The second bounded cohomology of a group with infinitely many ends ^{∗}

Koji Fujiwara ^{†} Mathematics Department, Keio University Yokohama, 223, Japan

              fujiwara@math.keio.ac.jp

Abstract

We study the second bounded cohomology of an amalgamated free product of groups, and an HNN extension of a group. As an application, we have a group with infinitely many ends has infinite dimensional second bounded cohomology.

Contents §1 Introduction §2 Quasi homomorphism and the counting function §3 Quasi homomorphisms of A ∗_{C} B §4 Choice of words for A ∗_{C} B §5 Proofs of Th 1 and Cor 1,2,3 §6 Quasi homomorphisms of A ∗_{C} φ §7 Choice of words for A ∗_{C} φ §8 Proofs of Th 2 and Th 3

∗1991 Mathematics Subject Classification, Primary: 20F32, Secondary: 55N20.

†This work was done when the author visited MSRI supported in part by NSF grant DMS-9022140 and a JSPS grant.

1 Introduction

Bounded cohomology was defined by F. Trauber for groups and by M. Gromov for spaces. We review the definition of bounded cohomology of a group G. Let

C_{b}^{k}(G; A) = {f : G^{k} → A | f has bounded range},

where A = Z or R. The boundary δ : C_{b}^{k}(G; A) → C_{b}^{k+1}(G; A) is given by

δf(g_{0}, …, g_{k}) = f(g_{1}, …, g_{k}) + ∑_{i=1}^{k}(−1)^{i}f(g_{0}, …, g_{i−1}g_{i}, … g_{k})
+(−1)^{k+1}f(g_{0}, …, g_{k−1}).

The cohomology of the complex {C_{b}^{k}(G; A), δ} is the bounded cohomology of G, denoted by H_{b}^{∗}(G; A). See [G], [I] as general references for the theory of bounded cohomology.

For any group G, the first bounded cohomology, H_{b}^{1}(G; A) is trivial. For an amenable group G, H_{b}^{n}(G; R) is trivial for any n. The first example of a group with non-trivial second bounded cohomology was given by R. Brooks, [B]. He showed a free group of rank greater than 1 has infinite dimensional second bounded cohomology. R.I. Grigorchuk investigated the structure of the second bounded cohomology of free groups, torus knot groups and surface groups, [Gr]. T. Yoshida [Y] and T. Soma [So1], [So2] studied the third bounded cohomology of surfaces and hyperbolic three manifolds. D.B.A. Epstein and the author showed a non-trivial word-hyperbolic group has infinite dimensional second bounded cohomology, [EF].

We state results on the second bounded cohomology of an amalgamated free product of groups:

Theorem 1 Let G = A ∗_{C} B. If |C\A/C| ≥ 3 and |B/C| ≥ 2, then the cardinality of the dimension of H_{b}^{2}(G; R) as a vector space over R is continuum.

Corollary 1 Let G = A ∗ B with A ≠ {1}, B ≠ {1}. If G ≠ Z_{2} ∗ Z_{2}, then the cardinality of the dimension of H_{b}^{2}(G; R) as a vector space over R is continuum.

Remark. (1) Cor 1 is a generalization of R. Brooks’ result on free groups.

(2) Since Z_{2} ∗ Z_{2} is amenable, H_{b}^{2}(Z_{2} ∗ Z_{2}; R) is trivial.

Corollary 2 Let G = A ∗_{C} B. If |A| = ∞, |C| < ∞, and |B/C| ≥ 2, then the cardinality of the dimension of H_{b}^{2}(G; R) as a vector space over R is continuum.

Corollary 3 Let G = A ∗_{C} B. If A is abelian, |A/C| ≥ 3, and |B/C| ≥ 2, then the cardinality of the dimension of H_{b}^{2}(G; R) as a vector space over R is continuum.

Example. P SL_{2}(Z) = Z_{2} ∗Z_{3} and SL_{2}(Z) = Z_{4} ∗_{Z_{2}} Z_{6} satisfy the assumption of Th 1. They are non-elementary word-hyperbolic groups too.

For an HNN extension of a group, we show:

Theorem 2 Let G = A ∗_{C} φ. If |A/C| ≥ 2, |A/φ(C)| ≥ 2, then then the cardinality of the dimension of H_{b}^{2}(G; R) as a vector space over R is continuum.

Remark. In Th 1 and 2, and Cor 1, 2 and 3, we don’t have to assume G is finitely generated.

We apply Th 1 and 2 to a group with infinitely many ends, namely, due to Stallings’ structure theorem [S], we have:

Theorem 3 If G is a finitely generated group with infinitely many ends, then the cardinality of the dimension of H_{b}^{2}(G; R) as a vector space over R is continuum.

To conclude, we state a conjecture. Conjecture. Let G be a group. If H_{b}^{2}(G; R) ≠ {0}, then H_{b}^{2}(G; R) is infinite dimensional as a vector space over R.

2 Quasi homomorphism and the counting function

We will review the counting function of a group w.r.t. a word. This was defined in [EF] as a generalization of R. Brooks’ counting function for free groups in [B]. Let G be a group with a (finite or infinite) set of generators and Γ(G) the Cayley graph. For a word w = x_{1}x_{2} … x_{n} in these generators, define |w| = n. Let \overline{w} be the element of G which is represented by the word w. Define w^{−1} = x_{n}^{−1} … x_{1}^{−1}. We sometimes identify a word w and the path starting at 1 and labeled by w in Γ(G). For a path α labeled by w, define |α| = |w| and \overline{α} = \overline{w}. For an element g in G, define |g| = inf_{α} |α|, where α ranges over all the paths with \overline{α} = g.

Let α be a finite path in Γ(G). Define |α|_{w} to be the maximal number of times that w can be seen as a subword of α without overlapping. We define

c_{w}(α) = sup_{α′}\{|α′|_{w} − (|α′| − |\overline{α}|)\} = |\overline{α}| − inf_{α′}(|α′| − |α′|_{w}),

where α′ ranges over all the paths with the same starting point as α and the same finishing point. If the infimum in the definition of c_{w}(α) is attained by α′, we say that α′ realizes c_{w} at α. We have |\overline{α}|/|w| ≥ c_{w}(α) ≥ 0. If α is a geodesic, then c_{w}(α) ≥ |α|_{w}.

We define h_{w} = c_{w} − c_{w^{−1}}. For each g in G, we choose γ_{g} to be a path from 1 to g and set h_{w}(g) = h_{w}(γ_{g}). Then h_{w}(g) does not depend on the choice of γ_{g}. Thus h_{w} ∈ C^{1}(G; Z).

