RIGIDITY OF GROPE GROUPS

MATIJA CENCELJ, KATSUYA EDA, AND ALEŠ VAVPETIČ

Abstract. We give a systematic definition of the fundamental groups of gropes, which we call grope groups. We show that there exists a nontrivial homomorphism from the minimal grope group M to another grope group G only if G is the free product of M with another grope group.

1. Introduction

Here we study groups whose classifying spaces are (open infinite) gropes (a recent short note on gropes in general is [10]). In algebra these groups first appeared in the proof of a lemma by Alex Heller [7] as follows. Let φ_{0} be a homomorphism from the free group F_{0} on one generator α to any perfect group P. Let

φ_{0}(α) = [p_{0}, p_{1}][p_{2}, p_{3}] ⋯ [p_{2n−2}, p_{2n−1}] ∈ P (∗)

then we can extend φ_{0} to a homomorphism φ_{1} of a (nonabelian) free group F_{1} on 2n generators β_{0}, …, β_{2n−1} by setting φ_{1}(β_{i}) = p_{i}. Note that φ_{0}(α_{0}) may have several different expressions as a product of commutators, so we may choose any; even if some of the elements p_{1}, …, p_{2n−1} coincide we let all elements β_{i} to be distinct. Now we repeat the above construction for every homomorphism φ_{1}|_{⟨β_{i}⟩} of the free group on one generator to P and thus obtain a homomorphism φ_{2} : F_{2} → P. Repeating the above construction we obtain a direct system of inclusions of free groups F_{1} → F_{2} → F_{3} → ⋯ and homomorphisms φ_{n} : F_{n} → P. The direct limit of F_{n} is a locally free perfect group D and every group obtained by the above construction is called a grope group (and its clasifying space is a grope). This construction shows therefore that every homomorphism from a free group on one generator to a perfect group P can be extended to a homomorphism from a grope group to P. Note that in case the perfect group P has the Ore property ([8], [6]) that every element in P is a commutator, in the above process (∗) we can make every generator in the chosen basis of F_{n} a single commutator of two basis elements of F_{n+1}. The group obtained in this way is the minimal grope group M. Clearly every grope group admits many epimorphisms onto M. In the sequel we show that M admits a nontrivial homomorphism to another grope group G only if the latter is the free product G ≅ M ∗ K where K is a grope group.

Date: July 29, 2006.

1991 Mathematics Subject Classification. (MSC 2000): 20F22, 20F12, 20F38 .

Key words and phrases. group, grope.

Supported in part by the Slovenian-Japanese research grant BI–JP/05-06/2, ARRS research program No. 0101-509, the ARRS research project of Slovenia No. J1–6128–0101–04 and the Grant-in-Aid for Scientific research (C) of Japan No. 16540125.

Gropes were introduced by Štan’ko [9]. They have an important role in geometric topology ([3], for more recent use in dimension theory see [5] and [4]). Their fundamental groups were used by Berrick and Casacuberta to show that the plus-construction in algebraic Ktheory is localization [2]. Recently [1] such a group has appeared in the construction of a perfect group with a nonperfect localization.

In the first part of the paper we give a systematic definition of grope groups and prove some technical lemmas. In the second part we prove that the minimal grope group admits nontrivial homomorphisms to almost no other grope group thus proving that there exist at least two distinct grope groups.

2. Systematic definition of grope groups and basic facts

For every positive integer n let \underline{n} = \{0, 1, …, n − 1\}. The set of non-negative integers is denoted by ℕ. We denote the set of finite sequences of elements of a set X by Seq(X) and the length of a sequence s ∈ Seq(X) by lh(s). The empty sequence is denoted by ∅.

For a non-empty set A let L(A) be the set {a, a^{−} : a ∈ A}, which we call the set of letters. We identify (a^{−})^{−} with a. Let W(A) = Seq(L(A)), which we call the set of words. For a word W ≡ a_{0} ⋯ a_{n}, define W^{−} ≡ a_{n}^{−} ⋯ a_{0}^{−}. We write W ≡ W′ for identity in W(A) while W = W′ for identity in the free group generated by A. For instance aa^{−} = ∅ but aa^{−} ≢ ∅. We adopt [a, b] = aba^{−1}b^{−1} as the definition of a commutator.

To describe all the grope groups we introduce some notation.

A grope frame S is a subset of Seq(ℕ) satisfying: ∅ ∈ S and for every s ∈ S there exists n > 0 such that \underline{2n} = \{i ∈ ℕ : si ∈ S\}.

For each grope frame S we induce formal symbols c_{s}^{S} for s ∈ S and define E_{m}^{S} = {c_{s}^{S} : lh(s) = m, s ∈ S} and a free group F_{m}^{S} = ⟨E_{m}^{S}⟩. Then define e_{m}^{S} : F_{m}^{S} → F_{m+1}^{S} by: e_{m}^{S}(c_{s}^{S}) = [c_{s0}^{S}, c_{s1}^{S}] ⋯ [c_{s 2k−2}^{S}, c_{s 2k−1}^{S}] where \underline{2k} = \{i ∈ ℕ : si ∈ S\}. Let G^{S} = \underrightarrow{lim}(F_{m}^{S}, e_{m}^{S} : m ∈ ℕ) and e_{mn}^{S} = e_{n−1}^{S} ⋯ e_{m}^{S} for m ≤ n and every such group G^{S} is a grope group.

For s ∈ S, s is binary branched, if \{i ∈ ℕ : si ∈ S\} = \underline{2}. Let S_{0} be a grope frame such that every s ∈ S_{0} is binary branched, i.e. S_{0} = Seq(\underline{2}). Then G^{S_{0}} = M is the so-called minimal grope group. Since e_{m}^{S} is injective, we frequently regard F_{m}^{S} is a subgroup of G^{S}.

