2000] 20F28, 20F06, 20F67 20E26, 20E22,

EVERY GROUP IS AN OUTER AUTOMORPHISM GROUP OF A FINITELY GENERATED GROUP

INNA BUMAGIN AND DANIEL T. WISE

Abstract. We show that every countable group Q is isomorphic to Out(N) where N is a finitely generated subgroup of a countable C′( \frac{1}{6} ) small-cancellation group G. Furthermore, when Q is finitely presented, we can choose G to be finitely presented and residually finite.

1. Introduction

While many groups arise as automorphisms of algebraic objects, it is not true that every group Q arises as Aut(N) for some group N. Indeed, it is well-known that a nontrivial cyclic group of odd order is not isomorphic to Aut(N) for any group N. The situation for Out(N) is markedly different and the question of whether a given group Q can be realized as an outer automorphism group of some group N has attracted some recent attention.

In [Mat89] it was shown that any group Q is isomorphic to Out(N) where N is the fundamental group of a graph of groups but the cardinality of the generating set of N is greater than the cardinality of the generating set of Q. The problem becomes more complicated if N is required to belong to a special class of groups. Any Q is isomorphic to Out(N) where N is simple [DGG01], or where N is torsion-free metabelian with trivial center [GP00]. Any countable Q is isomorphic to Out(N) where N is a locally finite p-group [DG93]. Finally, when Q is finite, N can be chosen to be the fundamental group of a closed hyperbolic 3-manifold [Koj88].

The first main result in this paper is Theorem 11 where we show that any countable Q is isomorphic to Out(N) where N is a finitely generated subgroup of a finitely generated C′(\frac{1}{6}) group G. The second main result is Theorem 15 where we show that if Q is finitely presented, N can be chosen to be a finitely generated subgroup of a finitely presented residually finite C′(\frac{1}{6}) group. We note that in analogy with Kojima’s result, when Q is finite, the group N produced by our construction is the fundamental group of a finite C′(\frac{1}{6}) complex.

The main idea is to use a variant of Rips’s construction [Rip82] in order to obtain a short exact sequence 1 → N → G → Q → 1, where G is a C′(\frac{1}{6}) group and N is finitely generated. As observed in [Wisb], the natural homomorphism Q → Out(N) is easily shown to be injective in this case. The main difficulty is to choose G so that there are no “surprise” outer automorphisms of N, and so Q → Out(N) is surjective. A criterion for establishing this surjectivity is given in Lemma 9 which is the central underlying technical result in this paper.

Date: October 9, 2002.

1991 Mathematics Subject Classification. [.

Key words and phrases. Outer automorphism groups, residually finite, small cancellation.

Research supported by NSERC grant.

We close this introduction by discussing the extent to which our results can be generalized. Since there are uncountably many finitely generated groups Q, it is not true that every such Q arises as Out(N) where N is a subgroup of a finitely presented group. Nevertheless we pose the following problem which would generalize our two main results:

Problem 1. Is every countable group Q isomorphic to Out(N) where N is finitely generated and residually finite?

2. Review of small-cancellation theory

A piece in a presentation ⟨a_{1}, a_{2}, ⋯ | R_{1}, R_{2}, … ⟩ is a word U such that US_{1} ≠ US_{2} and both (US_{1})^{±1} and (US_{2})^{±1} are cyclic permutations of relators.

The presentation satisfies C′(α) if for any piece U which is a subword of the cyclic word UR_{j}, we have |U| < α|R_{j}| where we use |X| to denote the length of the word X. We will also say that the set of words {R_{1}, R_{2}, … } in the free group ⟨a_{1}, a_{2}, ⋯ | −⟩ satisfies C′(α) if the presentation above satisfies C′(α). We say that G is a C′(α) group if G has a presentation that satisfies the C′(α) condition.

We will need the following well-known facts about the metric small cancellation groups whose presentations satisfy a C′(α) condition. The first theorem is the key result in small cancellation theory.

Proposition 2. [LS77, Chapter V, Theorem 4.4] If W is a freely reduced word which represents the identity element in a C′(\frac{1}{6}) group, then W contains a subword V such that V S is a conjugate of some relator or its inverse, and S is the concatenation of at most three pieces.

A proof of the following theorem for C′(\frac{1}{8}) groups, based on an idea of Lipschutz [Lip62] was given by Lyndon and Schupp [LS77, Chapter V, Theorem 10.1]. For C′(\frac{1}{6}) groups it was proved by Greendlinger [Gre60].