Let f ∈ C^{1}(G; R). If there exists a constant D < ∞ s.t.

|f(gh) − f(g) − f(h)| ≤ D,

for any g, h ∈ G, then we say f is a quasi homomorphism with defect D. Let f be a quasi homomorphism with defect D. Then |δf| ≤ D and δ(δf) = 0, thus δf ∈ Z_{b}^{2}(G; R), which defines [δf] ∈ H_{b}^{2}(G; R). Remark that we always have [δf] = 0 in H^{2}(G; R), however, we may have [δf] ≠ 0 in H_{b}^{2}(G; R) since f is not necessarily in C_{b}^{1}(G; R).

3 Quasi homomorphisms of A ∗_{C} B

Let G = A ∗_{C} B with |A/C| ≥ 2, |B/C| ≥ 2. Take the set {A ∪ B}\{1} as a set of generators of G and denote its Cayley graph by Γ(G). Note that if a generating set is infinite, then Γ is not locally finite.

If a word w = x_{1} … x_{n} satisfies n = 1 or x_{1}, x_{3}, … ∈ A\C (or B\C) and x_{2}, x_{4}, … ∈ B\C (or A\C, resp.), then we say w is reduced.

Lemma 3.1 A word w = x_{1} … x_{n} is reduced iff it is a geodesic in Γ.

Proof. Assume w is not reduced, then there exists a subword x_{i}x_{i+1} with x_{i}, x_{i+1} ∈ A(or x_{i}, x_{i+1} ∈ B). Then \overline{x_{i}x_{i+1}} ∈ A(or B resp.), thus |\overline{x_{i}x_{i+1}}| ≤ 1. Therefore x_{i}x_{i+1} is not a geodesic, hence w is not a geodesic.

On the other hand, assume w is not a geodesic. To show w is not reduced by contradiction, suppose w is reduced. Take a geodesic γ, hence reduced, s.t. \overline{w} = \overline{γ}. Then |w| = |γ| since reduced words representing a same element have same length. Thus w is a geodesic. This is a contradiction. We showed w is not reduced. q.e.d.

Lemma 3.2 Let w be a word and α a path. If w^{2} is reduced, then there is a geodesic which realizes c_{w} at α.

Proof. Since w^{2} is reduced, w is reduced. Let γ be a path which realizes c_{w} at α s.t. |γ|_{w} is minimal among all the realizing paths at α. We will show that γ is a geodesic. If |γ|_{w} = 0, then γ is a geodesic. Assume |γ|_{w} = n > 0. Write γ as

γ_{1}w_{1}γ_{2} … w_{n}γ_{n+1},

where w_{i} is a copy of w and γ_{i} may be an empty word. First, to show every γ_{i} is reduced by contradiction, suppose some γ_{i} is not reduced. Replace γ_{i} by a reduced word γ′_{i} with \overline{γ_{i}} = \overline{γ′_{i}}, then we have a new path

γ′ = γ_{1}w_{1}γ_{2} … γ′_{i} … w_{n}γ_{n+1},

which satisfies |γ′| < |γ|, |γ′|_{w} ≥ |γ|_{w} and \overline{γ} = \overline{γ′}. This contradicts that γ is a realizing path. Thus every γ_{i} is reduced. Next, in order to show that γ is reduced by contradiction, suppose not. Since w^{2} is reduced, there is a subword w_{i}γ_{i+1}w_{i+1} of γ which is not reduced and γ_{i+1} is not empty. Since w_{i}, w_{i+1}, γ_{i+1} are reduced, one of the following four cases is occurring.

(i) The last letter of w_{i} and the initial letter of γ_{i+1} are in A. (ii) The last letter of w_{i} and the initial letter of γ_{i+1} are in B. (iii) The last letter of γ_{i+1} and the initial letter of w_{i+1} are in A. (iv) The last letter of γ_{i+1} and the initial letter of w_{i+1} are in B.

We treat the case (i). Write

w_{i} = … b_{1}a_{1}, γ_{i+1} = a_{2}b_{2} …, a_{i} ∈ A, b_{i} ∈ B.

Rewrite the subword w_{i}γ_{i+1} in γ as

w_{i}γ_{i+1} = … b_{1}a_{1}a_{2}b_{2} … = … b_{1}a′b_{2} …,

Figure 1: Geodesic triangle. This illustrates Lemma 3.3.

where a′ = a_{1}a_{2} ∈ A. This gives a new word γ′ with |γ′| = |γ| − 1, |γ′|_{w} = |γ|_{w} −1. Since \overline{γ′} = \overline{γ} and |γ|−|γ|_{w} = |γ′|−|γ′|_{w}, γ′ is another realizing path with |γ′|_{w} < |γ|_{w}. This contradicts the choice of γ, thus γ is reduced. By Lemma 3.1, γ is a geodesic. A similar argument applies to the other three cases, and we omit it. q.e.d.

Lemma 3.3 Let α, β be paths starting at 1. We have

|c_{w}(α) − c_{w}(β)| ≤ 2|\overline{α^{−1}β}|.

Proof. Take realizing geodesics of c_{w} at α, α′, and at β, β′. Then c_{w}(α) = |α′|_{w}, c_{w}(β) = |β′|_{w}. Take a geodesic σ with \overline{α^{−1}β} = \overline{σ}.

Since the path α′σ satisfies \overline{α′σ} = \overline{β′},

|β′| − |β′|_{w} ≤ |α′σ| − |α′σ|_{w}.

Since |α′σ| = |α′| + |σ| and |α′σ|_{w} ≥ |α′|_{w} + |σ|_{w},

|β′| − |β′|_{w} ≤ |α′| + |σ| − |α′|_{w} − |σ|_{w}.

Thus

c_{w}(β) = |β′|_{w} ≥ |α′|_{w} + |β′| − |α′| − |σ| + |σ|_{w} ≥ c_{w}(α) − 2|σ|,

since |α′|_{w} = c_{w}(α), |β′| − |α′| ≥ −|σ|, |σ|_{w} ≥ 0. Similarly, c_{w}(α) ≥ c_{w}(β) − 2|σ|. Thus |c_{w}(α) − c_{w}(β)| ≤ 2|σ| = 2|\overline{α^{−1}β}|. q.e.d.

Lemma 3.4 Let α, β be paths starting at 1. We have

|h_{w}(α) − h_{w}(β)| ≤ 4|\overline{α^{−1}β}|.

Figure 2: Dividing geodesics. This illustrates Lemma 3.7. α = α_{1}α_{2}, γ = γ_{1}γ_{2}, σ = σ_{1}σ_{2}.

Proof. h_{w} = c_{w} − c_{w^{−1}} and Lemma 3.3. q.e.d.

Lemma 3.5

c_{w}(α) = c_{w^{−1}}(α^{−1}).