For a non-empty word W the head of W is the left most letter b of W, i.e. W ≡ bX for some word X, and the tail of W is the right most letter c of W, i.e. W ≡ Y c for some word Y. When AB ≡ W, we say that A is the head part of W and B is the tail part of W. For a word W ∈ W(E_{m}^{S}) and n ≥ m, we let e_{m,n}^{S}[W] be a word in W(E_{n}^{S}) defined as follows: e_{m,m}^{S}[W] ≡ W and e_{m,n+1}^{S}[W] is obtained by replacing every c_{t} in e_{m,n}^{S}[W] by

(P0) c_{t0}^{S}c_{t1}^{S}c_{t0}^{S−}c_{t1}^{S−} ⋯ c_{t 2k−2}^{S}c_{t 2k−1}^{S}c_{t 2k−2}^{S−}c_{t 2k−1}^{S−}

and every c_{t}^{S−} by

(P1) c_{t 2k−1}^{S}c_{t 2k−2}^{S}c_{t 2k−1}^{S−}c_{t 2k−2}^{S−} ⋯ c_{t1}^{S}c_{t0}^{S}c_{t1}^{S−}c_{t0}^{S−}

respectively.

We drop the superscript ^{S}, if no confusion can occur.

For a reduced word W ∈ W(E_{n}) with W ∈ F_{m} for m < n, let W_{0} ∈ W(E_{m}) such that e_{m,n}[W_{0}] ≡ W. (The existence of W_{0} is assured in Lemma 2.2.) A subword V of W is small, if there exists a letter c_{s} or c_{s}^{−} in W_{0} and i ∈ ℕ such that V is a subword of e_{m+1,n}[c_{si}] or e_{m+1,n}[c_{si}^{−}] respectively. (Note that being small depends on m. In the following usage of this notion m and n are always fixed in advance.)

Observation 1. Let n > m+1 and let W ≡ e_{m+1,n}[c_{s0}]. Suppose that X ∈ W(E_{n}) is a reduced word and X ∈ F_{m}. When W is a subword of X, W may appear in e_{m,n}[c_{s}] or e_{m,n}[c_{s}^{−}] and hence we cannot uniquely determine a successive letter to W in X. However, if X ≡ W Y for some Y, the head of Y is uniquely determined as c_{s10···0}. Also if we know the preceding letter to W, i.e. X ≡ ZW Y and we know the tail of Z which is c_{t10···0}^{−} or c_{t0···0}^{−} for some t or c_{s110···0}^{−}, the head of Y is uniquely determined. That is, the head is c_{s10···0}, if the preceding letter is c_{t10···0}^{−} or c_{t0···0}^{−} for some t and the head is c_{s110···0}, if the preceding letter is c_{s110···0}^{−}. (In the above we ignore i-digits for i > n.)

Observation 2. A letter c_{s0···0} ∈ W(E_{n}) for lh(s) = m possibly appears in e_{m,n}[W_{0}] in the following cases. When n = m + 1, c_{s0} appears once in e_{m,n}[c_{s}] and also once in e_{m,n}[c_{s}^{−}]. According to the increase of n, c_{s0···0} appears in many parts. c_{s0···0} appears 2^{n−m−1}-times in e_{m,n}[c_{s}] and also 2^{n−m−1}-times in e_{m,n}[c_{s}^{−}].

Lemma 2.1. For a word W ∈ W(E_{m}) and n ≥ m, e_{m,n}[W] is reduced, if and only if W is reduced. Lemma 2.2. For a reduced word V ∈ W(E_{n}) and n ≥ m, V ∈ F_{m} if and only if there exists W ∈ W(E_{m}) such that e_{m,n}[W] ≡ V.

Proof. The sufficiency is obvious. To see the other direction, let W be a reduced word in W(E_{m}) such that e_{m,n}[W] = V in F_{n}. By Lemma 2.1 e_{m,n}[W] is reduced. Since every element in F_{n} has a unique reduced word in W(E_{n}) presenting itself, we have e_{m,n}[W] ≡ V. □

Lemma 2.3. Let m < n and A be a non-empty word in W(E_{n}). Let X_{0}AY_{0} and X_{1}AY_{1} be reduced words in W(E_{n}) satisfying X_{0}AY_{0}, X_{1}AY_{1} ∈ F_{m}.

(1) If A is not small, X_{0}A ∉ F_{m} and X_{1}A ∉ F_{m}, then the heads of Y_{0} and Y_{1} are the same.

(2) Let X_{0} be an empty word. If A is not small and A ∉ F_{m}, the heads of Y_{0} and Y_{1} are the same.

(3) Let X_{0} and X_{1} be empty words. If A ∉ F_{m}, the heads of Y_{0} and Y_{1} are the same.

Proof. (1) Since X_{0}AY_{0} ∈ F_{m} but X_{0}A ∉ F_{m}, we have a letter c ∈ E_{m} ∪ E_{m}^{−} and words U_{0}, U_{1}, U_{2} such that U_{1} ≢ ∅, U_{2} ≢ ∅, X_{0}A ≡ U_{0}U_{1} and U_{1}U_{2} ≡ e_{m,n}[c]. Since A is not small, c and U_{0}, U_{1}, U_{2} are uniquely determined by A. Since the same thing holds for X_{1}AY_{1}, we have the conclusion by Observation 1 for n > m + 1. (The case for n = m + 1 is easier.)

(2) Since AY_{0} ∈ F_{m}, A ∉ F_{m} and A is not a small word, for any word B such that BA is reduced we have BA ∉ F_{m}. In particular X_{1}A ∉ F_{m} and the conclusion follows from (1).

(3) Since AY_{0} ∈ F_{m}, there are A_{0} and non-empty U_{0}, U_{1} such that A_{0} ∈ F_{m}, A ≡ A_{0}U_{0} and U_{0}U_{1} ≡ e_{m,n}[c] for some c ∈ E_{m} ∪ E_{m}^{−}. Since A ∉ F_{m}, the head of U_{1} is uniquely determined by A and hence the heads of Y_{0} and Y_{1} are the same (Observation 1). □

Lemma 2.4. Let m < n and A, X, Y in W(E_{n}) and AXA^{−}Y ∈ F_{m}. If AXA^{−}Y is reduced and A is not small, then AXA^{−} ∈ F_{m} and Y ∈ F_{m}.

