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Algorithmic recognition of infinite cyclic extensions

Bren Cavallo Department of Mathematics, CUNY Graduate Center, City University of New York. bcavallo@gradcenter.cuny.edu Jordi Delgado Departament de Matemàtiques, Universitat Politècnica de Catalunya. jorge.delgado@upc.edu Delaram Kahrobaei CUNY Graduate Center, PhD Program in Computer Science and NYCCT, Mathematics Department, City University of New York. dkahrobaei@gc.cuny.edu  and  Enric Ventura Departament de Matemàtiques, Universitat Politècnica de Catalunya. enric.ventura@upc.edu
(Date: August 4, 2025)
Abstract.

We prove that one cannot algorithmically decide whether a finitely presented \mathbb{Z}-extension admits a finitely generated base group, and we use this fact to prove the undecidability of the BNS invariant. Furthermore, we show the equivalence between the isomorphism problem within the subclass of unique \mathbb{Z}-extensions, and the semi-conjugacy problem for deranged outer automorphisms.

Key words and phrases:
extension, cyclic extension, decision problem, BNS invariant, undecidability
2010 Mathematics Subject Classification:
20E22, 20F10

In the present paper, we study algorithmic problems about recognition of certain algebraic properties among some families of group extensions. Indeed, we see that yet for the relatively easy family of \mathbb{Z}-extensions one can find positive and negative results, i.e., both solvable and unsolvable “recognition problems”.

For example, we prove that one cannot algorithmically decide whether a finitely presented \mathbb{Z}-extension admits a finitely generated base group. Even when the extension has a unique possible base group, it is not decidable in general whether this particular base group is finitely generated or not. As a consequence, we prove general undecidability for the Bieri–Neumann–Strebel invariant: there is no algorithm which, on input a finite presentation of a group GG and a character χ:G\chi\colon G\to\mathbb{R}, decides whether [χ][\chi] belongs to the BNS invariant of GG, [χ]Σ(G)[\chi]\in\Sigma(G), or not. Although this result seems quite natural, since this geometric invariant has long been agreed to be hard to compute in general (see for example [25, 28, 17, 18]), as far as we know, its undecidability does not seem to be contained in the literature. Following our study of recognition properties, we finally consider the isomorphism problem in certain classes of unique \mathbb{Z}-extensions, and prove that it is equivalent to the semi-conjugacy problem for the corresponding deranged outer automorphisms (see details in Section 8).

The structure of the paper is as follows. In Section 1 we state the recognition problems we are interested in. In Section 2 we introduce the most general framework for our study: (finitely presented) r\mathbb{Z}^{r}-extensions (denoted byr*\operatorname{-by-}\mathbb{Z}^{r}), unique r\mathbb{Z}^{r}-extensions (denoted !byr!\operatorname{-by-}\mathbb{Z}^{r}), as well as the subfamily of 𝖿𝗀byr\mathsf{fg}\operatorname{-by-}\mathbb{Z}^{r} groups, and will investigate the above problems for them. In Sections 3 and 4 we focus on the case r=1r=1 (i.e., infinite cyclic extensions) which will be the main target of the paper. The central result in Section 5 is 5.3, showing that the membership problem for 𝖿𝗀by\mathsf{fg}\operatorname{-by-}\mathbb{Z} (among other similar families) is undecidable, even within the class !by!\operatorname{-by-}\mathbb{Z}. As an application, Section 6 contains the undecidability of the BNS invariant (6.4). In Section 7 we search for “standard presentations” of 𝖿𝗀by\mathsf{fg}\operatorname{-by-}\mathbb{Z} groups (7.1). Finally, in Section 8 we characterize the isomorphism problem in the subclass of unique \mathbb{Z}-extensions by means of the so-called semi-conjugacy problem (a weakened version of the standard conjugacy problem) for deranged outer automorphisms (8.12).

1. Algorithmic recognition of groups

Algorithmic behavior of groups has been a very fundamental concern in Combinatorial and Geometric Group Theory since the very beginning of this branch of Mathematics in the early 1900’s. The famous three problems stated by Max Dehn in 1911 are prototypical examples of this fact: the Word, Conjugacy, and Isomorphism problems have been very influential in the literature along these last hundred years. Today, these problems (together with a great and growing collection of variations) are the center of what is known as Algorithmic Group Theory.

Dehn’s Isomorphism Problem is probably the paradigmatic example of what is popularly understood as “algorithmic recognition of groups”. Namely, let 𝒢𝖿𝗉\mathcal{G}_{\mathsf{fp}} be the family of finite presentations of groups. Then (with the usual abuse of notation of denoting in the same way a presentation and the presented group):

  • Isomorphism problem [ IP\operatorname{IP} ]: given two finite presentations G1,G2𝒢𝖿𝗉G_{1},G_{2}\in\mathcal{G}_{\mathsf{fp}}, decide whether they present isomorphic groups, G1G2G_{1}\simeq G_{2}, or not.

It is well known that, in this full generality, Dehn’s Isomorphism Problem is unsolvable, see for example [27]. So, a natural next step is to study what happens when we restrict the inputs to a certain subfamily 𝒢𝖿𝗉\mathcal{H}\subseteq\mathcal{G}_{\mathsf{fp}}:

  • Isomorphism problem within \mathcal{H} [ IP()]\operatorname{IP}(\mathcal{H})\,]: given two finite presentations H1,H2H_{1},H_{2}\in\mathcal{H}, decide whether they present isomorphic groups, H1H2H_{1}\simeq H_{2}, or not.

Since we are interested in groups, we will only consider families of presentations closed by isomorphism; in this way, the problems considered are actually about groups (although represented by finite presentations). The literature is full of results solving the isomorphism problem for more and more such subfamilies \mathcal{H} of 𝒢𝖿𝗉\mathcal{G}_{\mathsf{fp}}, or showing its unsolvability even when restricted to smaller and smaller subfamilies \mathcal{H}.

Another recognition aspect is that of deciding whether a given group satisfies certain property, i.e., whether it belongs to a certain previously defined family. For two arbitrary subfamilies ,𝒢𝒢𝖿𝗉\mathcal{H},\mathcal{G}\subseteq\mathcal{G}_{\mathsf{fp}}, we define the:

  • (Family) Membership problem for \mathcal{H} within 𝒢\mathcal{G} [ MP𝒢()\operatorname{MP}_{\mathcal{G}}(\mathcal{H}) ]: given a finite presentation G𝒢{G\in\mathcal{G}}, decide whether GG\in\mathcal{H} or not.

If 𝒢\mathcal{H}\subseteq\mathcal{G} and MP𝒢()\operatorname{MP}_{\mathcal{G}}(\mathcal{H}) is decidable we will also say that the inclusion 𝒢\mathcal{H}\subseteq\mathcal{G} is decidable. When the considered ambient family is the whole family of finitely presented groups (i.e., 𝒢=𝒢𝖿𝗉\mathcal{G}=\mathcal{G}_{\mathsf{fp}}) we will usually omit any reference to it and simply talk about the membership problem for \mathcal{H}, denoted MP()\operatorname{MP}(\mathcal{H}).

A classic undecidability result due to Adian [1, 2] and Rabin [29] (see also [27]) falls into this scheme. Namely, when \mathcal{H} is a Markov subfamily (i.e., a nonempty subfamily 𝒢𝖿𝗉\emptyset\neq\mathcal{H}\subseteq\mathcal{G}_{\mathsf{fp}} such that the subgroups of groups in \mathcal{H} do not completely cover 𝒢𝖿𝗉\mathcal{G}_{\mathsf{fp}}), then MP()\operatorname{MP}(\mathcal{H}) is not decidable. This turns out to include the impossibility of deciding membership for countless well known families of finitely presented groups (e.g., trivial, finite, abelian, nilpotent, solvable, free, torsion-free, simple, automatic, etc).

Note that MP()\operatorname{MP}(\mathcal{H}) being decidable is the same as saying that \mathcal{H} is a recursive set of finite presentations. And, even when MP()\operatorname{MP}(\mathcal{H}) is not decidable, we can still ask for a recursive enumeration of the elements in \mathcal{H}:

  • (Family) Enumeration problem for \mathcal{H} [ EP()\operatorname{EP}(\mathcal{H}) ]: enumerate all the elements in \mathcal{H}.

In many cases the considered subfamily 𝒢𝖿𝗉\mathcal{H}\subseteq\mathcal{G}_{\mathsf{fp}} entails a concept of “good” or “standard” presentation 𝒮\mathcal{S}\subseteq\mathcal{H}, for the groups presented. For example, if \mathcal{H} is the family of (finite presentations for) braid groups {Bnn2}\{B_{n}\mid n\geqslant 2\}, we can define the set of standard presentations 𝒮\mathcal{S}\subseteq\mathcal{H} to be those of the form

σ1,,σn1|σiσj=σjσi,|ij|>1σiσjσi=σjσiσj,|ij|=1;\left\langle\sigma_{1},\ldots,\sigma_{n-1}\,\left|\ \begin{array}[]{cc}\sigma_{i}\sigma_{j}=\sigma_{j}\sigma_{i},&|i-j|>1\\ \sigma_{i}\sigma_{j}\sigma_{i}=\sigma_{j}\sigma_{i}\sigma_{j},&|i-j|=1\end{array}\,\right.\hskip-2.0pt\right\rangle\,;

in this case, the family enumeration problem for 𝒮\mathcal{S} consists, on input an arbitrary finite presentation presenting a braid group, to compute its (unique) standard one for it, i.e., to recognize the number of strands nn.

2. Group extensions

Let GG and QQ be arbitrary groups. We say that GG is a group extension by QQ (or a QQ-extension) if GG can be homomorphically mapped onto QQ, i.e., if there exists a normal subgroup HGH\trianglelefteqslant G such that the quotient G/HG/H is isomorphic to QQ. Of course, this situation gives rise to the short exact sequence

1HGQ11\to H\to G\to Q\to 1

for some group HH, and we will say that GG is byQ*\operatorname{-by-}Q. One also says that such HH is a base group for the extension, and that GG is an extension of HH by QQ; so, if we want to specify the base group we will say that GG is HbyQH\operatorname{-by-}Q.

We remark that a given group extension by QQ may admit many, even non isomorphic, different base groups (see 8.2).

If HH is the only (as subset) normal subgroup of GG with quotient G/HG/H isomorphic to QQ, then we say that the QQ-extension is unique, or that GG is a unique extension by QQ; in the same vein as before, we will say that GG is !byQ!\operatorname{-by-}Q (or that GG is !HbyQ!H\operatorname{-by-}Q, if we want to specify who is the unique normal subgroup).

It will be convenient to extend this notation allowing to replace the groups HH and QQ by any group property (which we will usually write in 𝗌𝖺𝗇𝗌\mathsf{sans} typeface). Concretely, given two properties of groups, 𝖯1\mathsf{P}_{1}, 𝖯2\mathsf{P}_{2}, we say that a group GG is 𝖯1by𝖯2\mathsf{P}_{1}\operatorname{-by-}\mathsf{P}_{2} (resp., !𝖯1by𝖯2!\mathsf{P}_{1}\operatorname{-by-}\mathsf{P}_{2}) if it is HbyQH\operatorname{-by-}Q (resp., !HbyQ!H\operatorname{-by-}Q) for certain groups HH satisfying 𝖯1\mathsf{P}_{1}, and QQ satisfying 𝖯2\mathsf{P}_{2}. In this way we can easily refer to families of group extensions in terms of the behavior of their base and quotient groups. So, for example, a group GG is 𝖿𝗀byr\mathsf{fg}\operatorname{-by-}\mathbb{Z}^{r} if it is HbyrH\operatorname{-by-}\mathbb{Z}^{r} for some finitely generated group HH. And it is !𝖿𝗀byr!\mathsf{fg}\operatorname{-by-}\mathbb{Z}^{r} if it is !Hbyr!H\operatorname{-by-}\mathbb{Z}^{r} for some finitely generated group HH; i.e., if it has a unique normal subgroup with quotient isomorphic to r\mathbb{Z}^{r}, which happens to be finitely generated (not to be confused with GG having a unique finitely generated normal subgroup whose quotient is isomorphic to r\mathbb{Z}^{r}—and possibly some others infinitely generated as well).

When we need to add extra assumptions (i.e., satisfying some property 𝖯\mathsf{P}) on the elements of certain family 𝒢\mathcal{G}, we will denote the new family [𝒢]𝖯[\mathcal{G}]_{\mathsf{P}}. For example, [𝖺𝖻𝖾𝗅𝗂𝖺𝗇]𝖿𝗀[\mathsf{abelian}]_{\mathsf{fg}} denotes the family of finitely generated abelian groups, while [by]𝖿𝗉[*\operatorname{-by-}\mathbb{Z}]_{\mathsf{fp}} denotes the family of finitely presented extensions by \mathbb{Z}.

Recognizing r\mathbb{Z}^{r}-extensions (more concretely, solving the membership problem for the families byr*\operatorname{-by-}\mathbb{Z}^{r} and !byr!\operatorname{-by-}\mathbb{Z}^{r}) turns out to be very easy. Recall that, for a finitely generated group GG, its abelianization is always of the form Gab=rTG^{\operatorname{ab}}=\mathbb{Z}^{r}\oplus T, where rr is a nonnegative integer and TT is a finite abelian group (both canonically determined by GG and called, respectively, the (first) Betti number of GG, denoted r=b(G)r=\operatorname{b}(G), and the abelian torsion of GG).

Clearly, for a finitely generated group GG, b(G)\operatorname{b}(G) is the maximum rank for a free-abelian quotient of GG. Hence, we have the following straightforward characterizations.

Lemma 2.1.

Let GG be a finitely generated group, and kk a nonnegative integer. Then,

  1. (1)

    GG is byk*\operatorname{-by-}\mathbb{Z}^{k} if and only if b(G)k\operatorname{b}(G)\geq k;

  2. (2)

    GG is !byk!\operatorname{-by-}\mathbb{Z}^{k} if and only if b(G)=k\operatorname{b}(G)=k. ∎

Note that one can easily compute the Betti number of any group given by a finite presentation: just abelianize it (i.e., add as relators the commutators of any pair of generators in the presentation) and then apply the Classification Theorem for finitely generated abelian groups, which is clearly algorithmic. Thus, Lemma 2.1 immediately implies the decidability of the membership problem for these families of groups.

Corollary 2.2.

