Computability of finite quotients of finitely generated groups

Having one of CFQ, ReFQ or co-ReFQ is independent of a choice of a generating family.

Proof

Let 𝑆 and 𝑇 be two finite generating sets of a group 𝐺 (not necessarily of the same cardinality). Fix, for each 𝑠 in 𝑆, an expression s = t 1 α 1 ⁢ … ⁢ t k α k , with α i ∈ { - 1 , 1 } and t i ∈ T , that gives 𝑠 as a product of elements of 𝑇 or their inverses, and for each 𝑡 in 𝑇, an expression t = s 1 β 1 ⁢ s 2 β 2 ⁢ … ⁢ s k β k that describes 𝑡 in terms of the generators of 𝑆 and their inverses. For a finite group 𝐹 and a function 𝑓 from 𝑆 to 𝐹, define the function f ′ from 𝑇 to 𝐹 by

The function 𝑓 defines a homomorphism if and only if f ′ also defines a homomorphism φ ′ that satisfies φ ′ ⁢ ( s ) = f ⁢ ( s ) for 𝑠 in 𝑆. That last condition is an equality in 𝐹 that can be tested using the expressions s = t 1 α 1 ⁢ … ⁢ t k α k , even before it is known whether or not f ′ extends. Using this, all three properties, CFQ, ReFQ, co-ReFQ, can be seen to be independent of the chosen generating family of 𝐺. ∎

A subset 𝐴 of 𝐺 is said to be determinable in finite quotients of 𝐺 if there exists an algorithm that, given a morphism 𝜑 from 𝐺 onto a finite group 𝐹, can determine the image φ ⁢ ( A ) (i.e. solve the membership problem for φ ⁢ ( A ) in 𝐹).

A family ( A i ) i ∈ N of subsets of 𝐺 is uniformly determinable in finite quotients of 𝐺 if each A i is determinable in finite quotients of 𝐺, and if the algorithm that determines A i in finite quotients of 𝐺 depends recursively on 𝑖.

There exists a recursive subset of ℤ that is not determinable in finite quotients of ℤ.

Proof

See Lemma 33, disregarding the statement about the subset of ℤ being closed. ∎

There exists a subset of ℤ that is not re, but is determinable in finite quotients of ℤ.

Proof

For a function ℎ that grows faster than any recursive function, consider the enumeration 2 ⁢ h ⁢ ( 1 ) , 2 ⁢ h ⁢ ( 2 ) + 1 , 3 ⁢ h ⁢ ( 3 ) , 3 ⁢ h ⁢ ( 4 ) + 1 , 3 ⁢ h ⁢ ( 5 ) + 2 , 4 ⁢ h ⁢ ( 6 ) , …. This defines a set that is not re, but whose image in any quotient Z / n ⁢ Z of ℤ is all of Z / n ⁢ Z . ∎

A subset 𝐴 of a group 𝐺 is open in P ⁢ T ⁢ ( G ) if and only if, for any element 𝑎 of 𝐴, there is a morphism 𝜑 from 𝐺 onto a finite group 𝐹 such that φ - 1 ⁢ ( φ ⁢ ( a ) ) ⊆ A , i.e. such that if an element of 𝐺 has the same image as 𝑎 in 𝐹, it also belongs to 𝐴.

The closure B ¯ of a subset 𝐵 of 𝐺 is the biggest set of elements that satisfy the following condition: for any morphism 𝜑 from 𝐺 onto a finite group 𝐹, φ ⁢ ( B ¯ ) ⊆ φ ⁢ ( B ) .

Let 𝐴 and 𝐵 be disjoint subsets of 𝐺 that are determinable in finite quotients of 𝐺. Then McKinsey’s algorithm can be used to distinguish 𝐴 from 𝐵 if and only if the following two conditions hold:

Proof

This is a simple consequence of Fact 9. ∎

Let 𝐺 be a finitely generated group with ReFQ.

• If 𝐺 is residually finite, then it is co-re. If it is re and residually finite, it has solvable word problem.

• If 𝐺 is conjugacy separable, then there exists an algorithm that decides when two of its elements are not conjugate. If it is re and conjugacy separable, then it has solvable conjugacy problem.

• If 𝐺 is LERF, there is an algorithm that, given a tuple ( x 1 , … , x n , g ) of elements of 𝐺, stops exactly when 𝑔 does not belong to the subgroup of 𝐺 generated by ( x 1 , … , x n ) . If it is re and LERF, it has solvable generalized word problem.

Proof

All three points follow from Proposition 10, noticing that it provides a uniform algorithm, and using it respectively with

• 𝐴 and 𝐵 being singletons,

• 𝐴 and 𝐵 being conjugacy classes,

• 𝐴 being a finitely generated subgroup of 𝐺 and 𝐵 a singleton.