Proof. The head of the reduced word in W(E_{m}) for the element AXA^{−}Y is c_{s} or c_{s}^{−} for c_{s} ∈ E_{m}. According to c_{s} or c_{s}^{−}, A ≡ e_{m+1,n}[c_{s0}]Z or e_{m+1,n}[c_{sk}]Z for a non-empty word Z, where \underline{k + 1} = \{i ∈ ℕ : si ∈ S\} is even. Then A^{−} ≡ Z^{−}e_{m+1,n}[c_{s0}^{−}] or A^{−} ≡ Z^{−}e_{m+1,n}[c_{sk}^{−}] and hence AXA^{−} ∈ F_{m} and consequently Y ∈ F_{m}. □

Lemma 2.5. For e ≠ x ∈ F_{m}^{S} and u ∈ G^{S}, uxu^{−1} ∈ F_{m}^{S} implies u ∈ F_{m}^{S}.

Proof. There exists n ≥ m such that u ∈ F_{n}. Let W be a cyclically reduced word and V be a reduced word such that x = V W V^{−} in F_{m} and V W V^{−} is reduced. Then e_{m,n}(x) = e_{m,n}[V]e_{m,n}[W]e_{m,n}[V]^{−} and e_{m,n}[V] is reduced and e_{m,n}[W] is cyclically reduced by Lemma 2.1. Let U be a reduced word for u in F_{n}. Let k = lh(U). Then e_{m,n}(x^{2k+1}) = e_{m,n}[V]e_{m,n}[W]^{2k+1}e_{m,n}[V]^{−} and the right hand term is a reduced word. Hence the reduced word for ux^{k}u^{−} of the form Xe_{m,n}[W]Y, where Ue_{m,n}[V]e_{m,n}[W]^{k} = X and e_{m,n}[W]^{k}e_{m,n}[V]^{−}U^{−} = Y. Since ux^{k}u^{−1} ∈ F_{m}, X ∈ F_{m} and Y ∈ F_{m}. Now we have Ue_{m,n}[V] ∈ e_{m,n}(F_{m}) and hence U ∈ e_{m,n}(F_{m}), which implies the conclusion. □

Lemma 2.6. Let UW U^{−} be a reduced word in W(E_{n}). If UW U^{−} ∈ F_{m} and W is cyclically reduced, then U, W ∈ F_{m}.

Proof. If U is empty or n = m, then the conclusion is obvious. If U ∈ F_{m}, then W U^{−} ∈ F_{m} and so W ∈ F_{m}. Suppose that U is U ∉ F_{m}. Since UW U^{−}, UW^{−}U^{−} ∈ F_{m}, the head of W and that of W^{−} is the same by Lemma 2.3 (3), which contradicts that W is cyclically reduced. □

Lemma 2.7. Let XY and Y X be reduced words in W(E_{n}) for n ≥ m. If XY and Y X belong to F_{m}, then both of X and Y belong to F_{m}.

Proof. We may assume n > m. When n > m, the head of e_{m,n}[W] for a non-empty word W ∈ W(E_{m}) is c_{s0···0} or c_{sk0···0} where lh(s) = m and k + 1 = {i ∈ ℕ : si ∈ S} is even. (When n = m + 1, there appears no 0 ⋯ 0.) Since X^{−}Y^{−} ∈ F_{m} and X^{−}Y^{−} is reduced, the tail of X is of the form c_{s0···0}^{−} or c_{sk0···0}^{−}. We only deal with the former case. Suppose that X ∉ F_{m}. Since XY ∈ F_{m} and XY is reduced, X ≡ Ze_{m+1,n}[c_{s1}c_{s0}^{−}] for some Z. This implies X^{−} ≡ e_{m+1,n}[c_{s0}c_{s1}^{−}]Z^{−}, which contradicts that X^{−}Y^{−} ∈ F_{m} and X^{−}Y^{−} is reduced. Now we have X, Y ∈ F_{m}. □

Lemma 2.8. Let m < n and A, B, C in W(E_{n}) and e ≠ ABCA^{−}B^{−}C^{−} ∈ F_{m}. If ABCA^{−}B^{−}C^{−} is a reduced word and at least one of A, B, C is not small, then A, B, C ∈ F_{m}.

Proof. Since ABCA^{−}B^{−}C^{−} ≠ e, at most one of A, B, C is empty. When C is empty, the conclusion follows from Lemma 2.4 and the fact that BAB^{−}A^{−} is also reduced and BAB^{−}A^{−} ∈ F_{m}.

Now we assume that A, B, C are non-empty. If A is not small, then ABCA^{−} ∈ F_{m} and B^{−}C^{−} ∈ F_{m} by Lemma 2.4. Since BC is cyclically reduced, A ∈ F_{m} and BC ∈ F_{m} by Lemma 2.6. The conclusion follows from Lemma 2.7. In the case that C is not small, the argument is similar. The remaining case is when A and C are small. Then ABCA^{−}B^{−}C^{−} ∈ F_{m} and CBAC^{−}B^{−}A^{−} ∈ F_{m} imply A ≡ C, which contradicts the reducedness of ABCA^{−}B^{−}C^{−}. □

Lemma 2.9. Let m < n and A, B, C in W(E_{n}) and e ≠ ABCA^{−}B^{−}C^{−} ∈ F_{m}. If ABCA^{−}B^{−}C^{−} is a reduced word and A, B, C are small, then one of A, B, C is empty.

Assume C is empty. Then there exists c_{s} ∈ E_{m} such that s is binary branched and either

A ≡ e_{m+1,n}[c_{s0}] and B ≡ e_{m+1,n}[c_{s1}],

or

A ≡ e_{m+1,n}[c_{s1}] and B ≡ e_{m+1,n}[c_{s0}].

Proof. Since A, B, C are small, all the words A, B, C and their inverses must be subwords of e_{m+1,n}[c_{si}], i = 0, 1, or e_{m+1,n}[c_{si}^{−}], for an element c_{s} ∈ E_{m}, and in particular that either

ABCA^{−}B^{−}C^{−} = e_{m,n}(c_{s}) = e_{m+1,n}[c_{s0}c_{s1}c_{s0}^{−}c_{s1}^{−}]

or

ABCA^{−}B^{−}C^{−} = e_{m,n}(c_{s}^{−}) = e_{m+1,n}[c_{s1}c_{s0}c_{s1}^{−}c_{s0}^{−}],

where the left most and right most terms are reduced words. We remark that if the cardinality of {i ∈ ℕ : si ∈ S} were greater than 2, one of A, B, C would not be small; hence in our case s is binary branched.