For every k0k\geqslant 0, the membership problem for the families byk*\operatorname{-by-}\mathbb{Z}^{k} and !byk!\operatorname{-by-}\mathbb{Z}^{k} is decidable; i.e., there exists an algorithm which takes any finite presentation as input and decides whether the presented group is byk*\operatorname{-by-}\mathbb{Z}^{k} (resp., !byk!\operatorname{-by-}\mathbb{Z}^{k}) or not. ∎

Let us denote by πab:GGab\pi^{\operatorname{ab}}\colon G\twoheadrightarrow G^{\operatorname{ab}} the abelianization map, by πab:Gb(G)\pi^{\operatorname{ab}}_{*}\colon G\twoheadrightarrow\mathbb{Z}^{\operatorname{b}(G)} the abelianization map followed by the canonical projection onto the free-abelian part b(G)\mathbb{Z}^{\operatorname{b}(G)}, and let us also write Gab=πab(G)G^{\operatorname{ab}}_{*}=\pi^{\operatorname{ab}}_{*}(G). We collect here some elementary properties of the first Betti number which will be useful later.

Lemma 2.3.

Let GG be a finitely generated group. Then,

  1. (1)

    b(G)=b(Gab)=b(Gab)=rkGabrk(Gab)\operatorname{b}(G)=\operatorname{b}(G^{\operatorname{ab}})=\operatorname{b}(G^{\operatorname{ab}}_{*})=\operatorname{rk}{G^{\operatorname{ab}}_{*}}\leq\operatorname{rk}(G^{\operatorname{ab}}), with equality if and only if GabG^{\operatorname{ab}} is free-abelian;

  2. (2)

    for every subgroup HGabH\leqslant G^{\operatorname{ab}}, b(Gab/H)=b(Gab)b(H)\operatorname{b}(G^{\operatorname{ab}}/H)=\operatorname{b}(G^{\operatorname{ab}})-\operatorname{b}(H);

  3. (3)

    if Gab=H1HkG^{\operatorname{ab}}=H_{1}\oplus\cdots\oplus H_{k}, then b(Gab)=b(H1)++b(Hk)\operatorname{b}(G^{\operatorname{ab}})=\operatorname{b}(H_{1})+\cdots+\operatorname{b}(H_{k}). ∎

In addition, note that the kernels of both πab\pi^{\operatorname{ab}} and πab\pi^{\operatorname{ab}}_{*} are fully characteristic subgroups of GG (i.e., invariant under endomorphisms of GG). Hence, every ϕEnd(G)\phi\in\operatorname{End}(G) (resp., every ϕAut(G)\phi\in\operatorname{Aut}(G)) canonically determines endomorphisms ϕabEnd(Gab)\phi^{\operatorname{ab}}\in\operatorname{End}(G^{\operatorname{ab}}) and ϕabEnd(Gab){\phi^{\operatorname{ab}}_{*}\in\operatorname{End}(G^{\operatorname{ab}}_{*})} (resp., automorphisms ϕabAut(Gab)\phi^{\operatorname{ab}}\in\operatorname{Aut}(G^{\operatorname{ab}}) and ϕabAut(Gab)\phi^{\operatorname{ab}}_{*}\in\operatorname{Aut}(G^{\operatorname{ab}}_{*})). Of course, after choosing an abelian basis for r\mathbb{Z}^{r}, where r=b(G)r=\operatorname{b}(G), ϕab\phi^{\operatorname{ab}}_{*} can be thought of as an r×rr\times r square matrix over the integers (with determinant ±1\pm 1 if ϕ\phi is an automorphism of GG). In the following section we will relate certain properties of a \mathbb{Z}-extension with properties of its defining automorphism ϕ\phi.

3. \mathbb{Z}-extensions

We will concentrate now on infinite cyclic extensions, concretely in the family by*\operatorname{-by-}\mathbb{Z} and its subfamily !by!\operatorname{-by-}\mathbb{Z}. Let us describe them in a different way: since \mathbb{Z} is a free group, every short exact sequence of the form 1HG11\to H\to G\to\mathbb{Z}\to 1 splits, and so GG is a semidirect product of HH by \mathbb{Z}; namely GHαG\simeq H\rtimes_{\alpha}\mathbb{Z}, for some αAut(H)\alpha\in\operatorname{Aut}(H). Let us recall this well known construction in order to fix our notation.

Given an arbitrary group HH and an automorphism αAut(H)\alpha\in\operatorname{Aut}(H), define the semidirect product of HH by \mathbb{Z} determined by α\alpha as the group HαH\rtimes_{\alpha}\mathbb{Z} with underlying set H×H\times\mathbb{Z} and operation given by

(1) (h,m)(k,n)=(hαm(k),m+n),(h,m)\cdot(k,n)=(h\,\alpha^{m}(k),m+n),

for all h,kHh,k\in H, and m,nm,n\in\mathbb{Z}. Of course, h(h,0)h\mapsto(h,0) is a natural embedding of HH in HαH\rtimes_{\alpha}\mathbb{Z}, and we then have the natural short exact sequence

(2) 1HHα1.1\to H\to H\rtimes_{\alpha}\mathbb{Z}\to\mathbb{Z}\to 1.

Therefore, HαH\rtimes_{\alpha}\mathbb{Z} belongs to the family by*\operatorname{-by-}\mathbb{Z}. Recall that we can have HαKβH\rtimes_{\alpha}\mathbb{Z}\simeq K\rtimes_{\beta}\mathbb{Z}, with H=KH=K but αβ\alpha\neq\beta; or even with H≄KH\not\simeq K. We discuss this phenomena in Section 8 (see 8.3 and 8.2, respectively).

In the opposite direction, assume that GG is in the family by*\operatorname{-by-}\mathbb{Z}. Choose a homomorphism onto \mathbb{Z}, say ρ:G\rho\colon G\twoheadrightarrow\mathbb{Z}, and consider the short exact sequence given by

1HGρ1,1\to H\to G\stackrel{{\scriptstyle\rho}}{{\twoheadrightarrow}}\mathbb{Z}\to 1,

where H=kerρGH=\ker\rho\trianglelefteqslant G. Choose and denote by tt a preimage in GG of any of the two generators of \mathbb{Z} (note that choosing such tt is equivalent to choosing a split homomorphism for ρ\rho). Now consider the conjugation by tt in GG, say γt:GG\gamma_{t}\colon G\to G, gtgt1g\mapsto tgt^{-1}, and denote by αAut(H)\alpha\in\operatorname{Aut}(H) its restriction to HH (note that γt\gamma_{t} is an inner automorphism of GG, but α\alpha may very well not be inner as an automorphism of HH). By construction, we have

(3) th=α(h)t,th=\alpha(h)t,

for every hHh\in H. At this point, it is clear that every element from GG can be written in a unique way as htkht^{k}, for some hHh\in H and kk\in\mathbb{Z}. And—from (3)—the operation in GG can be easily understood by thinking that tt (respectively, t1t^{-1}) jumps to the right of elements from HH at the price of applying α\alpha (respectively, α1\alpha^{-1}):

(4) htmktn=hαm(k)tm+n.ht^{m}\cdot kt^{n}=h\,\alpha^{m}(k)\,t^{m+n}.

This is, precisely, the multiplicative version of (1). Hence, GHαG\simeq H\rtimes_{\alpha}\mathbb{Z}, the semidirect product of HH by \mathbb{Z} determined by α\alpha.

From this discussion it follows easily that, for any presentation of HH, say H=XRH=\langle\,X\!\mid\!R\,\rangle, and any αAut(H)\alpha\in\operatorname{Aut}(H), the semidirect product G=HαG=H\rtimes_{\alpha}\mathbb{Z} admits a presentation of the form

(5) X,t|R,txt1=α(x)(xX).\left\langle X,t\,\left|\ R\,,\,txt^{-1}=\alpha(x)\ (x\in X)\,\right.\hskip-2.0pt\right\rangle.

Note that (5) is a finite presentation if and only if the initial presentation for HH was finite. So, a group GG admits a finite presentation of type (5) if and only if GG is 𝖿𝗉by\mathsf{fp}\operatorname{-by-}\mathbb{Z}. This provides the notion of standard presentation in this context.

Definition 3.1.

A standard presentation for a 𝖿𝗉by\mathsf{fp}\operatorname{-by-}\mathbb{Z} group GG is a finite presentation of the form (5).

The previous discussion provides the following alternative descriptions for the family of finitely presented \mathbb{Z}-extensions. For any group HH, we have

[Hby]𝖿𝗉={Hα f.p.αAut(H)},[H\operatorname{-by-}\mathbb{Z}]_{\mathsf{fp}}=\{H\rtimes_{\alpha}\mathbb{Z}\text{ f.p.}\mid\alpha\in\operatorname{Aut}(H)\},

and then,

[by]𝖿𝗉\displaystyle[*\operatorname{-by-}\mathbb{Z}]_{\mathsf{fp}}\, ={G𝖿𝗉b(G)1}\displaystyle=\,\{\,G\ \mathsf{fp}\,\mid\operatorname{b}(G)\geqslant 1\,\}
={Hα𝖿𝗉H group, and αAut(H)}.\displaystyle=\,\{\,H\rtimes_{\alpha}\mathbb{Z}\ \mathsf{fp}\,\mid H\text{ group, and }\alpha\in\operatorname{Aut}(H)\,\}\,.
Remark 3.2.

Note that we have made no assumptions on the base group HH. Imposing natural conditions on it, we get the inclusions

(6) 𝖿𝗉by[𝖿𝗀by]𝖿𝗉[by]𝖿𝗉,\mathsf{fp}\operatorname{-by-}\mathbb{Z}\,\subseteq[\mathsf{fg}\operatorname{-by-}\mathbb{Z}]_{\mathsf{fp}}\,\subseteq\,[*\operatorname{-by-}\mathbb{Z}]_{\mathsf{fp}}\,,

which will be seen throughout the paper to be both strict.

The strictness of the second inclusion in (6) is a direct consequence of 5.2, while the strictness of the first one is proved below (we thank Conchita Martínez for pointing out the candidate group (7) in the subsequent proof).

Proposition 3.3.

The inclusion 𝖿𝗉by[𝖿𝗀by]𝖿𝗉\mathsf{fp}\operatorname{-by-}\mathbb{Z}\,\subseteq[\mathsf{fg}\operatorname{-by-}\mathbb{Z}]_{\mathsf{fp}} is strict. That is, there exist finitely presented \mathbb{Z}-extensions of finitely generated groups, which are not \mathbb{Z}-extensions of any finitely presented group.

Proof.

Let p<q<rp<q<r be three different prime numbers, and consider the additive group AA of the ring [1p,1q,1r]\mathbb{Z}[\frac{1}{p},\frac{1}{q},\frac{1}{r}], which is well known to be generated by X={1/pn}n{1/qn}n{1/rn}nX=\{1/p^{n}\}_{n\in\mathbb{N}}\cup\{1/q^{n}\}_{n\in\mathbb{N}}\cup\{1/r^{n}\}_{n\in\mathbb{N}}, but not finitely generated (for any given finite set of elements in AA, let kk be the biggest pp-exponent in the denominators, and it is easy to see that 1/pk+1A1/p^{k+1}\in A is not in the subgroup generated by them). Finally consider the following two commuting automorphisms α,β:AA\alpha,\beta\colon A\to A given by α:apra\alpha\colon a\mapsto\tfrac{p}{r}a, and β:aqra\beta\colon a\mapsto\tfrac{q}{r}a.

Our candidate GG is the (metabelian) semidirect product of AA by 2=t,s[t,s]\mathbb{Z}^{2}=\langle\,t,s\!\mid\![t,s]\,\rangle, with action tαt\mapsto\alpha, sβs\mapsto\beta, namely,

(7) G\displaystyle G =Aα,β2\displaystyle\,=\,A\rtimes_{\alpha,\,\beta}\mathbb{Z}^{2}
=A,t,s|ts=st,tat1=pra,sas1=qra.\displaystyle\,=\,\left\langle A,t,s\,\left|\ ts=st,\,\,tat^{-1}=\frac{p}{r}a,\,sas^{-1}=\frac{q}{r}a\,\right.\hskip-2.0pt\right\rangle.

(One has to be careful here with the notation: it is typically multiplicative for the nonabelian group GG, but additive for the abelian group AA, while α\alpha and β\beta are defined using products of rational numbers; beware, in particular, of the element 1A1\in\mathbb{Z}\subseteq A which is additive and, of course, nontrivial.)

It is easy to see that GG is generated by 1A1\in A, and t,s2t,s\in\mathbb{Z}^{2}: indeed, conjugating 1 by all powers of tt and ss we obtain, respectively, pn/rnp^{n}/r^{n} and qn/rnq^{n}/r^{n}, and then λnpn/rn+μnqn/rn=1/rn\lambda_{n}p^{n}/r^{n}+\mu_{n}q^{n}/r^{n}=1/r^{n} for appropriate integers λn,μn\lambda_{n},\mu_{n}, by Bezout’s identity; with the same trick and having rn/pnr^{n}/p^{n} and 1/rn1/r^{n}, we get 1/pnrn1/p^{n}r^{n} and so, 1/pn1/p^{n}; and similarly, one gets 1/qn1/q^{n}. Note that, in order to obtain all of AA, it is enough to get 1/pn1/p^{n}, 1/qn1/q^{n} and 1/rn1/r^{n} for nn big enough; this will be used later.

To see that GG is finitely presented, it is enough to use Theorem A(ii) in [3], which provides a precise condition for a finitely generated metabelian group to be finitely presented. This is a result, due to Bieri and Strebel, that later lead to the development of the so called Bieri–Neumann–Strebel theory (see  Section 6).

Note that the group GG is finitely generated and metabelian, having AA as an abelian normal subgroup with quotient 2\mathbb{Z}^{2}. We know that AA is not finitely generated as group; however, with 2\mathbb{Z}^{2} acting by conjugation, AA becomes a 2\mathbb{Z}^{2}-module, which is finitely generated by the exact same argument as in the previous paragraph. But even more: for all nontrivial valuation v:2v\colon\mathbb{Z}^{2}\to\mathbb{R}, AA is also finitely generated over at least one of the two monoids {(n,m)2v(n,m)0}\{(n,m)\in\mathbb{Z}^{2}\mid v(n,m)\geqslant 0\}, or {(n,m)2v(n,m)0}\{(n,m)\in\mathbb{Z}^{2}\mid v(n,m)\leqslant 0\}. This is because any such valuation has the form (n,m)αn+βm(n,m)\mapsto\alpha n+\beta m for some (0,0)(α,β)2(0,0)\neq(\alpha,\beta)\in\mathbb{Z}^{2}, and then it is routine to show that, starting with 1A1\in A, conjugating only either by those tnsmt^{n}s^{m} with αn+βm0\alpha n+\beta m\geqslant 0, or those with αn+βm0\alpha n+\beta m\leqslant 0, and adding, we can get all of AA (we leave the details to the reader). By Theorem A(ii) from [3], this implies that the group GG is finitely presented.