Proposition 3. Every torsion-element in a C′(\frac{1}{6}) small-cancellation group is a power of a conjugate of an element W, where W^{n} is a relator.

Greendlinger proved for C′(\frac{1}{8}) groups [Gre62], and later for C′(\frac{1}{6}) groups [Gri66], that in these groups, any two elements that commute necessarily belong to a cyclic subgroup. Truffault [Tru74] and Seymour [Sey74] independently, strengthened the result of Greendlinger and showed that in C′(\frac{1}{6}) groups, centralizers of non-trivial elements are cyclic. From this latter result we observe that:

Corollary 4. Let G be a C′(\frac{1}{6}) group, and let N be a non-cyclic subgroup of G. Then the centralizer of N in G is trivial.

Proof. Let N = ⟨h_{1}, h_{2}, ...⟩. Then C_{G}(N) ⊂ C_{G}(h_{1}) ∩ C_{G}(h_{2}) ∩ ... Let d ∈ C_{G}(h_{1}) ∩ C_{G}(h_{2}) ∩ ... hence h_{i} ∈ C_{G}(d) for all i. If d ≠ 1, then C_{G}(d) is cyclic so that H is cyclic, a contradiction. Thus, C_{G}(N) = {1}.

2.1. Construction of special small-cancellation words.

Definition 5. An x, y word is special if it is a positive word in x, y with no x^{2} or y^{3} subwords.

It is easy to provide sets of special words satisfying stringent small cancellation conditions. The following lemma provides us with examples of such sets; we will use these examples in our proof.

Lemma 6. Consider the infinite word W = xy(xy^{2})xy(xy^{2})^{2}xy(xy^{2})^{3} …

1. for each infinite sequence {n_{1}, n_{2}, … } of natural numbers, one can find an infinite family W_{∞} = {w_{1}, w_{2}, … } of words which satisfies the C′(\frac{1}{20}) condition, so that |w_{i}| ≥ n_{i} for each i.

2. for each positive integer m and a large enough integer p one can find a family V_{m} = \{v_{1}, v_{2}, …, v_{m}\} of words which satisfies the C′(\frac{1}{20}) condition, so that given a subset I of {1, 2, …, m}, the lengths of the elements of V_{m} are as follows: |v_{i}| = p for each i ∈ I and |v_{i}| = p + 1 for each i ∉ I.

Proof. To prove the first assertion, let

w_{i} = xy(xy^{2})^{k_{i}}xy(xy^{2})^{k_{i}+1} … xy(xy^{2})^{2k_{i}}, (1)

where k_{1} ≥ max\{60, \sqrt{\frac{2}{3}n_{1}}\} and k_{i} ≥ max\{2k_{i−1}, \sqrt{\frac{2}{3}n_{i}}\} be the words in W_{∞}. Obviously, w_{i} is a special x, y word of length |w_{i}| = \frac{9}{2}k_{i}^{2} + 5k_{i} + 1 > n_{i}, and the length of the maximal piece p_{i} = y(xy^{2})^{2k_{i}−1}xy(xy^{2})^{2k_{i}}xy of w_{i} equals |p_{i}| = 12k_{i} +2. Hence W_{∞} satisfies the C′(\frac{1}{20}) condition, and we have proven the first assertion.

To prove the second assertion, consider the finite subset

W_{m} = {w_{2}, w_{3}, …, w_{m+1}} ⊂ W_{∞}

chosen according to part (1) of this lemma (we set n_{i} = 1 for all i). We modify the words w_{i} ∈ W_{m} as follows. Let p > |w_{m+1}| be an integer number. A “correcting” exponent for i = 2, 3, …, m + 1, is the maximal integer c_{i} which does not exceed \frac{p−|w_{i}|}{2(k_{i}+1)}. Define

v_{i} = (xy)^{c_{i}}(xy^{2})^{k_{i}}(xy)^{c_{i}}(xy^{2})^{k_{i}+1} … (xy)^{c_{i}}(xy^{2})^{2k_{i}}x^{ϵ_{1}^{(i)}}(xy)^{ϵ_{2}^{(i)}}(xy^{2})^{ϵ_{3}^{(i)}},

where ϵ_{1}^{(i)}, ϵ_{3}^{(i)} ∈ {0, 1} and ϵ_{2}^{(i)} is a nonnegative integer, so that |v_{i}| = p. Note that the length of the correcting term x^{ϵ_{1}^{(i)}}(xy)^{ϵ_{2}^{(i)}}(xy^{2})^{ϵ_{3}^{(i)}} does not exceed 2(k_{i} + 1).