Proof. Clear from the definition of c_{w}. q.e.d.

Lemma 3.6

h_{w}(α) = −h_{w}(α^{−1}).

Proof. h_{w}(α^{−1}) = c_{w}(α^{−1}) − c_{w^{−1}}(α^{−1}) = c_{w^{−1}}(α) − c_{w}(α) = −h_{w}(α). q.e.d.

Lemma 3.7 Let α be a reduced path. If α = α_{1}α_{2}, then

|h_{w}(α) − h_{w}(α_{1}) − h_{w}(α_{2})| ≤ 10.

Proof. We will show

|c_{w}(α) − c_{w}(α_{1}) − c_{w}(α_{2})| ≤ 5.

Take realizing geodesics of c_{w} at α_{1}, γ_{1}, and at α_{2}, γ_{2}. We have |γ_{1}|_{w} = c_{w}(α_{1}), |γ_{2}|_{w} = c_{w}(α_{2}). Put γ = γ_{1}γ_{2}, then \overline{γ} = \overline{α}. Since α is reduced, it is a geodesic. Since |α| = |α_{1}| + |α_{2}| = |γ_{1}| + |γ_{2}| = |γ|, γ is a geodesic. Since γ is a geodesic with \overline{γ} = \overline{α}, c_{w}(α) ≥ |γ|_{w}. Thus

c_{w}(α) ≥ |γ|_{w} ≥ |γ_{1}|_{w} + |γ_{2}|_{w} = c_{w}(α_{1}) + c_{w}(α_{2}).

On the other hand, take a realizing geodesic σ at α, then c_{w}(α) = |σ|_{w}. Since α and σ are reduced, there exists subdivision of σ, σ = σ_{1}σ_{2} s.t. \overline{α_{1}^{−1}σ_{1}} = \overline{α_{2}σ_{2}^{−1}} = c, for some c ∈ C. Since |c| ≤ 1, from Lemma 3.3 and 3.5.

|c_{w}(α_{i}) − c_{w}(σ_{i})| ≤ 2, i = 1, 2.

Since σ_{1}, σ_{2} are geodesics, |σ_{i}|_{w} ≤ c_{w}(σ_{i}), i = 1, 2. Thus

c_{w}(α) = |σ|_{w} ≤ |σ_{1}|_{w} + |σ_{2}|_{w} + 1 ≤ c_{w}(σ_{1}) + c_{w}(σ_{2}) + 1 ≤ c_{w}(α_{1}) + c_{w}(α_{2}) + 5.

We showed |c_{w}(α)−c_{w}(α_{1})−c_{w}(α_{2})| ≤ 5. |c_{w^{−1}}(α)−c_{w^{−1}}(α_{1})−c_{w^{−1}}(α_{2})| ≤ 5 is similar. Since h_{w} = c_{w} − c_{w^{−1}}, we get |h_{w}(α) − h_{w}(α_{1}) − h_{w}(α_{2})| ≤ 10.

q.e.d.

Proposition 1 Let w be a word. We have

|δh_{w}| ≤ 78.

Proof. Let x, y be elements in G. δh_{w}(x, y) = h_{w}(x)+h_{w}(y)−h_{w}(xy). We will show |h_{w}(xy) − h_{w}(x) − h_{w}(y)| ≤ 78. Take geodesics, hence reduced paths, α, β and γ with \overline{α} = x, \overline{β} = y, \overline{γ} = xy. Since \overline{αβ} = \overline{γ} and α, β, γ are reduced, there exist subdivisions of α, β, γ,

α = α_{1}α_{2}α_{3}, β = β_{1}β_{2}β_{3}, γ = γ_{1}γ_{2}γ_{3},

s.t.

\overline{γ_{1}^{−1}α_{1}} = c_{1}, \overline{α_{3}β_{1}} = c_{2}, \overline{β_{3}γ_{3}^{−1}} = c_{3},

for some c_{1}, c_{2}, c_{3} ∈ C and \overline{α_{2}}, \overline{β_{2}}, \overline{γ_{2}} are (simultaneously) in A or B.

Since \overline{α_{2}} is in A or B, c_{w}(α_{2}), c_{w^{−1}}(α_{2}) ≤ |\overline{α_{2}}| ≤ 1. Thus |h_{w}(α_{2})| ≤ 2. Similarly, |h_{w}(β_{2})|, |h_{w}(γ_{2})| ≤ 2.

Using Lemma 3.7 twice for α = ((α_{1}α_{2})α_{3}),

|h_{w}(α) − h_{w}(α_{1}) − h_{w}(α_{2}) − h_{w}(α_{3})| ≤ 20,

thus |h_{w}(α) − h_{w}(α_{1}) − h_{w}(α_{3})| ≤ 22. Similarly,

|h_{w}(β) − h_{w}(β_{1}) − h_{w}(β_{3})| ≤ 22,
|h_{w}(γ) − h_{w}(γ_{1}) − h_{w}(γ_{3})| ≤ 22.

Figure 3: Dividing geodesic triangle. This illustrates Prop 1. c_{1}, c_{2}, c_{3} ∈ C. 9 Thus

|h_{w}(α) + h_{w}(β) − h_{w}(γ)|
≤ |h_{w}(α_{1}) + h_{w}(α_{3}) + h_{w}(β_{1}) + h_{w}(β_{3}) − h_{w}(γ_{1}) − h_{w}(γ_{3})| + 66
≤ |h_{w}(α_{1}) − h_{w}(γ_{1})| + |h_{w}(β_{1}) − h_{w}(α_{3}^{−1})| + |h_{w}(γ_{3}^{−1}) − h_{w}(β_{3}^{−1})| + 66
≤ 4(|c_{1}| + |c_{2}| + |c_{3}|) + 66 ≤ 78,

since |c_{i}| ≤ 1 and Lemma 3.4 and 3.6.

By definition, h_{w}(α) = h_{w}(x), h_{w}(β) = h_{w}(y), h_{w}(γ) = h_{w}(xy), thus

|h_{w}(x) + h_{w}(y) − h_{w}(xy)| ≤ 78.

q.e.d.

4 Choice of words for A ∗_{C} B

The goal of this section is to prove the following proposition.

Proposition 2 Let G = A ∗_{C} B with |C\A/C| ≥ 3 and |B/C| ≥ 2. Then there exist words w_{i}, 0 ≤ i < ∞ which satisfy the following conditions.

(1) For any i ≥ 0 and any n ≥ 1, h_{w_{i}}(w_{i}^{n}) = n.

(2) For any j > i ≥ 0 and any n ≥ 1, h_{w_{j}}(w_{i}^{n}) = 0.