Now consider the subgroup H=ts,AGH=\langle ts,A\rangle\leqslant G, which is clearly normal and produces a quotient G/H==zG/H=\mathbb{Z}=\langle\,z\!\mid\!-\,\rangle. Since

HAαβ=A,z|zaz1=pqr2aH\simeq\,A\rtimes_{\alpha\circ\beta}\mathbb{Z}\,=\,\left\langle A,z\,\left|\ zaz^{-1}=\frac{pq}{r^{2}}a\,\right.\hskip-2.0pt\right\rangle

is generated by 1,z1,z (by the same reason as above), we deduce that GG is both a \mathbb{Z}-extension of its finitely generated subgroup HH, and finitely presented; i.e., G[𝖿𝗀by]𝖿𝗉G\in[\mathsf{fg}\operatorname{-by-}\mathbb{Z}]_{\mathsf{fp}}.

It remains to see that G𝖿𝗉byG\notin\mathsf{fp}\operatorname{-by-}\mathbb{Z} (i.e., GG is not a \mathbb{Z}-extension of any finitely presented subgroup). We do not know whether this is true for every p,q,rp,q,r, but we shall prove it for particular values of the parameters; concretely for (p,q,r)=(2,3,5)(p,q,r)=(2,3,5).

It is easy to see that the derived subgroup GG^{\prime} is contained in AA. We shall prove that, when this inclusion is indeed an equality—for example, when (p,q,r)=(2,3,5)(p,q,r)=(2,3,5), as it is straighforward to see—then GG is not a \mathbb{Z}-extension of any finitely presented subgroup. That is, no normal subgroup NGN\trianglelefteqslant G with G/NG/N\simeq\mathbb{Z} can be finitely presented. In fact, let NGN\trianglelefteqslant G be such a subgroup. Then A=GNGA=G^{\prime}\trianglelefteqslant N\trianglelefteqslant G and, taking quotients by AA, we obtain 1N/AG/A=2=t,s1\trianglelefteqslant N/A\trianglelefteqslant G/A=\mathbb{Z}^{2}=\langle t,s\rangle. But G/NG/AN/A2N/A\mathbb{Z}\simeq G/N\simeq\frac{G/A}{N/A}\simeq\frac{\mathbb{Z}^{2}}{N/A}. So, it must be N/AN/A\simeq\mathbb{Z}.

Now, choose (n,m)(0,0)(n,m)\neq(0,0) such that N/AN/A is generated by tnsmAt^{n}s^{m}A (we can clearly assume n>0n>0); and deduce that NAφN\simeq A\ltimes_{\varphi}\mathbb{Z} with action φ:apnqmrn+ma\varphi\colon a\mapsto\frac{p^{n}q^{m}}{r^{n+m}}a. In particular, NN is finitely generated by an argument as above. Note also that the action of φ\varphi is by multiplication by a simplified fraction, say λ/μ\lambda/\mu, with λ\lambda and μ\mu both different from ±1\pm 1 (if m0m\geqslant 0 it is multiplication by pnqmrn+m\frac{p^{n}q^{m}}{r^{n+m}}; and if m<0m<0, it is multiplication by pnr|m|rnq|m|\frac{p^{n}r^{|m|}}{r^{n}q^{|m|}}).

Finally, let us apply again Theorem A(ii) in [3], now to the short exact sequence 1AN11\to A\to N\to\mathbb{Z}\to 1. The only nontrivial valuations \mathbb{Z}\to\mathbb{R} are 1±11\mapsto\pm 1, and it is easy to see that AA is not finitely generated neither as a +\mathbb{Z}^{+}-module (with finitely many elements one cannot obtain 1/λn1/\lambda^{n} for n0n\gg 0), nor as a \mathbb{Z}^{-}-module (with finitely many elements one cannot obtain 1/μn1/\mu^{n} for n0n\gg 0). Therefore, NN is not finitely presented, and the group GG is not a \mathbb{Z}-extension of any finitely presented subgroup, as we wanted to prove. ∎

4. Unique \mathbb{Z}-extensions

Recall that the family of unique \mathbb{Z}-extensions (i.e., groups having a unique normal subgroup with quotient \mathbb{Z}) which are finitely presented is denoted [!by]𝖿𝗉[!\operatorname{-by-}\mathbb{Z}]_{\mathsf{fp}}.

As seen in 2.1, the family [!by]𝖿𝗉[!\operatorname{-by-}\mathbb{Z}]_{\mathsf{fp}} consists precisely of those groups GG in [by]𝖿𝗉[*\operatorname{-by-}\mathbb{Z}]_{\mathsf{fp}} such that b(G)=1\operatorname{b}(G)=1. For finitely generated base groups HH, LABEL:prop:derangedcharacterization gives a quite simple characterization of this unicity condition in terms of the defining automorphism α\alpha. We first need a convenient description of the abelianization of a \mathbb{Z}-extension.

Lemma 4.1.

Let HH be an arbitrary group, and let αAut(H)\alpha\in\operatorname{Aut}(H). Then,

(8) (Hα)abHabim(αabid).(H\rtimes_{\alpha}\mathbb{Z})^{\operatorname{ab}}\,\simeq\,\frac{H^{\operatorname{ab}}}{\operatorname{im}(\alpha^{\operatorname{ab}}-\operatorname{id})}\,\oplus\,\mathbb{Z}\,.

Moreover, if HH is finitely generated, then so is HαH\rtimes_{\alpha}\mathbb{Z}, and

(9) (Hα)abk+1T,(H\rtimes_{\alpha}\mathbb{Z})^{\operatorname{ab}}\simeq\mathbb{Z}^{k+1}\oplus T,

where kk is the rank of ker(αabid)\ker(\alpha_{*}^{\operatorname{ab}}-\operatorname{id}), and TT is a finite abelian group.

Proof.

Let H=XRH=\langle\,X\!\mid\!R\,\rangle. Abelianizing Hα=X,tR,txit1=α(xi)(xiX)H\rtimes_{\alpha}\mathbb{Z}=\langle\,X,t\!\mid\!R,\,tx_{i}t^{-1}=\alpha(x_{i})\,\,\,(x_{i}\in X)\,\rangle, we get

(Hα)ab\displaystyle(H\rtimes_{\alpha}\mathbb{Z})^{\operatorname{ab}} =X,t|R,txit1=α(xi)(xiX),xixj=xjxi(xi,xjX),txi=xit(xiX)\displaystyle=\left\langle X,t\,\left|\ \begin{smallmatrix}R,\hfill\\ tx_{i}t^{-1}=\alpha(x_{i})\hfill&\hfill(x_{i}\in X),\\ x_{i}x_{j}=x_{j}x_{i}\hfill&\ \hfill(x_{i},x_{j}\in X),\\ tx_{i}=x_{i}t\hfill&\hfill(x_{i}\in X)\phantom{,}\end{smallmatrix}\,\right.\hskip-2.0pt\right\rangle
(10) X|R,xi=α(xi)(xiX),xixj=xjxi(xi,xjX)×t\displaystyle\simeq\,\left\langle X\,\left|\ \begin{smallmatrix}R,\hfill\\ x_{i}=\alpha(x_{i})\hfill&\hfill(x_{i}\in X),\\ x_{i}x_{j}=x_{j}x_{i}\hfill&\ \hfill(x_{i},x_{j}\in X)\end{smallmatrix}\,\right.\hskip-2.0pt\right\rangle\times\langle\,t\!\mid\!-\,\rangle
Habim(αabid)t.\displaystyle\simeq\,\frac{H^{\operatorname{ab}}}{\operatorname{im}(\alpha^{\operatorname{ab}}-\operatorname{id})}\,\oplus\,\langle\,t\!\mid\!-\,\rangle\,.

For the second part, suppose that HH is finitely generated. Then so is HabH^{\operatorname{ab}} and thus, using 2.3, we have

(11) b(Hα)\displaystyle\operatorname{b}(H\rtimes_{\alpha}\mathbb{Z}) =b(Habim(αabid))+1\displaystyle=\operatorname{b}\!\left(\frac{H^{\operatorname{ab}}}{\operatorname{im}(\alpha^{\operatorname{ab}}-\operatorname{id})}\right)+1
=b(Hab)b(im(αabid))+1\displaystyle=\operatorname{b}(H^{\operatorname{ab}})-\operatorname{b}(\operatorname{im}(\alpha^{\operatorname{ab}}-\operatorname{id}))+1
=b(Hab)b(im(αabid))+1\displaystyle=\operatorname{b}(H^{\operatorname{ab}}_{*})-\operatorname{b}(\operatorname{im}(\alpha_{*}^{\operatorname{ab}}-\operatorname{id}))+1
=b(ker(αabid))+1\displaystyle=\operatorname{b}(\ker(\alpha_{*}^{\operatorname{ab}}-\operatorname{id}))+1
=rk(ker(αabid))+1,\displaystyle=\operatorname{rk}\,(\ker(\alpha_{*}^{\operatorname{ab}}-\operatorname{id}))+1,

which is what we wanted to prove. ∎

This last result, combined with 2.1, provides a computable characterization for automorphisms defining !𝖿𝗀by!\mathsf{fg}\operatorname{-by-}\mathbb{Z} groups.

Proposition 4.2.

Let HH be an arbitrary group, and αAut(H)\alpha\in\operatorname{Aut}(H) such that the semidirect product HαH\rtimes_{\alpha}\mathbb{Z} is finitely generated. Then, the following conditions are equivalent:

  1. (1)

    HαH\rtimes_{\alpha}\mathbb{Z} is !by!\operatorname{-by-}\mathbb{Z};

  2. (2)

    b(Hα)=1\operatorname{b}(H\rtimes_{\alpha}\mathbb{Z})=1;

  3. (3)

    H=ker(πab)H=\ker(\pi^{\operatorname{ab}}_{*}) (i.e., HH is the full preimage of the torsion subgroup of (Hα)ab(H\rtimes_{\alpha}\mathbb{Z})^{\operatorname{ab}} under πab\pi^{\operatorname{ab}});

  4. (4)

    HH is a fully characteristic subgroup of HαH\rtimes_{\alpha}\mathbb{Z}.

Moreover, if HH is finitely generated, then the following additional condition is also equivalent:

  1. (5)

    αab\alpha^{\operatorname{ab}}_{*} has no nontrivial fixed points (equivalently, 11 is not an eigenvalue of αab\alpha^{\operatorname{ab}}_{*}, ker(αabid)=0\ker(\alpha^{\operatorname{ab}}_{*}-\operatorname{id})=0, or det(αabid)0\det(\alpha^{\operatorname{ab}}_{*}-\operatorname{id})\neq 0).

Proof.

[12[\ref{item:!-by-Z}\Leftrightarrow\ref{item:betti=1}]. This is precisely the content of 2.1(ii), for k=1k=1.

[23][\ref{item:betti=1}\Leftrightarrow\ref{item:H=preabelianization of torsion}]. This follows immediately from (4).

[34][\ref{item:H=preabelianization of torsion}\Rightarrow\ref{item:characteristic}]. This is clear, since the torsion subgroup of an abelian group is fully characteristic, and so is its full preimage.

[42][\ref{item:characteristic}\Rightarrow\ref{item:betti=1}]. By contradiction, suppose that HαtH\rtimes_{\alpha}\langle\,t\!\mid\!-\,\rangle has Betti number at least 22. Then, there exists an epimorphism ρ:Hαt2\rho\colon H\rtimes_{\alpha}\langle t\rangle\twoheadrightarrow\mathbb{Z}^{2}, and an element hHh\in H such that ρ(h)0\rho(h)\neq 0. Take a primitive element v2v\in\mathbb{Z}^{2} such that ρ(h)=λv\rho(h)=\lambda v for some λ\lambda\in\mathbb{Z}, λ0\lambda\neq 0. The subgroup v\langle v\rangle\simeq\mathbb{Z} is a direct summand of 2\mathbb{Z}^{2}, and the composition

Hαt𝜌2vHαtvt\begin{array}[]{ccccccl}H\rtimes_{\alpha}\langle t\rangle&\overset{\rho}{\twoheadrightarrow}&\mathbb{Z}^{2}&\twoheadrightarrow&\langle v\rangle&\hookrightarrow&H\rtimes_{\alpha}\langle t\rangle\\ &&&&v&\mapsto&t\end{array}

provides an endomorphism of HαH\rtimes_{\alpha}\mathbb{Z} mapping hHh\in H to tλt^{\lambda}, a contradiction with condition 4.

Finally, the equivalence [25][\ref{item:betti=1}\Leftrightarrow\ref{item:deranged}] is immediate from the equality (11). ∎

Definition 4.3.

We say that an automorphism αAut(H)\alpha\in\operatorname{Aut}(H) is deranged if one of (and thus all) the conditions 14\ref{item:!-by-Z}-\ref{item:characteristic} in 4.2 hold. Note that when the group HH is finitely generated, condition 5 provides further another equivalent definition of derangedness, this time expressed in terms of the automorphism. Note that in this last case, given αAut(H)\alpha\in\operatorname{Aut}(H) by images of generators, it is easy to check algorithmically whether α\alpha is deranged or not.

In particular, every automorphism of a group HH with b(H)=0\operatorname{b}(H)=0, i.e., with |Hab|<|H^{\operatorname{ab}}|<\infty, is (trivially) deranged. Note also that derangedness is, in fact, a property of outer automorphisms. The sets of deranged automorphisms and deranged outer automorphisms of a group HH will be denoted, respectively, Aut𝖽(H)\operatorname{Aut}_{\mathsf{d}}(H) and Out𝖽(H)\operatorname{Out}_{\mathsf{d}}(H).

Consequently, for any finitely generated group HH we have

[!Hby]𝖿𝗉={Hα𝖿𝗉αAut𝖽(H)},[!H\operatorname{-by-}\mathbb{Z}]_{\mathsf{fp}}=\{\,H\rtimes_{\alpha}\mathbb{Z}\ \mathsf{fp}\mid\alpha\in\operatorname{Aut}_{\mathsf{d}}(H)\,\},

and then,

[!𝖿𝗀by]𝖿𝗉\displaystyle[!\mathsf{fg}\operatorname{-by-}\mathbb{Z}]_{\mathsf{fp}}\, ={Hα𝖿𝗉H𝖿𝗀 and αAut𝖽(H)}\displaystyle=\,\{\,H\rtimes_{\alpha}\mathbb{Z}\ \mathsf{fp}\mid H\ \mathsf{fg}\text{ and }\alpha\in\operatorname{Aut}_{\mathsf{d}}(H)\,\}
(12) [!by]𝖿𝗉={G𝖿𝗉b(G)=1}.\displaystyle\subseteq\,[!\operatorname{-by-}\mathbb{Z}]_{\mathsf{fp}}\ =\ \{\,G\ \mathsf{fp}\mid\operatorname{b}(G)=1\}\,.