Property ReFQ is equivalent to having co-re membership problem for finite-index normal subgroups, that is, to having an algorithm that decides when an element is not in a given finite-index normal subgroup, and does not terminate otherwise.

Proof

Suppose first that 𝐺 has ReFQ, and let 𝑁 be a finite-index normal subgroup of 𝐺 generated by a family x 1 , … , x k . Finally, let 𝑔 be an element of 𝐺; we want to decide whether or not 𝑔 belongs to 𝑁. Enumerate the quotients ( F , f ) of 𝐺, and look for a finite quotient in which the image of 𝑔 is non-trivial, while the images of x 1 , … , x n are all trivial. If 𝑔 does not belong to 𝑁, such a quotient exists (the projection G → G / N ) , and this algorithm will terminate.

Now suppose 𝐺 has co-re membership problem for finite-index normal subgroups. Write ⟨ S | R ⟩ for a presentation of 𝐺. Let ( F , f ) be a finite group together with a function from 𝑆 to 𝐹. As 𝑓 does not necessarily define a morphism, we cannot yet apply Schreier’s method. But if F n is a free group with basis the 𝑛 generators of 𝐺, 𝑓 does define a morphism 𝜑 from F n to 𝐹, and thus we can find a family x 1 , … , x k of elements of F n that generate ker ⁡ ( φ ) . The finite group 𝐹 is thus given by the presentation ⟨ S | x 1 , … , x k ⟩ . (But x 1 , … , x k generate ker ⁡ φ as a group, and not only as a normal subgroup as would be guaranteed by any presentation of 𝐹 on the generators 𝑆.)

Now 𝑓 extends to a morphism if and only if 𝐹 satisfies all the relations of 𝐺, that is to say, if and only if the relations x 1 , … , x k imply the relations that appear in 𝑅, that is to say, if and only if ⟨ S | R , x 1 , … , x k ⟩ is just another presentation of 𝐹. But this is a presentation of G / N , where 𝑁 is the subgroup of 𝐺 generated by x 1 , … , x k . If 𝑓 does not extend to a morphism, G / N is a strict quotient of 𝐹.

Thus we can do the following: enumerate the elements of 𝐺, g 1 , g 2 , … . Then use the membership algorithm for 𝑁 (which, as we suppose, can only show something does not belong to 𝑁) to find elements that define different classes in G / N , that is, find g i 0 that does not belong to 𝑁; then g i 1 which is such that neither itself nor g i 0 ⁢ g i 1 - 1 belong to 𝑁, and g i 2 such that g i 2 , g i 0 ⁢ g i 2 - 1 and g i 1 ⁢ g i 2 - 1 do not belong to 𝑁, etc. If 𝐹 is a quotient of 𝐺, this method will yield card ⁡ ( F ) elements, at which point the algorithm has proven that 𝐹 is a quotient of 𝐺. Of course, if 𝐹 is not a quotient of 𝐺, it will never stop. ∎

For recursively presented groups, having CFQ and having solvable membership problem for finite-index normal subgroups are equivalent properties.

A group 𝐺, which admits a presentation ⟨ S | R ⟩ , has CFQ if the isomorphism problem is solvable for the following family of presentations: all finite presentations of finite groups, and all presentations of the form ⟨ S | R , R 1 ⟩ , where R 1 is a finite set of relations such that ⟨ S | R 1 ⟩ is finite.

There exists a recursive family of recursive presentations of finite groups, for which the isomorphism problem is unsolvable. Moreover, any two of those presentations differ only by a finite number of relations.

Any recursively presented group has co-ReFQ.

Proof

Let 𝐺 be a re group generated by 𝑆 of cardinality 𝑛. Let ( F , f ) be an 𝑛-marked finite group. For any relation 𝑟 of 𝐺, write

We again test the equality e = f ⁢ ( s 1 ) α 1 ⁢ … ⁢ f ⁢ ( s k ) α k . Since we suppose 𝐺 re, this can be carried out successively on all relations of 𝐺. If 𝑓 does not extend to a morphism, a relation true in 𝐺 but not in 𝐹 will eventually be found. ∎

If 𝐺 and 𝐻 both have one of CFQ, ReFQ, co-ReFQ, then so does their free product G * H .

Proof

Note that we have shown that those properties are independent of the generating family; thus we can show this using as a generating family of G * H the union of a generating family of 𝐺 and of one of 𝐻. The proof is then straightforward, as a function from that generating family to a finite group extends to a morphism of G * H if and only if both restrictions to 𝐺 and to 𝐻 extend as morphisms. ∎

ReFQ is inherited by finite-index subgroups.

