Lemma 5.2.2]** With the above definition, the natural epimorphism $\pi:E_{q}(G)\rightarrow G$ is a projective $q$-crossed $G$-module.

The relations of a consistent polycyclic presentation $F_{n}/R$ have the form:

• $g_{i}^{e_{i}}=g_{i+1}^{\alpha_{i,i+1}}...g_{n}^{\alpha_{i,n}}$ for $i\in I$,
• $g_{j}^{-1}g_{i}g_{j}=g_{j+1}^{\beta_{i,j,j+1}}...g_{n}^{\beta_{i,j,n}}$ para $j<i$,
• $g_{j}g_{i}g_{j}^{-1}=g_{j+1}^{\gamma_{i,j,j+1}}...g_{n}^{\gamma_{i,j,n}}$ para $j<i$ e $j\notin I$,

for some set $I\subseteq\{1,...,n\}$, certain exponents $e_{i}\in\mathbb{N}$ for $i\in I$, and $\alpha_{i,j}$, $\beta_{i,j,k}$, $\gamma_{i,j,k}\in\mathbb{Z},$ for all $i$, $j$ and $k$. To ease notation we shall write the defining relations of $G$ as relators, in the form $r_{1},....,r_{l}$. Thus, each relator $r_{j}$ is a word in the generators $g_{1},...,g_{n}$; that is, $r_{j}=r_{j}(g_{1},...,g_{n})$.

We now introduce $l$ new generators $t_{1},...,t_{l},$ one for each relator $r_{j}$, and define a new group $\epsilon(G)$ to be the group generated by $g_{1},...,g_{n},t_{1},...,t_{l}$, subject to the relators:

• (a)

$r_{i}(g_{1},...,g_{n})t_{i}^{-1},$ for $1\leq i\leq l$,

• (b)

$[t_{i},g_{j}],$ for $1\leq j\leq n$, $1\leq i\leq l$,

• (c)

$[t_{i},t_{j}],$ for $1\leq j<i\leq l$,

• (d)

$t_{i}^{q},$ for $1\leq i\leq l$.

Denote by $T_{q}$ the $q$-central subgroup of $\epsilon(G)$ generated by $\left\{t_{1},...,t_{l}\right\}.$ It follows directly from these relators that $\epsilon(G)$ is a $q$-central extension of $G$ by $T_{q}$.

The following Lemma asserts that the above relations give a polycyclic presentation of $E_{q}(G)$, possibly inconsistent.

Lemma 3.2**.**

Let $G$ be a polycyclic group given by a consistent polycyclic presentation $F_{n}/R$. Then we have: $\epsilon(G)\cong F_{n}/R^{q}[R,F_{n}]$, $T_{q}\cong R/R^{q}[R,F_{n}]$ and $\epsilon(G)/T_{q}\cong G$.

Proof.

It follows by relations $(a)$ above that $\epsilon(G)/T_{q}\cong G$, while by relations $(b)$, $(c)$ and $(d)$ we immediately see that $T_{q}$ is a $q$-central subgroup of $\epsilon(G).$ Define $\sigma:F_{n}\rightarrow\epsilon(G)$ given by $(g_{i})^{\sigma}=g_{i}$, $1\leqslant i\leqslant n$. Relations $(a)$ imply that $\sigma$ is an epimorphism and, since $\epsilon(G)$ is a $q$-central extension of $G,$ we have that $R^{q}[R,F_{n}]\leqslant Ker(\sigma)\leqslant R$. On the other side, there exists a well defined homomorphism from $\epsilon(G)$ to $\frac{F_{n}}{R^{q}[F_{n},R]}$, which is an epimorphism. Consequently, $Ker(\sigma)\leq R^{q}[R,F_{n}]\leq Ker(\sigma)$ and thus, $\epsilon(G)\cong\frac{F_{n}}{R^{q}[F_{n},R]}$. Therefore, $\epsilon(G)\cong E_{q}(G)$, where we get $T_{q}\cong\frac{R}{R^{q}[R,F_{n}]}$. ∎

By using an adaptation of the method described by Eick and Nickel in [9] (see also [22, p. 424]) we can determine a consistent polycyclic presentation for $E_{q}(G)$ from the (possibly inconsistent) polycyclic presentation given by Lemma 3.2. We then get a consistent polycyclic presentation for $E_{q}(G)$ in the generators $g_{1},...,g_{n},t_{1},...,t_{l}$ with the following relations:

• (1)

$r_{i}(g_{1},...,g_{n})t_{1}^{q_{i1}}...t_{l}^{q_{il}}$, for $1\leq i\leq l$;

• (2)

$[t_{i},g_{j}]$, for $1\leq i\leq n$, $1\leq j\leq l$;

• (3)

$[t_{i},t_{j}],$ for $1\leq j<i\leq l$;

• (4)

$t_{i}^{d_{i}},$ for $1\leq i\leq l,$ with $d_{i}\mid q,$

where $(q_{ij})_{1\leq i,j\leq l}$ is an appropriate invertible matrix over $\mathbb{Z}$. It may happen that $d_{i}=1$ for some $i\in\{1,...,l\}$. In this case the corresponding generator $t_{i}$ is redundant and can be removed.

Bellow we give a couple of simple examples in order to illustrate these results. The same examples will be used in subsequent sections.

Example 3.3**.**

First we consider the symmetric group $S_{3}$, given by the consistent polycyclic presentation

[TABLE]

According to the definition we have, say for $q=2$:

[TABLE]

where $t_{1},t_{2},t_{3}$ are central. Checking for consistency we find that $t_{2}=1$. Thus, a consistent polycyclic presentation of $E_{2}(S_{3})$ is

[TABLE]

Example 3.4**.**

In this second example we consider the infinite dihedral group, given by the following consistent polycyclic presentation:

[TABLE]

From this we get, for an arbitrary $q\geqslant 2$,

[TABLE]

Checking these relations for consistency we find that this presentation is consistent.

Now, from the polycyclic presentation of $E_{q}(G)$ given earlier we can determine a presentation for $\mathfrak{E}^{q}(G)$.

Lemma 3.5**.**

Let $\varsigma:F_{m}\rightarrow E_{q}(G)$ given by $(f_{i})\varsigma=w_{i},$ for $1\leq i\leq m$, where as before $w_{i}=w_{i}(g_{1},\ldots,g_{n})$ is a word in the generators $g_{1},\ldots,g_{n}.$ Then,

• (i)

$Ker(\varsigma)=[K,F_{m}]K^{q}$;

• (ii)

$\mathfrak{E}^{q}(G)\cong Im(\varsigma)/(S)\varsigma$.

Proof.