We only deal with the first case. Then ABC ≡ e_{m+1,n}[c_{s0}c_{s1}] and A^{−}B^{−}C^{−} ≡ e_{m+1,n}[c_{s0}^{−}c_{s1}^{−}]. In case A, B, C are non-empty, A is a proper subword of e_{m+1,n}[c_{s0}] or C is a proper subword of e_{m+1,n}[c_{s1}]. In either case A^{−}B^{−}C^{−} ≡ e_{m+1,n}[c_{s0}^{−}c_{s1}^{−}] does not hold. Hence one of A, B, C is empty. We may assume C is empty. Since A, B are small, A ≡ e_{m+1,n}[c_{s0}] and B ≡ e_{m+1,n}[c_{s1}]. □

3. Proof of Theorem 3.1

In this section we prove Theorem 3.1. The minimal grope group M = G^{S_{0}} admits a nontrivial homomorphism into a grope group G^{S}, if and only if there exists s ∈ S such that a frame {t ∈ Seq(ℕ) : st ∈ S} is equal to S_{0}.

It is easy to see that the condition on G^{S} in the above theorem is equivalent to G^{S} ≅ M ∗ K, where K is another grope group.

In our proof of Lemma 3.9 we analyze a reduction procedure of a word Y^{−}ABY X^{−}A^{−}B^{−}X where Y^{−}ABY and X^{−}A^{−}B^{−}X are reduced. Lemmas 3.2, 3.3, 3.4 and 3.5 show connections between our reduction steps in case at least one of X and Y is empty. Lemma 3.6 corresponds to the final step, i.e. when we have the reduced word. Lemmas 3.7 and 3.8 correspond to the case that X and Y are non-empty. In the following lemmas we assume m < n.

Lemma 3.2. Let A, B ∈ W(E_{n}) be non-empty reduced words such that ABA^{−}B^{−} ≠ e and AB and A^{−}B^{−} are reduced words. Then the following hold:

(1.1) If B ≡ B_{0}A, then B_{0} is non-empty, AB_{0} and A^{−}B_{0}^{−} are reduced words and AB_{0}A^{−}B_{0}^{−} = ABA^{−}B^{−}. In addition if AB_{0}, A^{−}B_{0}^{−} ∈ F_{m}, then AB, A^{−}B^{−} ∈ F_{m}.

(1.2) If A ≡ A_{0}B, then A_{0} is non-empty, A_{0}B and A_{0}^{−}B^{−} are reduced words and A_{0}BA_{0}^{−}B^{−} = ABA^{−}B^{−}. In addition if A_{0}B, A_{0}^{−}B^{−} ∈ F_{m}, then AB, A^{−}B^{−} ∈ F_{m}.

(1.3) If A ≡ A_{0}Z and B ≡ B_{0}Z for non-empty words A_{0} and B_{0} and B_{0}A_{0}^{−} is reduced, then A_{0}ZB_{0}A_{0}^{−}Z^{−}B_{0}^{−} is reduced and A_{0}ZB_{0}A_{0}^{−}Z^{−}B_{0}^{−} = ABA^{−}B^{−}. In addition if A_{0}, B_{0}, Z ∈ F_{m}, then AB, A^{−}B^{−} ∈ F_{m}.

Proof. We only show (1.1). The non-emptiness of B_{0} follows from ABA^{−}B^{−} ≠ e. Since AB and A^{−}B^{−} are reduced, AB_{0} and A^{−}B_{0}^{−} are cyclically reduced and hence the second statement follows from Lemma 2.7. □

Lemma 3.3. Let A, B, C ∈ W(E_{n}) be reduced words (possibly empty) such that ABCA^{−}B^{−}C^{−} ≠ e and AB and CA^{−}B^{−}C^{−} are reduced words. Then the following hold:

(2.1) If B ≡ B_{0}C^{−}, then AB_{0} and A^{−}CB_{0}^{−}C^{−} are reduced words and AB_{0}A^{−}CB_{0}^{−}C^{−} = ABCA^{−}B^{−}C^{−}. In addition if AB_{0}A^{−}, CB_{0}^{−}C^{−} ∈ F_{m}, then AB, CA^{−}B^{−}C^{−} ∈ F_{m}.

(2.2) If C ≡ B^{−}C_{0}, then AC_{0} and A^{−}B^{−}C_{0}^{−}B are reduced words and AC_{0}A^{−}B^{−}C_{0}^{−}B = ABCA^{−}B^{−}C^{−}. In addition if AC_{0}A^{−}, B^{−}C_{0}^{−}B ∈ F_{m}, then AB, CA^{−}B^{−}C^{−} ∈ F_{m}.

(2.3) If B ≡ B_{0}Z^{−} and C ≡ ZC_{0} for non-empty words B_{0} and C_{0} and B_{0}C_{0} is reduced, then AB_{0}C_{0}A^{−}ZB_{0}^{−}C_{0}^{−}Z^{−} is reduced and AB_{0}C_{0}A^{−}ZB_{0}^{−}C_{0}^{−}Z^{−} = ABCA^{−}B^{−}C^{−}. In addition if AB_{0}C_{0}A^{−}, ZB_{0}^{−}C_{0}^{−}Z^{−} ∈ F_{m}, then AB, CA^{−}B^{−}C^{−} ∈ F_{m}.

Proof. (2.1) The first proposition is obvious. Let B_{0} ≡ XB_{1}X^{−} for a cyclically reduced word B_{1}. Since (AX)B_{1}(AX)^{−}, (CX)B_{1}^{−}(CX)^{−} ∈ F_{m}, AX, CX, B_{1} ∈ F_{m} by Lemma 2.6. Now AB = (AX)B_{1}(CX)^{−} ∈ F_{m} and CA^{−}B^{−}C^{−} = (CX)(AX)^{−}(CB_{0}^{−}C^{−}) ∈ F_{m}. We see (2.2) similarly.