(3) For any i ≥ 0, \overline{w_{i}} ∈ [G, G].

Let G = A ∗_{C} B with |C\A/C| ≥ 3 and |B/C| ≥ 2. Let w and w′ be reduced paths starting at a same point and finishing at a same point,

w = a_{1}b_{1} … a_{n}b_{n},
w′ = a′_{1}b′_{1} … a′_{m}b′_{m},

where a_{1}, b_{n}, a′_{1}, b′_{m} may be empty.

Since \overline{w} = \overline{w′} and w and w′ are reduced, n = m and there exist c_{1}, … c_{n} and c′_{0}, c′_{1}, … c′_{n} in C s.t.

c′_{i−1}a_{i}c_{i}^{−1} = a′_{i}, c_{i}b_{i}c_{i}^{′−1} = b′_{i},

for 1 ≤^{∀} i ≤ n, where c′_{0} = c′_{n} = 1. See Figure 4.

Figure 4: Reduced paths w, w′ with common end points. \overline{w} = \overline{w′}.

Figure 5: Definition of P (v).

Let v be a subword of w,

v = a_{i}b_{i} … a_{j}b_{j},

where a_{i}, b_{j} may be empty. Define a subword of w′, denoted by P (v), by

P (v) = a′_{i}b′_{i} … a′_{j}b′_{j}.

Let v′ be a subword of w′. If v′ is a subword of P (v), we say v covers v′. If v′ = P (v), then we say v faces v′.

Since |C\A/C| ≥ 3, we can choose a_{1}, a_{2} ∈ A s.t. a_{1}, a_{2} ∈ A\C, and a_{2} ∉ Ca_{1}C. Take b ∈ B\C. Define words w_{i}, 0 ≤ i < ∞ by

w_{i} = (a_{1}b)^{10^{i}}(a_{1}^{−1}b^{−1})^{10^{i}}(a_{2}b)^{10^{i}}(a_{2}^{−1}b^{−1})^{10^{i}}
(a_{1}b)^{4·10^{i}}(a_{1}^{−1}b^{−1})^{4·10^{i}}(a_{2}b)^{4·10^{i}}(a_{2}^{−1}b^{−1})^{4·10^{i}}.

We write the subword (a_{1}b)^{4·10^{i}} by w_{i}(1, +) and the subword (a_{1}^{−1}b^{−1})^{4·10^{i}} by w_{i}(1, −).

Lemma 4.1 The words w_{i}, 0 ≤ i < ∞ satisfy the following conditions.

(1) For any i ≥ 0 and any n ≥ 1, w_{i}^{n} is reduced.

(2) For any i ≥ 0, |w_{i}| = 40 · 10^{i} and |w_{i}(1, ±)| = 8 · 10^{i}.

(3) For any i ≥ 0, \overline{w_{i}} ∈ [G, G].

Proof. (1) is clear from the choice of a_{1}, a_{2}, and b. (2) is obvious. For (3), use \overline{(a_{i}b)^{p}(a_{i}^{−1}b^{−1})^{p}} ∈ [G, G] for i = 1, 2 and any p ≥ 1. q.e.d.

Lemma 4.2 In any pair of paths having a same starting point and a same finishing point, the following conditions hold.

(1) a_{1} cannot face a_{2}.

(2) For any i ≥ 0, w_{i}^{2} cannot cover w_{i}^{−1}.

(3) For any i < j, w_{j}(1, +) cannot cover w_{i} and w_{j}(1, −)^{−1} cannot cover w_{i}.

(4) For any k > 0 and any i < j, w_{i}^{k} cannot cover w_{j} nor w_{j}^{−1}.

Proof. (1) If a_{1} faces a_{2} in some pair of paths, then there exist c_{1}, c_{2} ∈ C s.t. c_{1}a_{1}c_{2} = a_{2}, thus a_{2} ∈ Ca_{1}C, which contradicts our choice of a_{1} and a_{2}.

(2) Let W_{0} symbolize the order of the occurrence of elements labeled by a_{1}^{±1} or a_{2}^{±1} in w_{0} in the following way:

W_{0} = 1\overline{1}2\overline{2}1111\overline{1}\overline{1}\overline{1}\overline{1}2222\overline{2}\overline{2}\overline{2}\overline{2},

where 1 stands for a_{1}, 2 for a_{2}, \overline{1} for a_{1}^{−1}, and \overline{2} for a_{2}^{−1}. Here, remark that in W_{0}, we ignore b^{±1}’s of w_{0}. Let W_{0}^{−1} represent for w_{0}^{−1}:

W_{0}^{−1} = 2222\overline{2}\overline{2}\overline{2}\overline{2}1111\overline{1}\overline{1}\overline{1}\overline{1}2\overline{2}1\overline{1}.

In order to show the conclusion by contradiction, assume w_{0}^{2} covers w_{0}^{−1}. Then W_{0}^{2} covers W_{0}^{−1}. For each possible position for W_{0}^{2} covering W_{0}^{−1}, one can find some 1(or 2, \overline{1}, \overline{2}) in W_{0}^{−1} facing some 2 (or 1, \overline{2}, \overline{1} resp.) in W_{0}^{2}, see Figure 6. This means a_{1} is facing some a_{2}, which contradicts (1). Thus we showed (2) for i = 0. A similar argument works for w_{i} for any i ≥ 1, and we omit it.

(3) Observe that w_{j}(1, +) consists of a_{1} and b, while w_{i} contains a_{2}. If w_{j}(1, +) covers w_{i}, then each a_{2} in w_{i} faces some a_{1} in w_{j}(1, +), which contradicts (1). The same argument goes for w_{j}(1, −)^{−1}, and we omit it.

(4) Assume w_{i}^{k} covers w_{j} for some i < j and some k > 0. Then w_{j}(1, +) covers some w_{i} of w_{i}^{k} since 2|w_{i}| ≤ |w_{j}(1, +)|, which contradicts (3). Assume w_{i}^{k} covers w_{j}^{−1} for some i < j and some k > 0. Then w_{j}(1, −)^{−1} covers some w_{i} of w_{i}^{k}, which contradicts (3). q.e.d.