Proof

Let 𝐺 be a group, and 𝑁 a finite-index subgroup of 𝐺. An enumeration of the finite quotients of 𝐺 gives an enumeration of finite quotients of 𝑁, by restricting the homomorphisms to 𝑁. Not all quotients of 𝑁 need arise this way, but add the following: whenever a pair ( F , f ) is found that defines a quotient of 𝑁, list all quotients of the finite group 𝐹, and when a quotient F ⁢ → π ⁢ F 0 is found, add ( F 0 , π ∘ f ) to the list of quotients of 𝑁. We claim that all finite quotients of 𝑁 arise this way.

Let 𝑀 be a finite-index normal subgroup of 𝑁. Then 𝑀 is of finite index in 𝐺, it may not be normal in 𝐺, but it contains a normal subgroup M 0 which is both finite index and normal in 𝐺. Then G → G / M 0 restricts to a morphism N → N / M 0 , and M / M 0 is a normal subgroup in N / M 0 , and the quotient of N / M 0 by M / M 0 is, of course, N / M . ∎

Find a finitely presented residually finite group with a finitely generated subgroup that does not have CFQ.

Let 𝐺 be a finitely generated group. Let 𝐻 be a group obtained by adding finitely many relations and identities to 𝐺. If 𝐺 has any of ReFQ, co-ReFQ or CFQ, then so does 𝐻.

Proof

Let 𝐺 and 𝐻 be two finitely generated groups, with a morphism 𝜋 from 𝐺 onto 𝐻. Let ( F , f ) be a marked finite group. It is obvious that if 𝑓 extends to a homomorphism 𝜑 from 𝐻 onto 𝐹, then f ∘ π extends as well to a morphism 𝜓, which, in addition, satisfies ker ⁡ ( π ) ⊆ ker ⁡ ( ψ ) . On the other hand, if f ∘ π extends to a morphism 𝜓, and if ker ⁡ ( π ) is contained in ker ⁡ ( ψ ) , then 𝜓 factors through 𝜋 and 𝑓 will extend to a morphism. The diagram is the following:

Thus if 𝐺 has CFQ or ReFQ, we can reduce the finite quotient question of 𝐻 to a question about subgroups inclusion.

If ker ⁡ ( π ) is finitely generated as a normal subgroup, then the question “is ker ⁡ ( π ) contained in ker ⁡ ( ψ ) ?” can be solved in finite time, as it is solved by computing ψ ⁢ ( r ) for each 𝑟 in a generating family of ker ⁡ ( π ) . If ker ⁡ ( π ) is generated by an identity – that is a set of relations of the form v ⁢ ( x 1 , … , x k ) , where 𝑣 is a element of the free group on 𝑘 generators, and x 1 , … , x k take all possible values of G k – this question can also be answered because, to check whether or not an identity holds in a finite group, one only needs to check finitely many relations.

If 𝐺 is a co-ReFQ group, we can determine whether or not ker ⁡ ( π ) is contained in ker ⁡ ( ψ ) even without knowing if 𝜓 defines a morphism from 𝐺, and thus if ( F , f ) does not define a quotient of 𝐻, we will either prove that f ∘ π does not extend to a quotient of 𝐺, or that, even if it were to define a quotient, the inclusion of kernels would not hold. With this, all cases of Proposition 20 are covered. ∎

The fundamental group of a graph of groups with vertex groups that have CFQ (or ReFQ or co-ReFQ) and finitely generated edge groups also has CFQ (respectively ReFQ or co-ReFQ). This includes free products amalgamated over finitely generated subgroups and HNN extensions over finitely generated subgroups.

A direct product of groups that have one of CFQ, ReFQ or co-ReFQ, again has that property.

If two finitely generated groups 𝐺 and 𝐻 admit the same finitary image, then 𝐺 has any of CFQ, ReFQ, co-ReFQ if and only if 𝐻 shares the same property.

The wreath product H ≀ G of two finitely generated groups 𝐺 and 𝐻 has CFQ if and only if either 𝐻 has CFQ and is infinite, or if 𝐻 is finite and 𝐺 has CFQ.

Proof

Suppose that H ≀ G has CFQ. From the presentation of H ≀ G given above, it is easy to see that 𝐻 can be obtained as a quotient of H ≀ G by adding only finitely many relations to it. Thus, by Proposition 20, 𝐻 has CFQ.