(i). Notice that by definition $Im(\varsigma)$ covers $G\cong E_{q}(G)/T_{q}$ and hence $F_{m}/Ker(\varsigma)$ is a $q-$central extension of $G=F_{m}/K$. Thus, $[K,F_{m}]K^{q}\leq Ker(\varsigma)$. On the other hand, $F_{m}/[K,F_{m}]K^{q}$ is a polycyclic $q$-central extension of $G$ and, since by construction $E_{q}(G)$ is the largest $q$-central extension of $G$ with this property (by Proposition 2.6 it is a projective $q$-crossed $G-$module; see also [9, Lemma 3]), it follows that $E_{q}(G)$ contains $F_{m}/[K,F_{m}]K^{q}$ as a sub-factor via $\varsigma$. Thus, $Ker(\varsigma)=[F_{m},K]K^{q}$.

(ii). Now, by part $(i)$ we get that $Im(\varsigma)\cong\dfrac{F_{m}}{[F_{m},K]K^{q}}$ and, by definition, $\mathfrak{E}^{q}(G)=\dfrac{F_{m}}{S[F_{m},K]K^{q}}$. But $(S)\varsigma=\dfrac{S[K,F_{m}]K^{q}}{[K,F_{m}]K^{q}}$; consequently, $\dfrac{Im(\varsigma)}{(S)\varsigma}=\mathfrak{E}^{q}(G)$. ∎

Tuned in this way, in order to determine a presentation of $\mathfrak{E}^{q}(G)$ it suffices to determine generators for the subgroups $Im(\varsigma)$ and $(S)\varsigma$ of $E_{q}(G)$, since standard methods for polycyclic groups can be used in order to construct a consistent polycyclic presentation for the quotient $Im(\varsigma)/(S)\varsigma$ (see also [14, Chap. 8]).

Certainly, a set of generators for $Im(\varsigma)$ is given by $w_{1},...,w_{m}$. Let $s_{1},...,s_{k}$ be a set of defining relators for the finitely presented group $H=F_{m}/S$. Then, $(S)\varsigma$ is generated by $(s_{1})\varsigma,...,(s_{k})\varsigma$ as a subgroup, once $(S)\varsigma\leq T_{q}$ is central in $E_{q}(G)$. Thus, a set of generators for $(S)\varsigma$ can be determined by evaluating the relators $s_{1},...,s_{k}$ in $E_{q}(G)$.

4. Polycyclic presentations for the $q$-exterior square $G\wedge^{q}G$ and for the second homology group

$H_{2}(G,\mathbb{Z}_{q})$

According to Lemmas 2.5 and 3.2, we have the following:

Corollary 4.1**.**

$(i)$* $G\wedge^{q}G\cong E_{q}(G)^{\prime}E_{q}(G)^{q}$;

COROLLARY 4.1 $(ii)$ $H_{2}(G,\mathbb{Z}_{q})\cong\left(E_{q}(G)^{\prime}E_{q}(G)^{q}\right)\cap T_{q}.$*

Therefore, in order to obtain a presentation for the groups $G\wedge^{q}G$ and $H_{2}(G,\mathbb{Z}_{q})$, for a polycyclic group $G$ given by a consistent polycyclic presentation, we apply standard methods to determine presentations of subgroups of polycyclic groups (see for instance [14, Chap. 8]).

By the isomorphism given in Corollary 4.1, we obtain generators for $G\wedge^{q}G$ via $E_{q}(G)$.

Proposition 4.2**.**

The subgroup $(E_{q}(G))^{\prime}(E_{q}(G))^{q}$ of $E_{q}(G)$ is generated by the set

[TABLE]

Proof.

As $E_{q}(G)=\left\langle g_{1},...,g_{n},t_{1},...,t_{l}|r_{i}=t_{i},\left(t_{i}\text{ }q\text{-central}\right)\right\rangle$ is polycyclic, we immediately get that

[TABLE]

Now by a simple induction on $q$ we see that each power $g^{q},g\in E_{q}(G),$ is a word in the commutators $[g_{i},g_{j}]$ and in the $q$-th powers $g_{i}^{q}$ of the generators of $G$. ∎

Let $\pi:E_{q}(G)\to G$ be the natural epimorphism and choose a preimage $\widetilde{g}\in E_{q}(G)$ for each $g\in G$. By Proposition 2.6 we have an isomorphism $\beta:G\wedge^{q}G\to(E_{q}(G))^{\prime}(E_{q}(G))^{q}$ such that $(g\wedge h)^{\beta}=[\widetilde{g},\widetilde{h}]$ and $(\widehat{k})^{\beta}=(\widetilde{k})^{q}.$

Remark 4.3**.**

• $(i)$

As we have seen in the Introduction, $G$ acts naturally on $G\wedge^{q}G$ via $(g\wedge h)^{k}=g^{k}\wedge h^{k}$, $(\widehat{k})^{g}=\widehat{k^{g}}$ for all $g,h,k\in G$. In addition, this action is compatible with the isomorphism $\beta$ and $(g\wedge h)^{k}$ corresponds to $[\widetilde{g},\widetilde{h}]^{k}=[\widetilde{g^{k}},\widetilde{h^{k}}]$, while $(\widehat{k})^{g}$ corresponds to $(\widetilde{k}^{q})^{g}=\widetilde{(k^{g})}^{q}$. The image $w^{k}$ of an arbitrary element $w\in G\wedge^{q}G$ is obtained by writing $w$ as a product of $q$-th powers and commutators and then computing the action of $k$ upon each factor.

• $(ii)$

By construction, the map $\lambda:G\times G\times G\to G\wedge^{q}G,\;(g,h,k)\mapsto[\widetilde{g},\widetilde{h}](\widetilde{k})^{q}$ is a $q$-biderivation. Applying $\beta$ it corresponds to the $q$-biderivation $\lambda:G\times G\times G\to(E_{q}(G))^{\prime}(E_{q}(G))^{q},(g,h,k)\mapsto(g\wedge h)\widehat{k}$.

Note that we can determine the image of the action of $G$ and of the $q$-biderivation $\lambda$ (see 2.4) in the polycyclic presentation of $G\wedge^{q}G$ by using the above Remark.

Example 4.4** (Continuation of Example 3.3).**

We determine $S_{3}\wedge^{2}S_{3}$ by identifying it with the subgroup $\left(E_{2}(S_{3})\right)^{\prime}\left(E_{2}(S_{3})\right)^{2}=\left\langle[g_{1},g_{2}],g_{1}^{2},g_{2}^{2}\right\rangle=\left\langle w|w^{6}\right\rangle\cong C_{6},$ where $w=g_{2}^{2}t_{1}$.