We claim that V_{m} = \{v_{1}, v_{2}, …, v_{m}\} satisfies the C′(\frac{1}{20}) condition. Indeed, the values of c_{i} are all different, hence the maximal length of a piece p_{i} which is a subword of v_{i} is bounded by |p_{i}| ≤ |(xy^{2})^{2k_{i}−1}(xy)^{c_{i}}(xy^{2})^{2k_{i}}| < 2c_{i} + 12k_{i}. Since |v_{i}| > \frac{9}{2}k_{i}^{2} + 2c_{i}k_{i} + 2c_{i}, we have

\frac{|v_{i}|}{|p_{i}|} > \frac{\frac{9}{2}k_{i}^{2} + 2c_{i}k_{i} + 2c_{i}}{2c_{i} + 12k_{i}} = \frac{\frac{9}{4} \frac{k_{i}}{c_{i}} k_{i} + k_{i} + 1}{1 + 6\frac{k_{i}}{c_{i}}}.

It can be readily seen that \frac{|p_{i}|}{|v_{i}|} < \frac{1}{20} for all i > 1. The claim follows.

3. Outer automorphism groups and short exact sequences

The plan is to use the natural homomorphism Q → Out(N) which exists for any short exact sequence

1 → N → G → Q → 1 (2)

Specifically, G acts by conjugation on N, so there is a homomorphism ϕ̂ : G → Aut(N), and obviously ϕ̂(N) ⊂ Inn(N), so there is an induced homomorphism ϕ : Q → Aut(N)/ Inn(N) = Out(N). Our goal is as follows: for a given group Q, find G and N in (2) so that

ϕ : Q → Out(N) (3)

is an isomorphism.

3.1. The map ϕ is injective. We will prove that ϕ is injective by applying the following simple criterion which holds for an arbitrary extension (2).

Lemma 7. If the centralizer C_{G}(N) of N in G is contained in N, then ϕ is injective.

Proof. The kernel of ϕ is represented by elements g ∈ G which act on N like an inner automorphism, say conjugation by n ∈ N. Specifically, for each x ∈ N, we have x^{n} = x^{g} and so x^{ng^{−1}} = x. Thus ng^{−1} is in the centralizer of N, and so ng^{−1} ∈ N, and hence g ∈ N as claimed. Thus g represents a trivial element of Q.

In our construction, G will be a non-cyclic small cancellation group and N will be a 2-generated subgroup of G. Therefore, by Corollary 4, the centralizer of N is trivial, so it is obviously contained in N, and we have the following statement.

Lemma 8. If the group G in (2) is non-cyclic and C′(\frac{1}{6}), then ϕ is injective.

3.2. The map ϕ is surjective. Given a group Q presented by ⟨A | R⟩, where X = {a_{i}}_{i∈I} and R = {R_{j}}_{j∈J} , we form a presentation

⟨x, y, A | x^{a_{i}} = A_{i}, x^{a_{i}^{−1}} = B_{i}, y^{a_{i}} = C_{i}, y^{a_{i}^{−1}} = D_{i}, i ∈ I (4)

R_{j} = E_{j}, j ∈ J
x^{p}, y^{p}, xyxy^{2}xy^{3} … xy^{q}⟩.

Letting G denote the group presented above, we obtain a short exact sequence 1 → N → G → Q → 1 where N = ⟨x, y⟩. The first four types of relations guarantee that N is normal in G, and the R_{j} = E_{j} relations guarantee that G/N ≅ Q. The last three relations are added to facilitate the arguments determining Out(N) that we shall make below. Of course, a priori, there is no reason to conclude that N is nontrivial.

In order to obtain a group G with the desired properties, we use small cancellation techniques.

Lemma 9. If Presentation (4) for the group G in (2) satisfies the following conditions, then the map ϕ : Q → Out(N) is surjective.

1. Presentation (4) satisfies the C′(\frac{1}{11}) condition.

2. The relations of G are chosen so that A_{i}, B_{i}, C_{i}, D_{i} and E_{i} are special x, y words, and each such special word has length ≥ \frac{30}{31} of its corresponding relator.