Note that in (4) we wrote inclusion and not an equality because, in principle, it could happen that a finitely presented !by!\operatorname{-by-}\mathbb{Z} group has his unique normal subgroup with quotient isomorphic to \mathbb{Z} being not finitely generated. In the next section we shall construct such a group (see LABEL:cor:K*Z_isfg-by-Z_iff_K=1) showing that this inclusion is strict.

5. Undecidability results

Observe that if HH is finitely generated or finitely presented, then so is HαH\rtimes_{\alpha}\mathbb{Z} for every αAut(H)\alpha\in\operatorname{Aut}(H), i.e.,

𝖿𝗀by\displaystyle\mathsf{fg}\operatorname{-by-}\mathbb{Z} [by]𝖿𝗀,\displaystyle\,\subseteq\,[*\operatorname{-by-}\mathbb{Z}]_{\mathsf{fg}},
𝖿𝗉by\displaystyle\mathsf{fp}\operatorname{-by-}\mathbb{Z} [by]𝖿𝗉.\displaystyle\,\subseteq\,[*\operatorname{-by-}\mathbb{Z}]_{\mathsf{fp}}\,.

However, it is less obvious that the converse is not true in general: a semidirect product HαH\rtimes_{\alpha}\mathbb{Z} can be finitely presented, with HH not being finitely generated. Or, as was hinted few lines above, even worse: there do exist finitely presented \mathbb{Z}-extensions which are not \mathbb{Z}-extensions of any finitely generated group. In other words, the following inclusion is strict:

[𝖿𝗀by]𝖿𝗉[by]𝖿𝗉.[\mathsf{fg}\operatorname{-by-}\mathbb{Z}]_{\mathsf{fp}}\,\subset\,[*\operatorname{-by-}\mathbb{Z}]_{\mathsf{fp}}\,.

Indeed, this can happen even for unique \mathbb{Z}-extensions. This fact follows easily from the next lemma, showing that any free product KK*\mathbb{Z} has the form of a certain semidirect product.

Lemma 5.1.

Let K=XRK=\langle\,X\!\mid\!R\,\rangle be an arbitrary group with generators X={xj}jJX=\{x_{j}\}_{j\in J}, and consider the free product

iK=X(i)(i)|R(i)(i),\operatorname*{\scalebox{2.0}{\raisebox{-1.29167pt}{$\ast\hskip-0.3pt$}}}_{i\in\mathbb{Z}}K=\left\langle X^{(i)}\,(i\in\mathbb{Z})\,\left|\ R^{(i)}\,(i\in\mathbb{Z})\,\right.\hskip-2.0pt\right\rangle\,,

where X(i)|R(i)\left\langle X^{(i)}\,\left|\ R^{(i)}\,\right.\hskip-2.0pt\right\rangle (i)(i\in\mathbb{Z}) are disjoint copies of the original presentation for KK. Then,

(13) (iK)τK,\left(\operatorname*{\scalebox{2.0}{\raisebox{-1.29167pt}{$\ast\hskip-0.3pt$}}}_{i\in\mathbb{Z}}K\right)\rtimes_{\tau}\mathbb{Z}\ \simeq\ K*\mathbb{Z}\,,

where τ\tau is the automorphism of iK\operatorname*{\scalebox{2.0}{\raisebox{-1.29167pt}{$\ast\hskip-0.3pt$}}}_{i\in\mathbb{Z}}\,K defined by

(14) τ:xj(i)xj(i+1)(i,xj(i)X(i)).\tau\colon x_{j}^{(i)}\mapsto x_{j}^{(i+1)}\ \left(\forall i\in\mathbb{Z}\,,\,\forall x_{j}^{(i)}\in X^{(i)}\right).
Proof.

Naming tt the generator of \mathbb{Z}, we have

(16) (iK)τ\displaystyle\!\!\left(\operatorname*{\scalebox{2.0}{\raisebox{-1.29167pt}{$\ast\hskip-0.3pt$}}}_{i\in\mathbb{Z}}\,K\right)\rtimes_{\tau}\mathbb{Z} =t,X(i)(i)|R(i)(i),txj(i)t1=xj(i+1)(i,xj(i)X(i))\displaystyle\,=\,\left\langle t,X^{(i)}\ (i\in\mathbb{Z})\,\left|\ \!\!\begin{array}[]{l}R^{(i)}\ (i\in\mathbb{Z}),\\ t\,x_{j}^{(i)}\,t^{-1}=x_{j}^{(i+1)}\ \left(i\in\mathbb{Z},\,x_{j}^{(i)}\!\in X^{(i)}\right)\!\!\end{array}\,\right.\hskip-2.0pt\right\rangle
(19) t,X(i)(i)|R(0),txj(i)t1=xj(i+1)(i,xj(i)X(i))\displaystyle\,\simeq\,\left\langle t,X^{(i)}\ (i\in\mathbb{Z})\,\left|\ \!\!\begin{array}[]{l}R^{(0)},\\ t\,x_{j}^{(i)}\,t^{-1}=x_{j}^{(i+1)}\ \left(i\in\mathbb{Z},\,x_{j}^{(i)}\!\in X^{(i)}\right)\!\!\end{array}\,\right.\hskip-2.0pt\right\rangle
(20) t,X(0)|R(0)=K.\displaystyle\,\simeq\,\left\langle t,X^{(0)}\,\left|\ R^{(0)}\,\right.\hskip-2.0pt\right\rangle\,=\,K*\mathbb{Z}\,.

To see the last isomorphism, consider the maps from (20) to (19) given by

t\displaystyle t t,\displaystyle\mapsto t,
xj(0)\displaystyle x_{j}^{(0)} xj(0)(xj(0)X(0)),\displaystyle\mapsto x_{j}^{(0)}\ (x_{j}^{(0)}\in X^{(0)}),

and from (19) to (20) given by

t\displaystyle t t,\displaystyle\mapsto t,
xj(i)\displaystyle x_{j}^{(i)} tixj(0)ti(i,xj(i)X(i)).\displaystyle\mapsto t^{i}x_{j}^{(0)}t^{-i}\ (i\in\mathbb{Z},\ x_{j}^{(i)}\in X^{(i)}).

It is straightforward to see that they are both well-defined homomorphisms, and one inverse to the other. ∎

Corollary 5.2.

If KK is a group with finite abelianization (i.e., b(K)=0\operatorname{b}(K)=0), then the free product KK*\mathbb{Z} is a unique \mathbb{Z}-extension, and the following conditions are equivalent:

  1. (1)

    K is 𝖿𝗀byK*\mathbb{Z}\text{ is }\mathsf{fg}\operatorname{-by-}\mathbb{Z};

  2. (2)

    K is 𝖿𝗉byK*\mathbb{Z}\text{ is }\mathsf{fp}\operatorname{-by-}\mathbb{Z};

  3. (3)

    K is 𝖺𝖻𝖾𝗅𝗂𝖺𝗇byK*\mathbb{Z}\text{ is }\mathsf{abelian}\operatorname{-by-}\mathbb{Z};

  4. (4)

    K is 𝖿𝗂𝗇𝗂𝗍𝖾byK*\mathbb{Z}\text{ is }\mathsf{finite}\operatorname{-by-}\mathbb{Z};

  5. (5)

    K is 𝖿𝗋𝖾𝖾byK*\mathbb{Z}\text{ is }\mathsf{free}\operatorname{-by-}\mathbb{Z};

  6. (6)

    K=1K=1.

In particular, !𝖿𝗀by!\mathsf{fg}\operatorname{-by-}\mathbb{Z} is a strict subfamily of [!by]𝖿𝗉[!\operatorname{-by-}\mathbb{Z}]_{\mathsf{fp}} (and so, 𝖿𝗀by\mathsf{fg}\operatorname{-by-}\mathbb{Z} is a strict subfamily of [by]𝖿𝗉[*\operatorname{-by-}\mathbb{Z}]_{\mathsf{fp}}).

Proof.

Note that the abelianization of KK*\mathbb{Z} is (K)ab=Kab(K*\mathbb{Z})^{\operatorname{ab}}=K^{\operatorname{ab}}\oplus\mathbb{Z}, where |Kab|<|K^{\operatorname{ab}}|<\infty by hypothesis; therefore, b(K)=1\operatorname{b}(K*\mathbb{Z})=1. Thus, from 2.1(ii), KK*\mathbb{Z} is a unique \mathbb{Z}-extension, i.e., it contains a unique normal subgroup with quotient \mathbb{Z}. By 5.1, this unique normal subgroup is isomorphic to zK\operatorname*{\scalebox{2.0}{\raisebox{-1.29167pt}{$\ast\hskip-0.3pt$}}}_{z\in\mathbb{Z}}K, which is finitely generated (resp., finitely presented, abelian, finite, free) if and only if KK is trivial (the free case being true because b(K)=0\operatorname{b}(K)=0).

Taking KK to be a nontrivial finitely presented group with finite abelianization, we obtain that KK*\mathbb{Z} belongs to [!by]𝖿𝗉[!\operatorname{-by-}\mathbb{Z}]_{\mathsf{fp}} but not to !𝖿𝗀by\mathsf{fg}\operatorname{-by-}\mathbb{Z}. ∎

Next, inspired by a trick initially suggested by Maurice Chiodo, we will prove a stronger result. Not only the family !𝖿𝗀by\mathsf{fg}\operatorname{-by-}\mathbb{Z} is a strict subfamily of [!by]𝖿𝗉[!\operatorname{-by-}\mathbb{Z}]_{\mathsf{fp}}, but the membership problem between these two families is undecidable: it is impossible to decide algorithmically whether a given finitely presented unique \mathbb{Z}-extension is 𝖿𝗀by\mathsf{fg}\operatorname{-by-}\mathbb{Z} or not, i.e., whether its unique base group is finitely generated or not. To see this, we use a classic undecidability result: there is no algorithm which, on input a finite presentation, decides whether the presented group is trivial or not (see, for example, [27]).

Theorem 5.3.

For every group property 𝖯{𝖿𝗀,𝖿𝗉,𝖺𝖻𝖾𝗅𝗂𝖺𝗇,𝖿𝗂𝗇𝗂𝗍𝖾,𝖿𝗋𝖾𝖾}\mathsf{P}\in\{\mathsf{fg},\mathsf{fp},\mathsf{abelian},\mathsf{finite},\mathsf{free}\}, the membership problem for  𝖯by\mathsf{P}\operatorname{-by-}\mathbb{Z}  within  [!by]𝖿𝗉[!\operatorname{-by-}\mathbb{Z}]_{\mathsf{fp}} is undecidable.

In other words, there exists no algorithm which, on input a finite presentation of a group with Betti number 11, decides whether it presents a 𝖿𝗀by\mathsf{fg}\operatorname{-by-}\mathbb{Z} (resp., 𝖿𝗉by\mathsf{fp}\operatorname{-by-}\mathbb{Z}, 𝖺𝖻𝖾𝗅𝗂𝖺𝗇by\mathsf{abelian}\operatorname{-by-}\mathbb{Z}, 𝖿𝗂𝗇𝗂𝗍𝖾by\mathsf{finite}\operatorname{-by-}\mathbb{Z}, 𝖿𝗋𝖾𝖾by\mathsf{free}\operatorname{-by-}\mathbb{Z}) group or not.

Proof.

We will proceed by contradiction. Assume the existence of an algorithm, say 𝔄\mathfrak{A}, such that, given as input a finite presentation of a group with Betti number 11, outputs YES if it presents a 𝖯by\mathsf{P}\operatorname{-by-}\mathbb{Z} group, and NO otherwise.

Now, consider the following algorithm 𝔅\mathfrak{B} to check triviality: on input an arbitrary finite presentation K=XRK=\langle\,X\!\mid\!R\,\rangle:

  1. (1)

    abelianize KK and, using the Classification Theorem for finitely generated abelian groups, check whether KabK^{\operatorname{ab}} is trivial or not; if not, answer NO; otherwise KK is a perfect group and so, the new group KK*\mathbb{Z} has Betti number 1;

  2. (2)

    apply 𝔄\mathfrak{A} to the presentation X,tR\langle\,X,t\!\mid\!R\,\rangle, to decide whether KK*\mathbb{Z} is a 𝖯by\mathsf{P}\operatorname{-by-}\mathbb{Z} group or not.

According to 5.2, the output to step (ii) is YES if and only if KK is trivial. Hence, 𝔅\mathfrak{B} is deciding whether the given presentation XR\langle\,X\!\mid\!R\,\rangle presents the trivial group or not. This contradicts Adian–Rabin’s Theorem on the undecidability of the triviality problem. ∎

Of course, if the membership problem is not decidable within some family \mathcal{H}, it is also undecidable within any superfamily of \mathcal{H}. So, we immediately get the following consequence.

Corollary 5.4.

The membership problems for the families  𝖿𝗉by\mathsf{fp}\operatorname{-by-}\mathbb{Z} and 𝖿𝗀by\mathsf{fg}\operatorname{-by-}\mathbb{Z} are undecidable. ∎

As stated in the introduction, this is exactly the same as saying that the families (of finite presentations) 𝖿𝗉by\mathsf{fp}\operatorname{-by-}\mathbb{Z} and 𝖿𝗀by\mathsf{fg}\operatorname{-by-}\mathbb{Z} are not recursive. Note that none of these families is neither Markov nor co-Markov, and thus the two undecidability results in 5.4 are not contained in the classic ones due to Adian–Rabin. Indeed, any finitely presented group is a subgroup of some 𝖿𝗉by\mathsf{fp}\operatorname{-by-}\mathbb{Z} (and so, of some 𝖿𝗀by\mathsf{fg}\operatorname{-by-}\mathbb{Z}) group; therefore the families 𝖿𝗉by\mathsf{fp}\operatorname{-by-}\mathbb{Z} and 𝖿𝗀by\mathsf{fg}\operatorname{-by-}\mathbb{Z} are not Markov. On the other hand, every 𝖿𝗉\mathsf{fp} group embeds in some 22-generated simple group (see [27, Corollary 3.10]); since

𝗌𝗂𝗆𝗉𝗅𝖾𝗉𝖾𝗋𝖿𝖾𝖼𝗍¬(𝖿𝗀by)¬(𝖿𝗉by),\mathsf{simple}\ \Rightarrow\ \mathsf{perfect}\ \Rightarrow\ \neg(\mathsf{fg}\operatorname{-by-}\mathbb{Z})\ \Rightarrow\ \neg(\mathsf{fp}\operatorname{-by-}\mathbb{Z})\,,

the families 𝖿𝗉by\mathsf{fp}\operatorname{-by-}\mathbb{Z} and 𝖿𝗀by\mathsf{fg}\operatorname{-by-}\mathbb{Z} are not co-Markov either.

