Metabelian groups: Full-rank presentations, randomness and Diophantine problems

Let 𝐺 be a metabelian group given by a full-rank presentation G=⟨A∣R⟩M. Then there exist two finitely generated subgroups 𝐻 and 𝐾 of 𝐺 such that

1. 𝐻 is a free metabelian group of rank max⁡(|A|-|R|,0),

2. 𝐾 is a virtually abelian group with |R| generators, and its normal closure L=KG in 𝐺 is again virtually abelian,

3. G=⟨H,K⟩=L⁢H.

Let 𝐺 be a finitely generated metabelian group given by a full-rank presentation G=⟨A∣R⟩M such that |R|≤|A|-1. Then, in any direct decomposition of 𝐺, all but one of the direct factors are virtually abelian.

Let 𝐺 be a metabelian group given by a full-rank presentation G=⟨A∣R⟩M. Then the following hold.

1. If |R|≤|A|-2, then the ring of integers ℤ is e-interpretable in 𝐺, and the Diophantine problem in 𝐺 is undecidable.

2. If |R|≥|A|, then the Diophantine problem of 𝐺 is decidable (in fact, the first-order theory of 𝐺 is decidable).

For the case of deficiency 1 (i.e., m=n-1), decidability of the Diophantine problem over 𝐺 remains an interesting open problem. The recent work [15] proves decidability of the Diophantine problem in

Also some deficiency-1 presentations define cyclic groups, which have decidable Diophantine problem [7].

Let 𝐺 be a metabelian group given by a full-rank presentation G=⟨A∣R⟩M. If |A|-|R|=1, then the Diophantine problem in 𝐺 is decidable.

Theorem 1.6 ([11])

Let 𝑅 be a set of 𝑚 words of length ℓ in an alphabet

i.e., each word is obtained by successively concatenating randomly chosen letters from A±1 with uniform probability. Then M⁢(A,R) has full rank (that is, rank⁡(M⁢(A,R))=min⁡{n,m}) asymptotically almost surely as ℓ→∞.

Let ⟨A∣R⟩=⟨a1,…,an∣r1,…,rm⟩ be a presentation where all relators ri have length ℓ>0. Then the presentation ⟨A∣R⟩ has full rank asymptotically almost surely as ℓ tends to infinity.

A finite presentation ⟨A∣R⟩ is said to be in Smith normal form if the relation matrix M⁢(A,R) is in Smith normal form.

For any finite presentation ⟨A∣R⟩, there exists a finite presentation in Smith normal form ⟨A′∣R′⟩, with

such that, for any variety 𝒱, the groups G=⟨A∣R⟩V and G′=⟨A′∣R′⟩V are isomorphic. Moreover, such a presentation ⟨A′∣R′⟩ and an isomorphism G→G′ can be found algorithmically.

Proof

The presentation ⟨A′∣R′⟩ can be obtained by repeatedly applying Nielsen transformations on the tuples 𝐴 and 𝑅. Indeed, note that such a Nielsen transformation has the effect in the relation matrix M⁢(A,R) of adding or subtracting two columns or two rows, respectively. It is known (see [26]) that one can find a finite sequence of elementary row and column operations that transforms M⁢(A,R) into its Smith normal form. Performing the corresponding Nielsen transformation on ⟨A∣R⟩, one gets the required presentation ⟨A′∣R′⟩. The details of this procedure can be found in [11]. ∎

Let 𝐺 be a metabelian group given by a full-rank presentation in Smith normal form,

where ci∈[F⁢(A),F⁢(A)], αi∈Z∖{0} for all i=1,…,m, and m≤n. Then the following holds.

1. K=⟨a1,…,am⟩ is a virtually abelian group, and its normal closure L=KG in 𝐺 is again virtually abelian.

2. H=⟨am+1,…,an⟩ is a free metabelian group of rank n-m.

3. G=⟨H,K⟩=H⁢L.

Proof

We show first that K=⟨a1,…,am⟩ is virtually abelian. Note that K∩G′ is abelian. Set N=α1⁢⋯⁢αm. Then, for all g∈K, one has gN∈K∩G′. Hence, K/K∩G′ is finite, so 𝐾 is virtually abelian. Similarly, for L=KG, the subgroup L∩G′ is normal in 𝐿 and abelian. The quotient L/L∩G′ is abelian, of period 𝑁, and finitely generated (since L=K(L∩G′)), hence finite. It follows that 𝐿 is abelian-by-(finite abelian). Note that 𝐿 might not be finitely generated.