For (2.3) observe the following. Since the both B_{0} and C_{0} are nonempty, B_{0}C_{0} and B_{0}^{−}C_{0}^{−} are cyclically reduced. Hence, using Lemmas 2.6 and 2.7, we have (2.3). □

The next two lemmas are straightforward and we omit the proofs. Lemma 3.4. Let A, B, C ∈ W(E_{n}) be reduced words (possibly empty) such that ABA^{−}CB^{−}C^{−} ≠ e and AB and A^{−}CB^{−}C^{−} are reduced. Then the following hold: (3.1) If A ≡ A_{0}B, then A_{0}B and A_{0}^{−}CB^{−}C^{−} are reduced and A_{0}BA_{0}^{−}CB^{−}C^{−} =

ABA^{−}CB^{−}C^{−}. In addition if A_{0}BA_{0}^{−}, CB^{−}C^{−} ∈ F_{m}, then ABA^{−}, CB^{−}C^{−} ∈ F_{m}.

(3.2) If B ≡ B_{0}A, then AB_{0} and CA^{−}B_{0}^{−}C^{−} are reduced and AB_{0}CA^{−}B_{0}^{−}C^{−} = ABA^{−}CB^{−}C^{−}. In addition if AB_{0}, CA^{−}B_{0}^{−}C^{−} ∈ F_{m}, then ABA^{−}, CB^{−}C^{−} ∈ F_{m}.

(3.3) If B ≡ B_{0}Z and A ≡ A_{0}Z for non-empty words A_{0} and B_{0} and B_{0}A_{0}^{−} is reduced, then A_{0}ZB_{0}A_{0}^{−}CZ^{−}B_{0}^{−}C^{−} is reduced. In addition if A_{0}ZB_{0}A_{0}^{−}, CZ^{−}B_{0}^{−}C^{−} ∈ F_{m}, then ABA^{−}, CB^{−}C^{−} ∈ F_{m}.

Lemma 3.5. Let A, B, C ∈ W(E_{n}) be reduced words (possibly empty) such that ABA^{−}CB^{−}C^{−} ≠ e and A and BA^{−}CB^{−}C^{−} are reduced words. Then the following hold: (4.1) If A ≡ A_{0}B^{−}, A_{0} and BA_{0}^{−}CB^{−}C^{−} are reduced and A_{0}BA_{0}^{−}CB^{−}C^{−} =

ABA^{−}CB^{−}C^{−}. In addition if A_{0}BA_{0}^{−}, CB^{−}C^{−} ∈ F_{m}, then ABA^{−}, CB^{−}C^{−} ∈ F_{m}.

(4.2) If B ≡ A^{−}B_{0}, and B_{0}A^{−}CB_{0}^{−}AC^{−} is reduced and B_{0}A^{−}CB_{0}AC^{−} = ABA^{−}CB^{−}C^{−}. In addition if B_{0}A^{−}, CB_{0}^{−}AC^{−} ∈ F_{m}, then ABA^{−}, CB^{−}C^{−} ∈ F_{m}.

(4.3) If A ≡ A_{0}Z^{−} and B ≡ ZB_{0} for non-empty words A_{0}, B_{0} and A_{0}B_{0} is reduced, then A_{0}B_{0}ZA_{0}^{−}CB_{0}^{−}Z^{−}C^{−} is reduced and A_{0}B_{0}ZA_{0}^{−}CB_{0}^{−}Z^{−}C^{−} = ABA^{−}CB^{−}C^{−}. In addition if A_{0}B_{0}ZA_{0}^{−}, CB_{0}^{−}Z^{−}C^{−} ∈ F_{m}, then ABA^{−}, CB^{−}C^{−} ∈ F_{m}.

Lemma 3.6. Let A, B, C, D ∈ W(E_{n}) be reduced non-empty words.

(1) if ABA^{−}B^{−} is reduced and ABA^{−}B^{−} ∈ F_{m} and at least one of A, B is not small, then A, B ∈ F_{m};

(2) if ABCA^{−}B^{−}C^{−} is reduced and ABCA^{−}B^{−}C^{−} ∈ F_{m} at least one of A, B, C is not small, then A, B, C ∈ F_{m};

(3) if CABC^{−}DA^{−}B^{−}D^{−} is reduced and CABC^{−}DA^{−}B^{−}D^{−} ∈ F_{m}, then A, B, C, D ∈ F_{m}.

(4) if CAC^{−}DA^{−}D^{−} is reduced and CAC^{−}DA^{−}D^{−} ∈ F_{m}, then CAC^{−}, DA^{−}D^{−} ∈ F_{m}.

Proof. The statements (1) and (2) are paraphrases of Lemma 2.8.

(3) Let c be the head of C and d be the tail of D^{−}. Since c^{−} and d^{−} are contiguous, we have CABC^{−}, DA^{−}B^{−}D^{−} ∈ F_{m}. Since AB and A^{−}B^{−} are reduced and the both A and B are non-empty, AB is cyclically reduced. Now the conclusion follows from Lemmas 2.6 and 2.7.

(4) This follows from a reasoning in the proof of (3). □

Lemma 3.7. Let A^{−}B^{−} and X_{0}ABX_{0}^{−} be reduced words such that X_{0}AB ≡ BAX_{1} for some X_{1}. If lh(X_{0}) ≤ lh(B), then there exist A′, B′ such that lh(B′) < lh(B), (A′)^{−}(B′)^{−} and X_{0}A′B′X_{0}^{−} are reduced words, X_{0}A′B′ ≡ B′A′X_{1}, A^{−}B^{−}X_{0}ABX_{0}^{−} = (A′)^{−}(B′)^{−}X_{0}A′B′X_{0}^{−}, and A, B ∈ ⟨X_{0}, A′, B′⟩.

Proof. First we remark that lh(X_{0}) ≠ lh(B) since BX_{0}^{−} is reduced. Hence lh(B) > lh(X_{0}). If lh(B) = lh(X_{0}) + lh(A), then we have X_{0}A ≡ B ≡ AX_{1} and have the conclusion, i,e, A′ ≡ A and B′ ≡ ∅.