3. The exponent q is an integer satisfying q > 45.

4. The exponent p is a number satisfying p > 2q.

Proof. Pick α ∈ Aut(N). Since the only relators in G which are proper powers are x^{p} and y^{p}, by Proposition 3, the only torsion elements in G are conjugates of powers of x or y. Therefore, α(x) = g_{1}^{−1}z_{1}^{p_{1}}g_{1}, and α(y) = g_{2}^{−1}z_{2}^{p_{2}}g_{2} where each g_{i} ∈ G, each z_{i} ∈ {x, y}, and 0 ≤ p_{i} < p for each i = 1, 2.

The subgroup N has many more relations than just x^{p} and y^{p} and in particular, is not a free product. For instance W = xyxy^{2}xy^{3} … xy^{q} is a relation which does not hold between torsion elements in a free product of cyclic groups. We shall first prove that g_{1} = g_{2}, in other words we show that the image of α in Out(N) is conjugation by g_{1} (or g_{2}) pre-composed by a map θ that sends x to some power of x (or y), and sends y to some power of y (or x). We shall later prove that θ ∉ Aut(N) unless θ(x) = x and θ(y) = y. Our arguments show that in all the other cases the subgroup ⟨α(x), α(y)⟩ is isomorphic with ⟨α(x)⟩ ∗ ⟨α(y)⟩ and therefore cannot be isomorphic to N.

We argue as follows. We may assume that the words g_{1}^{−1}z_{1}^{p_{1}}g_{1} and g_{2}^{−1}z_{2}^{p_{2}}g_{2} are freely reduced. Suppose that g_{1} ≠ g_{2}. If g_{1}g_{2}^{−1} = z_{1}^{q_{1}}z_{2}^{q_{2}} for some integers q_{1}, q_{2}, then g_{1} = z_{1}^{q_{1}}z_{2}^{q_{2}}g_{2}, hence g_{1}^{−1}z_{1}^{p_{1}}g_{1} = g_{2}^{−1}z_{2}^{−q_{2}}z_{1}^{p_{1}}z_{2}^{q_{2}}g_{2}. We see that α is conjugation by g_{2} pre-composed by conjugation by z_{2}^{q_{2}} and by the map θ; in this case, we can proceed to the next step. Therefore, we may assume that g_{1}g_{2}^{−1} ≠ z_{1}^{q_{1}}z_{2}^{q_{2}} for any integers q_{1}, q_{2}. Consider the free group F = ⟨x̃, ỹ, ã_{1}, ã_{2}, ⋯ | −⟩ and the homomorphism η : F → G that sends x̃ to x, ỹ to y and ã_{i} to a_{i} for each i ∈ I. Pick the shortest g̃, z̃_{1}, z̃_{2} ∈ F so that η(g̃) = g_{1}g_{2}^{−1}, η(z̃_{i}) = z_{i}, and consider the subgroup K̃ of F generated by z̃_{1} and g̃z̃_{2}g̃^{−1}. We claim that the image K = η(K̃) in G is a free product of two cyclic subgroups and therefore cannot be isomorphic to N. Indeed, consider the graph Γ (see [Sta83]) illustrated in Figure 1 corresponding to the generators of K̃. The graph Γ is represented by the bold part of Figure 1. It has two closed paths (labelled by z̃_{1}^{p_{1}} and z̃_{2}^{p_{2}}), which are Figure 1. α(N) is the fundamental group of the bold graph. The thin path represents a relator V S

connected by an arc labelled by g̃. In the product g̃z̃_{2}g̃^{−1}, absorb the last z̃_{2} syllable of g̃ if necessary, and change the base point of Γ if g̃ starts with z̃_{1}. By the above observations, we can assume that g̃ starts with a letter different from z̃_{1} and ends by a letter different from z̃_{2}. If the intersection K̃ ∩ker(η) is not trivial, then by Proposition 2, there is k̃ ∈ K̃ that contains a subword V such that V S is a conjugate of some relator or its inverse, and S is the concatenation of at most three pieces. In particular, this large part V of a relator is represented by a path in the graph. Unless V is either x^{n} or y^{n} for some 0 < n < p, the word V contains a long sequence of alternating powers of x and y. Therefore, the path V necessarily travels through the entire arc g̃. We distinguish the following two cases.

1. V traverses g̃ at most once (Figure 1).

2. V traverses g̃ more than once. In the first case, g̃ is not a geodesic. Indeed, add an arc labelled by S to Γ so as to obtain the relator U = V S as the label of a closed path in the resulting graph. Since V traverses g̃ at most once, the complement of g̃ in R consists of at most z_{1}^{2}, z_{2}^{2}, S and two pieces of U. Therefore,

|g̃| ≥ |U| − |S| − 4 − \frac{2}{11}|U| > |U| − 4 − \frac{5}{11}|U| > \frac{5}{11}|U|

the last inequality holds for U > 44. Hence, g̃ can be replaced by the shorter path which consists of S and two pieces.