• (i)

By Remark $(i)$ the image of $(g_{1}\wedge g_{2})^{g_{1}}$ in the consistent polycyclic presentation of $S_{3}\wedge^{2}S_{3}$ corresponds to the element $[g_{1},g_{2}]^{g_{1}}$ of $\left(E_{2}(S_{3})\right)^{\prime}\left(E_{2}(S_{3})\right)^{2}$. In turn, evaluating this element using the relations of $\left(E_{2}(S_{3})\right)^{\prime}\left(E_{2}(S_{3})\right)^{2}$ we obtain the element

[TABLE]

• (ii)

Analogously, the image of $(g_{1},g_{2},g_{1})\lambda$ in the consistent polycyclic presentation of $S_{3}\wedge^{2}S_{3}$ corresponds to the element $[g_{1},g_{2}]g_{1}^{2}$ of $\left(E_{2}(S_{3})\right)^{\prime}\left(E_{2}(S_{3})\right)^{2}$ and thus, using the relations of $\left(E_{2}(S_{3})\right)^{\prime}\left(E_{2}(S_{3})\right)^{2}$, we evaluate this element to get

[TABLE]

Example 4.5** (Continuation of Example 3.4).**

Now we determine $D_{\infty}\wedge^{2}D_{\infty}$ by identifying it with the subgroup

[TABLE]

where $w_{1}=g_{2}^{2},w_{2}=t_{1},w_{3}=t_{2}$.

• (i)

Analogous to the previous example, by Remark $(i)$ the image of $(g_{1}\wedge g_{2})^{g_{1}}$ in the consistent polycyclic presentation of $D_{\infty}\wedge^{2}D_{\infty}$ corresponds to the element $[g_{1},g_{2}]^{g_{1}}$ in $\left(E_{q}(D_{\infty})\right)^{\prime}\left(E_{q}(D_{\infty})\right)^{2},$ that is, to

[TABLE]

• (ii)

Similarly, the image of $(g_{1},g_{2},g_{1})\lambda$ corresponds to the element $[g_{1},g_{2}]g_{1}^{2}$ of $\left(E_{q}(D_{\infty})\right)^{\prime}\left(E_{q}(D_{\infty})\right)^{2}$, which results in

[TABLE]

5. The $q$-exterior center of a polycyclic group

Our next step is to show that we can easily determine the $q$-exterior center of a polycyclic group $G$ given by a consistent polycyclic presentation, using a consistent polycyclic presentation for $E_{q}(G)$ and standard methods for polycyclic groups (see [14, Chap. 8]). These techniques also extend those found in [9] for the exterior center (case $q=0$).

Theorem 5.1**.**

Let $G$ be a polycyclic group and $\pi:E_{q}(G)\to G$ the natural epimorphism. Then, $Z_{q}^{\wedge}(G)=\left(Z(E_{q}(G))\right)\pi$.

Proof.

For each $g\in G$ let $\widetilde{g}$ be a pre-image of $g$ in $E_{q}(G)$ under the epimorphism $\pi$, i.e., $(\widetilde{g})\pi=g$. Now, $[\widetilde{g},\widetilde{a}]=1$ for all $a\in G$ if, and only if, $[\widetilde{g},x]=1$ for all $x\in E_{q}(G)$. Indeed, given $x\in E_{q}(G)$ then $x^{\pi}\in G$ and by assumption $[\widetilde{g},\widetilde{x^{\pi}}]=1$. On the other hand, $\left(\widetilde{x^{\pi}}\right)^{\pi}=x^{\pi}$. Thus, $\left(\widetilde{x^{\pi}}\right)^{-1}x\in Ker(\pi)\leq Z(E_{q}(G))$. Therefore, we have

$1=[\widetilde{x^{\pi}}^{-1}x,\widetilde{g}]$ $=[\widetilde{x^{\pi}}^{-1},\widetilde{g}]^{x}[x,\widetilde{g}]$ $=[\widetilde{x^{\pi}},\widetilde{g}]^{-(\widetilde{x^{\pi}})^{-1}x}[x,\widetilde{g}]$ $=[\widetilde{x^{\pi}},\widetilde{g}][x,\widetilde{g}]$ $=[x,\widetilde{g}]$.

Conversely, given $a\in G$ we have $\widetilde{a}\in E_{q}(G)$. By assumption $[x,\widetilde{g}]=1$ for all $x\in E_{q}(G)$ and, in particular, for $x=\widetilde{a}$. Thus, $[\widetilde{a},\widetilde{g}]=1$, for all $a\in G$.

Now, remind that the map $\beta:G\wedge^{q}G\to(E_{q}(G))^{\prime}(E_{q}(G))^{q}$ given by $(g\wedge h)\beta=[\widetilde{g},\widetilde{h}]$ and $(\widehat{k})\beta=\widetilde{k}^{q}$ is an isomorphism, and thus we get:

• $Z_{q}^{\wedge}(G)=\left\{g\in G|1=g\wedge a\in G\wedge^{q}G\hskip 5.69046pt,\;\forall\;a\in G\right\}$
• $=\left\{g\in G|[\widetilde{g},\widetilde{a}]=1\hskip 5.69046pt,\;\forall\;a\in G\right\}$ (by using $\beta$)
• $=\left\{g\in G|[\widetilde{g},x]=1,\;\forall\;x\in E_{q}(G)\right\}$
• $=\left\{g\in G|\widetilde{g}\in Z(E_{q}(G))\right\}$
• $=\left(Z(E_{q}(G))\right)\pi$.

∎

Thus, by Theorem 5.1, the $q$-exterior center of a group $G$ given by a consistent polycyclic presentation can be easily determined: first we determine a polycyclic presentation for $E_{q}(G)$ and its corresponding natural epimorphism $\pi:E_{q}(G)\to G.$ Then we compute the center $Z(E_{q}(G))$ using standard methods for polycyclically presented groups (see [holt, Chap. 8]) and, finally, we apply $\pi$ to obtain $Z_{q}^{\wedge}(G)=\left(Z(E_{q}(G))\right)\pi$.

Example 5.2** (Continuation of Example 3.3).**

It follows from the consistent polycyclic presentation of $E_{2}(S_{3})$ that $Z\left(E_{2}(S_{3})\right)=\left\langle t_{1},t_{3}\right\rangle$. Thus, $Z_{2}^{\wedge}\left(S_{3}\right)=1$ and so, as one should expect, $S_{3}$ is $2$-capable. In fact, the group $Q$ given by

[TABLE]

has center $Z(Q)=<a^{2}>=Z_{2}(Q),$ of order 2, and $S_{3}\cong Q/Z(Q)$.

Example 5.3** (Continuation of Example 3.4).**

It follows from the polycyclic presentation of $E_{q}(D_{\infty})$ that $Z_{q}\left(E_{q}(D_{\infty})\right)=\left\langle t_{1},t_{2}\right\rangle$. Thus, $Z^{\wedge}_{q}\left(D_{\infty}\right)=1;$ hence, $D_{\infty}$ is $q$-capable for all $q\geq 0$. Indeed, the group

[TABLE]

has center $Z(Q)=<a^{2}>=Z_{q}(Q),$ of order $q$ and $D_{\infty}\cong Q/Z(Q)$.