6. Implications for the BNS invariant

Since the early 1980’s, in a series of papers by R. Bieri, W. Neumann, and R. Strebel (see [3, 4]), several gradually more general invariants—called Sigma (or BNS) Invariants—have been introduced to deal with finiteness conditions for presentations of groups. Concretely in [4], they present an invariant that characterizes those normal subgroups of a finitely generated group GG that are finitely generated and contain the commutator [G,G][G,G] of GG. Over the years, this theory has been reformulated in more geometric terms (for a modern version see the survey [32]). Below, we recall this construction and characterization, and discuss some implications of our undecidability results from Section 5.

For a finitely generated group GG, consider the real vector space Hom(G,)\operatorname{Hom}(G,\mathbb{R}) of all homomorphisms χ:G\chi\colon G\to\mathbb{R} (from GG to the additive group of the field of real numbers), which we call characters of GG. Note that, since \mathbb{R} is abelian and torsion-free, any character χ\chi must factor through πab\pi^{\operatorname{ab}}_{*} (abelianizing and then killing the torsion), i.e.,

χ:GπabGabGab.\chi\colon G\stackrel{{\scriptstyle\pi^{\operatorname{ab}}}}{{\twoheadrightarrow}}G^{\operatorname{ab}}\twoheadrightarrow G^{\operatorname{ab}}_{*}\rightarrow\mathbb{R}\,.

Thus, Hom(G,)=Hom(r,)=r\operatorname{Hom}(G,\mathbb{R})=\operatorname{Hom}(\mathbb{Z}^{r},\mathbb{R})=\mathbb{R}^{r}, where r=b(G)r=\operatorname{b}(G). We will consider the set of nontrivial characters modulo the equivalence relation given by positive scaling:

(21) χ1χ2λ>0 s.t. χ2=λχ1.\chi_{1}\sim\chi_{2}\ \Leftrightarrow\ \exists\lambda>0\text{ s.t. }\chi_{2}=\lambda\chi_{1}\,.

They form the so-called character sphere of GG, denoted S(G)=Hom(G,)/\operatorname{S}(G)=\operatorname{Hom}(G,\mathbb{R})^{*}/\sim which, equipped with the quotient topology, is homeomorphic to the unit Euclidean sphere of dimension r1{r-1} (through the natural identification of each ray emanating from the origin with its unique point of norm 1).

For example, if GG is not by*\operatorname{-by-}\mathbb{Z} (i.e., if b(G)=0)\operatorname{b}(G)=0), then Hom(G,)={0}\operatorname{Hom}(G,\mathbb{R})=\{0\} and the character sphere is empty (so, for this class of groups the BNS theory will be vacuous). More interestingly, if GG is !by!\operatorname{-by-}\mathbb{Z} (i.e., b(G)=1\operatorname{b}(G)=1), then the character sphere of GG is a set of just two points, namely S(G)={+1,1}\operatorname{S}(G)=\{+1,-1\}. Similarly, if b(G)=2,3,\operatorname{b}(G)=2,3,\ldots, then S(G)\operatorname{S}(G) is the unit circle in 2\mathbb{R}^{2}, the unit sphere in 3\mathbb{R}^{3}, and so on.

For any given (equivalence class of a) nontrivial character χ\chi, consider now the following submonoid of GG, called the positive cone of χ\chi:

(22) Gχ={gGχ(g)0}=χ1([0,+)),G_{\chi}=\{g\in G\mid\chi(g)\geq 0\}=\chi^{-1}([0,+\infty))\,,

to be thought of as the full subgraph of the Cayley graph Γ(G,X)\Gamma(G,X) determined by the vertices in GχG_{\chi} (once a set of generators XX is fixed). The Sigma invariant Σ(G)\Sigma(G) can then be defined as follows (we note that this is not the original definition given in [4], but a more geometrically appealing one, which was not noticed to be equivalent until several years later, see [26, Theorem 3.19]).

Definition 6.1.

Let G=XG=\langle X\rangle be a finitely generated group, and Γ(G,X)\Gamma(G,X) its Cayley graph. Then the set

(23) Σ(G)={[χ]S(G)Gχ is connected}S(G)\Sigma(G)=\left\{\,[\chi]\in\operatorname{S}(G)\mid G_{\chi}\text{ is connected}\right\}\subseteq\operatorname{S}(G)

does not depend on the choice of the finite generating set XX (see [32]), and is called the (first) Sigma—or BNSinvariant of GG.

Interestingly, this notion is quite related with commutativity. The extreme examples are free and free-abelian groups, for which it is easy to see that the BNS invariants are, respectively, the empty set and the full character sphere: Σ(Fr)=\Sigma(F_{r})=\emptyset, for r2r\geqslant 2; and Σ(r)=S(r)\Sigma(\mathbb{Z}^{r})=\operatorname{S}(\mathbb{Z}^{r}), for r1r\geqslant 1.

The set of characters vanishing on a certain subgroup HGH\leqslant G determine the following subsphere

S(G,H)={[χ]S(G)χ(H)=0}S(G),\operatorname{S}(G,H)=\{[\chi]\in\operatorname{S}(G)\mid\chi(H)=0\}\subseteq\operatorname{S}(G)\,,

which happens to contain interesting information about HH itself.

Theorem 6.2 (Bieri–Neumann–Strebel, [4]).

Let HH be a normal subgroup of a finitely generated group GG with G/HG/H abelian. Then, HH is finitely generated if and only if S(G,H)Σ(G)\operatorname{S}(G,H)\subseteq\Sigma(G). In particular, the commutator subgroup [G,G][G,G] is finitely generated if and only if Σ(G)=S(G)\Sigma(G)=\operatorname{S}(G). ∎

Note that if GG is HbyH\operatorname{-by-}\mathbb{Z}, then S(G,H)={[πH],[πH]}\operatorname{S}(G,H)=\{[\pi_{H}],-[\pi_{H}]\}, where πH:GG/H\pi_{H}\colon G\twoheadrightarrow G/H\simeq\mathbb{Z} is the canonical projection modulo HH. In this case, 6.2 tells us that

(24) H is 𝖿𝗀[πH],[πH]Σ(G).H\text{ is }\mathsf{fg}\ \Leftrightarrow\ [\pi_{H}],-[\pi_{H}]\in\Sigma(G)\,.

It follows an interesting characterization of 𝖿𝗀by\mathsf{fg}\operatorname{-by-}\mathbb{Z} groups.

Proposition 6.3.

A finitely generated group GG is 𝖿𝗀by\mathsf{fg}\operatorname{-by-}\mathbb{Z} if and only if its BNS invariant contains a pair of antipodal points; i.e.,

(25) G is 𝖿𝗀by[χ]S(G) s.t. [χ],[χ]Σ(G).G\text{ is }\mathsf{fg}\operatorname{-by-}\mathbb{Z}\ \Leftrightarrow\ \exists[\chi]\in\operatorname{S}(G)\text{ s.t. }[\chi],-[\chi]\in\Sigma(G)\,.
Proof.

The implication to the right is clear from (24).

The implication to the left is also clear after making sure that we can always choose such a character χ\chi with cyclic image (i.e., such that rkχ(G)=1\operatorname{rk}_{\mathbb{Z}}\chi(G)=1). To see this, we observe that, given a nontrivial character χ:G\chi\colon G\to\mathbb{R}, one has rkχ(G)=1\operatorname{rk}_{\mathbb{Z}}\chi(G)=1 if and only if there exists λ>0\lambda>0 such that λχ\lambda\chi has integral image, λχ:G\lambda\chi\colon G\twoheadrightarrow\mathbb{Z}\subseteq\mathbb{R}. In other words, rank-one characters correspond, precisely, to those points in the sphere S(G)\operatorname{S}(G) which are projections of integral (or rational) points from r{0}\mathbb{R}^{r}\setminus\{0\}. Thus, rank-one characters form a dense subset of S(G)\operatorname{S}(G). This, together with the fact that Σ(G)\Sigma(G) is an open subset of S(G)\operatorname{S}(G) (see [32, Theorem A3.3]) allows us to deduce, from the hypothesis, the existence of a pair of antipodal points of rank one. ∎

As a corollary, and using 5.3, we obtain the main result in this section: the BNS invariant is not uniformly decidable (even for groups with Betti number 11).

Theorem 6.4.

There is no algorithm such that, given a finite presentation of a group GG (with Betti number 11), and a character [χ]S(G)[\chi]\in\operatorname{S}(G), decides whether [χ][\chi] belongs to Σ(G)\Sigma(G) or not.

Proof.

Given a finite presentation of a !by!\operatorname{-by-}\mathbb{Z} group GG (so b(G)=1\operatorname{b}(G)=1, and S(G)\operatorname{S}(G) has two points), we can abelianize and construct the unique two characters ±π:G\pm\pi\colon G\twoheadrightarrow\mathbb{Z}. Assuming the existence of an algorithm like the one in the statement, we could algorithmically decide whether π,π\pi,-\pi both belong to Σ(G)\Sigma(G) or not, i.e., according to 6.3, whether GG is 𝖿𝗀by\mathsf{fg}\operatorname{-by-}\mathbb{Z} or not. This contradicts 5.3. ∎

We note that, in the case of a one-relator group G=a,brG=\langle\,a,b\!\mid\!r\,\rangle, K. Brown provided an interesting algorithm for deciding whether a given character χ:G\chi\colon G\to\mathbb{R} belongs to Σ(G)\Sigma(G) or not, by looking at the sequence of χ\chi-images of the prefixes of the relation rr (assumed to be in cyclically reduced form); see [8]. Later, N. Dunfield, J. Button and D. Thurston found applications of this result to 33-manifold theory; see [14, 9, 15].

7. Recursive enumerability of presentations

A standard presentation of a given 𝖿𝗉by\mathsf{fp}\operatorname{-by-}\mathbb{Z} group GG has been defined as a finite presentation of the form

X,t|R(X),txt1=α(x)(xX),\left\langle X,t\,\left|\ R(X)\,,\,txt^{-1}=\alpha(x)\ (x\in X)\,\right.\hskip-2.0pt\right\rangle,

where α\alpha is an automorphism of XR\langle\,X\!\mid\!R\,\rangle. It is natural to ask for an algorithm to compute one—or all—standard presentations for such a group GG, since this algorithm will provide explicit computable ways to think GG as a semidirect product (i.e., an explicit base group H=XRH=\langle\,X\!\mid\!R\,\rangle, and an explicit automorphism α\alpha, such that GHαG\simeq H\rtimes_{\alpha}\mathbb{Z}).

We have seen that membership for 𝖿𝗉by\mathsf{fp}\operatorname{-by-}\mathbb{Z} is undecidable (5.4). However, given a finite presentation for a 𝖿𝗉by\mathsf{fp}\operatorname{-by-}\mathbb{Z} group GG, we can use Tietze transformations to obtain a recursive enumeration of all the finite presentations for GG. In the following proposition we provide a (brute force) filtering process which extracts from it a recursive enumeration of all the standard ones.

Proposition 7.1.

Given a finite presentation of a 𝖿𝗉by\mathsf{fp}\operatorname{-by-}\mathbb{Z} group GG, the set of standard presentations for GG is recursively enumerable.

Proof.

Let PP be the finite presentation given (of a 𝖿𝗉by\mathsf{fp}\operatorname{-by-}\mathbb{Z} group GG). We will start enumerating all finite presentations of GG by successively applying to PP chains of elementary Tietze transformations in all possible ways. This process is recursive and eventually visits all finite presentations for GG (all standard presentations among them).

Now, it will be enough to construct a recognizing subprocess 𝔖\mathfrak{S} which, applied to any finite presentation PP^{\prime} for GG, if PP^{\prime} is in standard form it halts and returns PP^{\prime}, and if not it halts returning “NO, PP^{\prime} is not standard”, or works forever. Having 𝔖\mathfrak{S}, we can keep following the enumeration of all finite presentations PP^{\prime} for GG via Tieze transformations and, for each one, start and run in parallel the recognizing process 𝔖\mathfrak{S} for it; we maintain all of them running in parallel (some of them possibly forever), and at the same time we keep opening new ones, simultaneously aware of the possible halts (each one killing one of the parallel processes and possibly outputting a genuine standard presentation for GG).

So, we are reduced to design such a recognizing process 𝔖\mathfrak{S}. For a given finite presentation PP^{\prime} of GG, let us perform the following steps:

  1. (1)

    Check whether PP^{\prime} matches the scheme

    (26) X,t|R,txit1=wi(xiX),\left\langle X,t\,\left|\ R\,,\,tx_{i}t^{-1}=w_{i}\,\,\,(x_{i}\in X)\,\right.\hskip-2.0pt\right\rangle,

    where X={x1,,xn}X=\{x_{1},\ \ldots,x_{n}\} and R={r1,,rm}R=\{r_{1},\ldots,r_{m}\} are finite, and the wiw_{i}’s and rjr_{j}’s are all (reduced) words on XX. If PP does not match this scheme, then halt and answer “NO, PP^{\prime} is not standard”; otherwise go to the next step.

  2. (2)

    With PP^{\prime} being of the form (26), consider the group H=XR=F(X)/RH=\langle\,X\!\mid\!R\,\rangle=F(X)/\left\langle\!\left\langle R\right\rangle\!\right\rangle and let us try to check whether the map xiwix_{i}\mapsto w_{i} extends to a well-defined homomorphism α:HH\alpha\colon H\to H. For this, we must check whether α(rj)=1\alpha(r_{j})=1 in HH or not (but caution! we cannot assume in general a solution to the word problem for HH). Enumerate and reduce the elements in R\left\langle\!\left\langle R\right\rangle\!\right\rangle and check whether, for every relator rj(x1,,xn)Rr_{j}(x_{1},\ldots,x_{n})\in R, the word rj(w1,,wn)r_{j}(w_{1},\ldots,w_{n}) appears in the enumeration. If this happens for all j=1,,mj=1,\ldots,m, then go to the next step (with PP^{\prime} being of the form

    (27) X,t|R,txit1=α(xi)(xiX),\left\langle X,t\,\left|\ R\,,\,tx_{i}t^{-1}=\alpha(x_{i})\,\,\,(x_{i}\in X)\,\right.\hskip-2.0pt\right\rangle,

    where αEnd(F(X)/R)\alpha\in\operatorname{End}(F(X)/\!\left\langle\!\left\langle R\right\rangle\!\right\rangle) ).