Now we show that ⟨am+1,…,an⟩ is a free metabelian group of rank n-m. Assume that n-m≥2. By Romanovskii’s aforementioned result [23] (see the discussion after the statement of Theorem 1.1), we know that there exists a subset A1⊆A with |A1|=|A|-|R| such that ⟨A1⟩ is free metabelian freely generated by A1. We claim that A1={am+1,…,an}. Indeed, otherwise, there exists ai∈A1 such that ait∈G′ for some t∈Z∖{0}. Note that |A1|≥2, so there is aj∈A1 with i≠j. It follows then that [ait,[ai,aj]]=1, a contradiction with the fact that A1 freely generates ⟨A1⟩ as a free metabelian group.

If n-m=1, then the map an→1 and ai→0 for all i=1,…,n-1 gives rise to a homomorphism G→Z. Hence, an has infinite order in 𝐺, and ⟨an⟩ is a free metabelian group of rank 1. We note in passing that, in the case when m=n-1, the element of 𝐴 that generates an infinite cyclic group (i.e., a free metabelian group of rank 1) is not necessarily unique, e.g., in

both a1 and a2 generate an infinite cyclic subgroup. ∎

Proof of Theorem 1.1

Let 𝐺 be a metabelian group given by a full-rank presentation G=⟨A∣R⟩M. By Proposition 2.2, one can find algorithmically another presentation ⟨A′∣R′⟩M=⟨a1′,…,an′∣r1′,…,rm′⟩M of 𝐺 which is in Smith normal form.

If m≤n, then Lemma 2.3 applied to the presentation ⟨A′,∣R′⟩ gives subgroups 𝐻 and 𝐾 with the required properties. From these, we obtain the required subgroups in 𝐺 by inverting the isomorphism ⟨A∣R⟩=⟨A′∣R⟩.

On the other hand, if m>n, then 𝐺 is a quotient of the full-rank metabelian group with zero deficiency G1=⟨a1′,…,an′∣r1′,…,rn′⟩ (since M⁢(A′,R′) is in Smith normal form, this is a full-rank presentation), and by the previous case, it follows that G1 is virtually abelian. Since 𝐺 is a quotient of G1, 𝐺 is also virtually abelian. ∎

Note that if the presentation in Lemma 2.3 is in Smith normal form, but not of full rank, then, by the Generalized Freiheitssatz [23], the free subgroup H=⟨A0⟩ of 𝐺 exists, but there might not exist corresponding subgroups 𝐾 and 𝐿 for 𝐻. Indeed, let

Then a3,a4 generate a free metabelian group of rank 2, but G/HG is free metabelian of rank 2, so there is not a virtually abelian subgroup 𝐿 such that G=H⁢L.

Finally, note that Theorem 1.1 or Lemma 2.3 may not reveal fully the structure of 𝐺 and how it is related to the subgroups 𝐻 and 𝐾. For example, consider

Lemma 2.3 tells us that ⟨a3,a4⟩ is freely generated by a3,a4 and ⟨a1,a2⟩ is a virtually abelian group. On the other hand, it is easy to check that

and 𝐺 is the free metabelian product of two Baumslag–Solitar groups, which is not clear directly from the decomposition G=H⁢L.

Proof of Theorem 1.2

We showed in [11] that if 𝐻 is a finitely generated nilpotent group of class c≥2 given by a finite full rank presentation H=⟨A∣R⟩Nc with |R|≤|A|-1, then, in any direct decomposition of 𝐻, all but one of the direct factors are finite. We will use this fact in our proof. Now let 𝐺 be a finitely generated metabelian group given by a full-rank presentation G=⟨A∣R⟩M such that |R|≤|A|-1. Assume G=G1×⋯×Gk for some k≥2 and some subgroups Gi, i=1,…,k. Let 𝜋 be the natural projection of 𝐺 onto H=G/γ3⁢(G). The quotient 𝐻 admits the full-rank presentation ⟨A∣R⟩N2, so, by the result mentioned above, all but one, say π⁢(Gk), of the groups π⁢(G1),…,π⁢(Gk) are finite. Hence, ker⁡π∩Gi has finite index in Gi for all i=1,…,k-1. However, ker⁡π is abelian since ker⁡π=γ3⁢(G)≤[G,G]. ∎

A set D⊂Mm is called definable by systems of equations in ℳ, or e-definable in ℳ, if there exists a finite system of equations, say

in the language of ℳ such that, for any tuple a∈Mm, one has that a∈D if and only if the system ΣD⁢(a,y) on variables 𝐲 has a solution in ℳ. In this case, ΣD is said to e-define𝐷 in ℳ.