W_{0}^{2}: 1 \overline{1} 2 ¯2 1 1 1 1 \overline{1} \overline{1} \overline{1} \overline{1} 2 2 2 2 \overline{2} \overline{2} \overline{2} \overline{2} 1 \overline{1} 2 \overline{2} 1 1 1 1 \overline{1} \overline{1} \overline{1} \overline{1} 2 2 2 2 \overline{2} \overline{2} \overline{2} \overline{2} 2• 2 2 2 \overline{2} \overline{2} \overline{2} \overline{2} 1 1 1 1 \overline{1} \overline{1} \overline{1} \overline{1} 2 \overline{2} 1 \overline{1}• 2 2 2 2• \overline{2} \overline{2} \overline{2} \overline{2}• 1 1 1 1• \overline{1} \overline{1} \overline{1} \overline{1}• 2 \overline{2} 1 \overline{1} 2 2 2• 2• \overline{2} \overline{2} \overline{2}• \overline{2}• 1 1 1• 1• \overline{1} \overline{1} \overline{1}• \overline{1}• 2 \overline{2} 1 \overline{1} 2 2• 2• 2• \overline{2} \overline{2}• \overline{2}• \overline{2}• 1 1• 1• 1• \overline{1} \overline{1}• \overline{1}• \overline{1}• 2 \overline{2} 1 \overline{1} 2• 2• 2• 2• \overline{2}• \overline{2}• \overline{2}• \overline{2}• 1• 1• 1• 1• \overline{1}• \overline{1}• \overline{1}• \overline{1}• 2• \overline{2}• 1• \overline{1}• 2• 2• 2• 2 \overline{2}• \overline{2}• \overline{2}• \overline{2} 1• 1• 1• 1 \overline{1}• \overline{1}• \overline{1}• \overline{1} 2 \overline{2} 1 \overline{1} 2• 2• 2 2 \overline{2}• \overline{2}• \overline{2} \overline{2} 1• 1• 1 1 \overline{1}• \overline{1}• \overline{1} \overline{1} 2 \overline{2} 1 \overline{1} 2• 2 2 2 \overline{2}• \overline{2} \overline{2} \overline{2} 1• 1 1 1 \overline{1}• \overline{1} \overline{1} \overline{1} 2 \overline{2} 1 \overline{1} 2 2 2 2 \overline{2} \overline{2} \overline{2} \overline{2} 1 1 1 1 \overline{1} \overline{1} \overline{1} \overline{1}• 2• \overline{2} 1 \overline{1} 2 2 2 2 \overline{2} \overline{2} \overline{2} \overline{2} 1 1 1 1 \overline{1} \overline{1} \overline{1}• \overline{1} 2• \overline{2} 1 \overline{1} 2 2 2 2 \overline{2} \overline{2} \overline{2} \overline{2} 1 1 1 1 \overline{1} \overline{1}• \overline{1} \overline{1} 2• \overline{2} 1 \overline{1} 2 2 2 2 \overline{2} \overline{2} \overline{2} \overline{2} 1 1 1 1• \overline{1}• \overline{1} \overline{1} \overline{1} 2• \overline{2}• 1 \overline{1} 2 2 2 2 \overline{2} \overline{2} \overline{2} \overline{2} 1 1 1• 1 \overline{1} \overline{1} \overline{1} \overline{1} 2 \overline{2}• 1 \overline{1} 2 2 2 2 \overline{2} \overline{2} \overline{2} \overline{2} 1 1• 1 1 \overline{1} \overline{1} \overline{1} \overline{1} 2 \overline{2}• 1 \overline{1} 2 2 2 2 \overline{2} \overline{2} \overline{2} \overline{2}• 1• 1 1 1 \overline{1} \overline{1} \overline{1} \overline{1} 2 \overline{2}• 1• \overline{1} 2 2 2 2 \overline{2} \overline{2} \overline{2}• \overline{2} 1 1 1 1 \overline{1} \overline{1} \overline{1} \overline{1} 2 \overline{2} 1• \overline{1} 2 2 2 2 \overline{2} \overline{2}• \overline{2} \overline{2} 1 1 1 1 \overline{1} \overline{1} \overline{1} \overline{1} 2 \overline{2} 1• \overline{1} 2 2 2 2• \overline{2}• \overline{2} \overline{2} \overline{2} 1 1 1 1 \overline{1} \overline{1} \overline{1} \overline{1} 2 \overline{2} 1• \overline{1}• 2 2 2• 2 \overline{2} \overline{2} \overline{2} \overline{2} 1 1 1 1 \overline{1} \overline{1} \overline{1} \overline{1} 2 \overline{2} 1 \overline{1}• 2 2• 2 2 \overline{2} \overline{2} \overline{2} \overline{2} 1 1 1 1 \overline{1} \overline{1} \overline{1} \overline{1} 2 \overline{2} 1 \overline{1}•

Figure 6: W_{0}^{2} cannot cover W_{0}^{−1}. The first row is W_{0}^{2}. The second to the last rows describe all the possible positions for W_{0}^{−1}. We put • below illegal pairs of (1, 2) or (\overline{1}, \overline{2}). In each position for W_{0}^{−1}, we find at least one such pair, which is a contradiction.

Figure 7: w_{i}^{2} cannot cover w_{i}^{−1}.

Lemma 4.3 c_{w_{i}} and c_{w_{i}^{−1}} satisfy the following conditions.

(1) For any n ≥ 1 and any i ≥ 0, c_{w_{i}}(w_{i}^{n}) = n.

(2) For any n ≥ 1 and any i ≥ 0, c_{w_{i}^{−1}}(w_{i}^{n}) = 0.

(3) For any j > i ≥ 0 and any n ≥ 1, c_{w_{j}}(w_{i}^{n}) = 0.

(4) For any j > i ≥ 0 and any n ≥ 1, c_{w_{j}^{−1}}(w_{i}^{n}) = 0.

Proof. (1) Since w_{i}^{n} is reduced, w_{i}^{n} is a geodesic. Thus c_{w_{i}}(w_{i}^{n}) ≥ |w_{i}^{n}|_{w_{i}} = n. On the other hand, c_{w_{i}}(w_{i}^{n}) ≤ |\overline{w_{i}^{n}}|/|w_{i}| = n. Thus c_{w_{i}}(w_{i}^{n}) = n.

(2) To show c_{w_{i}^{−1}}(w_{i}^{n}) = 0 by contradiction, assume c_{w_{i}^{−1}}(w_{i}^{n}) > 0. Take a realizing geodesic of c_{w_{i}^{−1}} at w_{i}^{n}, α. Then |α|_{w_{i}^{−1}} > 0. Fix a subword labeled by w_{i}^{−1} in α. Since w_{i}^{n} and α are reduced, there is a subword of w_{i}^{n} labeled by w_{i}^{2} which covers the w_{i}^{−1} in α. But w_{i}^{2} cannot cover w_{i}^{−1} by Lemma 4.2(2), a contradiction. See Figure 7.

(3) To show the claim by contradiction, assume c_{w_{j}}(w_{i}^{n}) > 0 for some i < j and n. Take a realizing geodesic α of c_{w_{j}} at w_{i}^{n}. Then |α|_{w_{j}} > 0. Fix some subword labeled by w_{j} in α.