6. A consistent polycyclic presentation for

$\nu^{q}(G)/\Delta^{q}(G)$

As seen in Sec. 4, we can determine a consistent polycyclic presentation $F_{r}/U$ for the $q$-exterior square $G\wedge^{q}G$ in the generator $w_{1},...,w_{r}$ and relators, say $u_{1},...,u_{s}$. From such a presentation we will determine a consistent polycyclic presentation for the group $\nu^{q}(G)/\Delta^{q}(G)$. Remember that $\nu^{q}(G)/\Delta^{q}(G)\cong(G\wedge^{q}G)\rtimes(G\times G)$.

According to Remark $(i)$, we can determine the image of the $q$-biderivation $\lambda:G\times G\times G\to G\wedge^{q}G$: $(g,h,1)\mapsto(g\wedge h)$ and $(1,1,k)\mapsto\widehat{k}$ in the consistent polycyclic presentation we obtained for $G\wedge^{q}G$. Analogously, we can construct the natural action of $G$ on the presentation found for $G\wedge^{q}G$, which is given by $(g\wedge h)^{x}=g^{x}\wedge h^{x}$, $(\widehat{k})^{x}=\widehat{(k^{x})}$.

Recall that we are given a consistent polycyclic presentation of group $G$; as before,

$G=\langle g_{1},\ldots,g_{n}\mid r_{1},\ldots,r_{l}\rangle.$

Definition 6.1**.**

Define $\tau^{q}(G)$ to be the group generated by $g_{1},...,g_{n},g_{1}^{\varphi},...,g_{n}^{\varphi},w_{1},...,w_{r}$, subject to the following defining relations:

• (1)

$r_{i}(g_{1},...,g_{n})=1$, for $1\leq i\leq l$;

• (2)

$r_{i}(g_{1}^{\varphi},...,g_{n}^{\varphi})=1$, for $1\leq i\leq l$;

• (3)

$u_{i}(w_{1},...,w_{r})=1,$ for $1\leq i\leq s$;

• (4)

$g_{i}^{-1}g_{j}^{\varphi}g_{i}=g_{j}^{\varphi}\left((g_{i},g_{j},1)\lambda\right)^{-1}$, for $1\leq i,j\leq n$,

$g_{i}g_{j}^{\varphi}g_{i}^{-1}=g_{j}^{\varphi}\left((g_{i}^{-1},g_{j},1)\lambda\right)^{-1},$ for $1\leq i,j\leq n$ $i\notin I;$

• (5)

$g_{j}^{-1}w_{i}g_{j}=w_{i}^{g_{j}}$, for $1\leq i\leq r$, $1\leq j\leq n,$

$g_{j}w_{i}g_{j}^{-1}=w_{i}^{g_{j}^{-1}}$, for $1\leq i\leq r$, $1\leq j\leq n$, $j\notin I$,

$g_{j}^{-\varphi}w_{i}g_{j}^{\varphi}=w_{i}^{g_{j}}$,for $1\leq i\leq r$, $1\leq j\leq n$,

$g_{j}^{\varphi}w_{i}g_{j}^{-\varphi}=w_{i}^{g_{j}^{-1}}$, for $1\leq i\leq r$, $1\leq j\leq n$, $j\notin I$.

Notice that we can compute the right hand side of the relations $(4)$ and $(5)$ as words in $w_{1},...,w_{r}$ (see Remark $(i)$).

Theorem 6.2**.**

Let $W\leq\tau^{q}(G)$ be the subgroup $\langle w_{1},...,w_{r}\rangle$. Then we have:

• (i)

$W$* is a normal subgroup of $\tau^{q}(G)$ and $\tau^{q}(G)/W\cong G\times G$;*

• (ii)

The presentation of $\tau^{q}(G)$ in definition 6.1 is a consistent polycyclic presentation;

• (iii)

$W\cong G\wedge^{q}G$;

• (iv)

The map $\psi:\nu^{q}(G)\to\tau^{q}(G)$ defined by $(g_{i})\psi=g_{i}$, $(g_{i}^{\varphi})\psi=g_{i}^{\varphi}$ and $(\widehat{k})\psi=(1,1,k)\lambda$, for all $1\leq i\leq n$ and all $k\in G,$ extends to a well defined homomorphism (also denoted by $\psi$) such that $\operatorname{Ker}\psi=\Delta^{q}(G)$.

Proof.

The proof is mainly based on a careful analysis of the sets of defining relations (1) – (5) of $\tau^{q}(G)$, as established in the Definition 6.1. The relations $(5)$ tell us that $W$ is in fact a normal subgroup of $\tau^{q}(G)$. The relations $(1),(2)$ and $(4)$ imply that $\tau^{q}(G)/W\cong G\times G^{\varphi}$. In addition, relations $(3)$ show that $W$ is a factor of $G\wedge^{q}G$. Thus, $\tau^{q}(G)$ satisfies the exact sequence

[TABLE]

Now, the relations $(5)$ imply that $G\times G^{\varphi}$ acts by conjugation on $W,$ in the same way as $G\times G$ acts naturally on $G\wedge^{q}G$. In particular, we get that $[w,g]=w^{-1}w^{g}$ and, analogously, $[w,h^{\varphi}]=w^{-1}w^{h^{\varphi}}$, for all words $w$ in $w_{1},...,w_{r}$, all words $g$ in $g_{1},...,g_{n}$ and all words $h^{\varphi}$ in $g_{1}^{\varphi},...,g_{n}^{\varphi}$. Furthermore, the definition of a $q$-biderivation and relations $(4)$ imply that $[g,h^{\varphi}]=(g,h,1)\lambda$, for all words $g$ in $g_{1},...,g_{n}$ and $h^{\varphi}$ in $g_{1}^{\varphi},...,g_{n}^{\varphi}$.

Part $(i)$ then follows directly from the above considerations.

$(ii).$ The relations $(1)$ – $(5)$ already have the form of a polycyclic presentation. Thus, it remains to check them for consistency. Well, all consistency relation in the generators $g_{1},...,g_{n}$ is satisfied, once relations $(1)$ come from a consistent polycyclic presentation of $G$. Analogously, for the relations $(2)$ and $(3)$; they say that all consistency in the generators $g_{1}^{\varphi},...,g_{n}^{\varphi}$ and $w_{1},...,w_{r}$ are also satisfied. Besides that, if a consistency relation involves one generator of the $w_{1},...,w_{r}$, then it is satisfied, once $G\times G^{\varphi}$ acts on $W$ likewise $G\times G$ acts naturally on $G\wedge^{q}G$. Therefore, the bottom line is really to check the consistency relations in $g_{1},...,g_{n},g_{1}^{\varphi},...,g_{n}^{\varphi}$, involving mixed generators $g_{i}$ e $g_{j}^{\varphi}$. They are:

**: **

$g_{k}^{\varphi}(g_{j}g_{i})=(g_{k}^{\varphi}g_{j})g_{i}$ para $j>i$;

**: **

$g_{k}^{\varphi}(g_{j}^{\varphi}g_{i})=(g_{k}^{\varphi}g_{j}^{\varphi})g_{i}$ para $k>j$;

**: **

$((g_{j}^{\varphi})^{e_{j}})g_{i}=(g_{j}^{\varphi})^{e_{j}-1}(g_{j}^{\varphi}g_{i})$ para $j\in I$;

**: **

$g_{j}^{\varphi}(g_{i}^{e_{i}})=(g_{j}^{\varphi}g_{i})g_{i}^{e_{i}-1}$ para $i\in I$;

**: **

$g_{j}^{\varphi}=(g_{j}^{\varphi}g_{i}^{-1})g_{i}$ para $i\notin I$.

