On the residual finiteness and other properties of (relative) one-relator groups

A relative one-relator presentation has the form P =where X is a set, H is a group, and R is a group word on X and H. We show that if the group word on X obtained from R by deleting all the terms from H has what we call the unique max-min property, then the group defined by P is residually finite if and only if H is residually finite (Theorem 1). We apply this to obtain new results concerning the residual finiteness of (ordinary) one-relator groups (Theorem 4). We also obtain results concerning the conjugacy problem for one-relator groups (Theorem 5), and results concerning the relative asphericity of presentations of the form P (Theorem 6).


Introduction
The question of when one-relator groups are residually finite is still open. In the torsion-free case there are well-known examples of groups which are not residually finite, namely the Baumslag-Solitar/Meskin groups [4], [15]: where U, V do not generate a cyclic subgroup of the free group on x, and |l| = |m|, |l|, |m| > 1. On the other hand, there are some examples which are known to be residually finite. For instance, it was shown in [3] that if where U, V are positive words on an alphabet x and the exponent sum of x in UV −1 is 0 for each x ∈ x, or if where U, V are (not necessarily positive) words on x such that no letter x ∈ x appears in both U and V , then G = x; W is residually finite. In the torsion case there is the well-known open question: Question 1 [2], [5, Question OR1] Is every one-relator group with torsion residually finite? Question 1 is known to be true when G = x; W n where W is a positive word and n > 1 [9] (see also [19]). In [20], Wise obtains further related results, summed up by his "Quasi-Theorem 1.3": If W is sufficiently positive, and W n is sufficiently small cancellation, then G is residually finite.
A related open question is: Question 2 [5,Question OR6], [11,Question 8.68] If a torsion-free one-relator group G 1 = x; W is residually finite, then is G n = x; W n also residually finite for n > 1?
(Of course, if Question 1 is true, then Question 2 is trivially true.) It was shown in [1] that Question 2 holds true when W has the form (1) or (2). Here, amongst other things, we tackle Question 2 by considering relative presentations. A relative presentation has the form where H is a group and r is a set of expressions of the form The word is called the x-skeleton of R. We do not require that the x-skeleton is reduced or cyclically reduced. The group G = G(P) defined by P is the quotient of H * F (where F is the free group on x) by the normal closure of the elements of H * F represented by the expressions R ∈ r. The composition of the canonical imbedding H → H * F with the quotient map H * F → G is called the natural homomorphism, denoted by ν : H → G (or simply H → G).
As is normal, we will often abuse notation and write G = x, H; r , or G ∼ = x, H; r . When r consists of a single element R, then we have the one-relator relative presentation P = x, H; R .
Heuristically, G = G(P) should be governed by the "shape" of the x-skeleton of R and the algebraic properties of H.
Here we introduce the unique max-min property for the "shape" of W . (Words of the form (1) are a very special case.) For a group H, denote by M H the class of one-relator relative presentations of the form (5), where W has the unique max-min property.
is residually finite if and only if H is residually finite.
We can deduce from this Theorem 2 (Substitution Theorem). Let K be a one-relator group given by an ordinary presentation y, z; S(y, z) , and let P = x, H; R be an M H -presentation. Then the group given by the relative presentation x, y, H; S(y, R) is residually finite if and only if H and K are residually finite.
We can give the proof of this straightaway. Consider the M H * K -presentation P = x, H * K; Rz −1 . By Theorem 1, L = G(P) is residually finite if and only if H * K is residually finite, which is equivalent to requiring that both H and K are residually finite (using results discussed in [12] p417). Now note that L ∼ = x, y, z, H; S(y, z), Rz −1 ∼ = x, y, H; S(y, R) .
In particular, taking K to be defined by z; z n (n > 1) we have: Theorem 3. If G = x, H; R is a residually finite M H -group, then the group G n = x, H; R n (n > 1) is also residually finite.
Now take H to be a free group Φ. Then M Φ -groups are one-relator groups. Since Φ is residually finite ( [12],p116 or p417), we obtain the following theorem concerning residual finiteness of one-relator groups.
Theorem 4 Every M Φ -group G = x, Φ; R is a residually finite one-relator group. Moreover, if K = y, z; S(y, z) is a one-relator group, then the one-relator group K = x, y, Φ; S(y, R) is residually finite if and only if K is residually finite. In particular, G n = x, Φ; R n (n > 1) is residually finite.
The solution of the conjugacy problem for one-relator groups with torsion has been solved by B.B.Newman [16]. However, for the torsion-free case the problem is still open [5, Question O5].
Theorem 5 Every M Φ -group (Φ a finitely generated free group) has solvable conjugacy problem. Also, such groups have solvable power conjugacy problem.
(Two elements c, d of a group are said to be power conjugate if some power of c is conjugate to some power of d.) Other aspects of relative presentations (and in particular, one-relator relative presentations) have been studied intensively, particularly asphericity. Recall [6] that a relative presentation P is aspherical (more accurately, diagrammatically aspherical) if every spherical picture over P contains a dipole. Under a weaker condition on "shape" (the unique min property, or equivalently the unique max property) we can prove: Theorem 6 Let P be a relative presentation as in (5), where W has the unique min property. Then P is aspherical.
It then follows from [6] (see Corollary 1 of Theorem 1.1, Theorem 1.3, and Theorem 1.4) that for the group G = G(P) we have: (i) the natural homomorphism H → G is injective; (ii) every finite subgroup of G is contained in a conjugate of H; (iii) for any left ZG-module A, and any right ZG-module B, for all n ≥ 3.

