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BY 4.0 license Open Access Published by De Gruyter October 1, 2020

Obstruction to a Higman embedding theorem for residually finite groups with solvable word problem

Emmanuel Rauzy
From the journal Journal of Group Theory


We prove that, for a finitely generated residually finite group, having solvable word problem is not a sufficient condition to be a subgroup of a finitely presented residually finite group. The obstruction is given by a residually finite group with solvable word problem for which there is no effective method that allows, given some non-identity element, to find a morphism onto a finite group in which this element has a non-trivial image. We also prove that the depth function of this group grows faster than any recursive function.


It is well known that there can be no Higman embedding theorem for recursively presented finitely generated residually finite groups, that is to say, not all finitely generated recursively presented residually finite groups embed into finitely presented residually finite groups. Indeed, a theorem of McKinsey [9, 11, 3] states that all finitely presented residually finite groups have solvable word problem, while on the other hand several recursively presented residually finite groups are known that fail to have solvable word problem: for instance, one example was constructed by Meskin in [10], and one by Dyson in [4].

It was unknown whether or not the condition of having solvable word problem is sufficient for such embeddings to exist. For instance, in the article [7], the authors ask whether “unsolvability of the word problem is the only obstacle” to embed recursively presented residually finite groups into finitely presented residually finite groups. We answer that question negatively.

Theorem 1

There exists a finitely generated residually finite group with solvable word problem, that does not embed in any finitely presented residually finite group.

Call a group 𝐺effectively residually finite if there is an algorithm that takes a word 𝑤 on the generators of 𝐺 as input and, if we in 𝐺, produces a morphism 𝜑 from 𝐺 to a finite group 𝐹 that satisfies φ(w)e. What we actually prove here is the following theorem.

Theorem 2

There exists a finitely generated residually finite group with solvable word problem, that is not effectively residually finite.

Theorem 1 then follows from the following facts.

Fact 3

A finitely presented residually finite group is effectively residually finite.

Fact 4

A finitely generated subgroup of an effectively residually finite group is itself effectively residually finite.

Fact 3 follows from McKinsey’s algorithm. It comes from the fact that a finitely presented group has computable finite quotients; see [12]. Fact 4 is straightforward.

In the first section, Dyson’s groups are introduced, and they are used to reduce the proof of Theorem 2 to a problem about subsets of ℤ. In the second section, this problem is solved. In the last section, we show that the group constructed in order to prove Theorem 2 has a depth function that grows faster than any recursive function.

Following [4], throughout, recursively presented groups will be called re groups (for recursively enumerable), and groups in which there is an algorithm that recognizes non-trivial elements will be called co-re groups. A group has solvable word problem if and only if it is re and co-re.

1 Dyson’s groups

The author already used Dyson’s groups to investigate the property of having computable finite quotients [12]. What we construct here is a strengthening of the result obtained in that article; we will include here all definitions but omit the proofs that already appear there.

Dyson’s groups are amalgamated products of two lamplighter groups.

The lamplighter group 𝐿 is the wreath product of ℤ and Z/2Z, noted ZZ/2Z, which is by definition the semi-direct product ZZZ/2Z, where ℤ acts on ZZ/2Z by permuting the indices. It admits the following presentation:


The element aiεa-i of 𝐿 corresponds to the element of ZZ/2Z with only one non-zero coordinate in position iZ. We call it ui. Consider another copy L^ of the lamplighter group; with an isomorphism, we write gg^. For each subset 𝒜 of ℤ, define L(A) to be the amalgamated product of 𝐿 and L^, with ui=aiεa-i identified with u^i=a^iε^a^-i for each 𝑖 in 𝒜. It has the following presentation:


Whether the group L(A) is residually finite depends on whether the set 𝒜 is closed in the profinite topology on ℤ. The profinite topology on a group 𝐺, which we denote by PT(G), is the topology defined by taking cosets of finite index normal subgroups as a basis for the open sets. Thus a subset 𝒜 of ℤ is open in PT(Z) if and only if, for every 𝑛 in 𝒜, there exists an integer 𝑝 such that n+pZA. The profinite topology on ℤ was rediscovered in 1955 by Furstenberg in [5] (the profinite topology on an arbitrary group was defined in 1950 in [6]), where it is used to give an elegant proof of the existence of infinitely many primes, and it was proven in [8] that this topology comes from a metric, which is given by the following formula:


The norm x is thus the reciprocal of the greatest integer 𝑛 with the property that 1,2,,n all divide 𝑥. A sequence converges to 0 in that topology if and only if, for any integer 𝑘, there exists a point beyond which 𝑘 divides all terms of the sequence. For instance, n! goes to 0 as 𝑛 goes to infinity. Define a function 𝜃 on the natural numbers by θ(n)=lcm{1,2,3,,n}. θ(n) is the smallest non-zero natural number such that θ(n)1n. The closed ball of radius 1n and centered in 𝑥, which is the set {yZ,d(x,y)1n}, is simply the set x+θ(n)Z. It is in fact also open. Call B¯(x,r) the closed ball centered in 𝑥 and of radius 𝑟, and B(x,r) the corresponding open ball (but note that the latter is not the interior of the former). The distance 𝑑 is in fact ultrametric: for 𝑥, 𝑦 and 𝑧 integers, one has d(x,z)max(d(x,y),d(y,z)). This implies that each point of a ball can be taken as its center, and thus that if two balls intersect, one is contained in the other.