If lh(B) < lh(X_{0}) + lh(A), we have k > 0 and A_{0}, A_{1} such that B ≡ X_{0}A_{0}A_{1}, A ≡ (A_{0}A_{1})^{k}A_{0}, and A_{1} is non-empty. (We remark that A_{0} may be empty.) Let A′ ≡ A_{0} and B′ ≡ A_{1}. Since lh(X_{0}) + lh(A) = lh(B) + (k − 1)lh(A_{0}A_{1}) + lh(A_{0}), we have B ≡ A_{1}A_{0}X_{1}. Let A′ ≡ A_{0} and B′ ≡ A_{1}, then we have the conclusion.

If lh(B) > lh(X_{0}) + lh(A), we have k > 0 and B_{0}, B_{1} such that B_{0}B_{1} ≡ X_{0}A, B ≡ (B_{0}B_{1})^{k}B_{0}, and B_{1} is non-empty. (We remark that B_{0} may be empty.) Since lh(B_{1}B_{0}) = lh(AX_{1}), we have B_{1}B_{0} ≡ AX_{1}. Now B ≡ X_{0}A(B_{0}B_{1})^{k−1}B_{0} ≡ (B_{0}B_{1})^{k−1}B_{0}AX_{1} holds. Let A′ ≡ A and B′ ≡ (B_{0}B_{1})^{k−1}, then we have the conclusion. □

In Lemma 3.7 we have A^{−}B^{−}X_{0}ABX_{0}^{−} = X_{1}X_{0}^{−} = (A′)^{−}(B′)^{−}X_{0}A′B′X_{0}^{−}. Lemma 3.8. Let A, B, X, Y ∈ W(E_{n}) be reduced words (possibly empty) such that X and Y are non-empty, Y^{−}A^{−}B^{−}Y X^{−}ABX ≠ e, Y^{−}A^{−}B^{−}Y and X^{−}ABX are reduced words, and the reduced word of Y^{−}A^{−}B^{−}Y X^{−}ABX is cyclically reduced.

If Y^{−}A^{−}B^{−}Y X^{−}ABX ∈ F_{m}, then

(1) Y^{−}A^{−}B^{−}Y, X^{−}ABX ∈ F_{m}, or

(2) Y^{−}A^{−}B^{−}Y X^{−}ABX is equal to c_{s} or c_{s}^{−} for some s such that lh(s) = m and s is binary branched.

Proof. If Y X^{−} is reduced, then Y^{−}A^{−}B^{−}Y X^{−}ABX is cyclically reduced. By an argument analyzing the head and the tail of Y^{−} and X we can see Y^{−}A^{−}B^{−}Y, X^{−}ABX ∈ F_{m}.

Otherwise, in the cancellation of Y^{−}A^{−}B^{−}Y X^{−}ABX the leftmost Y^{−} or the rightmost X is deleted. Since Y^{−}A^{−}B^{−}Y X^{−}ABX ≠ e and lh(Y^{−}A^{−}B^{−}Y) = 2lh(Y) + lh(AB) and lh(X^{−}ABX) = 2lh(X) + lh(AB), lh(X) ≠ lh(Y). We suppose that lh(X) > lh(Y), i.e. the head of Y^{−} is deleted. Then we have X ≡ ZY for a non-empty word Z.

We first analyze a reduced word of A^{−}B^{−}Z^{−}ABZ, where A^{−}B^{−} is deleted. The head part of Z^{−}AB is BA. Applying Lemma 3.7 for X_{0} ≡ Z^{−} and X_{1} repeatedly, we have reduced words A_{0} and B_{0} such that Z^{−}A_{0}B_{0}Z is reduced, Z^{−}A_{0}B_{0} ≡ B_{0}A_{0}X_{1} for some X_{1}, A_{0}^{−}B_{0}^{−}Z^{−}A_{0}B_{0}Z = A^{−}B^{−}Z^{−}ABZ, A, B ∈ ⟨Z, A_{0}, B_{0}⟩ and lh(B_{0}) < lh(Z).

It never occurs that the both A_{0} and B_{0} are empty, but one of A_{0} and B_{0} may be empty. If B_{0} = ∅, interchange the role of A_{0} and B_{0} and by Lemma 3.7 we can assume B_{0} is non-empty and lh(B_{0}) < lh(Z).

First we deal with the case A_{0} is empty. Since the left most B_{0}^{−} is deleted in the reduction of B_{0}^{−}Z^{−}B_{0}Z, we have non-empty Z_{0} such that Z ≡ Z_{0}B_{0}^{−} and have a reduced word Z_{0}^{−}B_{0}Z_{0}B_{0}^{−} with Z_{0}^{−}B_{0}Z_{0}B_{0}^{−} = B_{0}^{−}Z^{−}B_{0}Z. Since the left most Y^{−} is deleted in the reduction of Y^{−}B_{0}^{−}Z^{−}B_{0}ZY and Z_{0}^{−}B_{0}Z_{0}B_{0}^{−}Y is reduced, Z_{0}^{−}B_{0}Z_{0}B_{0}^{−} is cyclically reduced and hence the reduced word of Y^{−}A^{−}B^{−}Y X^{−}ABX is a cyclical transformation of Z_{0}^{−}B_{0}Z_{0}B_{0}^{−}. By the fact that Y is the head part of B_{0}^{−}Z^{−}B_{0}ZY, Y is of the form (Z_{0}^{−}B_{0}Z_{0}B_{0}^{−})^{k}Y_{0} where Y_{0}Y_{1} ≡ Z_{0}^{−}B_{0}Z_{0}B_{0}^{−} for some non-empty Y_{1} and k ≥ 0.