Consider for example the first of these relations. Supposing that $g_{i}^{-1}g_{j}g_{i}=r_{ij}(g_{1},...,g_{n})=r_{ij}$ in the defining relations of $G$ and using the fact that $\lambda$ is a $q$-biderivation, we get:

**: **

$g_{k}^{\varphi}(g_{j}g_{i})=(g_{j}g_{i})(g_{k}^{\varphi})^{g_{j}g_{i}}$

**: **

$=(g_{i}r_{ij})\left(g_{k}^{\varphi}\left(\left(g_{j}g_{i},g_{k},1\right)\lambda\right)^{-1}\right)$,

**: **

$(g_{k}^{\varphi}g_{j})g_{i}=\left(g_{j}g_{k}^{\varphi}\left(\left(g_{j},g_{k},1\right)\lambda\right)^{-1}\right)g_{i}$

**: **

$=g_{j}g_{k}^{\varphi}g_{i}\left(\left(\left(g_{j},g_{k},1\right)\lambda\right)^{-1}\right)^{g_{i}}$

**: **

$=g_{j}g_{i}(g_{k}^{\varphi})^{g_{i}}\left(\left(g_{j},g_{k},1\right)\lambda^{g_{i}}\right)^{-1}$

**: **

$=g_{i}r_{ij}g_{k}^{\varphi}\left(\left(g_{i},g_{k},1\right)\lambda\right)^{-1}\left(\left(g_{j}^{g_{i}},g_{k}^{g_{i}},1\right)\lambda\right)^{-1}$

**: **

$=g_{i}r_{ij}g_{k}^{\varphi}\left(\left(g_{j}^{g_{i}},g_{k}^{g_{i}},1\right)\lambda\left(g_{i},g_{k},1\right)\lambda\right)^{-1}$

**: **

$=g_{i}r_{ij}g_{k}^{\varphi}\left(\left(g_{j}g_{i},g_{k},1\right)\lambda\right)^{-1}$.

The other consistency relations can be checked by similar calculations. Thus we obtain that $\tau^{q}(G)$ is given by a consistent polycyclic presentation.

Part $(iii)$ follows from $(ii)$ and from the theory of polycyclic presentations (see for instance [14, Sec. 8.3]), once $W$, as a subgroup of $\tau^{q}(G)$, has a consistent polycyclic presentation in the generators $w_{1},...,w_{r}$ and relations $u_{1},...,u_{s}$. Therefore, $W\cong G\wedge^{q}G$.

$iv)$ Since $\lambda$ is a $q$-biderivation, all relations of $\nu^{q}(G)$ hold in $\tau^{q}(G)$. Thus, $\psi:\nu^{q}(G)\to\tau^{q}(G)$ is an epimorphism, once $g_{i},g_{j}^{\varphi}\in Im(\psi)$ for all $1\leq i,j\leq n$, and if a word $w_{i}\in W\cong G\wedge^{q}G$ is a product of commutators and $q$-th powers, then $w_{i}\in Im(\psi)$, for all $1\leq i\leq r$. Consequently, $Im(\psi)=\tau^{q}(G)$. Besides that, $([g,h^{\varphi}])\psi=[g,h^{\varphi}]=(g,h,1)\lambda$ and $(\widehat{k})\psi=(1,1,k)\lambda$, for all words, $g$ in generators the $g_{1},...,g_{n}$, $h^{\varphi}$ in $g_{1}^{\varphi},...,g_{n}^{\varphi}$, and all $k\in G$. Therefore, the map induced by $\psi$ on the subgroup $\Upsilon^{q}(G)$ coincides with the map $\delta:\Upsilon^{q}(G)\to G\wedge^{q}G$, by construction. We then get the following commutative diagram:

\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\Upsilon^{q}(G)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\delta}$$\textstyle{\nu^{q}(G)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\psi}$$\textstyle{G\times G^{\varphi}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\cong}$$\textstyle{1}$$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{W\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\tau^{q}(G)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{G\times G^{\varphi}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{1.}

It follows that $Ker(\psi)\leq\Upsilon^{q}(G)$ and, consequently, $Ker(\psi)=Ker(\delta)=\Delta^{q}(G)$. This completes the proof. ∎

Example 6.3** (Continuation of Example 3.3).**

According to the above result, the following polycyclic presentation is a presentation of $\tau^{2}(S_{3})=\nu^{2}(S_{3})/\Delta^{2}(S_{3})$ as the group generated by $g_{1},g_{2},g_{1}^{\varphi},g_{2}^{\varphi},w$ subject to the following relations:

• (1)

$g_{1}^{2}=1$, $g_{1}^{-1}g_{2}g_{1}=g_{2}^{-1}$, $g_{2}^{3}$,

• (2)

$(g_{1}^{\varphi})^{2}=1$, $(g_{1}^{\varphi})^{-1}g_{2}^{\varphi}g_{1}^{\varphi}=(g_{2}^{\varphi})^{-1}$, $(g_{2}^{\varphi})^{3}$,

• (3)

$w^{6}=1$,

• (4)

$g_{1}^{-1}g_{1}^{\varphi}g_{1}=g_{1}^{\varphi}$*, *

$g_{1}^{-1}g_{2}^{\varphi}g_{1}=g_{2}^{\varphi}w^{2}$*, *

$g_{2}^{-1}g_{1}^{\varphi}g_{2}=g_{1}^{\varphi}w^{4}$*, *

$g_{2}^{-1}g_{2}^{\varphi}g_{2}=g_{2}^{\varphi}$,

• (5)

$g_{1}^{-1}wg_{1}=w^{5}$*, *

$g_{2}^{-1}wg_{2}=w$*, *

$(g_{1}^{\varphi})^{-1}wg_{1}^{\varphi}=w^{5}$*, *

$(g_{2}^{\varphi})^{-1}wg_{2}\varphi=w$.