Suppose additionally that 𝐻 is finite. A finite group 𝐹, marked by a function 𝑓, is a quotient of 𝐺 if and only if the function 𝑔, defined as the identity on the generators of 𝐻 and as 𝑓 on those of 𝐺, which thus sends a generating family of H ≀ G to one of H ≀ F , can be extended to a morphism. But, because 𝐻 is finite, H ≀ F is also finite, and thus it can be determined whether or not it is a quotient of H ≀ G . This proves that 𝐺 has CFQ.

For the converse, suppose first that 𝐻 is finite and that 𝐺 has CFQ. Because 𝐻 is finite, a presentation for H ≀ G can be obtained by adding finitely many relations to a presentation of G * H , which has CFQ because both 𝐺 and 𝐻 have. Thus Proposition 20 applies again.

Suppose now that 𝐻 is infinite and has CFQ. In a finite quotient of H ≀ G , some non-trivial element of 𝐻 must necessarily have a trivial image. But it is easy to see from the presentation of H ≀ G given above that, for any element h ≠ 1 of 𝐻, the relation h = 1 together with the relations of H ≀ G will always also imply that [ g 1 , g 2 ] = 1 for any pair ( g 1 , g 2 ) of elements of 𝐺. This implies that any finite quotient of H ≀ G is in fact a finite quotient of H ≀ G ab , where G ab denotes the abelianization G / [ G , G ] of 𝐺. It thus suffices to prove that if 𝐻 has CFQ and 𝐺 is abelian, then the wreath product H ≀ G also has CFQ.

Consider a finite group 𝐹 together with a function 𝑓 that goes from a generating family of H ≀ G to 𝐹. To decide whether or not 𝑓 can be extended to a morphism, one can first check whether or not the restriction f H of 𝑓 to a generating family of 𝐻 extends as a morphism from 𝐻 onto a subgroup F H of 𝐹, using the fact that 𝐻 has CFQ. If it does not, then 𝑓 cannot be extended to a group morphism. If it does, one can compute a presentation of F H . The finite group 𝐹 is thus a quotient of H ≀ G if and only if it is a quotient of F H ≀ G . But 𝐺 is finitely presented, being abelian, and F H is finite; thus this wreath product is finitely presented, and a finite presentation of it can be obtained from the presentation of F H . This presentation can in turn be used to decide whether or not 𝐹 is a quotient of F H ≀ G , and thus of H ≀ G .

(Note that this last part of the proof is non-constructive: we showed that H ≀ G admits a finite quotient algorithm because H ≀ G ab does, and that H ≀ G ab has CFQ because G ab is finitely presented, but we do not claim that a finite presentation for G ab can be effectively found, and thus the existence of an algorithm that recognizes the quotients of H ≀ G remains abstract.) ∎

L-presented groups have CFQ.

Theorem 29 ([5, Theorem 2.1])

There exists a finitely presented group 𝐺 in which there is no algorithm to decide which elements have trivial image in every finite quotient.

There exists a finitely presented group 𝐺 whose finitary image is not recursively presented.

There exists a group 𝐻 which is finitely presentable as a residually finite group, but which has unsolvable word problem.

Proof of Theorem 3

By Proposition 22, a group which is finitely presented as a residually finite group must have CFQ. Thus if such a group has unsolvable word problem, it satisfies the criteria required by Theorem 3. A group, finitely presented as a residually finite group, with unsolvable word problem exists by Theorem 29. ∎

L ⁢ ( A ) n is finitely presented and residually finite. It is the amalgamated product of two copies of the finite wreath product Z / n ⁢ Z ≀ Z / 2 ⁢ Z , and it admits the following presentation:

Proof

The given presentation is obtained from the presentation of L ⁢ ( A ) , adding relations a n and a ^ n , and simplifying the relations as can be done. It then follows from that presentation that L ⁢ ( A ) n is an amalgamated product of two copies of the finite wreath product Z / n ⁢ Z ≀ Z / 2 ⁢ Z . Finally, it is well known that an amalgamated product of finite groups is residually finite (see [17]). ∎

Let 𝒜 be a subset of ℤ.

1. L ⁢ ( A ) is re, co-re or has solvable word problem if and only if 𝒜 is respectively re, co-re or recursive.

2. L ⁢ ( A ) is residually finite if and only if 𝒜 is closed in P ⁢ T ⁢ ( Z ) .

3. L ⁢ ( A ) has CFQ if and only if the function which to 𝑛 associates A mod n is recursive.

Proof

We prove all three points in order.

