An uncountable family of finitely generated residually finite groups

Let 𝐹 be a group with an isomorphic copy F ¯ . Let H ≤ F and H ¯ ≤ F ¯ be corresponding subgroups. The double of 𝐹 along 𝐻 is the amalgamated product G = F * H = H ¯ F ¯ .

A sequence s = ( s n ) n = 0 ∞ is multiplying if s n and s n + 1 / s n are positive integers for all 𝑛, and lim ⁡ s n = ∞ .

Let 𝐻 be a closed subgroup of a residually finite group 𝐹. Then the double F * H = H ¯ F ¯ is also residually finite.

The based core at 𝑥 of a connected graph 𝐵 is the smallest connected subgraph containing 𝑥 and all closed cycles of 𝐵.

If 𝔰 is multiplying, then H s is a closed subgroup of 𝐹.

Proof

As above, every finitely generated subgroup of a free group is closed. Hence, it suffices to show that H s is the intersection of a sequence of finitely generated subgroups of 𝐹.

For each k ≥ 0 , define S k ≤ F as follows:

To show H s ⊆ ⋂ S k , it suffices to show that b m ⁢ a s m ⁢ b - m ∈ S k for m , k ≥ 0 . Fix m , k ≥ 0 . If m < k , then b m ⁢ a s m ⁢ b - m ∈ S k by definition. If m ≥ k , then let m = q ⁢ k + r for integers q , r ≥ 0 with 0 ≤ r < k . Let f = s m / s r . Since 𝔰 is multiplying, 𝑓 is a positive integer. Thus,

It remains to show that ⋂ S k ⊆ H s . Consider w ∈ ⋂ S k . Since lim ⁡ s n = ∞ , we may choose 𝑘 with k > 2 ⁢ | w | and s k - | w | > | w | . The based core 𝐶 of the covering space associated with S k is shown in Figure 1. The graph 𝐶 is the union of a based length 𝑘 cycle of 𝑏-edges, and a length s i cycle of 𝑎-edges attached at its 𝑖-th vertex for 0 ≤ i < k .

Figure 1

The core 𝐶 of the based covering space associated with S k .

Let 𝜔 be the immersed loop in 𝐶 labelled by 𝑤. Then 𝜔 is shorter than the a s k - m (red) loops for 1 ≤ m ≤ | w | , because s k - | w | > | w | . Thus, 𝜔 cannot traverse an edge of those (red) loops. Therefore, 𝜔 is loop immersed in the subgraph of 𝐶 obtained by removing those (red) loops.

Moreover, 𝜔 cannot traverse an edge of b | w | at the bottom of the graph since 𝜔 cannot backtrack.

Figure 2

The based graph C ′ after removing the (red) loops a s k - m for 1 ≤ m ≤ | w | and the bottom b | w | path in Figure 1. The base point is bold.

Therefore, 𝜔 is immersed in the subgraph C ′ obtained by discarding edges which 𝜔 cannot traverse as explained above. See Figure 2. Since C ′ is a based subgraph of the based core of the covering space associated with H s , we see that w ∈ H s . ∎

It is not difficult to show that G s is residually a finite 𝑝-group if and only if all elements of 𝔰 are powers of 𝑝. Furthermore, G s fails to be residually finite without the requirement that 𝔰 is nonbounded.

Let 𝔰 be multiplying. Let 𝑚 be a positive integer. Let

Then

where ⌊ a , b ⌋ = gcd ⁡ ( a , b ) .

Proof

Recall that G s = F * H s = H ¯ s F ¯ . Images of subgroups in G s → G ^ s are hatted,

We shall compute H 2 ⁢ ( G ^ s ) via the Mayer–Vietoris sequence for amalgamated products. Firstly, we claim G ^ s decomposes as a double as follows:

• G ^ s ≅ Q * K = K ¯ Q ¯ , where Q = F / ⟨ ⟨ a m ⟩ ⟩ = F ^ and Q ¯ = F ¯ / ⟨ ⟨ a ¯ m ⟩ ⟩ = F ¯ ^

• and where K = im ⁡ ( H s ) under F → Q , and K ¯ = im ⁡ ( H ¯ s ) under F ¯ → Q ¯ .

The map G ^ s → Q * K = K ¯ Q ¯ is induced by F → Q and F ¯ → Q ¯ . The retraction maps G s → F and G s → F ¯ project to retractions G ^ s → Q and G ^ s → Q ¯ . Moreover, the retractions G s → F and G s → F ¯ project to G ^ s → F ^ and G ^ s → F ¯ ^ , respectively. The inclusions Q ↪ G ^ s and Q ↪ G ^ s induce Q * K = K ¯ Q ¯ → G ^ s , which provides the inverse. Thus, G ^ s ≅ Q * K = K ¯ Q ¯ .