Max-min property
Let x be an alphabet. A weight function on x is a function Let W be a word on x as in (4). Given a weight function θ, we then have the function (where φ(0) = 0 since the empty sum is taken to be 0). We will say that the weight function is admissible for W if φ(r) = 0. For visual purposes, it is useful to extend φ to a piecewise linear function φ : [0, r] → R, so that the graph of φ in the interval [j − 1, j] is the straight line segment joining the points (j − 1, φ(j − 1)), (j, φ(j)) (0 < j ≤ r). We will informally refer to this graph as "the graph of W " (with respect to θ).
A word W as in (4) will be said to have the unique max-min property if for some admissible strict weight function θ, the graph of W has a unique maximum and a unique minimum. To be precise, we require that, for some admissible strict weight function, and some k, l ∈ {1, 2, . . . , r}, we have φ(j) < φ(k) for all j ∈ {1, 2, . . . , r} − k and φ(j) > φ(l) for all j ∈ {1, 2, . . . , r} − {l}. We also require that x k = x k+1 and x l = x l+1 (subscripts modulo r). This amounts to requiring that W is "reduced at the unique maximum and minimum", that is, (subscripts modulo r). For at the maximum and minimum we must have either x j = x j+1 , or x j = x j+1 and ε j = −ε j+1 (j = k, l). If the two letters occurring at the unique maximum are not disjoint from the two letters occurring at the unique minimum (i.e. {x k , x k+1 } ∩ {x l , x l+1 } is not empty), then we will say that W has the strong unique max-min property.
A word W as in (4) will be said to have the unique min property if for some strict weight function θ, the graph of W has a unique minimum (but not necessarily a unique maximum). The unique max property is defined similarly, but is not really of interest because replacing θ by −θ will convert this property to the unique min property.
We let M 1 H (respectively S 1 H ) denote the subclass of M H consisting of relative presentations of the form (5) for which W has the unique max-min property (respectively, the strong unique max-min property) with respect to the weight function Proof. Let G = x, H; R with R as in (3), and suppose W = x ε 1 1 x ε 1 2 . . . x εr 2 has the unique max-min property with respect to some strict weight function θ : x → Z. We can assume θ(x) > 0 for all x. For if θ(x) < 0 then we can replace x by x −1 . Let LetĜ = x, H;R , whereR is obtained from R by replacing each occurrence of y ±1 by (y 1 y 2 . . . y θ(y) ) ±1 (y ∈ y). It is easy to see that thex-skeletonŴ ofR has the unique max-min property with respect to 1 :x → Z. (The graph ofŴ is obtained from that of W by "stretching" along the horizontal axis.) Moreover, G is embedded intoĜ, for we have the retraction ρ with section µ: LetR be obtained from R as follows. For each y ∈ y, replace all occurrences of y ±1 by (y 1 y 2 ) ±1 , and replace all occurrences of a ±1 (respectively, b ±1 , c ±1 , d ±1 ) by (ea) ±1 (respectively, (be) ±1 , (ec) ±1 , (de) ±1 ). LetĜ = x, H;R , and letŴ be the word obtained fromR by deleting all terms from H. The graph ofŴ under the weight function 1 :x → Z is the graph of W magnified by a factor of 2, and e occurs at the unique maximum and the unique minimum. Moreover, G is embedded intoĜ for we have the retraction ρ with section µ:Ĝ Remark 1 Note that in both the above proofs we have µν =ν, where ν : H → G, ν : H →Ĝ are the natural homomorphims. Thus ifν is injective then so is ν.
Remark 2 Note also from the proof of the above two lemmas we get that every M Hgroup is a retract of an S 1 H -group.