Since α and w_{i}^{n} are reduced, there exists a subword labeled by w_{i}^{k} in w_{i}^{n} which covers the w_{j} in α. But w_{i}^{k} cannot cover w_{j} by Lemma 4.2(4), a contradiction.

(4) Similar to (3). q.e.d. Now we are in a position to prove Proposition 2.

Proof of Proposition 2. Let w_{i}, 0 ≤ i < ∞ be the set of words defined before in this section. We will show they are the words we want. Since h_{w_{i}} = c_{w_{i}} − c_{w_{i}^{−1}}, h_{w_{i}}(w_{i}^{n}) = n by Lemma 4.3(1) and (2). By Lemma 4.3(3) and (4), h_{w_{j}}(w_{i}^{n}) = 0 for i < j. For a homomorphism ϕ : G → R, ϕ(\overline{w_{i}}) = 0 since \overline{w_{i}} ∈ [G, G] by Lemma 4.1(3). q.e.d.

5 Proofs of Th 1 and Cor 1,2,3

Proof of Th 1. Let w_{i}, 0 ≤ i < ∞ be the words in Prop 2. By Prop 1, all the cocycles δh_{w_{i}} have the same bound. It follows that if (a_{i})_{i} ∈ l^{1}, then ∑_{i} a_{i}δh_{w_{i}} is also a well-defined cocycle. We get a real linear map

ω : l^{1} → H_{b}^{2}(G; R),

which sends (a_{i})_{i} to the cohomology class of ∑_{i} a_{i}δh_{w_{i}}. In order to show ω is injective, suppose ω((a_{i})) = 0. Then

δ(∑_{i=0}^{∞} a_{i}h_{w_{i}}) = δb

for some b ∈ C_{b}^{1}(G; R). This means

∑_{i} a_{i}h_{w_{i}} − b = ϕ,

for some homomorphism ϕ : G → R. Applying this to \overline{w_{0}^{n}} ∈ G, we find

a_{0}n − b(\overline{w_{0}^{n}}) = ϕ(\overline{w_{0}^{n}}) = 0,

for all n ≥ 1 by Prop 2. Since b is bounded, a_{0} = 0. Similarly, a_{i} = 0 for all i ≥ 1. Thus ω is injective. It is well-known that the cardinality of the dimension of l^{1} as a vector space is continuum. q.e.d. Proof of Cor 1. W.l.o.g., |A| ≥ 3. Since C = {1}, |C\A/C| = |A| ≥ 3. Apply Th 1. q.e.d. Proof of Cor 2. Since |C| < ∞ and |A| = ∞, |C\A/C| = ∞. Apply Th 1.

q.e.d.

Proof of Cor 3. Since A is abelian, |C\A/C| = |A/C| ≥ 3. Apply Th 1. q.e.d.

6 Quasi homomorphisms of A ∗_{C} φ

Let G = A∗_{C} φ =< A, t; c = t^{−1}φ(c)t, ^{∀}c ∈ C > with |A/C| ≥ 2, |A/φ(C)| ≥ 2. Let g be

g = a_{1}t^{n_{1}}a_{2}t^{n_{2}} … a_{I}t^{n_{I}}a_{I+1},

with a_{i} ∈ A and n_{i} ≠ 0, where a_{1}, a_{I+1} may be empty. Suppose g satisfies the following two conditions (i) and (ii) for 1 ≤^{∀} i ≤ I − 1.

(i) If n_{i} > 0, n_{i+1} < 0, then a_{i+1} ∉ C. (ii) If n_{i} < 0, n_{i+1} > 0, then a_{i+1} ∉ φ(C). Then we say g is reduced.

The following fact is known as Britton’s lemma.

Lemma 6.1 (Britton, [LS]) Suppose 1 ≤ I. If g is reduced, then g ≠ 1 in G.

As an application of Britton’s lemma, we have:

Lemma 6.2 Let

g = a_{1}t^{n_{1}}a_{2}t^{n_{2}} … a_{I}t^{n_{I}}a_{I+1}, a_{i} ∈ A
h = b_{1}t^{m_{1}}b_{2}t^{m_{2}} … b_{J}t^{m_{J}} b_{J+1}, b_{j} ∈ A

be reduced with n_{i} = ±1, m_{j} = ±1. If g = h in G, then

I = J, n_{1} = m_{1}, …, n_{I} = m_{I},

and

a_{I+1}b_{I+1}^{−1}, t^{n_{I}}a_{I+1}b_{I+1}^{−1}t^{−m_{I}}, a_{I}t^{n_{I}}a_{I+1}b_{I+1}^{−1}t^{−m_{I}}b_{I}^{−1}, …
…, t^{n_{1}}a_{2}t^{n_{2}}a_{3} … a_{I+1}b_{I+1}^{−1} … b_{3}^{−1}t^{−m_{2}}b_{2}^{−1}t^{−m_{1}}

are in C or φ(C), especially, in A.

Proof. W.l.o.g., we assume J ≤ I. h^{−1} = b_{J+1}^{−1}t^{−m_{J}} b_{J}^{−1}t^{−m_{J−1}} … b_{1}^{−1} is reduced since h is reduced. Put

d_{I+1} = a_{I+1}b_{J+1}^{−1},

then d_{I+1} ∈ A and

gh^{−1} = (a_{1}t^{n_{1}}a_{2}t^{n_{2}} … a_{I}t^{n_{I}})d_{I+1}(t^{−m_{J}} b_{J}^{−1}t^{−m_{J−1}} … b_{1}^{−1})

in G. Since gh^{−1} = 1 in G, the expression on the right hand side is not reduced by Britton’s lemma. But, since a_{1}t^{n_{1}}a_{2}t^{n_{2}} … a_{I}t^{n_{I}} and t^{−m_{J}} b_{J}^{−1}t^{−m_{J−1}} … b_{1}^{−1} are reduced, we have

n_{I} = m_{J} = 1(or − 1), d_{I+1} ∈ C(φ(C), resp.),

and t^{n_{I}}d_{I+1}t^{−m_{J}} ∈ φ(C)(C, resp.).

Put

c_{I} = t^{n_{I}}d_{I+1}t^{−m_{J}} , d_{I} = a_{I}c_{I}b_{J}^{−1}.

Clearly d_{I} ∈ A and

gh^{−1} = (a_{1}t^{n_{1}}a_{2}t^{n_{2}} … a_{I−1}t^{n_{I−1}})d_{I}(t^{−m_{J−1}}b_{J−1}^{−1}t^{−m_{J−2}} … b_{1}^{−1})

in G. By Britton’s lemma, the expression on the right hand side is not reduced. Thus

n_{I−1} = m_{J−1} = 1(or − 1),

d_{I} ∈ C(φ(C), resp.), and c_{I−1} = t^{n_{I−1}}d_{I}t^{−m_{J−1}} ∈ φ(C)(C, resp.).