In the second case, observe that our choice of the relations implies that V cannot travel through the entire arc g̃ more than twice. Indeed, each occurrence of g̃ in V is followed by an occurrence of g̃^{−1}. Since both x and y appear in the relators only with positive exponents, and g̃ is nontrivial, g̃ necessarily contains ã_{i}^{±1} for some i. The structure of relators implies that

V = g̃_{1}z_{i_{1}}^{±2}g̃^{±1}z_{i_{2}}^{±2}g̃^{∓1}z_{i_{1}}^{±2}g̃_{2}

where g̃_{1} and g̃_{2} are subpaths of g̃. Observe that g̃, g̃_{1} and g̃_{2} are pieces. Altogether we have |V | < \frac{4}{11}|U| + 6 and since 6 < \frac{1}{11}|U|, we obtain the following inequality |V | < \frac{5}{11}|U| which contradicts Proposition 2. Therefore, each k̃ ∈ K̃ ∩ ker(η) contains more than half of x^{r} or y^{r}. Let w̃ ∈ F be a word of length at least two in the free product ⟨z̃_{1}⟩ ∗ ⟨g̃z̃_{2}g̃^{−1}⟩, and let w̃ = w̃_{1} … w̃_{s} be its normal form. Our arguments imply that η(w̃) is nontrivial in G unless η(w̃_{i}) = 1_{G} for each i = 1, …, s. The claim follows.

We have shown that α(x) = g_{1}^{−1}z_{1}^{p_{1}}g_{1}, and α(y) = g_{1}^{−1}z_{2}^{p_{2}}g_{1}. We now show that p_{1} = p_{2} = 1. We are choosing shortest representatives subject to 0 < p_{i} < p. Notice that z_{1} ≠ z_{2}, for otherwise α(N) is cyclic and since by Lemma 10 below, N is not cyclic, α is not an automorphism. First, suppose that z_{1} = x, so that z_{2} = y. In other words, we suppose that α sends x to a power of x and y to a power of y (up to conjugacy). The image of each relator of G under α is necessarily trivial. Let W denote the last relator in Presentation 4. Consider

W′ = g_{1}α(W)g_{1}^{−1} = x^{p_{1}}y^{p_{2}}x^{p_{1}}y^{2p_{2}} … x^{p_{1}}y^{qp_{2}}.

Since α(W) =_{G} 1, we have W′ =_{G} 1, and Proposition 2 implies that W′ contains more than half of a relator. Since we have chosen shortest representatives for α(x) and α(y), the word W′ does not contain more than half of the relator x^{p}. The other relators are words with no x^{n} subwords for |n| > 1; therefore, p_{1} = 1. Note that if we started with a negative value of p_{1}, then we would actually have p+p_{1} = 1, so that p_{1} = −(p−1). But then g_{1}^{−1}xg_{1} should be the shortest representative for α(x). Furthermore, in all cyclic conjugates of A_{i}^{±1}, B_{i}^{±1}, C_{i}^{±1}, D_{i}^{±1} and E_{i}^{±1} the generator y appears with exponents ±1 and ±2. Assume that the exponents of x and y in W′ are positive. For each m with 1 < m < q, we have that (m − 1)p_{2}, mp_{2}, (m + 1)p_{2} ≡_{p} 1 or 2. So either p_{2} ≡_{p} 0 (which is impossible) or 2p_{2} ≡_{p} 0. In this latter case, we also have that p_{2} ≡_{p} ±1, therefore p = 2, but our construction implies that p ≫ 2. Hence, we have to use W when shortening W′. In W, each two sequential powers of y differ by 1, and their values increase, hence p_{2} = 1.

Now, assume z_{1} = y so that α(x) = g_{1}^{−1}y^{p_{1}}g_{1}, and α(y) = g_{1}^{−1}x^{p_{2}}g_{1}. Then

W′ = g_{1}α(W)g_{1}^{−1} = y^{p_{1}}x^{p_{2}}y^{p_{1}}x^{2p_{2}} … y^{p_{1}}x^{qp_{2}} = 1.