Remark 3
The referee has brought my attention to the work of K.S.Brown [8], which is concerned with whether a homomorphism χ from a one-relator group B = x; W (|x| ≥ 2, W as in (4) and cyclically reduced) onto Z has finitely generated kernel. Such a homomorphism is induced by a weight function θ which is admissible for W . However, since θ need not be strict, it is necessary to interpret the max-min property more widely.
Thus the unique maximum could be a "plateau": ie, for some k ∈ {1, 2, . . . , r} we could have φ(k) = φ(k + 1) and φ(j) < φ(k) for all j ∈ {1, 2, . . . , r} − {k, k + 1} (subscripts modulo r). Similarly, the unique minimum could be a "reverse plateau". Then according to Brown [8], as restated in Theorem 2.2 of [13], ker χ is finitely generated if and only if |x| = 2, and W has the unique max-min property in the above sense with respect to the corresponding weight function. In our work we could also allow non-strict weight functions. However, for the most part this can be avoided. For example, if the unique maximum is a plateau with x k = x k+2 then we could transform it to a genuine maximum by deleting x k+1 from x and replacing H by H * x k+1 . However, if the unique maximum is a plateau with x k = x k+2 then some of our arguments need to be modified, which we leave as an exercise for the reader.
In particular we have the (closed) paths (n, R).
There is an obvious action of Z on the above graph of groups, with i ∈ Z acting on vertices by i · n = i + n (n ∈ Z), and on the edges and vertex groups by i.(n, z) = (i + n, z) (n ∈ Z, z ∈ x ±1 ∪ H). This action of course extends to paths. Thus (i, α) = i.(0, α). In particular, (i, R) = i.(0, R), so Z acts onP.
If we regard P as a 2-complex of groups with a single vertex o, edges x ε (x ∈ x, ε = ±1), vertex group H, and defining path R, then we have a mapping of 2-complexes of groups ρ :P −→ P . This induces a homomorphism ρ * : π 1 (P, 0) −→ π 1 (P, o) = G which is injective, and Imρ * = K. This can easily be proved by adapting the standard arguments of covering space theory for ordinary 2-complexes (see for example [17] pp 157-159), to this relative situation.

Proof of Theorem 1
Since residual finiteness is closed under taking subgroups, it follows from Lemmas 1 and 2 and the Remark 1 at the end of §2 that it suffices to prove Theorem 1 for S 1 H -groups. We will make use of the following results: (a) A free product F * B, where F is a free group, is residually finite if and only if B is residually finite; (b) An infinite cyclic extension of a finitely generated group L is residually finite if and only if L is residually finite. (The first of these follows from results on p417 of [12]; the second is a special case of Theorem 7, p29 of [14]. ) We can assume x is finite. For if not let x ′ be the set of letters occurring in R. Then G is isomorphic to G ′ * Ψ where G ′ ∼ = x ′ , H; R , and Ψ is the free group on x − x ′ . So by (a) above, it is enough to work with G ′ .
Let G be defined by an S 1 H presentation as in (5), with e ∈ x occurring at both the unique maximum and the unique minimum of the graph of W under the weight function θ = 1. We denote the maximum and minimum values of φ W by M, m respectively. Note that m ≤ 0 ≤ M and m < M.
We first deal with the trivial case when M − m = 1. Then up to cyclic permutation and inversion, R = eha −1 h ′ , where a ∈ x − {e}, h, h ′ ∈ H. Thus G = Φ * H, where Φ is the free group on x − {e}, so the theorem holds by (a) above.
We have the epimorphism Also, we have the homomorphism Then ψη = id Z , so G is a semidirect product K ⋊ Z, where K = ker ψ, and with the action of n ∈ Z on K being induced by conjugation by f n . The fundamental group ofP (at the vertex 0), as in §3, is isomorphic to K.
We will obtain a relative presentation for K by collapsing a maximal tree. The edges (n, f ) ±1 form a maximal tree T in Γ. Let R n be the word on by deleting all edges from T which occur in (n, R) and replacing all terms (i, is a relative presentation for K. Moreover, since the edges in T constitute an orbit under the action of Z on our graph of groups, the action of Z on K is given by the automorphism Now consider the HNN-extension K of K given by the relative presentation Q = (n, x) (n ∈ Z, x ∈ x, x = f ), * n∈Z H n , s; R n (n ∈ Z) s(n, x)s −1 = (n + 1, x) (n ∈ Z, x ∈ x, x = e, f ), The automorphism µ of K can be extended to an automorphism µ of K by defining µ(s) = s. Then G = K ⋊ µ Z can be embedded into G = K ⋊ µ Z. By our assumption, up to cyclic permutation and inversion, (0, R) will have the form where h, h ′ ∈ H, ε = ±1, a, b ∈ x − {e}, and each term (i, z) occurring in the paths γ 0 , δ 0 is such that both its initial and terminal vertices lie in the range m + 1, m + 2, . . . , M − 1. Then R 0 = (M − 1, e)α 0 (m, e) ε β 0 where α 0 , β 0 do not contain any occurrence of (i, e) ±1 with i ≤ m or i ≥ M − 1. More generally, for n ∈ Z R n = (n + M − 1, e)α n (n + m, e) ε β n where α n , β n do not contain any occurrence of (i, e) ±1 with i ≤ n + m or i ≥ n + M − 1. Let F 0 be the free group on Then there is a homomorphism K → H * F 0 defined as follows: and (inductively), for k = 0, 1, 2, . . .
This homomorphism is actually an isomorphism. The inverse is defined by Thus G is an infinite cyclic extension of the group F 0 * H.