  3. (3)

    With PP^{\prime} being of the form (27), let us try to check now whether α\alpha is bijective, looking by brute force for its eventual inverse: enumerate all possible nn-tuples (v1,,vn)(v_{1},\ldots,v_{n}) of reduced words on XX and for each one, check simultaneously whether rj(v1,,vn)=1r_{j}(v_{1},\ldots,v_{n})=1 in HH for all j=1,,mj=1,\ldots,m (i.e., whether β:HH\beta\colon H\to H, xivix_{i}\mapsto v_{i} is a well-defined endomorphism of HH) and whether vi(w1,,wn)=1v_{i}(w_{1},\ldots,w_{n})=1 and wi(v1,,vn)=1w_{i}(v_{1},\ldots,v_{n})=1 in HH, for all i=1,,ni=1,\ldots,n (i.e., whether αβ=βα=id\alpha\beta=\beta\alpha=\operatorname{id} and so αAut(H)\alpha\in\operatorname{Aut}(H)). We do this in a similar way as in the previous step: enumerate the normal closure R\left\langle\!\left\langle R\right\rangle\!\right\rangle (an infinite process) and wait until all the mentioned words appear in the enumeration. When this happens (if so), halt the process and output PP^{\prime} as a standard presentation for GG.

For any given PP^{\prime}, step (i) finishes in finite time and either rejects PP^{\prime}, or recognizes that PP^{\prime} is of the form (26) and sends the control to step (ii). Now step (ii) either works forever, or it halts recognizing PP^{\prime} of the form (27) and sending the control to step (iii) (note that, by construction, it is guaranteed that if PP^{\prime} is really in standard form then α\alpha is a well-defined endomorphism of HH and step (ii) will eventually halt in finite time). Finally, the same happens in step (iii): it either works forever, or it halts recognizing that PP^{\prime} is in standard form (again by construction, it is guaranteed that if PP^{\prime} is really in standard form, then α\alpha is bijective and step (iii) will eventually catch its inverse and halt in finite time).

Process 𝔖\mathfrak{S} is built, and this concludes the proof. ∎

We remark that we can apply the previous algorithm to an arbitrary finite presentation PP of a (arbitrary) group GG: if GG is a 𝖿𝗉by\mathsf{fp}\operatorname{-by-}\mathbb{Z} group the process will enumerate all its standard presentations, while if GG is not 𝖿𝗉by\mathsf{fp}\operatorname{-by-}\mathbb{Z} the process will work forever outputting nothing. So, we can successively apply—in parallel—the previous algorithm to any enumerable family \mathcal{H} of presentations to obtain an enumeration of all standard 𝖿𝗉by\mathsf{fp}\operatorname{-by-}\mathbb{Z} presentations within \mathcal{H}. Taking =𝒢𝖿𝗉\mathcal{H}=\mathcal{G}_{\mathsf{fp}}, we get an enumeration of all standard 𝖿𝗉by\mathsf{fp}\operatorname{-by-}\mathbb{Z} presentations.

Corollary 7.2.

The set of standard presentations of 𝖿𝗉by\mathsf{fp}\operatorname{-by-}\mathbb{Z} groups is recursively enumerable. ∎

Applying all possible Tietze transformations to every standard presentation outputted by this procedure, we obtain an enumeration of all finite presentations of 𝖿𝗉by\mathsf{fp}\operatorname{-by-}\mathbb{Z} groups. This enriches 5.4 in the following way.

Corollary 7.3.

The set of finite presentations of 𝖿𝗉by\mathsf{fp}\operatorname{-by-}\mathbb{Z} groups is recursively enumerable but not recursive. ∎

8. On the isomorphism problem for unique \mathbb{Z}-extensions

Let us consider now problems of the first kind mentioned in Section 1: isomorphism problems within families of the form [𝖯by]𝖿𝗉[\mathsf{P}\operatorname{-by-}\mathbb{Z}]_{\mathsf{fp}}.

To begin with, we combine 5.1 with the following one to see that a \mathbb{Z}-extension can have non isomorphic base groups. The proof is just a direct writing of the corresponding presentations.

Lemma 8.1.

Let HH be an arbitrary group, and ϕAut(H)\phi\in\operatorname{Aut}(H). Then,

(28) (Hϕ)×(H×)Φ,(H\rtimes_{\phi}\mathbb{Z})\times\mathbb{Z}\,\simeq\,(H\times\mathbb{Z})\rtimes_{\Phi}\mathbb{Z}\,,

where ΦAut(H×)\Phi\in\operatorname{Aut}(H\times\mathbb{Z}) is defined by (h,t)(ϕ(h),t)(h,t)\mapsto(\phi(h),t). ∎

Corollary 8.2.

Isomorphic \mathbb{Z}-extensions can have nonisomorphic base groups, even of different type. More precisely, there exist a finitely presented group HH, a non finitely generated group HH^{\prime}, and automorphisms αAut(H)\alpha\in\operatorname{Aut}(H) and βAut(H)\beta\in\operatorname{Aut}(H^{\prime}), such that HαHβH\rtimes_{\alpha}\mathbb{Z}\simeq H^{\prime}\rtimes_{\beta}\mathbb{Z}. In particular, HHHHH\rtimes\mathbb{Z}\,\simeq\,H^{\prime}\rtimes\mathbb{Z}\ \nRightarrow\ H\simeq H^{\prime}.

Proof.

Let KK be any nontrivial finitely presented group. Consider H=KH=K*\mathbb{Z}, which is also finitely presented, and H=(iK)×H^{\prime}=\left(\operatorname*{\scalebox{2.0}{\raisebox{-1.29167pt}{$\ast\hskip-0.3pt$}}}_{i\in\mathbb{Z}}K\right)\times\mathbb{Z}, which is not finitely generated. Combining (13) and (28), we get

H×=(K)×((iK)τ)×((iK)×)T=HT,H\times\mathbb{Z}=\left(K*\mathbb{Z}\right)\times\mathbb{Z}\ \simeq\,\left(\left(\operatorname*{\scalebox{2.0}{\raisebox{-1.29167pt}{$\ast\hskip-0.3pt$}}}_{i\in\mathbb{Z}}K\right)\rtimes_{\tau}\mathbb{Z}\right)\times\mathbb{Z}\ \simeq\ \left(\left(\operatorname*{\scalebox{2.0}{\raisebox{-1.29167pt}{$\ast\hskip-0.3pt$}}}_{i\in\mathbb{Z}}K\right)\times\mathbb{Z}\right)\rtimes_{T}\mathbb{Z}=H^{\prime}\rtimes_{T}\mathbb{Z}\,,

where τAut(iK)\tau\in\operatorname{Aut}(\operatorname*{\scalebox{2.0}{\raisebox{-1.29167pt}{$\ast\hskip-0.3pt$}}}_{i\in\mathbb{Z}}\,K) is the automorphism (14) defined in 5.1, and TAut(H)T\in\operatorname{Aut}(H^{\prime}) the corresponding one according to LABEL:lem:rtimes_times_=_timesrtimes. The result follows taking α=idH\alpha=\operatorname{id}_{H} and β=T\beta=T. ∎

So, there is considerable flexibility in describing cyclic extensions as semidirect products. Even fixing the base group, this flexibility persists within the possible defining automorphisms. For example, one can easily see that HγH×H\rtimes_{\gamma}\mathbb{Z}\simeq H\times\mathbb{Z}, for every inner automorphism γInn(H)\gamma\in\operatorname{Inn}(H). A bit more generally, the following is a folklore lemma which is straightforward to prove (see [6]).

Lemma 8.3.

Let HH be an arbitrary group, and let α,βAut(H)\alpha,\beta\in\operatorname{Aut}(H). If β=γξα±1ξ1\beta=\gamma\xi\alpha^{\pm 1}\xi^{-1} for some γInnH\gamma\in\operatorname{Inn}H and some ξAut(H)\xi\in\operatorname{Aut}(H), then HαHβH\rtimes_{\alpha}\mathbb{Z}\simeq H\rtimes_{\beta}\mathbb{Z}. ∎

The existence of such γInnH\gamma\in\operatorname{Inn}H and ξAut(H)\xi\in\operatorname{Aut}(H) is exactly the same as [β][\beta] being conjugate to [α]±1[\alpha]^{\pm 1} in Out(H)\operatorname{Out}(H). This condition turns out to have some protagonism along the rest of the paper, making convenient to have a general shorthand terminology for it.

Definition 8.4.

Let GG be an arbitrary group. A pair of elements g,hGg,h\in G are said to be semi-conjugate if gg is conjugate to either hh or h1h^{-1}; we denote this situation by gh±1g\sim h^{\pm 1}.

With this terminology, 8.3 states that, when the defining automorphisms α,βAut(H)\alpha,\beta\in\operatorname{Aut}(H) are semi-conjugate in Out(H)\operatorname{Out}(H), then the corresponding semidirect products HαH\rtimes_{\alpha}\mathbb{Z} and HβH\rtimes_{\beta}\mathbb{Z} are isomorphic. Note also the following necessary condition: by 4.2, α\alpha is deranged if and only if b(Hα)=1\operatorname{b}(H\rtimes_{\alpha}\mathbb{Z})=1 so, in order for HαH\rtimes_{\alpha}\mathbb{Z} and HβH\rtimes_{\beta}\mathbb{Z} to be isomorphic, a necessary condition is that α\alpha and β\beta are either both simultaneously deranged, or both not deranged.

Apart from this, not much is known in general about characterizing or deciding when two \mathbb{Z}-extensions of a given group are isomorphic. In [6], Bogopolski–Martino–Ventura proved that, when the base group HH is free of rank 22, the converse to 8.3 also holds, providing a quite neat characterization of isomorphism within the family of F2byF_{2}\operatorname{-by-}\mathbb{Z} extensions and (using the decidability of the conjugacy problem in Out(F2)\operatorname{Out}(F_{2}), see [5]) a positive solution to the isomorphism problem within this family of groups.

Theorem 8.5 (Bogopolski–Martino–Ventura, [6]).

Let α,βAut(F2)\alpha,\beta\in\operatorname{Aut}(F_{2}). Then,

(29) F2αF2β[α] and [β] are semi-conjugate in Out(F2).F_{2}\rtimes_{\alpha}\mathbb{Z}\simeq F_{2}\rtimes_{\beta}\mathbb{Z}\ \Leftrightarrow\ [\alpha]\text{ and }[\beta]\text{ are semi-conjugate in }\operatorname{Out}(F_{2}).

In particular, the isomorphism problem within the family F2byF_{2}\operatorname{-by-}\mathbb{Z} is decidable. ∎

However, in this same paper, a counterexample was given (suggested by W. Dicks) to see that this equivalence is not true for free groups of higher rank, where the situation is, in general, much more complicated. The example is the following: consider the free group of rank 3, F3=a,b,cF_{3}=\langle\,a,b,c\!\mid\!\,\,\rangle, and the automorphisms α:F3F3\alpha\colon F_{3}\to F_{3} given by abcb1ab2c3a\mapsto b\mapsto c\mapsto b^{-1}ab^{-2}c^{3}, and β:F3F3\beta\colon F_{3}\to F_{3} by abca1b2cb1a\mapsto b\mapsto c\mapsto a^{-1}b^{2}cb^{-1}. It happens that F3αF3βF_{3}\rtimes_{\alpha}\mathbb{Z}\simeq F_{3}\rtimes_{\beta}\mathbb{Z} (see [6] for details), while α\alpha and β\beta are not semi-conjugate in Out(F3)\operatorname{Out}(F_{3}) because they abelianize to two 3×33\times 3 matrices of determinants, respectively, 1 and -1. As far as we know, the isomorphism problem for FrbyF_{r}\operatorname{-by-}\mathbb{Z} groups is open for r3r\geq 3.

The goal of the present section is to prove that an equivalence like (29) still holds, but under kind of an orthogonal condition: rather than restricting the base group to be F2F_{2}, we will leave HH arbitrary finitely generated, and impose conditions on the defining automorphism. Note that such an equivalence reduces the isomorphism problem in the family of restricted extensions, to the conjugacy problem in the corresponding family of outer automorphisms of the base group (or even to a weaker problem, if semi-conjugacy is not algorithmically equivalent to conjugacy).

This context strongly suggests defining the semi-conjugacy problem much in the same way that the standard conjugacy problem, and asking for the relationship between them. We state both problems together in order to make the comparison clear.

Definition 8.6.

Let XR\langle\,X\!\mid\!R\,\rangle be a finite presentation for a group GG. Then:

  • Conjugacy Problem for GG [ CP(G)\operatorname{CP}(G) ]: given two words u,vu,v in X±X^{\pm}, decide whether they represent conjugate elements in GG (uGvu\sim_{G}v) or not.

  • Semi-conjugacy Problem for GG [ 12CP(G)\textstyle\frac{1}{2}\!\operatorname{CP}(G) ]: given two words u,vu,v in X±X^{\pm}, decide whether they represent semi-conjugate elements in GG (uGv±1u\sim_{G}v^{\pm 1}) or not.

Question 1.

Is there a (finitely presented) group with decidable semi-conjugacy problem but undecidable conjugacy problem?

This question looks quite tricky. Of course, if two elements g,hGg,h\in G are not semi-conjugate, then they are not conjugate either. But if gh1g\sim h^{-1}, it is not clear how this information can help, in general, to decide whether ghg\sim h or not; in this sense the answer to the question seems reasonable to be negative. But, on the other hand, the two algorithmic problems are so close that it seems hard to construct a counterexample.

In our case, the condition demanded for the defining automorphisms is derangedness (see 4.3). The first observation is the following: suppose HαKβH\rtimes_{\alpha}\mathbb{Z}\simeq K\rtimes_{\beta}\mathbb{Z} for some groups HH, KK, and some deranged automorphisms αAut(H)\alpha\in\operatorname{Aut}(H) and βAut(K)\beta\in\operatorname{Aut}(K). Then, by construction, HH and KK are respectively, the unique normal subgroups with quotient isomorphic to \mathbb{Z}. Hence HKH\simeq K and, after expressing β\beta in terms of the generators of HH, we can think that both α,βAut(H)\alpha,\beta\in\operatorname{Aut}(H). The next step is to show that, under the derangedness condition, HαHβH\rtimes_{\alpha}\mathbb{Z}\simeq H\rtimes_{\beta}\mathbb{Z} implies that [α],[β]Out(H)[\alpha],[\beta]\in\operatorname{Out}(H) are semi-conjugate. To see this, we need to analyze how homomorphisms between unique \mathbb{Z}-extensions look like.