Example 6.4** (Continuation of Example 3.4).**

Again, according to Theorem 6.2 we find that $\tau^{2}(D_{\infty})=\nu^{2}(D_{\infty})/\Delta^{2}(D_{\infty})$ has the polycyclic presentation in the generators $g_{1},g_{2},g_{1}^{\varphi},g_{2}^{\varphi},w_{1},w_{2},w_{3}$ subject to the relations:

• (1)

$g_{1}^{2}=1$, $g_{1}^{-1}g_{2}g_{1}=g_{2}^{-1}$,

• (2)

$(g_{1}^{\varphi})^{2}=1$, $(g_{1}^{\varphi})^{-1}g_{2}^{\varphi}g_{1}^{\varphi}=(g_{2}^{\varphi})^{-1}$,

• (3)

$w_{1}^{-1}w_{2}w_{1}=w_{2}$, $w_{2}^{2}$,

$w_{1}^{-1}w_{3}w_{1}=w_{3}$, $w_{3}^{2}$,

$w_{2}^{-1}w_{3}w_{2}=w_{3}$,

• (4)

$g_{1}^{-1}g_{1}^{\varphi}g_{1}=g_{1}^{\varphi}$*, *

$g_{1}^{-1}g_{2}^{\varphi}g_{1}=g_{2}^{\varphi}w_{1}^{-1}w_{3}$*, *

$g_{2}^{-1}g_{1}^{\varphi}g_{2}=g_{1}^{\varphi}w_{1}w_{3}$*, *

$g_{2}^{-1}g_{2}^{\varphi}g_{2}=g_{2}^{\varphi}$,

$g_{2}g_{1}^{\varphi}g_{2}^{-1}=g_{1}^{\varphi}w_{1}^{-1}w_{3}$*, *

$g_{2}g_{2}^{\varphi}g_{2}^{-1}=g_{2}^{\varphi}$,

• (5)

$g_{1}^{-1}w_{1}g_{1}=w_{1}^{-1}$*,           $g_{1}^{-1}w_{2}g_{1}=w_{2}$,      $g_{1}^{-1}w_{1}g_{1}=w_{3}$, *

$g_{2}^{-1}w_{1}g_{2}=w_{1}$*,             $g_{2}^{-1}w_{2}g_{2}=w_{2}$,      $g_{2}^{-1}w_{3}g_{2}=w_{3}$, *

$g_{2}w_{1}g_{2}^{-1}=w_{1}$*,             $g_{2}w_{2}g_{2}^{-1}=w_{2}$,      $g_{2}w_{3}g_{2}^{-1}=w_{3}$, *

$(g_{1}^{\varphi})^{-1}w_{1}g_{1}^{\varphi}=w_{1}^{-1}$*,       $(g_{1}^{\varphi})^{-1}w_{2}g_{1}^{\varphi}=w_{2}$,        $(g_{1}^{\varphi})^{-1}w_{1}g_{1}^{\varphi}=w_{3}$, *

$(g_{2}^{\varphi})^{-1}w_{1}g_{2}^{\varphi}=w_{1}$*,        $(g_{2}^{\varphi})^{-1}w_{2}(g_{2}^{\varphi})=w_{2}$,       $(g_{2}^{\varphi})^{-1}w_{3}(g_{2}^{\varphi})=w_{3}$, *

$g_{2}^{\varphi}w_{1}(g_{2}^{\varphi})^{-1}=w_{1}$,        $g_{2}^{\varphi}w_{2}(g_{2}^{\varphi})^{-1}=w_{2}$,        $g_{2}^{\varphi}w_{3}(g_{2}^{\varphi})^{-1}=w_{3}$.

7. A polycyclic presentation for

$\nu^{q}(G)$

We can now use the consistent polycyclic presentation of $\tau^{q}(G)$ in place of $F_{n}/R$ and the finite presentation of $\nu^{q}(G)$ in place of $F_{m}/S$. Note that the epimorphism $\psi:\nu^{q}(G)\to\tau^{q}(G)$ has the required form. Thus we get the following

Theorem 7.1**.**

$\nu^{q}(G)=\mathfrak{E}^{q}(\tau^{q}(G))$.

Proof.

We have $\tau^{q}(G)=F_{n}/R$, $\nu^{q}(G)=F_{m}/S$ and the epimorphism $\psi:\nu^{q}(G)\to\tau^{q}(G)$ with the kernel $Ker(\psi)=K/S$. By definition, $\mathfrak{E}^{q}(\tau^{q}(G))\cong F_{m}/K^{q}[K,F_{m}]S$. By Theorem 6.2 the group $\nu^{q}(G)$ is a $q$-central extension of $\tau^{q}(G)$. Thus, $[K,F_{m}]K^{q}\leq S$. Since by Lemma 3.5 $\mathfrak{E}^{q}(\tau^{q}(G))=F_{m}/K^{q}[K,F_{m}]S$, it follows that $\mathfrak{E}^{q}(\tau^{q}(G))=F_{m}/S=\nu^{q}(G),$ as desired. ∎

Notice that we used a finite presentation for $\nu^{q}(G)$, to which we guarantee finiteness only in the case that $G$ is finite. On the other hand, the epimorphism $\psi:\nu^{q}(G)\to\tau^{q}(G)$ does not depend on the finiteness of the presentation of $\nu^{q}(G)$ and so we can consider that epimorphism. If $G$ is an infinite polycyclic group then by definition $\nu^{q}(G)$ is given by an infinite presentation, say $F/S$, where $F$ is a free group on the generators of $\nu^{q}(G)$, which we denote by $X$, of infinite rank, and where $S$ is the normal closure of the relations (13)—(18). On the other hand, $\nu^{q}(G)$ is polycyclic and so it has a finite polycyclic presentation $\left\langle X_{0}|S_{0}\right\rangle$, where $X_{0}\subseteq X$ and $S$ is the normal closure of $S_{0}$, $\bar{S_{0}}=S$. Thus, we can use the results in Lemma 3.5 and it suffices to prove that the image of $\varsigma$ is generated by the elements $g_{1},...,g_{n},g_{1}^{\varphi},...,g_{n}^{\varphi},\widehat{g_{i}},...,\widehat{g_{n}},$ and $(S)\varsigma$ is generated by the defining relations of $\nu^{q}(G)$ evaluated only on the polycyclic generators of $G$.

Proposition 7.2**.**

Consider the subgroup $L$ of $E_{q}\left(\tau^{q}(G)\right)$ given by

[TABLE]

Then, $Im(\varsigma)=L$. In addition, $(S)\varsigma$ is generated by the defining relations of $\nu^{q}(G)$ in the polycyclic generators of $G$ and $G^{\varphi}$ in $E_{q}\left(\tau^{q}(G)\right)$.

Proof.