Of course, the distance 𝑑 is effective: 𝑑 is a recursive function. This implies that both the closed and open balls of PT(Z) are recursive sets.

Call an open set 𝑂effectively open (in PT(Z)) if there is an effective procedure that, given any number 𝑛 in 𝑂, computes some integer 𝑝 such that n+pZ is contained in 𝑂. A set is effectively closed if its complement is effectively open.

We can now state the properties of the group L(A) that will allow us to prove Theorem 2.

Proposition 5

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 PT(Z).

  3. L(A) is effectively residually finite if and only if 𝒜 is co-re and effectively closed in PT(Z).

(1) and (2) were proved by Dyson in [4]. See also [12]. We prove here only the third point.


Suppose first that L(A) is effectively residually finite. Then L(A) is co-re, and by (1), 𝒜 is co-re as well. Let 𝑥 be an integer in the complement of 𝒜 such that uxu^x-11 in L(A). By our hypothesis, we can effectively find a morphism 𝜑 from L(A) to a finite group 𝐹, with φ(uxu^x-1)1 in 𝐹. Call 𝑁 the product of the orders of the images of 𝑎 and a^ in 𝐹. We then claim that A(x+NZ)=. Indeed, if it were not the case, there would exist an integer 𝑘 such that x+kNA, that is to say, such that ux+kN=u^x+kN in L(A). But then, this would imply


Because it can similarly be proved that φ(u^x)=φ(u^x+kN), this contradicts the assumption that φ(uxu^x-1)1. Thus 𝒜 does not meet x+NZ.

The proof of the converse follows closely the proof of (2) given in [12] (which differs from the original proof of Dyson), as one only needs to see that the hypothesis that 𝒜 is effectively closed in PT(Z) is enough to effectively carry out that proof. As we actually only use the first implication of (3) in the proof of Theorem 2, no further details are given here. ∎

From Proposition 5, to prove Theorem 2, it suffices to build 𝒜 with the following properties: 𝒜 is recursive, 𝒜 is closed, but not effectively so.

2 Construction in ℤ

Lemma 6

There exists a recursive subset 𝒜 of ℤ, closed in PT(Z), but not effectively so.


We construct a set ℬ, which will be the complement of the announced 𝒜. Thus it has to be recursive and open but not effectively open.

Call pn the 𝑛-th prime number. Define a sequence (tn)n0 by tn=p1pn. This sequence is defined so that pk divides tn if and only if kn. Note also that tn divides tn+1.

Consider an effective enumeration of all Turing machines: M1 is the first machine, M2 is the second, …. We will build ℬ as a disjoint union of open sets Xn, each Xn being a neighborhood of tn defined thanks to a run of the 𝑛-th Turing machine Mn. If this machine does not halt, Xn is a closed ball centered at tn of radius 1tn+1. If it halts, it is a finite union of balls, one of which is centered at tn, the radius of which depends of the number of steps needed for Mn to halt. Thanks to this, information of the form “Xn contains a ball of radius 𝑟 centered in tn” will translate into “if Mn halts, it does so in less than 𝑁 steps”, where 𝑁 can be computed from 𝑟.

Initialize Xn={tn}. Call rn=1tn+1 and m=θ(tn+1) so that the closed ball B¯(tn,rn) is the set tn+mZ. Start a run of the machine Mn.

After each computation step of Mn, note 𝑘 the number of steps already done in the computation and add, to Xn, tn+km, tn-km, as well as open balls centered in those numbers that are contained in B¯(tn,rn) and do not contain tn. That is, we replace Xn by


Because the distance 𝑑 is ultrametric, both balls B(tn+km,12d(tn,tn+km)) and B(tn-km,12d(tn,tn-km)) are contained in B¯(tn,rn).

If, at some point, the machine Mn halts, Xn consists of tn and of finitely many open balls centered at points tn±km. By construction, a point from Xn{tn} is at distance at least infk{12d(tn,tn±km)} from tn. This infimum can be computed; call it 𝑟. Then compute the smallest natural number 𝑦 such that d(tn,y)<r, and call r the distance d(tn,y). We then add to Xn the ball B(un,r). This implies that Xn cannot contain any ball of center tn and of radius strictly greater than r because it does not contain 𝑦. In particular, any ball centered in tn that contains one of the elements of the form tn±km that were added to Xn is not contained in Xn.