If Y_{0} is empty, we have Y^{−}A^{−}B^{−}Y X^{−}ABX = Z_{0}^{−}B_{0}Z_{0}B_{0}^{−}. If one of Z_{0} and B_{0} is not small, then Z_{0}, B_{0} ∈ F_{m} by Lemma 2.8 and we have Y^{−}A^{−}B^{−}Y, X^{−}ABX ∈ F_{m} by Lemma 3.7 and the fact Y = (Z_{0}^{−}B_{0}Z_{0}B_{0}^{−})^{k}. Otherwise, i.e., when of Z_{0} and B_{0} are small, Y^{−}A^{−}B^{−}Y X^{−}ABX = Z_{0}^{−}B_{0}Z_{0}B_{0}^{−} is equal to c_{s} or c_{s}^{−} for some s such that lh(s) = m and s is binary branched by Lemma 2.9.

If Y_{0} ≡ Z_{0}^{−}, Y_{0} ≡ Z_{0}^{−}B_{0} or Y_{0} ≡ Z_{0}^{−}B_{0}Z_{0}, the argument is similar to the case that Y_{0} is empty. Otherwise Y_{0} cut short Z_{0}^{−}, B_{0}, Z_{0} or B_{0}^{−}. Since arguments are similar, we only deal with the case that Y_{0} ≡ Z_{0}^{−}B_{1} where B_{1}B_{2} ≡ B_{0} for non-empty B_{1} and B_{2}. Then Y^{−}A^{−}B^{−}Y X^{−}ABX = B_{2}Z_{0}B_{2}^{−}B_{1}^{−}Z_{0}^{−}B_{1} and hence B_{2}Z_{0}B_{2}^{−}, B_{1}^{−}Z_{0}^{−}B_{1} ∈ F_{m} by Lemma 3.6 (4). Let Z_{1} be a cyclically reduced word such that Z_{0} ≡ U^{−}Z_{1}U. Then Z_{1}, B_{2}U^{−}, UB_{1} ∈ F_{m} by Lemma 2.6. Now

Y^{−}Z_{0}Y = B_{1}^{−}Z_{0}(B_{1}B_{2}Z_{0}^{−}B_{2}^{−}B_{1}^{−}Z_{0})^{k}Z_{0}(Z_{0}^{−}B_{1}B_{2}Z_{0}B_{2}^{−}B_{1}^{−})^{k}Z_{0}^{−}B_{1}
= (B_{1}^{−}Z_{0}B_{1}B_{2}Z_{0}^{−}B_{2}^{−})^{k}B_{1}^{−}Z_{0}B_{1}(B_{2}Z_{0}B_{2}^{−}B_{1}^{−}Z_{0}^{−}B_{1})^{k}
Y^{−}B_{0}Y = B_{1}^{−}Z_{0}(B_{1}B_{2}Z_{0}^{−}B_{2}^{−}B_{1}^{−}Z_{0})^{k}B_{1}B_{2}(Z_{0}^{−}B_{1}B_{2}Z_{0}B_{2}^{−}B_{1}^{−})^{k}Z_{0}^{−}B_{1}
= B_{1}^{−}Z_{0}B_{1}(B_{2}Z_{0}^{−}B_{2}^{−}B_{1}^{−}Z_{0}B_{1})^{k}B_{2}Z_{0}^{−}B_{1}(B_{2}Z_{0}B_{2}^{−}B_{1}^{−}Z_{0}^{−}B_{1})^{k}.

Hence Y^{−}Z_{0}Y, Y^{−}B_{0}Y ∈ F_{m}. Since Z = Z_{0}B_{0}^{−} and A, B ∈ ⟨Z, B_{0}⟩, we have Y^{−}ABY, X^{−}A^{−}B^{−}X ∈ F_{m}.

Next we suppose that A_{0} is non-empty. We have k > 0 and A_{1} and A_{2} such that Z^{−} ≡ B_{0}A_{1}A_{2}, A_{0} ≡ (A_{1}A_{2})^{k}A_{1}, X_{1} ≡ A_{2}A_{1}B_{0}. Since X^{−}AB ≡ UX_{1} for some U and X^{−}ABZ is reduced, X_{1}Z ≡ A_{2}A_{1}B_{0}A_{2}^{−}A_{1}^{−}B_{0}^{−} is a reduced word. By the assumption a reduced word of Y^{−}A_{2}A_{1}B_{0}A_{2}^{−}A_{1}^{−}B_{0}^{−}Y is cyclically reduced and A_{2}A_{1}B_{0}A_{2}^{−}A_{1}^{−}B_{0}^{−}Y is reduced, hence X_{1}Z ≡ A_{2}A_{1}B_{0}A_{2}^{−}A_{1}^{−}B_{0}^{−} is cyclically reduced and the reduced word of Y^{−}A_{2}A_{1}B_{0}A_{2}^{−}A_{1}^{−}B_{0}^{−}Y is given by a cyclical transformation of A_{2}A_{1}B_{0}A_{2}^{−}A_{1}^{−}B_{0}^{−}. Hence Y ≡ (A_{2}A_{1}B_{0}A_{2}^{−}A_{1}^{−}B_{0}^{−})^{k}Y_{0} where k ≥ 0 and A_{2}A_{1}B_{0}A_{2}^{−}A_{1}^{−}B_{0}^{−} ≡ Y_{0}Y_{1} for some Y_{1}.

For instance the reduced word of Y^{−}A_{2}A_{1}B_{0}A_{2}^{−}A_{1}^{−}B_{0}^{−}Y is of the form B_{0}A_{2}^{−}A_{1}^{−}B_{0}^{−}A_{2}A_{1} or B_{2}A_{2}^{−}A_{1}^{−}B_{2}^{−}B_{1}^{−}A_{2}A_{1}B_{1} where B_{0} ≡ B_{1}B_{2}. By Lemma 3.6 (4) or (3) respectively we conclude A_{1}, A_{2}, B_{0} ∈ F_{m} or A_{1}, A_{2}, B_{1}, B_{2} ∈ F_{m} which implies Y^{−}ABY, X^{−}A^{−}B^{−}X ∈ F_{m}. □

Lemma 3.9. For every grope group G^{S} the following hold:

If e ≠ [u, v] ∈ F_{m} and at least one of u and v does not belong to F_{m}, then [u, v] is conjugate to c_{s} or c_{s}^{−} in F_{m} for some s such that lh(s) = m and s is binary branched.