Apply x^{p} whenever it is possible. As we showed above, we still remain with different exponents of x, for instance each three sequential exponents of x in W′ are different. In the rest of the relations x appears with the exponent 1 only, therefore large subwords of these relations cannot occur in W′. Hence by Proposition 2, W′ ≠ 1, therefore, α(W) ≠ 1, a contradiction.

In conclusion, we have shown that α(x) = g_{1}^{−1}xg_{1} and α(y) = g_{1}^{−1}yg_{1}, so α is induced by conjugation by an element of G as claimed.

Lemma 10. If a group G has a presentation satisfying the conditions of Lemma 9, then the subgroup N = ⟨x, y⟩ is not cyclic so that G is not cyclic.

Proof. The freely reduced word xyx^{−1}y^{−1} is too short to contain a large part of any relation. Therefore by Proposition 2, we conclude that x and y do not commute, and so N cannot be cyclic, hence G is not cyclic.

3.3. Countable groups are outer automorphism groups of finitely generated groups. We apply Lemma 9 to prove the following theorem.

Theorem 11. Every countable group Q is isomorphic with Out(N) where N is a finitely generated subgroup of a countable C′(\frac{1}{6}) group.

Proof. Let Q = ⟨a_{1}, a_{2}, ⋯ | R_{1}, R_{2}, … ⟩. We consider the group G presented by (4). We choose relations in Presentation (4) so that G satisfies the C′(\frac{1}{11}) condition. Also, we have the short exact sequence (2) where N = ⟨x, y⟩. Our goal is to show that the map ϕ : Q → Out(N) is an isomorphism.

In order to define the relations of G, we define an auxiliary sequence of integers {n_{j}} by n_{j} = 30|R_{i}| for j = 5i and n_{j} = 1 for j ≠ 5i, i = 1, 2, …, ∞ and use this sequence to choose a sequence of words W_{∞} = {w_{1}, w_{2}, … } according to Lemma 6; the family W_{∞} satisfies the C′(\frac{1}{20}) condition. Let

V = {A_{1}, B_{1}, C_{1}, D_{1}, E_{1}, A_{2}, B_{2}, C_{2}, D_{2}, E_{2}, … }

be an ordered set of symbols. We assign values to the symbols from V according to the natural one-to-one correspondence between V and W_{∞}, so that A_{1} = w_{1}, B_{1} = w_{2}, C_{1} = w_{3} etc. In particular, our choice of the values of E_{i} implies that |E_{i}| ≥ 30|R_{i}|, hence the length of a maximal piece of the relation R_{i}^{−1}E_{i} is bounded above by (\frac{1}{30} + \frac{1}{20})|w_{5i}| > |R_{i}|+|p_{i}|. We conclude that the family \{R_{i}^{−1}E_{i} | i = 1, 2, …, ∞\} satisfies the C′(\frac{1}{11}) condition. Let q = 45, and let p be a number bigger than twice the maximal exponent of y that occurs in the relations A_{i}, B_{i}, C_{i}, D_{i}, E_{i} and xyxy^{2}xy^{3} … xy^{45}; for instance p = 97. Clearly, the presentation of G satisfies the C′(\frac{1}{11}) condition, in particular, G is a C′(\frac{1}{6}) group. Furthermore, by Lemma 10, G is not cyclic. Therefore, ϕ is injective by Lemma 8. It is readily seen that the presentation of G satisfies the conditions of Lemma 9, hence ϕ is surjective.

4. Finitely presented groups are Outer automorphism groups of finitely generated residually finite groups

We begin this section with the following lemma which will be used to prove the residual finiteness of our groups.

Lemma 12. Let G be presented by ⟨x, y, a_{1}, … a_{r} | V_{1}, …, V_{n}⟩, where V_{1}, …, V_{n} are reduced and cyclically reduced words. Let ψ : ⟨x, y, a_{1}, …, a_{k}⟩ → ℤ be such that ψ(x) = ψ(y) = 1, and ψ(a_{i}) = 0 for each i. Suppose that x and y only appear positively in the relators V_{j}. Suppose that ψ(V_{j}) is constant and ψ(V_{j}) > \frac{8}{9}|V_{j}| and suppose that the presentation for G satisfies the C′(\frac{1}{9}) condition. Then G is residually finite.

Lemma 12 will be deduced from a result of [Wisa] which we shall state after the following definition.