We define d_{I−1}, d_{I−2}, …, d_{I−J+2} and c_{I−1}, c_{I−2}, …, c_{I−J+1} inductively by

c_{i} = t^{n_{i}}d_{i+1}t^{−m_{J−I+i}}, d_{i} = a_{i}c_{i}b_{J−I+i}^{−1}.

See Figure 8.

Using Britton’s lemma repeatedly, we have

n_{I−J+j} = m_{j}, 1 ≤^{∀} j ≤ J

and

d_{I+1}, d_{I}, …, d_{I−J+2}, c_{I}, c_{I−1}, …, c_{I−J+1} ∈ C or φ(C).

Put

d_{I−J+1} = a_{I−J+1}c_{I−J+1}b_{1}^{−1}.

Clearly d_{I−J+1} ∈ A. To complete the proof, it suffices to show J = I. In order to show it by contradiction, suppose J < I. Then

gh^{−1} = a_{1}t^{n_{1}}a_{2} … t^{n_{I−J−1}}a_{I−J}t^{n_{I−J}} d_{I−J+1}

Figure 8: This illustrates Lemma 6.2. d_{i}, c_{i} ∈ A for I − J + 1 ≤^{∀} i ≤ I. If J < I, then it contradicts Britton’s lemma.

in G and the expression on the right hand side is reduced, which contradicts Britton’s lemma, since gh^{−1} = 1 in G. We got I = J. q.e.d.

We take {t} ∪ A\{1} as a set of generators of G and write the Cayley graph of G w.r.t. this set by Γ.

Lemma 6.3 If a path α is a geodesic in Γ, then it is reduced.

Proof. If a path α is not reduced, then we can make it shorter using tct^{−1} = φ(c), hance α is not a geodesic. q.e.d. Remark. Compare Lemma 6.3 with Lemma 3.1. A reduced path is not always a geodesic in this case. For example, the left hand side of φ(c)^{−1}tc = t for c ∈ C\{1} is reduced but not a geodesic.

Lemma 6.4 Let α be a path and w a word. If w^{2} is reduced, then there is a reduced path β which realizes c_{w} at α.

Proof. Take a path β which realizes c_{w} at α s.t. |β|_{w} is minimal among all the realizing paths. Then a similar argument to the proof of Lemma 3.2 shows β is reduced. We omit the detail. q.e.d.

Lemma 6.5 Let α, β be paths starting at 1. We have

|c_{w}(α) − c_{w}(β)| ≤ 2|\overline{α^{−1}β}|.
|h_{w}(α) − h_{w}(β)| ≤ 4|\overline{α^{−1}β}|.

Proof. Similar to Lemma 3.3 and 3.4. q.e.d.

Lemma 6.6 Let α be a reduced path. If α = α_{1}α_{2}, then

|h_{w}(α) − h_{w}(α_{1}) − h_{w}(α_{2})| ≤ 10.

Proof. Similar to Lemma 3.7. q.e.d.

Proposition 3

|δh_{w}| ≤ 78.

Proof. The outline is similar to the proof of Proposition 1. Let x, y be elements in G. We show |h_{w}(xy) − h_{w}(x) − h_{w}(y)| ≤ 78. Take reduced paths α, β, γ with \overline{α} = x, \overline{β} = y, \overline{γ} = xy. Since \overline{αβ} = \overline{γ} and α, β, γ are reduced, by Lemma 6.1 and 6.2, there exist subdivisions of α, β, γ,

α = α_{1}α_{2}α_{3}, β = β_{1}β_{2}β_{3}, γ = γ_{1}γ_{2}γ_{3},

s.t. \overline{γ_{1}^{−1}α_{1}} = c_{1}, \overline{α_{3}β_{1}} = c_{2}, \overline{β_{3}γ_{3}^{−1}} = c_{3}, for some c_{1}, c_{2}, c_{3} ∈ C ∪ φ(C), and \overline{α_{2}}, \overline{β_{2}}, \overline{γ_{2}} ∈ A. By Lemma 6.6,

|h_{w}(α) − h_{w}(α_{1}) − h_{w}(α_{2}) − h_{w}(α_{3})| ≤ 20.

Since |h_{w}(α_{2})| ≤ 2, |h_{w}(α) − h_{w}(α_{1}) − h_{w}(α_{3})| ≤ 22. Similarly,

|h_{w}(β) − h_{w}(β_{1}) − h_{w}(β_{3})| ≤ 22, |h_{w}(γ) − h_{w}(γ_{1}) − h_{w}(γ_{3})| ≤ 22.

By a similar argument to the proof of Prop 1,

|h_{w}(x) + h_{w}(y) − h_{w}(xy)| = |h_{w}(α) + h_{w}(β) − h_{w}(γ)| ≤ 78.

q.e.d.

7 Choice of words for A ∗_{C} φ

Let G = A ∗_{C} φ with |A/C| ≥ 2, |A/φ(C)| ≥ 2.

Lemma 7.1 Let w be

w = t^{n_{1}}a_{1}t^{n_{2}}a_{2} … t^{n_{I}}a_{I},

s.t. a_{i} ∈ A\{1}, n_{i} ∈ Z\{0} for 1 ≤^{∀} i ≤ I. We denote the set of the following conditions (1.1), …, (1.4) by Condition I and (2.1), …, (2.4) by Condition II. (1.1) 0 < n_{1}, n_{3}, n_{5}, . . .. (1.2) 0 > n_{2}, n_{4}, n_{6}, . . .. (1.3) a_{1}, a_{3}, a_{5}, … ∉ C. (1.4) a_{2}, a_{4}, a_{6}, … ∉ φ(C). (2.1) 0 > n_{1}, n_{3}, n_{5}, . . .. (2.2) 0 < n_{2}, n_{4}, n_{6}, . . .. (2.3) a_{1}, a_{3}, a_{5}, … ∉ φ(C). (2.4) a_{2}, a_{4}, a_{6}, … ∉ C.

If either the Condition I or II holds, then w is a geodesic in Γ.