If L ⁢ ( A ) is re or co-re, then clearly so is 𝒜, as 𝑛 belongs to 𝒜 if and only if u n = u ^ n in L ⁢ ( A ) , which proves one direction of (1). It is clear that if 𝒜 is re, the presentation of L ⁢ ( A ) given above is re as well. Suppose now that 𝒜 is co-re. We can enumerate the complement of 𝒜, and thus enumerate elements of the form

where α i , β i ∈ Z , x 1 , … , x k and y 1 , … , y k are elements of the base groups of 𝐿 and L ^ that have null components corresponding to indices in 𝒜, and 𝑧 is any element in 𝐿 or in L ^ . The elements written this way are exactly the elements in normal form for the amalgamated product L ⁢ ( A ) . Recall that the normal form in an amalgamated product (see [16]) necessitates two choices of transversals, here one for L / ( ⨁ A Z / 2 ⁢ Z ) and one for L ^ / ( ⨁ A Z / 2 ⁢ Z ) , and that an element in normal form is a consecutive product of an element of one transversal, followed by one of the other, etc., terminated by any element of one of these groups. But elements a α ⁢ x , with 𝛼 in ℤ and 𝑥 in the base group with null components on 𝒜, indeed form the most natural transversal for L / ( ⨁ A Z / 2 ⁢ Z ) . Thus any non-trivial element 𝑔 of L ⁢ ( A ) is equal to exactly one element in this enumeration. Ideally, we would then enumerate words that give the identity in L ⁢ ( A ) , and listing words that can be obtained by concatenating a word in normal form to a word that defines the identity would give the desired enumeration. Since L ⁢ ( A ) is not supposed re, we cannot directly enumerate this set of trivial words, but we will over-approximate it by a re set. For 𝑤 as in (∗), denote by B w the set of all indices that appear in elements x i or y i of the base groups ( B w can be empty). The over-approximation consists of the words corresponding to the identity in L ⁢ ( Z ∖ B w ) . Note that 𝑤 is also in normal form in L ⁢ ( Z ∖ B w ) , thus non-trivial there. Of course, B w is finite; thus Z ∖ B w is re; thus, as we already remarked, we can enumerate words (in 𝑎, a ^ , 𝜀, ε ^ ) that correspond to the identity in L ⁢ ( Z ∖ B w ) . Then, for any such word w 1 , the product w ⁢ w 1 corresponds to a non-identity element of L ⁢ ( Z ∖ B w ) , thus to a non-identity element in L ⁢ ( A ) since, as A ⊆ ( Z ∖ B w ) , L ⁢ ( Z ∖ B w ) satisfies more relations than L ⁢ ( A ) . Thus enumerating products w ⁢ w 1 with w 1 = e in L ⁢ ( Z ∖ B w ) will only yield non-identity elements in L ⁢ ( A ) . Moreover, every element of the form w ⁢ w 2 , where w 2 is a word that is the identity in L ⁢ ( A ) , will arise this way, again because L ⁢ ( Z ∖ B w ) satisfies more relations than L ⁢ ( A ) .

Because the algorithm that enumerates relations in L ⁢ ( Z ∖ B w ) depends recursively on 𝑤, this process can be applied simultaneously to all words 𝑤 in normal form, giving an enumeration of all words that correspond to non-identity elements of L ⁢ ( A ) , thus proving that L ⁢ ( A ) is co-re.

Finally, 𝒜 is recursive if and only if it is both re and co-re, if and only if L ⁢ ( A ) has solvable word problem. This ends the proof of (1).

We first sketch a conceptually simple proof of (2). Recall that the group L ⁢ ( A ) is the double of the lamplighter group 𝐿 over the subgroup H A generated by elements of the form a i ⁢ ε ⁢ a - i for 𝑖 in 𝒜. Statement (2) is implied by the two following arguments:

• the double of a residually finite group 𝐺 over a subgroup 𝐻 is itself residually finite if and only if 𝐻 is separable in 𝐺;

• the subgroup H A is separable in 𝐿 if and only if 𝒜 is separable in ℤ.

We now give another proof of (2), which can be easily rendered effective. We in fact prove slightly more than (2): for a subset 𝒜 of ℤ, the finitary image L ⁢ ( A ) f of L ⁢ ( A ) is L ⁢ ( A ¯ ) , where A ¯ denotes the closure of 𝒜 in P ⁢ T ⁢ ( Z ) . First we show that if 𝑛 belongs to A ¯ , then in any finite quotient ( F , f ) of L ⁢ ( A ) , the images of u n and u ^ n are the same in 𝐹. Let ( F , f ) be some finite quotient of L ⁢ ( A ) . Let 𝑝 and p ′ be the orders of f ⁢ ( a ) and f ⁢ ( a ^ ) in 𝐹. Then, because 𝑛 belongs to A ¯ , n + p ⁢ p ′ ⁢ Z must meet with 𝒜, as it is a neighborhood of 𝑛. Thus we have 𝑘 such that n + p ⁢ p ′ ⁢ k ∈ A , that is, such that u n + p ⁢ p ′ ⁢ k = u ^ n + p ⁢ p ′ ⁢ k in L ⁢ ( A ) . Then, in 𝐹 (we omit to write the homomorphism onto 𝐹),

This shows that L ⁢ ( A ) f is a quotient of L ⁢ ( A ¯ ) . It is then sufficient to see that L ⁢ ( A ¯ ) is residually finite to see that L ⁢ ( A ) f = L ⁢ ( A ¯ ) .