Definition 13. [Wisa][∥ψ∥, peak(V), δ(V)] Fix a homomorphism ψ: ⟨a_{1}, …, a_{r} | −⟩ → ℤ. Let ∥ψ∥ = max{|ψ(a_{i})| : 1 ≤ i ≤ r}. Let V be a word in {a_{1}^{±1}, …, a_{r}^{±1}}. Let subwords(V) denote the set of subwords of the cyclic word V, so each element of subwords(V) is a subword of some cyclic permutation of V. We define peak(V) to be max{−ψ(U) : U ∈ subwords(V)}. Note that peak(V) ≥ 0 because subwords(V) contains the empty word. The number peak(V) bounds the extent to which V can backtrack relative to ψ. For example, if ψ is the exponent sum homomorphism then ∥ψ∥ = 1, and if V = a^{−1}b^{3}a^{−2}b^{3}aba^{−3} then ψ(V) = 2 and peak(V) = 4. Finally we define δ(V) to be 0 or 1 depending upon whether ψ(V) is even or odd.

We quote the following from [Wisa, Thm 3.6]:

Theorem 14. The group ⟨a_{1}, …, a_{r} | V_{1}, …, V_{s}⟩ is residually finite provided that the following conditions hold:

1. Each V_{i} is reduced and cyclically reduced and the presentation satisfies C′(\frac{1}{n}).

2. There is a homomorphism ψ: ⟨a_{1}, …, a_{r} |⟩ → ℤ such that ψ(V_{1}) = ψ(V_{2}) = ⋯ = ψ(V_{s}) ≥ 1.

3. There exists ρ such that ∥ψ∥ ≤ ρ + 1, and ρ ≥ peak(V_{i}) and ψ(V_{i}) > 4ρ + 1 for each i.

4. For each i there are constants λ_{i} and ϵ_{i} such that λ_{i}|U| + ϵ_{i} ≥ |ψ(U)| for each U ∈ subwords(V_{i}).

5. For each i we let δ(V_{i}) be 0 or 1 in case ψ(V_{i}) is even or odd. Finally, the following equation is satisfied for each i:

\frac{ψ(V_{i}) − δ(V_{i})}{2} − 2ρ − ϵ_{i} ≥ \frac{4λ_{i}}{n} |V_{i}|

Now we can prove Lemma 12.

Proof. We deduce the statement of Lemma 12 from Theorem 14. Conditions (1) and (2) of Theorem 14 are a part of our assumptions. Note that peak(V_{j}) = 0 and ∥ψ∥ = 1, hence if we set ρ = 0, then (3) will hold. ψ(V_{j}) is equal to the sum of the lengths of the subwords of V_{j} formed by x and y, because x and y only appear positively in the relators V_{j} and the exponents of a_{i} do not affect the value of ψ(V_{j}). Hence, for λ_{i} = 1 and ϵ_{i} = 0 inequality (4) holds. Plug the values of ρ, λ_{i}, ϵ_{i} and n = 9 into equation (5) and obtain:

\frac{ψ(V_{i}) − δ(V_{i})}{2} ≥ \frac{4}{9}|V_{i}|

which follows immediately from our assumptions. Hence by Theorem 14, G is residually finite.

The following theorem is the main result of this section.

Theorem 15. Every finitely presented group Q is isomorphic to Out(N) where N is a finitely generated subgroup of a finitely presented residually finite C′(\frac{1}{11}) group.

Proof. Given a group Q presented by ⟨a_{1}, a_{2}, …, a_{k} | R_{1}, R_{2}, …, R_{t}⟩, we form a presentation of G as follows (cf. (4)):

⟨x, y, a_{1}, a_{2}, …, a_{k} | x^{a_{i}} = A_{i}, x^{a_{i}^{−1}} = B_{i}, y^{a_{i}} = C_{i}, y^{a_{i}^{−1}} = D_{i}, (5)

R_{i} = E_{i}, x^{p}, y^{p}, W ⟩

where A_{i}, B_{i}, C_{i}, D_{i}, E_{i} and W are positive words in x, y, and p is a sufficiently large number (which we specify below). Thus we obtain a short exact sequence 1 → N → G → Q → 1 (cf. (2)) where N = ⟨x, y⟩. Define a homomorphism ψ: G → ℤ by ψ(x) = ψ(y) = 1 and ψ(a_{j}) = 0 for each j (cf. Lemma 12). Therefore, for each g ∈ G which we consider a word in the alphabet of the generators of G, the value ψ(g) is the sum of the exponents of x and y which occur in g.