By definition of $L$, to show that $Im(\varsigma)=L$ it suffices to show that $(1,1,k)\lambda\in L$ for all $k\in G$. Let’s prove this by induction on the number of polycyclic generators of $G$. If $n=1$ then $G=<g_{1}>$ and so $k=g_{1}^{\alpha}$, for some $\alpha\in\mathbb{Z}$. Thus, $(1,1,k)\lambda=(1,1,g_{1}^{\alpha})\lambda$. For $\alpha\geq 2$, using relation 24 in Definition 2.3 we have

$(1,1,k)\lambda=(1,1,g_{1}^{2})\lambda$

$=\underbrace{(1,1,g_{1})\lambda}_{\in L}\underbrace{\displaystyle\prod_{i=1}^{q-1}\left(\left((g_{1},g_{1}^{-i},1)\lambda\right)^{g_{1}^{q-1-i}}\right)}_{=[g_{1},g_{1}^{-i\varphi}]\in L}\underbrace{(1,1,g_{1})\lambda}_{\in L}$ $\in L$.

By assuming it for $\alpha-1$ then, analogously,

$(1,1,k)\lambda=(1,1,g_{1}^{\alpha})\lambda$

$=(1,1,g_{1})\lambda\displaystyle\prod_{i=1}^{q-1}\left(\left((g_{1},g_{1}^{-i(n-1)},1)\lambda\right)^{g_{1}^{q-1-i}}\right)(1,1,g_{1}^{n-1})\lambda$ $\in L$.

In addition, again by the very definition of $\lambda$, as above, we obtain $(1,1,1)\lambda=1$ and $(1,1,k^{-1})\lambda=\left(\prod_{i=1}^{q-1}\left((x,x^{i},1)\lambda\right)\right)^{-1}\left((1,1,x)\lambda\right)^{-1},$ which are elements of $L$. This completes the case $n=1$.

Suppose $n\geq 1$ and that our assertion is true for $n-1$. If $k=g_{1}^{\alpha_{1}}...g_{n}^{\alpha_{n}}$ then, the same argument used above gives:

$(1,1,k)\lambda=(1,1,g_{1}^{\alpha_{1}}...g_{n}^{\alpha_{n}})\lambda$

$=(1,1,g_{1}^{\alpha_{1}})\lambda\displaystyle\prod_{i=1}^{q-1}\left(\left((g_{1}^{\alpha_{1}},(g_{2}^{\alpha_{2}}...g_{n}^{\alpha_{n}})^{-i},1)\lambda\right)^{(g_{1}^{\alpha_{1}})^{q-1-i}}\right)(1,1,g_{2}^{\alpha_{2}}...g_{n}^{\alpha_{n}})\lambda$ $\in L$

Thus, $Im(\varsigma)=L$. ∎

Therefore, having obtained a consistent polycyclic presentation of $\tau^{q}(G)$, we can extend it by adding new ($q$-central) generators $t_{i}$, one for each relator $r_{i}$ of $\tau^{q}(G)$, and changing each relator $r_{i}$ by $r_{i}t_{i}^{-1}$. Then, we evaluate the consistency relations among the relators of $\nu^{q}(G)$ in this new presentation and apply Lemma 3.5(ii).

The following result can by used in order to reduce the number of new generators added and the number of relators evaluated in this process.

Lemma 7.3**.**

It is redundant to add new generators corresponding to relations (1) and (2) in the definition of $\tau^{q}(G)$. If these generators are not introduced, then it is redundant evaluate the relators (1) and (2) in the definition of $\nu^{q}(G)$.

Proof.

The relators (1) and (2) in the definition of $\nu^{q}(G)$ coincide with the relators (1) and (2) in the definition of $\tau^{q}(G)$. Therefore, if we add new generators corresponding to those relators in (1) and (2) of Definition 6.1 and then we evaluate the relators (1) and (2) in the definition of $\nu^{q}(G)$, then as a result we obtain the corresponding generators. This means that the corresponding generators are eliminated in the process of the constructing the factor group as described in Lemma 3.5(b). This proves the result. ∎

Example 7.4** (Continuation of Example 3.3).**

We compute a polycyclic presentation of $\nu^{2}(S_{3})$ as a central extension $\mathfrak{E}^{2}(\tau^{2}(S_{3}))$ of $\tau^{2}(S_{3})$. There is a lot of calculations to get such a presentation (by hand) and so we’ll omit the details. We obtain a polycyclic presentation for $\nu^{2}(S_{3})$ in the generators $g_{1},g_{2},g_{1}^{\varphi},g_{2}^{\varphi},w,t$ and defining relations given by:

• (1)

$g_{1}^{2}=1$, $g_{1}^{-1}g_{2}g_{1}=g_{2}^{-1}$, $g_{2}^{3}$,

• (2)

$(g_{1}^{\varphi})^{2}=1$, $(g_{1}^{\varphi})^{-1}g_{2}^{\varphi}g_{1}^{\varphi}=(g_{2}^{\varphi})^{-1}$, $(g_{2}^{\varphi})^{3}$,

• (3)

$w^{6}=t$, $t^{2}$, $t$-central,

• (4)

$g_{1}^{-1}g_{1}^{\varphi}g_{1}=g_{1}^{\varphi}w^{6}$*, *

$g_{1}^{-1}g_{2}^{\varphi}g_{1}=g_{2}^{\varphi}w^{8}$*, *

$g_{2}^{-1}g_{1}^{\varphi}g_{2}=g_{1}^{\varphi}w^{4}$*, *

$g_{2}^{-1}g_{2}^{\varphi}g_{2}=g_{2}^{\varphi}$,

• (5)

$g_{1}^{-1}wg_{1}=w^{5}$*, *

$g_{2}^{-1}wg_{2}=w$*, *

$g_{1}^{-1}wg_{1}=w^{5}$*, *

$g_{2}^{-1}wg_{2}=w$.

From this we get the $2$-tensor square $S_{3}\otimes^{2}S_{3}\cong<w>\leq\nu^{2}(S_{3})$, that is, $S_{3}\otimes^{2}S_{3}\cong\mathbb{Z}_{12}$. In addition, we immediatly find that $\Delta^{2}(S_{3})\cong\mathbb{Z}_{2}$.

Example 7.5**.**

Here we compute a polycyclic presentation for $\nu^{3}(D_{\infty})$ as the group generatoed by $g_{1},g_{2}$, $g_{1}^{\varphi},g_{2}^{\varphi}$, $w_{1},w_{2}$ subject to the relations:

• (1)

$g_{1}^{2}=1$, $g_{1}^{-1}g_{2}g_{1}=g_{2}^{-1}$,

• (2)

$(g_{1}^{\varphi})^{2}=1$, $(g_{1}^{\varphi})^{-1}g_{2}^{\varphi}g_{1}^{\varphi}=(g_{2}^{\varphi})^{-1}$,

• (3)

$w_{1}^{2}$, $w_{1}^{-1}w_{2}w_{1}=w_{2}^{-1}$,

• (4)