Definition 8.7.

Let GG be a group, and HH a subgroup of GG. An endomorphism ΦEnd(G)\Phi\in\operatorname{End}(G) is called HH-stable if it leaves HH invariant as a subgroup. The collection of all HH-stable endomorphisms of GG form a submonoid denoted EndH(G)End(G)\operatorname{End}_{H}(G)\leqslant\operatorname{End}(G). In a similar way, the collection of all HH-stable automorphisms of GG form a subgroup denoted AutH(G)Aut(G)\operatorname{Aut}_{H}(G)\leqslant\operatorname{Aut}(G).

A general description of the HH-stable endomorphisms and automorphisms of infinite-cyclic extensions of HH follows.

Proposition 8.8.

Let HH be a group generated by X={xiiI}X=\{x_{i}\mid i\in I\}, and let α,βAut(H)\alpha,\beta\in\operatorname{Aut}(H). Then, any homomorphism from HαH\rtimes_{\alpha}\mathbb{Z} to HβH\rtimes_{\beta}\mathbb{Z} mapping HH to HH is of the form

(30) Φϵ,ϕ,h0:HαHβ,xiϕ(xi)th0tϵ\begin{array}[]{rcl}\Phi_{\epsilon,\phi,h_{0}}\colon H\rtimes_{\alpha}\mathbb{Z}&\to&H\rtimes_{\beta}\mathbb{Z},\\ x_{i}&\mapsto&\phi(x_{i})\\ t&\mapsto&h_{0}\,t^{\epsilon}\end{array}

where ϵ\epsilon\in\mathbb{Z}, h0Hh_{0}\in H, and ϕEnd(H)\phi\in\operatorname{End}(H) are such that γh0βϵϕ=ϕα\gamma_{h_{0}}\beta^{\epsilon}\phi=\phi\alpha.

Furthermore, Φϵ,ϕ,h0\Phi_{\epsilon,\phi,h_{0}} is an isomorphism if and only if ϵ=±1\epsilon=\pm 1 and ϕAut(H)\phi\in\operatorname{Aut}(H). Thus, HH-stable automorphisms of HαH\rtimes_{\alpha}\mathbb{Z} are precisely

(31) AutH(Hα)\displaystyle\operatorname{Aut}_{H}(H\rtimes_{\alpha}\mathbb{Z}) ={Φϵ,ϕ,h0ϵ=±1,h0H,ϕAut(H) s.t. γh0αϵϕ=ϕα}.\displaystyle=\left\{\,\Phi_{\epsilon,\phi,h_{0}}\mid\epsilon=\pm 1,\,h_{0}\in H,\,\phi\in\operatorname{Aut}(H)\text{ s.t. }\gamma_{h_{0}}\alpha^{\epsilon}\phi=\phi\alpha\,\right\}.
Proof.

Let Φ:HαHβ\Phi\colon H\rtimes_{\alpha}\mathbb{Z}\to H\rtimes_{\beta}\mathbb{Z} be a homomorphism leaving HH invariant, and let us denote by ϕ:HH\phi\colon H\to H its restriction to HH. Write Φ(t)=h0tϵ\Phi(t)=h_{0}\,t^{\epsilon} for some h0Hh_{0}\in H and ϵ\epsilon\in\mathbb{Z}. Applying Φ\Phi to both sides of the relation tht1=α(h)tht^{-1}=\alpha(h) in the domain, we get

h0βϵϕ(h)h01=h0tϵϕ(h)tϵh01=Φ(tht1)=Φ(α(h))=ϕα(h),h_{0}\cdot\beta^{\epsilon}\phi(h)\cdot h_{0}^{-1}=h_{0}t^{\epsilon}\cdot\phi(h)\cdot t^{-\epsilon}h_{0}^{-1}=\Phi(tht^{-1})=\Phi(\alpha(h))=\phi\alpha(h)\,,

for all hHh\in H. Hence, γh0βϵϕ=ϕα\gamma_{h_{0}}\beta^{\epsilon}\phi=\phi\alpha and Φ=Φϵ,ϕ,h0\Phi=\Phi_{\epsilon,\phi,h_{0}} has the desired form.

Assume now that Φϵ,ϕ,h0\Phi_{\epsilon,\phi,h_{0}} is an isomorphism (in particular, ϕ:HH\phi\colon H\to H is injective). Then we must have ϵ=±1\epsilon=\pm 1, otherwise tt would not be in the image. On the other hand, since HHαH\trianglelefteqslant H\rtimes_{\alpha}\mathbb{Z}, we have that

ϕ(H)=Φϵ,ϕ,h0(H)Hβ=Φϵ,ϕ,h0(Hα)=ϕ(H),h0tϵ,\phi(H)=\Phi_{\epsilon,\phi,h_{0}}(H)\trianglelefteqslant H\rtimes_{\beta}\mathbb{Z}=\Phi_{\epsilon,\phi,h_{0}}(H\rtimes_{\alpha}\mathbb{Z})=\langle\phi(H),\,h_{0}t^{\epsilon}\rangle\,,

and so, any element of HβH\rtimes_{\beta}\mathbb{Z} can be written in the form ϕ(h)(h0tϵ)k\phi(h)(h_{0}t^{\epsilon})^{k}, for some hHh\in H and kk\in\mathbb{Z}; and it belongs to HH if and only if k=0k=0. Thus, Φ(H)=H\Phi(H)=H and ϕAut(H)\phi\in\operatorname{Aut}(H). For the converse, it is clear that ϵ=±1\epsilon=\pm 1 and ϕAut(H)\phi\in\operatorname{Aut}(H) implies that Φϵ,ϕ,h0\Phi_{\epsilon,\phi,h_{0}} is an isomorphism. The final statement follows immediately. ∎

Note that 4.2 states precisely that EndH(Hα)=End(Hα)\operatorname{End}_{H}(H\rtimes_{\alpha}\mathbb{Z})=\operatorname{End}(H\rtimes_{\alpha}\mathbb{Z}) if and only if α\alpha is deranged. This fact, together with the previous description provides a characterization of isomorphic deranged extensions in terms of semi-conjugacy.

Corollary 8.9.

Let HH and KK be two arbitrary groups, and let αAut(H)\alpha\in\operatorname{Aut}(H) and βAut(K)\beta\in\operatorname{Aut}(K) be two deranged automorphisms. Then,

(32) HαKβHK and [α][β]±1 in Out(H),H\rtimes_{\alpha}\mathbb{Z}\simeq K\rtimes_{\beta}\mathbb{Z}\ \Leftrightarrow\ H\simeq K\text{ and }[\alpha]\sim[\beta^{\prime}]^{\pm 1}\text{ in }\operatorname{Out}(H)\,,

where β=ψ1βψAut(H)\beta^{\prime}=\psi^{-1}\beta\psi\in\operatorname{Aut}(H), and ψ:HK\psi\colon H\to K is any isomorphism.

Proof.

For any isomorphism ψ:HK\psi\colon H\to K, it is clear that Kβ=ψ(H)βHψ1βψK\rtimes_{\beta}\mathbb{Z}=\psi(H)\rtimes_{\beta}\mathbb{Z}\simeq H\rtimes_{\psi^{-1}\beta\psi}\mathbb{Z}. Hence, the statement is equivalent to saying

HαHβ[α][β]±1 in Out(H),H\rtimes_{\alpha}\mathbb{Z}\simeq H\rtimes_{\beta}\mathbb{Z}\ \Leftrightarrow\ [\alpha]\sim[\beta]^{\pm 1}\text{ in }\operatorname{Out}(H)\,,

for α,βAut(H)\alpha,\beta\in\operatorname{Aut}(H). The implication to the left is a general fact (see 8.3), and the implication to the right is a direct consequence of 8.8: since α\alpha and β\beta are deranged, any isomorphism from HαH\rtimes_{\alpha}\mathbb{Z} to HβH\rtimes_{\beta}\mathbb{Z} must map HH to HH and so, must be of the form Φϵ,ϕ,h0\Phi_{\epsilon,\phi,h_{0}} for some ϵ=±1\epsilon=\pm 1, h0Hh_{0}\in H, and ϕAut(H)\phi\in\operatorname{Aut}(H) satisfying γh0βϵϕ=ϕα\gamma_{h_{0}}\beta^{\epsilon}\phi=\phi\alpha. Hence, [α][β]±1[\alpha]\sim[\beta]^{\pm 1} in Out(H)\operatorname{Out}(H). ∎

We are now ready to prove the main result in this section: for any family \mathcal{H} of finitely presented groups with decidable isomorphism problem, we characterize when the family [!by]𝖿𝗉[!\mathcal{H}\operatorname{-by-}\mathbb{Z}]_{\mathsf{fp}} has again decidable isomorphism problem, in terms of a certain variation of the conjugacy problem for outer automorphisms of groups in \mathcal{H}.

Note that 8.9 clearly insinuates a link between the isomorphism problem for deranged extensions, and the semi-conjugacy problem for deranged outer automorphisms of the base group. However, there is a subtlety at this point: the supposed algorithm solving the isomorphism problem will receive the input (the compared groups) as finite presentations of the \mathbb{Z}-extensions. From those, we know how to compute suitable base groups H,KH,K, and automorphisms α,β\alpha,\beta (see 7.1), but this last ones are given by images of the generators in the starting presentations, and not as words in some presentation of the corresponding automorphism groups, which would be the appropriate inputs for the standard conjugacy problem there.

So, in general, one must distinguish between these two close but not necessarily identical situations. As before, we state both problems together to emphasize the difference between them.

Definition 8.10.

Let XR\langle\,X\!\mid\!R\,\rangle be a presentation for a group GG, YS\langle\,Y\!\mid\!S\,\rangle a presentation for Aut(G)\operatorname{Aut}(G), and assume |X|<|X|<\infty. Then:

  • (Standard) conjugacy problem for Aut(G)\operatorname{Aut}(G) [ CP(Aut(G))\operatorname{CP}(\operatorname{Aut}(G)) ]: given two automorphisms α,βAut(G)\alpha,\beta\in\operatorname{Aut}(G) as words in the presentation of Aut(G)\operatorname{Aut}(G), decide whether α\alpha and β\beta are conjugate to each other in Aut(G)\operatorname{Aut}(G).

  • Aut\operatorname{Aut}-conjugacy problem for GG [ CPG(Aut(G))\operatorname{CP}_{\!\scriptscriptstyle{G}}(\operatorname{Aut}(G)) ]: given two automorphisms α,βAut(G)\alpha,\beta\in\operatorname{Aut}(G) by images of (the finitely many) generators XX, decide whether α\alpha and β\beta are conjugate to each other in Aut(G)\operatorname{Aut}(G).

Similarly, we define the Out\operatorname{Out}-conjugacy problem [ CPG(Out(G))\operatorname{CP}_{\!\scriptscriptstyle{G}}(\operatorname{Out}(G)) ], the Aut\operatorname{Aut}-semi-conjugacy problem [ 12CPG(Aut(G))\textstyle\frac{1}{2}\!\operatorname{CP}_{\!\scriptscriptstyle{G}}(\operatorname{Aut}(G)) ], and the Out\operatorname{Out}-semi-conjugacy problem [ 12CPG(Out(G))\textstyle\frac{1}{2}\!\operatorname{CP}_{\!\scriptscriptstyle{G}}(\operatorname{Out}(G)) ] for GG (in contrast with the standard CP(Out(G))\operatorname{CP}(\operatorname{Out}(G)), 12CP(Aut(G))\textstyle\frac{1}{2}\!\operatorname{CP}(\operatorname{Aut}(G)), and 12CP(Out(G))\textstyle\frac{1}{2}\!\operatorname{CP}(\operatorname{Out}(G))).

Note that, in general, these pairs of problems are similar but not identical: from the algorithmic point of view it could be very different to have an automorphism of GG given as the collection of images of a finite set of generators of GG, or as a word (composition of generators for Aut(G)\operatorname{Aut}(G)). Consider, for example, the Baumslag–Solitar group G=BS(2,4)G=\operatorname{BS}(2,4), which is finitely generated, but whose automorphism group Aut(G)\operatorname{Aut}(G) is known to be not finitely generated (see [11]).

However, knowing in advance a finite set of generators for Aut(G)\operatorname{Aut}(G) (respectively, Out(G)\operatorname{Out}(G)) as images of generators of GG, these two kinds of problems turn out to be equivalent.

Proposition 8.11.

Let XR\langle\,X\!\mid\!R\,\rangle be a presentation for a group GG, X={x1,,xn}X=\{x_{1},\ldots,x_{n}\}, and {ui,ji=1,,n,j=1,,N}\{u_{i,j}\mid i=1,\ldots,n,\,\,j=1,\ldots,N\} a finite set of words in X±X^{\pm} such that {αj:xiui,jj=1,,N}\{\alpha_{j}\colon x_{i}\mapsto u_{i,j}\mid j=1,\ldots,N\} is a well defined finite family of automorphisms generating Aut(G)\operatorname{Aut}(G). Then,

(33) CPG(Aut(G)) is decidableCP(Aut(G)) is decidable .\operatorname{CP}_{\!\scriptscriptstyle{G}}(\operatorname{Aut}(G))\text{ is decidable}\ \Leftrightarrow\ \operatorname{CP}(\operatorname{Aut}(G))\text{ is decidable\,}.

The same is true replacing conjugacy by semi-conjugacy, and/or Aut\operatorname{Aut} by Out\operatorname{Out}.

Proof.

Suppose that GG has decidable Aut\operatorname{Aut}-conjugacy problem. Given two automorphisms α,βAut(G)\alpha,\beta\in\operatorname{Aut}(G) as words on the αi\alpha_{i}’s, say α=a(α1,,αN)\alpha=a(\alpha_{1},\ldots,\alpha_{N}) and β=b(α1,,αN)\beta=b(\alpha_{1},\ldots,\alpha_{N}), we can compute the corresponding compositions of αj\alpha_{j}’s and obtain explicit expressions for α(xi)\alpha(x_{i}) and β(xi)\beta(x_{i}) in terms of XX, for i=1,,ni=1,\ldots,n. Now, applying the solution to the Aut\operatorname{Aut}-conjugacy problem for GG we decide whether α\alpha and β\beta are conjugate to each other in Aut(G)\operatorname{Aut}(G).