Observe that, in the notation above, if D⊂Mm is e-definable, then it is definable in ℳ by the formula ∃y⁢ΣD⁢(x,y). Such formulas are called positive primitive, or pp-formulas. Hence, e-definable subsets are sometimes called pp-definable. On the other hand, in number theory, such sets are usually referred to as Diophantine. And yet, in algebraic geometry, they can be described as projections of algebraic sets.

An algebraic structure A=(A;f,…,r,…,c,…) is called e-interpretable in another algebraic structure ℳ if there exists n∈N, a subset D⊆Mn and an onto map (called the interpreting map) ϕ:D↠A such that the following holds.

1. 𝐷 is e-definable in ℳ.

2. For every function f=f⁢(x1,…,xn) in the language of 𝒜, the preimage by 𝜙 of the graph of 𝑓, i.e., the set {(x1,…,xk,xk+1)∣ϕ⁢(xk+1)=f⁢(x1,…,xk)}, is e-definable in ℳ.

3. For every relation 𝑟 in the language of 𝒜, and also for the equality relation = in 𝒜, the preimage by 𝜙 of the graph of 𝑟 is e-definable in ℳ.

Let 𝒜 be e-interpretable in ℳ with an interpreting map ϕ:D↠A (in the notation of Definition 3.3). Then, for every finite system of equations S⁢(x) in 𝒜, there exists a finite system of equations S*⁢(y,z) in ℳ such that if (b,c) is a solution to S*⁢(y,z) in ℳ, then b∈D and ϕ⁢(b) is a solution to S⁢(x) in 𝒜. Moreover, any solution 𝐚 to S⁢(x) in 𝒜 arises in this way, i.e., a=ϕ⁢(b) for some solution (b,c) to S*⁢(y,z) in ℳ for some i=1,…,k. Furthermore, there is a polynomial time algorithm that constructs the system S*⁢(y,z) when given a system S⁢(x).

If 𝒜 is e-interpretable in ℳ, then D⁢(A) is reducible to D⁢(M). Consequently, if D⁢(A) is undecidable, then D⁢(M) is undecidable as well.

E-interpretability is a transitive relation, i.e., if A1 is e-interpretable in A2, and A2 is e-interpretable in A3, then A1 is e-interpretable in A3.

Proposition 3.7 ([10])

Let 𝐻 be a normal subgroup of a group 𝐺. If 𝐻 is e-definable in 𝐺 (as a set), then the natural map π:G→G/H is an e-interpretation of G/H in 𝐺. Consequently, D⁢(G/H) is reducible to D⁢(G).

Theorem 3.8 ([11])

Let 𝐺 be a finitely generated nonabelian nilpotent group admitting a full rank presentation of deficiency at least 2 (i.e., there are at least two more generators than relations). Then the ring of integers ℤ is e-interpretable in 𝐺, and in particular, the Diophantine problem of 𝐺 is undecidable.

Let 𝐺 be a group, and let 𝐻 be a normal verbal subgroup of 𝐺 with finite verbal width. Then the quotient G/H is e-interpretable in 𝐺.

Proof

By the argument above, the subgroup 𝐻 is e-definable in 𝐺. Now the result follows from Proposition 3.7. ∎

Proof of Theorem 1.3

Let G=⟨A∣R⟩M be a full rank presentation of a metabelian group 𝐺. To prove (1), note first that, since γ3⁢(G)≥[G′,G′], the 2-nilpotent quotient G/γ3⁢(G) admits a presentation ⟨A∣R⟩N2 in the variety N2 of nilpotent groups of class at most 2. In particular, the group G/γ3⁢(G) has a full-rank presentation of deficiency at least 2. By Theorem 3.8, the ring ℤ is e-interpretable in G/γ3⁢(G). It is known that, in a finitely generated metabelian group, any verbal subgroup has finite width [25]; in particular, the subgroup γ3⁢(G) has finite width in 𝐺. Hence, by Proposition 3.9, the group G/γ3⁢(G) is e-interpretable in 𝐺. It follows from transitivity of e-interpretations that ℤ is e-interpretable in 𝐺, so the Diophantine problem in 𝐺 is undecidable.

To prove the second statement of the theorem, observe that if |R|≥|A|, then, by Theorem 1.1, the group 𝐺 is virtually abelian. Hence, the Diophantine problem in 𝐺 is decidable (see [7]). ∎