To calculate H 1 ⁢ ( H ^ s ) , we first determine H ^ s through the following sequence of isomorphisms. The first holds by declaring a ^ , b ^ to be the images of a , b . The second holds by the Normal Form Theorem for free products [9, Theorem IV.1.2]. The third holds for each factor as s j has order m / ⌊ s j , m ⌋ in Z / m ⁢ Z .

Therefore, as homology is summable over free products, we have

To compute H 2 ⁢ ( F ^ ) , we construct an aspherical complex 𝑋 with π 1 ⁢ X ≅ F ^ , so H 2 ⁢ ( F ^ ) = H 2 ⁢ ( X ) (see [2, Proposition 2.4.1]). Let X 2 be the standard 2-complex of ⟨ a , b ∣ b m ⟩ . The aspherical complex 𝑋 is formed by adding appropriate 𝑛-cells to X 2 for n ≥ 3 .

∂ 2 : C 2 ⁢ ( X ) → C 1 ⁢ ( X ) is injective since the attaching map of the 2-cell has degree 𝑚. Thus,

The following exact sequence is deduced from the Mayer–Vietoris sequence for amalgamated products [2, Proposition 2.7.7]:

Applying equations (4.1), (4.3) and (4.4), we obtain the following exact sequence:

The map ϕ : H 1 ⁢ ( H ^ s ) → H 1 ⁢ ( F ^ ) ⊕ H 1 ⁢ ( F ¯ ^ ) is induced by H ^ s ↪ F ^ and H ¯ ^ s ↪ F ¯ ^ . By (4.2), ϕ ⁢ ( [ b ^ j ⁢ a ^ s j ⁢ b ^ - j ] ) = ( [ a ^ s j ] , [ a ¯ ^ s j ] ) . Since { [ b ^ j ⁢ a ^ s j ⁢ b ^ - j ] : j ≥ 0 } generates H 1 ⁢ ( H ^ s ) , we have im ⁡ ( ϕ ) = ⌊ m , s 0 ⌋ ⁢ Z / m ⁢ Z ≅ Z m / ⌊ m , s 0 ⌋ as s 0 ∣ s j for j ≥ 0 . Therefore, by the first isomorphism theorem, we reach the conclusion

An element g ∈ G is loose if C = ⟨ g ⟩ is a maximal cyclic subgroup and there exists a maximal cyclic subgroup C ′ such that C ∩ C ′ is a proper finite index subgroup of 𝐶.

If s ≠ ( 1 ) , then g ∈ G s is loose if and only if 𝑔 is conjugate to a ± 1 or a ¯ ± 1 .

Proof

Choose 𝑝 with s p > 1 . Let C = ⟨ b p ⁢ a ⁢ b - p ⟩ and C ′ = ⟨ b ¯ p ⁢ a ¯ ⁢ b ¯ - p ⟩ . Then C ∩ C ′ = ⟨ b p ⁢ a s p ⁢ b - p ⟩ = ⟨ b ¯ p ⁢ a ¯ s p ⁢ b ¯ - p ⟩ is a proper finite index subgroup of 𝐶 and C ′ . Hence, b p ⁢ a ⁢ b - p and b ¯ p ⁢ a ¯ ⁢ b ¯ - p are loose and thus 𝑎 and a ¯ are loose, since looseness is preserved by conjugation.

For the other direction, consider the action of G s on the Bass–Serre tree. Suppose ⟨ g ⟩ ≠ ⟨ k ⟩ are maximal cyclic subgroups with ⟨ g ⟩ ∩ ⟨ k ⟩ infinite. Note that either both are hyperbolic or both are elliptic. Indeed, if 𝑔 is hyperbolic, then 𝑘 is hyperbolic since k n ∈ ⟨ g ⟩ for some 𝑛.

Suppose 𝑔 and 𝑘 are hyperbolic, then since ⟨ g ⟩ and ⟨ k ⟩ are commensurable, they stabilize the same axis in the tree and hence act on the axis in the same way. By maximality of ⟨ g ⟩ and ⟨ k ⟩ , we have ⟨ g ⟩ = ⟨ k ⟩ , which contradicts our assumption.

Suppose 𝑔 and 𝑘 are elliptic. Let 𝑢 and 𝑣 be the vertices stabilized by 𝑔 and 𝑘, and let 𝐴 be the geodesic from 𝑢 to 𝑣. Note that u ≠ v since free groups have no loose subgroups. Note that the pointwise-stabilizer of 𝐴 is ⟨ g ⟩ ∩ ⟨ k ⟩ .

Let A = e 1 ⁢ e 2 ⁢ ⋯ ⁢ e m , with vertices u = u 0 , u 1 , … , u m = v . Since 𝑔 stabilizes u 0 , there exists ℓ such that 𝑔 stabilizes u ℓ but not e ℓ + 1 . Thus, 𝑔 is in the vertex group associated to u ℓ but not in the edge group associated to e ℓ . However, g n is in the edge group associated to e ℓ for some n > 0 .