Conversely, suppose Aut(G)\operatorname{Aut}(G) has decidable conjugacy problem, and we are given two automorphisms α,βAut(G)\alpha,\beta\in\operatorname{Aut}(G) by the images of the xix_{i}’s, say α(xi)\alpha(x_{i}) and β(xi)\beta(x_{i}), i=1,,ni=1,\ldots,n. We will express α\alpha and β\beta as compositions of the αj\alpha_{j}’s, and then apply the assumed solution to the conjugacy problem for Aut(G)\operatorname{Aut}(G) to decide whether α\alpha and β\beta are conjugate to each other, or not. We can do this by a brute force enumeration of all possible formal reduced words ww on α1,,αN\alpha_{1},\ldots,\alpha_{N} and, for each one, computing the tuple (w(x1),,w(xn))(w(x_{1}),\ldots,w(x_{n})) and trying to check whether it equals (α(x1),,α(xn))(\alpha(x_{1}),\ldots,\alpha(x_{n})), or (β(x1),,β(xn))(\beta(x_{1}),\ldots,\beta(x_{n})) (following a brute force enumeration of the normal closure R\left\langle\!\left\langle R\right\rangle\!\right\rangle, like in the proof of 7.1).

The proofs of the other versions of the statement are completely analogous. For the conjugacy problems we need to add another brute force search layer enumerating all possible conjugators; we leave details to the reader. ∎

After this proposition we can prove the main result in this section.

Theorem 8.12.

Let \mathcal{H} be a family of finitely presented groups with decidable isomorphism problem. Then, the isomorphism problem of !by!\mathcal{H}\operatorname{-by-}\mathbb{Z} is decidable if and only if the Out𝖽\operatorname{Out}_{\mathsf{d}}-semi-conjugacy problem of HH is decidable for every HH in \mathcal{H}; i.e.,

IP(!by) decidable12CPH(Out𝖽(H)) decidable, H.\operatorname{IP}(!\mathcal{H}\operatorname{-by-}\mathbb{Z})\text{ decidable}\ \Leftrightarrow\ \textstyle\frac{1}{2}\!\operatorname{CP}_{\!\scriptscriptstyle{H}}(\operatorname{Out}_{\mathsf{d}}(H))\text{ decidable, }\forall H\in\mathcal{H}\,.
Proof.

Suppose that every HH\in\mathcal{H} has decidable Out\operatorname{Out}-semi-conjugacy problem for deranged inputs. Given finite presentations of two groups GG and GG^{\prime} in !by!\mathcal{H}\operatorname{-by-}\mathbb{Z}, we run 7.1 to compute standard presentations for them, and extract finite presentations for base groups and defining automorphisms (say HH and αAut(H)\alpha\in\operatorname{Aut}(H) for GG, and KK and βAut(K)\beta\in\operatorname{Aut}(K) for GG^{\prime}, respectively). We have G=HαG=H\rtimes_{\alpha}\mathbb{Z} and G=KβG^{\prime}=K\rtimes_{\beta}\mathbb{Z} and, by hypotheses, α\alpha and β\beta are deranged. Furthermore, since b(G)=b(G)=1\operatorname{b}(G)=\operatorname{b}(G^{\prime})=1, HH and KK are the unique normal subgroups of GG and GG^{\prime}, respectively, with quotient \mathbb{Z}; hence, H,KH,K\in\mathcal{H}.

Now we apply the isomorphism problem within \mathcal{H} to the obtained presentations for HH and KK, and decide whether they are isomorphic as groups. If H≄KH\not\simeq K then, by 8.9, G≄GG\not\simeq G^{\prime} and we are done. Otherwise, we construct an explicit isomorphism ψ:HK\psi\colon H\to K (by a brute force search procedure like the ones above), we compute β=ψ1βψAut(H)\beta^{\prime}=\psi^{-1}\beta\psi\in\operatorname{Aut}(H), and we apply our solution to the Out\operatorname{Out}-semi-conjugacy problem for HH\in\mathcal{H} to the inputs α\alpha and β\beta^{\prime}, (which are deranged, by construction). The output on whether [α][\alpha] and [β][\beta^{\prime}] are or are not semi-conjugate in Out(H)\operatorname{Out}(H) is the final answer we are looking for (again by 8.9).

For the converse, assume that the isomorphism problem is decidable in the family !by!\mathcal{H}\operatorname{-by-}\mathbb{Z}, and fix a finite presentation XR\langle\,X\!\mid\!R\,\rangle for a group HH\in\mathcal{H}. Given two deranged automorphisms α,βOut(H)\alpha,\beta\in\operatorname{Out}(H) via images of the generators xiXx_{i}\in X, build the corresponding standard presentations for HαH\rtimes_{\alpha}\mathbb{Z} and HβH\rtimes_{\beta}\mathbb{Z} (which are groups in !by!\mathcal{H}\operatorname{-by-}\mathbb{Z}, by construction) and apply the assumed solution to the isomorphism problem for this family to decide whether they are isomorphic or not. By LABEL:cor:isomorphicderanged_extensions, the answer is affirmative if and only if [α][\alpha] and [β][\beta] are semi-conjugate in Out(H)\operatorname{Out}(H). ∎

We apply now 8.12 to special families of groups with decidable isomorphism problem. Some of these corollaries are already known in the literature; other methods provide alternative approaches.

Taking \mathcal{H} to be a single group HH, we get the following result.

Corollary 8.13.

Let HH be a finitely presented group. Then the isomorphism problem is decidable within the family !Hby!H\operatorname{-by-}\mathbb{Z} if and only if HH has decidable Out\operatorname{Out}-semi-conjugacy problem for deranged inputs. In particular, if |Out(H)|<|\operatorname{Out}(H)|<\infty, then !Hby!H\operatorname{-by-}\mathbb{Z} has decidable isomorphism problem. ∎

Taking \mathcal{H} to be the families of finite, finitely generated abelian, or polycyclic groups, 8.12 provides well-known results, since the obtained extensions turn out to be subfamilies of that of virtually-polycyclic groups for which the isomorphism problem is known to be decidable (see [30]).

For the family of Braid groups =𝖻𝗋𝖺𝗂𝖽={Bnn2}\mathcal{B}=\mathsf{braid}=\{B_{n}\mid n\geqslant 2\}, the specially simple structure of its outer automorphism group allows us to state the isomorphism problem within the family by\mathcal{B}\operatorname{-by-}\mathbb{Z}.

Corollary 8.14.

The isomorphism problem is decidable within the family by\mathcal{B}\operatorname{-by-}\mathbb{Z}.

Proof.

It is well known that, for every n2n\geq 2, Out(Bn)={id¯,ι¯}\operatorname{Out}(B_{n})=\{\overline{\operatorname{id}},\overline{\iota}\}, where ι:BnBn\iota\colon B_{n}\to B_{n} is the automorphism given by σiσi1\sigma_{i}\mapsto\sigma_{i}^{-1} (see [16]). Then, (since automorphisms in the same inner class induce the same \mathbb{Z}-extension, by 8.3), we have

(34) by={Bn×,n2}{Bnι,n2}.\mathcal{B}\operatorname{-by-}\mathbb{Z}=\{B_{n}\times\mathbb{Z},n\geq 2\}\cup\{B_{n}\rtimes_{\iota}\mathbb{Z},n\geq 2\}\,.

We claim that they are all pairwise nonisomorphic. Those in the first term of the union cannot be isomorphic to those in the second one because id\operatorname{id} is not deranged, while ι\iota is (in other words, b(Bn×)=2\operatorname{b}(B_{n}\times\mathbb{Z})=2 while b(Bnι)=1\operatorname{b}(B_{n}\rtimes_{\iota}\mathbb{Z})=1). Two deranged extensions BnιB_{n}\rtimes_{\iota}\mathbb{Z} and BmιB_{m}\rtimes_{\iota}\mathbb{Z} can only be isomorphic if their base groups are, and this happens only when n=mn=m (this can be seen, for example, by observing that the center Z(Bn)\textrm{Z}(B_{n}) is generated by the full twist (σ1σ2σn1)n(\sigma_{1}\sigma_{2}\cdots\sigma_{n-1})^{n} and so, the abelianization of Bn/Z(Bn)B_{n}/\textrm{Z}(B_{n}) is cyclic of order n(n1)n(n-1)). Finally, the same argument shows that Bn×Bm×B_{n}\times\mathbb{Z}\simeq B_{m}\times\mathbb{Z} if and only if n=mn=m.

Thus, the isomorphism problem within by\mathcal{B}\operatorname{-by-}\mathbb{Z} is decidable: given two finite presentations of groups G1G_{1} and G2G_{2} in by\mathcal{B}\operatorname{-by-}\mathbb{Z}, explore the two trees of Tietze transformations until finding standard presentations for them, i.e., until recognizing their number of strands, say nn and mm. Now G1G2G_{1}\simeq G_{2} if and only if b(G1)=b(G2)\operatorname{b}(G_{1})=\operatorname{b}(G_{2}) and n=mn=m. ∎

Finally, let us consider the case of finitely generated free groups, =[𝖿𝗋𝖾𝖾]𝖿𝗀={Fnn0}\mathcal{F}=[\mathsf{free}]_{\mathsf{fg}}=\{F_{n}\mid n\geqslant 0\}. To start with, the isomorphism problem for \mathcal{F} is decidable like in the case of Braid groups (since FnFmn=mF_{n}\simeq F_{m}\,\Leftrightarrow\,n=m). A solution to the conjugacy problem in Out(Fn)\operatorname{Out}(F_{n}) was announced by M. Lustig in the preprints [22, 23]. Although this project is not completed (and there is no published version yet), it is believed that Out(Fn)\operatorname{Out}(F_{n}) has decidable conjugacy problem. However, at this moment we can only say to have firm complete solutions for some classes of outer automorphisms:

  1. (1)

    the case of rank 2 is easily decidable because Out(F2)GL2()\operatorname{Out}(F_{2})\simeq\operatorname{GL}_{2}(\mathbb{Z});

  2. (2)

    for finite-order elements of Out(Fn)\operatorname{Out}(F_{n}) an algorithm to solve the conjugacy problem follows from results of S. Krstić (see [19]);

  3. (3)

    J. Los and, independently, Z. Sela solved the conjugacy problem in Out(Fr)\operatorname{Out}(F_{r}) for irreducible inputs, see [31, 21, 24];

  4. (4)

    for Dehn twist automorphisms, the conjugacy problem has been solved by Cohen–Lustig, see [10];

  5. (5)

    finally, Krstić–Lustig–Vogtmann solved the conjugacy problem in Out(Fn)\operatorname{Out}(F_{n}) for linearly growing automorphisms, i.e., for roots of Dehn twists, see [20].

If the conjugacy problem in Out(Fn)\operatorname{Out}(F_{n}) were decidable in general, we could deduce from 8.12 that the isomorphism problem for the family ![𝖿𝗋𝖾𝖾]𝖿𝗀by![\mathsf{free}]_{\mathsf{fg}}\operatorname{-by-}\mathbb{Z} is decidable as well. By the moment, we can only restrict our attention to the above mentioned subsets of Out(Fn)\operatorname{Out}(F_{n}), where the conjugacy problem is firmly known to be decidable, and we obtain the isomorphism problem for the corresponding subfamilies (see the proof of 8.12).

Corollary 8.15.

If the conjugacy problem for Out(Fn)\operatorname{Out}(F_{n}) is decidable, then the isomorphism problem within the family ![𝖿𝗋𝖾𝖾]𝖿𝗀by![\mathsf{free}]_{\mathsf{fg}}\operatorname{-by-}\mathbb{Z} is also decidable. ∎

Corollary 8.16.

The isomorphism problem within the following families is decidable:

  1. (1)

    !F2by!F_{2}\operatorname{-by-}\mathbb{Z};

  2. (2)

    {FnααOut(Fn) deranged and finite order}\{F_{n}\rtimes_{\alpha}\mathbb{Z}\mid\alpha\in\operatorname{Out}(F_{n})\text{ deranged and finite order}\,\};

  3. (3)

    {FnααOut(Fn) deranged and irreducible}\{F_{n}\rtimes_{\alpha}\mathbb{Z}\mid\alpha\in\operatorname{Out}(F_{n})\text{ deranged and irreducible}\,\};

  4. (4)

    {FnααOut(Fn) deranged and linearly growing}\{F_{n}\rtimes_{\alpha}\mathbb{Z}\mid\alpha\in\operatorname{Out}(F_{n})\text{ deranged and linearly growing}\,\}. ∎

It is worth mentioning that our approach is somehow opposite to that taken by Dahmani in [12]. In this interesting preprint the author solves the conjugacy problem for atoroidal automorphisms of FnF_{n}. An automorphism αOut(Fn)\alpha\in\operatorname{Out}(F_{n}) is atoroidal if no proper power of α\alpha fixes any nontrivial conjugacy class (note that this notion is similar in spirit to our notion of derangedness, though they do not coincide). Brinkmann proved in [7] that FnαF_{n}\rtimes_{\alpha}\mathbb{Z} is hyperbolic if and only if α\alpha is atoroidal. And α,βOut(Fn)\alpha,\beta\in\operatorname{Out}(F_{n}) are conjugate to each other if and only if FnαF_{n}\rtimes_{\alpha}\mathbb{Z} is isomorphic to FnβF_{n}\rtimes_{\beta}\mathbb{Z} with an automorphism mapping FnF_{n} to FnF_{n}, and tt to an element of the form wt1wt^{1} (i.e., with an stable and positive automorphism in our language, see the proof of 8.9). Then, Dahmani uses a variation of the celebrated solution to the general isomorphism problem for hyperbolic groups (see [31, 13]) to determine whether FnαF_{n}\rtimes_{\alpha}\mathbb{Z} and FnβF_{n}\rtimes_{\beta}\mathbb{Z} are isomorphic through an isomorphism of the above type, and so deciding whether the atoroidal automorphisms α\alpha and β\beta are conjugated to each other in Out(Fn)\operatorname{Out}(F_{n}). Our approach has been the opposite: we have used the conjugacy problem in Out(Fn)\operatorname{Out}(F_{n}) (more precisely, those particular cases where it is known to be decidable) to solve the isomorphism problem in the corresponding families of unique \mathbb{Z}-extensions.


Acknowledgements

We thank Ha Lam for interesting conversations during an early stage of the development of this paper. We also thank Maurice Chiodo for suggesting a nice trick used in 5.3; and Conchita Martínez for pointing out the counterexample used in 3.3.

The second named author thanks the support of Universitat Politècnica de Catalunya through the PhD grant number 81–727. The third named author was partially supported by a PSC-CUNY grant from the CUNY research foundation, the City Tech foundation, the ONR (Office of Naval Research) grant N000141210758 and N00014-15-1-2164, and AAAS (American Association for The Advancement of Science) grant 71527-0001. Finally, the second and fourth named authors acknowledge partial support from the Spanish Government through grant number MTM2014-54896-P.


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