To specify the relators of G, let W_{∞} = {w_{1}, w_{2}, … } be the family of x, y words chosen according to part (1) of Lemma 6, and let m = 4k + t,. Let a positive integer p satisfy the following inequalities: p > |w_{m+1}| and p > 30 max_{1≤i≤t} |R_{i}|; also we assume that p = \frac{q(q+3)}{2} for some integer q. Let I = {1, 2, …, t}, and let V_{m} be the family of words obtained from the finite subset W_{m} of W_{∞}, according to part (2) of Lemma 6. Fix a one-to-one correspondence between the finite set of variables {A_{j}, B_{j}, C_{j}, D_{j}, E_{i}} and the finite set of words V_{m}(x, y) = {v_{1}, …, v_{m}} and assign values to A_{j}, B_{j}, C_{j}, D_{j} and E_{j} according to this correspondence. The only requirement is that the following condition is satisfied: |A_{j}| = |B_{j}| = |C_{j}| = |D_{j}| = p and |E_{j}| = p + 1. Let

W = xyxy^{2}xy^{3} … xy^{q}

where q is the integer chosen above (it is easy to see that q ≫ 45). Obviously, this presentation of G satisfies the conditions of Lemma 12, hence G is residually finite. Moreover, the presentation of G satisfies the conditions of Lemma 9, hence ϕ is surjective. By Lemma 10, G is not cyclic, hence by Lemma 8, ϕ is injective. Therefore, Q is isomorphic to Out(N) where N is the subgroup of G generated by x and y.

In conclusion, we note that if N is finitely generated and residually finite then Aut(N) is residually finite [Bau63].

References

[Bau63] Gilbert Baumslag. Automorphism groups of residually finite groups. J. London Math. Soc., 38:117–118, 1963.

[DG93] Manfred Dugas and Rüdiger Göbel. On locally finite p-groups and a problem of Philip Hall’s. J. Algebra, 159(1):115–138, 1993.

[DGG01] Manfred Droste, Michèle Giraudet, and Rüdiger Göbel. All groups are outer automorphism groups of simple groups. J. London Math. Soc. (2), 64(3):565– 575, 2001.

[GP00] Rüdiger Göbel and Agnes T. Paras. Outer automorphism groups of metabelian groups. J. Pure Appl. Algebra, 149(3):251–266, 2000.

[Gre60] Martin Greendlinger. On Dehn’s algorithms for the conjugacy and word problems, with applications. Comm. Pure Appl. Math., 13:641–677, 1960.

[Gre62] Martin Greendlinger. A class of groups all of whose elements have trivial centralizers. Math. Z., 78:91–96, 1962.

[Gri66] M. Grindlinger. The problem of conjugacy and coincidence with an anticenter in the theory of groups. Sibirsk. Mat. Z̆., 7:785–803, 1966.

[Koj88] Sadayoshi Kojima. Isometry transformations of hyperbolic 3-manifolds. Topology Appl., 29(3):297–307, 1988.

[Lip62] Seymour Lipschutz. On powers of elements in S-groups. Proc. Amer. Math. Soc., 13:181–186, 1962.

[LS77] Roger C. Lyndon and Paul E. Schupp. Combinatorial group theory. Springer-Verlag, Berlin, 1977. Ergebnisse der Mathematik und ihrer Grenzgebiete, Band 89.

[Mat89] Takao Matumoto. Any group is represented by an outerautomorphism group. Hiroshima Math. J., 19(1):209–219, 1989.

[Rip82] E. Rips. Subgroups of small cancellation groups. Bull. London Math. Soc., 14(1):45–47, 1982.

[Sey74] J. E. Seymour. Conjugate powers and unique roots in certain small cancellation groups. PhD thesis, University of Illinois, 1974.

[Sta83] John R. Stallings. Topology of finite graphs. Invent. Math., 71(3):551–565, 1983.

[Tru74] Bernard Truffault. Centralisateurs des éléments dans les groupes de Greendlinger. C. R. Acad. Sci. Paris. Sér. A, 279:317–319, 1974.

[Wisa] Daniel T. Wise. Residual finiteness of one-relator groups with torsion. J. London Math. Soc. (2). To appear.

[Wisb] Daniel T. Wise. A residually finite version of Rips’s construction. Bull. London Math. Soc. To appear.

Dept. of Math., McGill University, Montreal, Quebec, Canada H3A 2K6

   E-mail address: bumagin@math.mcgill.ca

Dept. of Math., McGill University, Montreal, Quebec, Canada H3A 2K6

   E-mail address: wise@math.mcgill.ca