We suppose that 𝒜 is closed to omit the closure notation. Let 𝑤 be a non-identity element of L ⁢ ( A ) , and write it in normal form w = a α 1 ⁢ x 1 ⁢ a ^ ⁢ y 1 β 1 ⁢ … ⁢ a α k ⁢ x k ⁢ a ^ ⁢ y k β k ⁢ z , as in the proof of (1).

Suppose first that the normal form is the trivial one: w = z with 𝑧 in 𝐿 or in L ^ . Then 𝑤 is non-trivial in the quotient of L ⁢ ( A ) obtained by identifying the two copies 𝐿 and L ^ of the lamplighter group (i.e. ⟨ L ⁢ ( A ) | a = a ^ , ε = ε ^ ⟩ ), which is just the lamplighter group itself, which is residually finite.

We can now suppose the normal form has several terms. Each x i is an element of ⨁ Z ∖ A Z / 2 ⁢ Z , that is to say, a product ∏ u k i , j with k i , j ∉ A . Because 𝒜 is closed, for each such k i , j , there is p i , j that satisfies ( k i , j + p i , j ⁢ Z ) ∩ A = ∅ . Similarly, for each x ^ i , introduce integers p i , j ′ , j = 1 , 2 , … . Write 𝑁 for the product ∏ p i , j ⁢ ∏ p i , j ′ . (It is 1 if, for all 𝑖, x i = x ^ = i e .) We claim that 𝑤 is non-trivial in L ⁢ ( A ) N . Indeed, 𝑁 was chosen so that, for any ( i , j ) , k i , j (or its remainder modulo 𝑁) does not belong to A mod N . This implies that 𝑤 is also in normal form in L ⁢ ( A ) N (by Lemma 30), and thus non-trivial there.

Again by Lemma 30, L ⁢ ( A ) N is residually finite, so we have proven that L ⁢ ( A ) is residually residually finite, which of course is the same as residually finite.

Finally, we prove (3). Suppose A mod n depends recursively on 𝑛. Let ( F , f ) be a finite group together with a function 𝑓 from { a , a ^ , ε , ε ^ } to 𝐹. To determine whether or not 𝑓 defines a homomorphism, compute the orders of f ⁢ ( a ) and of f ⁢ ( a ^ ) , and let 𝑛 be their product. If 𝑓 extends to a morphism, this morphism factors through the projection π : L ⁢ ( A ) → L ⁢ ( A ) n . By Lemma 30, a finite presentation for L ⁢ ( A ) n can be found from the computation of A mod n . It can then be determined in finite time from this presentation whether 𝑓 defines a homomorphism from L ⁢ ( A ) n to 𝐹 or not.

Suppose now that L ⁢ ( A ) has CFQ. Let 𝑛 be a natural number. To compute A mod n , consider all possible presentations for L ⁢ ( A ) n : for B ⊂ { 0 , … , n - 1 } , define the presentation ∏ B ,

All these presentations define residually finite groups, and because they are finitely presented, they have CFQ. Then L ⁢ ( A ) n also has CFQ because it is obtained from L ⁢ ( A ) by adding two relations; thus we can start to enumerate the quotients of L ⁢ ( A ) n . Also start enumerating the quotients of all groups given by the presentations ∏ B for B ⊂ { 0 , … , n - 1 } . Those 2 n lists are all different (because, as the presentations ∏ B give residually finite groups, a list contains a finite group in which the images of a j ⁢ ε ⁢ a - j and a ^ ε ^ j a ^ - j differ if and only if 𝑗 does not belong to 𝐵), and only one corresponds to the list of quotients of L ⁢ ( A ) n . It can be determined, in a finite number of steps, which of those lists corresponds to L ⁢ ( A ) n , and thus which presentation ∏ B gives a presentation of L ⁢ ( A ) n , and then one can conclude that B = A mod n . ∎

There exists a non-recursive subset 𝒜 of ℤ, closed in P ⁢ T ⁢ ( Z ) , for which A mod n depends recursively on 𝑛.