$g_{1}^{-1}g_{1}^{\varphi}g_{1}=g_{1}^{\varphi}$*, *

$g_{1}^{-1}g_{2}^{\varphi}g_{1}=g_{2}^{\varphi}w_{2}^{-2}$*, *

$g_{2}^{-1}g_{1}^{\varphi}g_{2}=g_{1}^{\varphi}w_{2}^{2}$*, *

$g_{2}^{-1}g_{2}^{\varphi}g_{2}=g_{2}^{\varphi}$,

$g_{2}g_{1}^{\varphi}g_{2}^{-1}=g_{1}^{\varphi}w_{2}^{-2}$*, *

$g_{2}g_{2}^{\varphi}g_{2}^{-1}=g_{2}^{\varphi}$,

• (5)

$g_{1}^{-1}w_{1}g_{1}=w_{1}$*,                            $g_{1}^{-1}w_{2}g_{1}=w_{2}^{-1}$, *

$g_{2}^{-1}w_{1}g_{2}=w_{1}w_{2}^{2}$,                        $g_{2}^{-1}w_{2}g_{2}=w_{2}$,

$g_{2}w_{1}g_{2}^{-1}=w_{1}w_{2}^{-2}$,                      $g_{2}w_{2}g_{2}^{-1}=w_{2}$,

$(g_{1}^{\varphi})^{-1}w_{1}g_{1}^{\varphi}=w_{1}$,                      $(g_{1}^{\varphi})^{-1}w_{2}g_{1}^{\varphi}=w_{2}^{-1}$,

$(g_{2}^{\varphi})^{-1}w_{1}g_{2}^{\varphi}=w_{1}w_{2}^{2}$,                  $(g_{2}^{\varphi})^{-1}w_{2}(g_{2}^{\varphi})=w_{2}$,

$g_{2}^{\varphi}w_{1}(g_{2}^{\varphi})^{-1}=w_{1}w_{2}^{-2}$,                 $g_{2}^{\varphi}w_{2}(g_{2}^{\varphi})^{-1}=w_{2}$.

According to this presentation we find that $D_{\infty}\otimes^{3}D_{\infty}\cong D_{\infty}$.

Notice that the computation of a presentation of $\nu^{q}(G)$ becomes relatively simple if the group $G$ is $q$-perfect, according to Theorem 7.6 bellow. We shall continue using the same notation as before. More specifically, let $F_{n}/R$ be consistent polycyclic presentation for the polycyclic group $G$ in the generators $g_{1},...,g_{n}$, relators $r_{1},...,r_{l}$, and index set $I$. Let $F_{r}/U$ be a consistent polycyclic presentation for $G\wedge^{q}G$ in the generators $w_{1},...,w_{r}$ and relators $u_{1},...,u_{s}$, as found in Sec. 4. We determine the image of the $q$-biderivation $\lambda:G\times G\times G\to G\wedge^{q}G$: $(g,h,1)\mapsto(g\wedge h)$ e $(1,1,k)\mapsto\widehat{k}$ in the consistent polycyclic presentation obtained for $G\wedge^{q}G$ and construct the natural action of $G$ on that presentation found for $G\wedge^{q}G$ (as defined before: $(g\wedge h)^{x}=g^{x}\wedge h^{x}$, $(\widehat{k})^{x}=\widehat{k^{x}}$, according to Remark $(i)$).

Theorem 7.6**.**

Let $G$ be a polycyclic group given as above. If $G$ is $q$-perfect, then the group $\nu^{q}(G)$ is the group generated by $g_{1},...,g_{n},g_{1}^{\varphi},...,g_{n}^{\varphi}$, $w_{1}...,w_{r}$, subject to the defining relations

• (1)

$r_{i}(g_{1},...,g_{n})=1$* for $1\leq i\leq l$,*

• (2)

$r_{i}(g_{1}^{\varphi},...,g_{n}^{\varphi})=1$* para $1\leq i\leq l$,*

• (3)

$u_{i}(w_{1},...,w_{r})=1$* para $1\leq i\leq s$,*

• (4)

$g_{i}^{-1}g_{j}^{\varphi}g_{i}=g_{j}^{\varphi}\left\{(g_{i},g_{j},1)\lambda\right\}^{-1}$* para $1\leq i,j\leq n$,*

$g_{i}g_{j}^{\varphi}g_{i}^{-1}=g_{j}^{\varphi}\left\{(g_{i}^{-1},g_{j},1)\lambda\right\}^{-1}$* para $1\leq i,j\leq n$ $i\notin I$*

• (5)

$g_{j}^{-1}w_{i}g_{j}=w_{i}^{g_{j}}$*, para $1\leq i\leq r$, *$1\leq j\leq n$

$g_{j}w_{i}g_{j}^{-1}=w_{i}^{g_{j}^{-1}}$*, para $1\leq i\leq r$, $1\leq j\leq n$, *$j\notin I$

$g_{j}^{-\varphi}w_{i}g_{j}^{\varphi}=w_{i}^{g_{j}}$*, para $1\leq i\leq r$, *$1\leq j\leq n$

$g_{j}^{\varphi}w_{i}g_{j}^{-\varphi}=w_{i}^{g_{j}^{-1}}$, para $1\leq i\leq r$, $1\leq j\leq n$, $j\notin I$.

Proof.

In effect, according to Definition 6.1 the above presentation is the same as that of $\tau^{q}(G)$. By Theorem 6.2, $\tau^{q}(G)\cong\nu^{q}(G)/\Delta^{q}(G)$. If $G$ is $q$-perfect, then we have $\Delta^{q}(G)=1$ and so $\tau^{q}(G)\cong\nu^{q}(G)$. Consequently, the given presentation is a presentation of $\nu^{q}(G)$. ∎

7.1. A polycyclic presentation for the $q$-tensor square of a polycyclic group

By all we have seen, a method for determining a consistent polycyclic presentation for the $q$-tensor square $G\otimes^{q}G$ from a given consistent polycyclic presentation of $G$ consists of:

Algorithm 7.7**.**

• $a)$

Determine a consistent polycyclic presentation for $G\wedge^{q}G$.

• $b)$

Determine a consistent polycyclic presentation for $\tau^{q}(G)$.

• $c)$

Determine a consistent polycyclic presentation for $\nu^{q}(G)$.

• $d)$

Determine a consistent polycyclic presentation for the subgroup $\Upsilon^{q}(G)$ of $\nu^{q}(G)$.

Step $(a)$ is a direct application of the method for computing a central extension in Sec. 3. If $G=F_{n}/R$ a consistent polycyclic presentation of $G$, then we can determine a consistent polycyclic presentation for $E_{q}(G)=F_{n}/[F_{n},R^{q}]R^{q}$ and we get $G\wedge^{q}G$ as the subgroup $(E_{q}(G))^{\prime}(E_{q}(G))^{q}$. Step $(b)$ is thus a direct application of the method developed in Sec. 4. Step $(c)$ is obtained by another application of the method for computing a central extension in Sec. 3, in order to compute $\mathfrak{E}^{q}(\tau^{q}(G))$ which, by Theorem 7.1, is isomorphic to $\nu^{q}(G)$. Finally, step $(d)$ is an application of standad methods for compting presentations of subgroups of polycycliclly presented groups.