Proof. We treat the case that the Condition I holds. The other case is similar. Under this condition, clearly w is reduced. Let γ be a geodesic with \overline{w} = \overline{γ}. Then by Lemma 6.3, γ is reduced. By Lemma 6.2, we have

γ = b_{0}τ_{1}b_{1}τ_{2}b_{2} … τ_{I}b_{I},

s.t. b_{i} ∈ A\{1} for 1 ≤ i ≤ I − 1, and b_{0}, b_{I} ∈ A\{1} or is empty, and

τ_{i} = tb_{i,1}tb_{i,2} … tb_{i,n_{i}−2}tb_{i,n_{i}−1}t, if i is odd,
τ_{i} = t^{−1}b_{i,1}t^{−1}b_{i,2} … t^{−1}b_{i,n_{i}−2}t^{−1}b_{i,n_{i}−1}t^{−1}, if i is even,

where b_{i,j} ∈ A\{1} or is empty, for 1 ≤^{∀} i ≤ I and 1 ≤^{∀} j ≤ n_{i} − 1.

We will show b_{I} is not empty. To show this by contradiction, assume b_{I} is empty. Then

wγ^{−1} = … ta_{I}t^{−1} …, if I is odd,
wγ^{−1} = … t^{−1}a_{I}t …, if I is even.

Since w and γ^{−1} are reduced, wγ^{−1} is reduced by (1.3) if I is odd or by (1.4) if I is even. Then it follows from Britton’s lemma that \overline{wγ^{−1}} ≠ 1, a contradiction. We got b_{I} is not empty. Thus

|γ| ≥ ∑_{i=1}^{I} |τ_{i}| + I ≥ ∑_{i=1}^{I} |n_{i}| + I = |w|.

We have the first inequality because b_{1}, …, b_{I} are not empty, and the second one since the number of t’s in τ_{i} is n_{i} for 1 ≤^{∀} i ≤ I. Since γ is a geodesic, we have |γ| = |w|, hence w is a geodesic. q.e.d.

Lemma 7.2 Take g ∈ A\C and h ∈ A\φ(C) and fix them. Let w_{i}, 0 ≤ i < ∞, be words s.t.

w_{i} = t^{10^{i}}gt^{−10^{i}}ht^{10^{i}}g^{−1}t^{−10^{i}}h^{−1}t^{2·10^{i}}gt^{−2·10^{i}}ht^{3·10^{i}}g^{−1}t^{−3·10^{i}}h^{−1}.

Then the words satisfy the following conditions.

(1) For any i ≥ 0 and any n ≥ 1, c_{w_{i}}(w_{i}^{n}) = n.

(2) For any i ≥ 0 and any n ≥ 1, c_{w_{i}^{−1}}(w_{i}^{n}) = 0.

(3) For any j > i ≥ 0 and any n ≥ 1, c_{w_{j}^{±1}}(w_{i}^{n}) = 0.

(4) For any i ≥ 0, \overline{w_{i}} ∈ [G, G].

Proof. (1) By Lemma 7.1, w_{i}^{n} is a geodesic. Thus c_{w_{i}}(w_{i}^{n}) = n.

(2) In order to show c_{w_{0}^{−1}}(w_{0}^{n}) = 0 for any n ≥ 1 by contradiction, suppose c_{w_{0}^{−1}}(w_{0}^{n}) > 0 for some n. By Lemma 6.4, take a reduced path α s.t. \overline{α} = \overline{w_{0}^{n}} and |α|_{w_{0}^{−1}} > 0. We describe the order of t’s and t^{−1}’s appearing in w_{0} by W_{0},

W_{0} = + − + − + + − − + + + − − −,

where + stands for t, − for t^{−1} and we ignore g^{±1}, h^{±1} in w_{0}. Then W_{0}^{n} represents w_{0}^{n}. Let Λ be the order of t’s and t^{−1}’s in α. Applying Lemma 6.2 for α and w_{0}^{n}, we get W_{0}^{n} = Λ. Λ contains W_{0}^{−1} as a subpiece since |α|_{w_{0}^{−1}} > 0, where

W_{0}^{−1} = + + + − − − + + − − + − + − .

But it is easy to check that W_{0}^{n} cannot contain W_{0}^{−1} as a subset, thus we get a contradiction. Hence c_{w_{0}^{−1}}(w_{0}^{n}) = 0. Similarly, we can show that c_{w_{i}^{−1}}(w_{i}^{n}) = 0 for any i ≥ 1 and any n ≥ 1.

(3) Let W_{i}, 1 ≤ i < ∞ be the order of t’s and t^{−1}’s in w_{i}. To show c_{w_{j}}(w_{i}^{n}) = 0 for all n ≥ 1 and all j > i > 0 by contradiction, assume c_{w_{j}}(w_{i}^{n}) > 0 for some n ≥ 1 and some j > i > 0. Then, by our assumption, W_{j} must be contained in W_{i}^{n} as a subword. This is a contradiction since W_{j} has 3 · 10^{j} consecutive t’s, but W_{i} has at most 3 · 10^{i} consecutive t’s. Thus we have c_{w_{j}}(w_{i}^{n}) = 0. c_{w_{j}^{−1}}(w_{i}^{n}) = 0 is similar.

(4) Set T = t^{10^{i}}, then

w_{i} = [T, g][gh, [T, g^{−1}] ][T, g^{−1}][g, h][T^{2}, g][gh, [T^{3}, g^{−1}] ][T^{3}, g^{−1}][g, h],

thus \overline{w_{i}} ∈ [G, G]. q.e.d.

Proposition 4 Let G = A ∗_{C} φ with |A/C| ≥ 2, |A/φ(C)| ≥ 2. There exist words w_{i}, 0 ≤ i < ∞ which satisfy the following conditions.

(1) For any i ≥ 0 and any n ≥ 1, h_{w_{i}}(w_{i}^{n}) = n.

(2) For any j > i ≥ 0 and any n ≥ 1, h_{w_{j}}(w_{i}^{n}) = 0.

(3) For any i ≥ 0, \overline{w_{i}} ∈ [G, G].

Proof. The words w_{i}, 0 ≤ i < ∞ in Lemma 7.2 clearly satisfy (1), (2) and (3). q.e.d.

8 Proofs of Th 2 and Th 3

Proof of Th 2. We can show Th 2 using Prop 3 and Prop 4 as we showed Th 1 using Prop 1 and Prop 2. We omit the detail. q.e.d. Proof of Th 3. By J. Stallings’ structure theorem [S], G is either

(1) A ∗_{C} B with |C| < ∞, |A/C| ≥ 3 and |B/C| ≥ 2, or

(2) A ∗_{C} φ with |C| < ∞ and |A/C| ≥ 2.

In the case (1), if |A| = ∞ or |B| = ∞, then we have the conclusion by Cor 2. If |A| < ∞ and |B| < ∞, then G is word-hyperbolic. Since G has infinitely many ends, it is non-elementary. It is known that the conclusion of Th 3 holds for a non-elementary word-hyperbolic group, [EF]. In the case (2), since |A/C| ≥ 2 and |C| < ∞, we have |A/φ(C)| ≥ 2. Apply Th 2.

q.e.d.

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