Of course, if the machine Mn does not halt, Xn will be the whole ball B¯(tn,rn).

This ends the definition of Xn, and ℬ is defined as the union Xn. This union is disjoint because, by the choice of the radius rn, any element of Xn is divisible by pn, but none is divisible by pn+1. We now prove that ℬ defined this way satisfies all three properties that appear in the statement of this lemma.

ℬ is clearly open, because each Xn is open, whether or not the machine Mn halts.

ℬ is a recursive set. It is obviously recursively enumerable because it was defined by an effective enumeration. To see that it is also co-re, let 𝑥 be an integer; we want to decide whether 𝑥 belongs to ℬ. By looking at the prime decomposition of 𝑥, one can find up to one 𝑛 such that 𝑥 might belong to Xn. Because Xn is always contained in B¯(tn,rn), if d(tn,x)>rn, 𝑥 cannot be in Xn. Otherwise, it belongs to B¯(tn,rn)=tn+mZ, and we can find 𝑘 such that x=tn+km. Then, if a run of Mn lasts more then 𝑘 steps, automatically, 𝑥 will belong to Xn. On the other hand, if Mn stops in less than 𝑘 steps, Xn can be determined explicitly as a finite union of open balls, and thus the question of whether 𝑥 belongs to Xn can be settled. Because the problem “does Mn halt in more then 𝑘 steps” is a computable one, in either case, we will be able to determine whether or not 𝑥 belongs to ℬ.

Finally, ℬ is not effectively open. Suppose we have an algorithm that gives, for 𝑥 in ℬ, an integer 𝑘 such that x+kZ is contained in ℬ. Applying it to tn, we can find a radius 𝑟 such that B(tn,r) is contained in ℬ. We will show that this information implies a new information of the form: if the machine Mn halts, then it halts in less than 𝑁 steps. This would of course allow one to solve the halting problem; thus such an algorithm does not exist.

Indeed, we have seen that if Mn halts in 𝑁 steps, Xn cannot contain any ball centered in tn that contains an element of the form tn±km, with kN. Turning this around, computing 𝑁 such that tn+mN belongs to B(tn,r) (for instance, N=θ(1r+1)), the information “B(tn,r) is contained in ℬ” implies that either Mn does not halt, or it halts in less then 𝑁 steps.

This ends the proof of Lemma 6. ∎

What we do not know yet is whether the condition of being recursively presented and effectively residually finite, which is necessary to be a subgroup of a finitely presented residually finite group, is also sufficient. For instance, the strictly stronger condition of having computable finite quotients is not known to be unnecessary (see [12]).

3 Non-recursive depth function

In [1], Bou-Rabee introduced the residual finiteness growth function, or depth function, ρG, of a residually finite group 𝐺. To a natural number 𝑛, ρG associates the smallest number 𝑘 such that, for any non-trivial element of length at most 𝑛 in 𝐺, there exists a finite quotient of 𝐺 of order at most 𝑘 such that the image of this element in that quotient is non-trivial.

A finitely generated subgroup 𝐻 of a finitely generated residually finite group 𝐺 must have a depth function ρH that grows more slowly than that of 𝐺 (see [1]). Because it is easy to see that a finitely presented residually finite group always has a recursive growth function, a subgroup of a finitely presented residually finite group must have its depth function bounded above by a recursive function. Note that, in [7], it was shown that finitely presented residually finite groups can have arbitrarily large recursive depth function. It was also known that the depth function of a residually finite group could grow arbitrarily fast (see [2]).

Note that an effectively residually finite group with solvable word problem always has its depth function bounded above by a recursive function.

Proposition 7

Let 𝒜 be the subset of ℤ given in Lemma 6. The depth function ρL(A) of L(A) cannot be smaller than a recursive function.


Suppose there exists a recursive function 𝑓 such that ρL(A)f. Then, for each 𝑛 which is not in 𝒜, unu^n-1 is a non-identity element of L(A), and thus it has a non-trivial image in a finite quotient 𝐹 of size at most f(4n+2) (because un and u^n are of word length 2n+1). Now the orders of 𝑎 and a^ in 𝐹 both divide f(4n+2)!; thus 𝐹 is a quotient of the quotient of L(A) given by adding to it the two relations


Thus, in this group as well, unu^n-1 is a non-identity element, and we know that this implies that n+f(4n+2)!Z does not meet 𝒜 (see the proof of Proposition 5). This shows that 𝒜 is effectively closed, contradicting our hypothesis. ∎

  1. Communicated by: Pavel A. Zalesskii


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Received: 2020-02-22
Revised: 2020-09-16
Published Online: 2020-10-01
Published in Print: 2021-05-01

© 2020 Rauzy, published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.