Proof

We construct a set ℬ, which will be the complement of the desired 𝒜. Thus it has to be open, re but not co-re, and for any 𝑎 and 𝑏, the question “is a + b ⁢ Z a subset of ℬ?” has to be solvable in a finite number of steps. Indeed, 𝑎 belongs to A mod b if and only if a + b ⁢ Z meets 𝒜, if and only if a + b ⁢ Z is not a subset of the complement of 𝒜.

Let p n be the 𝑛-th prime number. Define two sequences ( x n ) n ≥ 0 and ( y n ) n ≥ 1 by the following:

These sequences have the following properties:

• for any 𝑛, x n ∣ x n + 1 and y n ∣ y n + 1 ;

• for any integer 𝑏, there is some (computable) 𝑛 such that b ∣ x n and b ∣ y n ;

• p k divides x n if and only if k ≥ n , and p k divides y n if and only if k > n ;

• for integers k , k ′ , n , n ′ , with k ≤ n and k ′ ≤ n ′ , y k + x n ⁢ Z and y k ′ + x n ′ ⁢ Z are disjoint if and only if k ≠ k ′ , and otherwise one is a subset of the other.

Consider a recursive function 𝑓 whose image is re but not co-re. Assume that, for any 𝑛, 1 ≤ f ⁢ ( n ) ≤ n (it is easy to see that such a function exists). Then we define ℬ as the union,

Since 𝑓 is a recursive function, ℬ is re. It is not co-re, however, because y m belongs to ℬ if and only if 𝑚 belongs to the image of 𝑓 (this follows directly from the properties of the sequences ( x n ) n ≥ 0 and ( y n ) n ≥ 1 ). The set ℬ is open because it is defined as a union of open sets.

All that is left to see is that we can decide, for integers 𝑎 and 𝑏, whether or not a + b ⁢ Z is a subset of ℬ. Suppose that a < b . If a = 0 , then 0 ∈ a + b ⁢ Z , but 0 ∉ B ; thus a + b ⁢ Z is not a subset of ℬ. If 𝑎 is non-zero, no element of a + b ⁢ Z is divisible by 𝑏. Thus, because of the second property of the sequences ( x n ) n ≥ 0 and ( y n ) n ≥ 1 quoted above, there exists 𝑁 such that if N ≤ k ≤ n , then a + b ⁢ Z ∩ y k + x n ⁢ Z = ∅ . Thus a + b ⁢ Z is a subset of ℬ if and only if it is a subset of the set B N , defined by

Define a pseudo-inverse 𝑔 of 𝑓 by g ⁢ ( m ) = inf ⁡ { n , f ⁢ ( n ) = m } . Because we chose 𝑓 such that, for any 𝑛, f ⁢ ( n ) ≤ n , for any 𝑚, g ⁢ ( m ) ≥ m . If 𝑚 is not in the image of 𝑓, put g ⁢ ( m ) = ∞ . The set B N can be expressed as the disjoint union,

Because x k ∣ x g ⁢ ( k ) , B N is contained in the set C N , defined by

It can be determined whether or not a + b ⁢ Z is contained in C N because the sequences ( x n ) n ≥ 1 and ( y n ) n ≥ 1 can be computed. If a + b ⁢ Z is not contained in C N , then it is not contained in B N either.

If it is contained in C N , then a + b ⁢ Z is contained in B N if and only if, for each 𝑘, a + b ⁢ Z ∩ y k + x k ⁢ Z is contained in B N . But, because B N and C N are disjoint unions, a + b ⁢ Z ∩ y k + x k ⁢ Z is contained in B N if and only if it is contained in y k + x g ⁢ ( k ) ⁢ Z . (If 𝑘 is not in Im ⁡ ( f ) and g ⁢ ( k ) = ∞ , by convention, we have y k + x g ⁢ ( k ) ⁢ Z = { y k } .) Now this question can be effectively answered. If a + b ⁢ Z ∩ y k + x k ⁢ Z is empty, there is nothing to do. Otherwise, it is of the form t + lcm ⁡ ( b , x k ) ⁢ Z . Enumerate f ⁢ ( 1 ) , f ⁢ ( 2 ) , … , f ⁢ ( lcm ⁡ ( b , x k ) ) . Either 𝑘 is in that list, in which case g ⁢ ( k ) can be computed and the question “is a + b ⁢ Z ∩ y k + x k ⁢ Z contained in y k + x g ⁢ ( k ) ⁢ Z ?” can be settled, or 𝑘 does not appear in the enumeration, which shows that g ⁢ ( k ) is greater than lcm ⁡ ( b , x k ) . In this last case, as x g ⁢ ( k ) is greater than g ⁢ ( k ) , y k + x g ⁢ ( k ) ⁢ Z cannot contain any set of the form t + lcm ⁡ ( b , x k ) ⁢ Z . ∎

Alternate proof of Theorem 3

It follows from Proposition 31 that the group L ⁢ ( A ) , where 𝒜 is the set constructed in Lemma 32, satisfies the requirements of Theorem 3. ∎

There exists a recursive subset 𝒜 of ℤ, closed in P ⁢ T ⁢ ( Z ) , for which A mod n does not depend recursively on 𝑛.