Remark 4
Note that by sending s to the generator 1 ∈ Z ⊂ G = K ⋊ µ Z, we obtain a retraction of G onto G (with section induced by the inclusion of K into K).
We can now complete the proof. Clearly the natural homomorphism from H into G is injective (and is thus injective into G). Hence if H is not residually finite then neither is G. It remains to show that if H is residually finite then so is G (and thus G). Case 1. If H is finitely generated then the result holds straight away by (a) and (b) above.
Case 2. Suppose that H is not finitely generated. For any homomorphism θ from H to a group H θ we obtain an induced homomorphism from G = (F 0 * H) ⋊ µ Z to G θ = (F 0 * H θ ) ⋊ µ Z which acts as θ on H and acts as the identity on F 0 and Z.

Proof of Theorem 5
Lemma 3 Let C be a group which is a retract of a group B. If B has solvable conjugacy (or power conjugacy) problem, then so does C.
Proof. By assumption we have maps B Thus the result follows. Now it is shown in [7] that infinite cyclic extensions of finitely generated free groups have solvable conjugacy, and power conjugacy, problem. By Remarks 2, 4, every M Φgroup is a retract of such a group.

Proof of Theorem 6
We will assume familiarity with the terminology in § §1.2, 1.4 of [6]. As in Lemma 1, we can assume that θ(x) > 0 for all x. We can extend θ to any word U = y ε 1 1 y ε 2 2 . . . y εs s , (s > 0, y i ∈ x, ε i = ±1, 1 ≤ i ≤ s) by θ(U) = s i=0 ε i θ(y i ). Let P be a based connected spherical picture (with at least one disc) over P, with global basepoint O, and basepoint O ∆ for each disc ∆. (Note that since R is not periodic, there will be just one basepoint for each disc.) We will also choose, for each region R, a point O R in the interior of R.
We can relabel P to obtain a pictureP overP as follows: (a) For each region R, choose a tranverse path γ R from O to O R , and let U R (a word on x) be the label on the path γ R . Then the potential q(R) of R is θ(U R ). (This is independent of the choice of path γ R , since θ(W ) = 0.) (b) For an arc tranversely labelled x ∈ x say, relabel it by (q(R), x) where R is the region where the tranverse arrow on the arc begins.
(c) For a corner of a disc, with label h ∈ H say, relabel the corner by (q, h), where q is the potential of the region in which the corner occurs.
For a disc ∆, let q ∆ be the potential of the region containing O ∆ . Then in the relabelled picture, ∆ will be labelled by the path (q ∆ , R).
Let Θ be a minimal disc, that is, a disc such that q Θ ≤ q ∆ for all discs ∆. Let m be the minimum value of φ θ W , and let e be one of the two distinct letters occurring at the unique minimum. Then in the path (0, R) there is a unique edge labelled (m, e). Now Θ is labelled by (q Θ , R) inP, and thus there is a unique edge labelled (m + q Θ , e) incident with Θ. This arc must intersect another disc Θ ′ , which must also be labelled by (q Θ , R), but with the opposite orientation. Thus we obtain a dipole inP where Θ, Θ ′ are the discs of the dipole. Reverting to P, this dipole inP gives rise to a dipole in P.