It thus suffices to show that H < F has the property that if g r ∈ H for some 𝑟 but g ∉ H , then 𝑔 is conjugate into ⟨ a ⟩ . Equivalently, we show that if a path 𝑆 in the graph associated to 𝐻 has a power S m which is closed, then either 𝑆 is closed or 𝑆 is a conjugate of a power of 𝑎.

Observe that the 𝑏-exponent sum of 𝑆 equals 0, for otherwise, no S r is closed. Assume without loss of generality that 𝑆 is cyclically reduced, and moreover, if 𝑆 is not a power of 𝑎, then 𝑆 starts with 𝑏. We then deduce that 𝑆 starts on the 𝑏 ray, for otherwise, S r embeds in a tree attached to the interior of an 𝑎-loop. If 𝑆 ends on the 𝑏 ray, then 𝑆 is closed since the 𝑏-exponent sum of 𝑆 is 0. Finally, if 𝑆 ends in the interior of an 𝑎-loop, then as above, S r - 1 embeds in a tree attached to the interior of an 𝑎-loop, so S r is not closed. ∎

Each G s is finitely generated and residually finite. Moreover, we have G s ≇ G t when s ≠ t .

Proof

Let 𝔰 and 𝔱 be multiplying sequences. Suppose ϕ : G s → G t is an isomorphism. Denote generators for G s as a , b , a ¯ , b ¯ and that for G t as a ′ , b ′ , a ¯ ′ , b ¯ ′ . Denote A m = ⟨ ⟨ a m , a ¯ m ⟩ ⟩ ⊆ G s and A m ′ = ⟨ ⟨ a m , a ¯ m ⟩ ⟩ ⊆ G t .

By Proposition 5.2, loose elements a ± 1 , a ¯ ± 1 can be mapped to conjugates of ( a ′ ) ± 1 or ( a ¯ ′ ) ± 1 . For s 0 = 1 , since a = a ¯ , so ϕ ⁢ ( A m ) = A m ′ for all 𝑚. For s 0 > 1 , by abelianization of G s and G t , ϕ ⁢ ( a ) is conjugate to ( a ′ ) ± 1 and ϕ ⁢ ( a ¯ ) is conjugate to ( a ¯ ′ ) ± 1 , or ϕ ⁢ ( a ) is conjugate to ( a ¯ ′ ) ± 1 and ϕ ⁢ ( a ¯ ) is conjugate to ( a ′ ) ± 1 . Therefore, ϕ ⁢ ( A m ) = A m ′ for all 𝑚.

Suppose sequences s ≠ t first differ at index 𝑘, i.e. s j = t j for j < k and s k ≠ t k . Without loss of generality, let s k > t k . Using s j ∣ s k for j ≤ k , by Lemma 4.1,

If k = 0 , then Z ⌊ s 0 , t 0 ⌋ = Z t 0 . Thus, H 1 ⁢ ( G s / A s k ) ≇ H 1 ⁢ ( G t / A s k ′ ) . This is impossible.

If k ≥ 1 , then ⌊ s k , t k ⌋ < s k . Thus, ⊕ j = k ∞ Z s k / ⌊ s k , t j ⌋ ≇ 1 , so

This is impossible. ∎

A group 𝐺 generated by a finite set 𝑆 has solvable word problem if there is a computer program that can determine whether a word in 𝑆 represents 1 G . A subgroup 𝐻 of 𝐺 has solvable membership problem if there is a computer program that determines whether or not a word in 𝑆 represents an element in 𝐻. A sequence 𝔰 is computable if there is a computer program that outputs s n for input 𝑛. Unsolvable means not solvable and uncomputable means not computable.

There are uncountably many finitely generated residually finite groups with unsolvable word problem.

Proof

Consider groups G s . We claim that if 𝔰 is an uncomputable sequence, then H s has unsolvable membership problem. Assume there is a program that determines whether ( b n ⁢ a p ⁢ b - n ) ∈ H s . Then enumerating 𝑝 gives a program that outputs s n on input 𝑛.

We now show G s has unsolvable word problem if H s has unsolvable membership problem. Assume there is a program that determines whether a word is trivial in G s . Consider the involution ϕ : G s → G s with ϕ ⁢ ( a ) = a ¯ and ϕ ⁢ ( b ) = b ¯ . Then determining whether the word w ⁢ ϕ ⁢ ( w ) - 1 represents 1 G for any word 𝑤 in { a , b } solves the membership problem for 𝐻.

It thus suffices to prove that there are uncountably many uncomputable multiplying sequences 𝔰. There are countably many computable sequences since there are countably many (finite) computer programs. However, there are uncountably many multiplying sequences, e.g. let s n + 1 / s n ∈ { 2 , 4 } . ∎