Proof

Let p n be the 𝑛-th prime number. Fix some effective enumeration M 1 , M 2 , … of all Turing machines. Consider the following process. Start running all those machines simultaneously, as is done to show that the halting problem is re. While running calculations on the 𝑛-th machine, at each new step in the computation, produce a new power of p 2 ⁢ n : p 2 ⁢ n , p 2 ⁢ n 2 , p 2 ⁢ n 3 , … . If the computation on this machine stops after 𝑘 steps, end the list p 2 ⁢ n , p 2 ⁢ n 2 , … , p 2 ⁢ n k already produced with p 2 ⁢ n + 1 k + 1 .

Write 𝒜 for the set of all prime powers obtained this way. Then 𝒜 is obviously re, as it was defined by an effective enumeration process. It is even recursive. Indeed, for a number 𝑥, if 𝑥 is not the power of a prime, then 𝑥 is not in 𝒜. If it is the power of a prime of even index, say x = p 2 ⁢ n k , then 𝑥 belongs to 𝒜 if and only if the 𝑛-th Turing machine does not stop in less than 𝑘 calculations steps. This question can be effectively settled. Similarly, if 𝑥 is the power of a prime of odd index, x = p 2 ⁢ n + 1 k , then 𝑥 belongs to 𝒜 if and only if the 𝑛-th Turing machine stops in exactly 𝑘 calculations steps; this also can be determined.

Of course, A mod m does not depend recursively on 𝑚. Indeed, the question “does 0 belong to A mod p 2 ⁢ n + 1 ?” is, by construction, equivalent to “does the 𝑛-th Turing machine halt?”.

Finally, we show that 𝒜 is a closed set, which is equivalent to finding, for any 𝑥 not in 𝒜, a number 𝑦 such that x + y ⁢ Z does not meet 𝒜. If 𝑥 has several prime divisors, then x + x ⁢ Z works because any element of it has several prime divisors. If 𝑥 is the power of a prime of even index, x = p 2 ⁢ n k , and 𝑥 is not in 𝒜, it must be that the 𝑛-th Turing machine stops in strictly less than 𝑘 steps. Thus the only elements in 𝒜 that are multiples of p 2 ⁢ n will have a valuation in p 2 ⁢ n lower than 𝑘. Thus x + x ⁢ Z will also work. The last case is if 𝑥 is the power of a prime of odd index, x = p 2 ⁢ n + 1 k . In this case, we claim that x + p 2 ⁢ n + 1 ⁢ x ⁢ Z does not meet 𝒜. Indeed, 𝑥 is the only power of p 2 ⁢ n + 1 contained in x + p 2 ⁢ n + 1 ⁢ x ⁢ Z ; all other elements of it have at least two different prime divisors. ∎

Proof of Theorem 2

By Proposition 31, for a subset 𝒜 of ℤ, Dyson’s group L ⁢ ( A ) satisfies the requirements of Theorem 2 provided that 𝒜 is closed, recursive and that the function A mod n is not computable. Lemma 33 provides such a set. ∎

There exists a group with solvable word problem and recursive depth function, that does not have CFQ.

Proof

It appears clearly in the proof of Lemma 33 that the constructed 𝒜 is effectively closed: if 𝑥 does not belong to it, then some 𝑦 such that x + y ⁢ Z does not meet 𝒜 can effectively be found. Going back to the proof of the first point of Proposition 31, it appears that, given a non-identity element 𝑤 of L ⁢ ( A ) , the recursiveness of 𝒜 allows one to compute its normal form. Then, in the proof of the second point of that same proposition, it appears that, from this normal form and the effective closeness of 𝒜, some integer 𝑁 can be effectively found such that 𝑤 is non-trivial in L ⁢ ( A ) N . A presentation of L ⁢ ( A ) N cannot necessarily be found, but there are 2 N possible finite presentations for it, all of them with recursive depth function. Taking the supremum of those depth functions allows one to compute a recursive upper bound to the depth function of L ⁢ ( A ) .

Although it is not clear whether that depth function is recursive or not, by taking the direct product of the group L ⁢ ( A ) with a finitely presented group with recursive depth function greater than that of L ⁢ ( A ) , which exists by [14], one obtains a group which still has solvable word problem and uncomputable finite quotients, and whose depth function is recursive. ∎