Proof. We have n > m such that u, v ∈ F_{n}. It suffices to show the lemma in case that the reduced word for [u, v] is cyclically reduced. For, suppose that we have the conclusion of the lemma in the indicated case. Let [u, v] ∈ F_{m} and [u, v] = XY X^{−} where XY X^{−} is a reduced word and Y is cyclically reduced. Then we have [X^{−}uX, X^{−}vX] = X^{−}[u, v]X = Y. On the other hand X, Y ∈ F_{m} by Lemma 2.6. By the assumption at least one of X^{−}uX and X^{−}vX does not belong to F_{m}. Since [u, v] is conjugate to Y in F_{m}, we have the conclusion.

Let u, v ∈ F_{n} such that [u, v] ≠ e and the reduced word for [u, v] is cyclically reduced. There exist a cyclically reduced non-empty word V_{0} ∈ W(E_{n}) and a reduced word X ∈ W(E_{n}) such that v = X^{−}V_{0}X and the word X^{−}V_{0}X is reduced. Let U_{0} be a reduced word for uX^{−}. Since V_{0} is a cyclically reduced word, at least one of U_{0}V_{0} and V_{0}U_{0}^{−} is reduced. When U_{0}V_{0} is reduced, there exist k ≥ 0 and reduced words Y, A, B such that Y^{−}ABY is reduced, U_{0} ≡ Y^{−}AV_{0}^{k} and V_{0} ≡ BA. When V_{0}U_{0}^{−} is reduced, there exist k ≥ 0 and reduced words Y, A, B such that Y^{−}ABY is reduced, U_{0} ≡ Y^{−}A(V_{0}^{−})^{k} and V_{0} ≡ BA. In the both bases uvu^{−1} = Y^{−}ABY and v = X^{−}BAX. We remark that AB and BA are cyclically reduced.

We analyze a reduction procedure of Y^{−}ABY X^{−}A^{−}B^{−}X in the following. (Case 0): X and Y are empty.

In this case the both A and B are non-empty and corresponds to Lemma 3.2. Using (1.1) and (1.2) alternately and (1.3) possibly as the last step we obtain a reduced word of ABA^{−}B^{−}. If the reduced word XY ZX^{−}Y^{−}Z^{−} satisfies that one of X, Y, Z is not small, by (1) and (2) of Lemma 3.6 and applying Lemma 3.2 repeatedly we can see A, B ∈ F_{m}. Otherwise, one of X, Y, Z is empty and [u, v] = c_{s} or [u, v] = c_{s}^{−} for some binary branched s with lh(s) = m by Lemma 2.9. (Case 1): Y is empty, but X is non-empty. (Case 2): X is empty, but Y is non-empty.

In these cases arguments are symmetric, we only deal with (Case 1). There is possibility that one of A and B may be empty, though at least one of A and B is non-empty. We assume that A is non-empty. We trace Lemmas 3.3, 3.4, 3.5 to get a reduced word of ABX^{−}A^{−}B^{−}X. Then we apply one of (2), (3) and (4) of Lemma 3.6 to the reduced word and applying Lemma 3.2 repeatedly we get a reduced word. Then we have A, B ∈ F_{m}, which implies u, v ∈ F_{m}, or [u, v] = c_{s} etc. as in (Case 0). (Case 3): The both X and Y are non-empty.

Only in this case we use the assumption that the reduced word of Y^{−}ABY X^{−}A^{−}B^{−}X is cyclically reduced. By Lemma 3.8 we have the conclusion. □

Lemma 3.10. Let F be a free group generated by C and c, d ∈ C be distinct elements. If [c, d] = [u, v] for u, v ∈ F, then neither u nor v belongs to the commutator subgroup of F.

Proof. Since c, d are generators, [c, d] ∉ [F, [F, F]] and the conclusion follows. □

Lemma 3.11. Let F be a free group generated by B and b_{0}, b_{1} ∈ B be distinct. If c, d ∈ {b, b^{−} : b ∈ B} and [b_{0}, b_{1}] = [x^{−1}cx, y^{−1}dy] for x, y ∈ F, then c, d ∈ {b_{0}, b_{0}^{−}, b_{1}, b_{1}^{−}} and moreover c ∈ {b_{0}, b_{0}^{−}} iff d ∈ {b_{1}, b_{1}^{−}} and c ∈ {b_{1}, b_{1}^{−}} iff d ∈ {b_{0}, b_{0}^{−}}.

Proof. Using a canonical projection to ⟨b_{0}, b_{1}⟩ we easily see that c, d ∈ {b_{0}, b_{0}^{−}, b_{1}, b_{1}^{−}}. To see the remaining part it suffices to show that if c = b_{0}, and d = b_{0} or b_{0}^{−}, then [b_{0}, b_{1}] ≠ [x^{−1}cx, y^{−1}dy] for any x, y.

We show that b_{0}b_{1}b_{0}^{−}b_{1}^{−} is not cyclically equivalent to the reduced word for [x^{−1}cx, y^{−1}dy]. For this purpose we may assume x = e. We only deal with d = b_{0}. We have a reduced word Y such that y^{−1}b_{0}y = Y^{−}b_{0}Y and Y^{−}b_{0}Y is reduced. (Note that y = Y may not hold.) The head of Y is not b_{0} nor b_{0}^{−}, since Y^{−}b_{0}Y is reduced. When the tail of Y is b_{0} or b_{0}^{−}, we choose n ≥ 0 so that Y ≡ Zb_{0}^{n} or Y ≡ Z(b_{0}^{−})^{n} respectively and n is maximal. Then Z is non-empty. Now b_{0}Z^{−}b_{0}Zb_{0}^{−}Z^{−}b_{0}^{−}Z is a cyclically reduced word which is cyclically equivalent to b_{0}Y^{−}b_{0}Y b_{0}^{−}Y^{−}b_{0}^{−}Y. Since b_{0}Z^{−}b_{0}Zb_{0}^{−}Z^{−}b_{0}^{−}Z is not cyclically equivalent to b_{0}b_{1}b_{0}^{−}b_{1}^{−}, we have the conclusion. □

Proof of Theorem 3.1. Let h : G^{S_{0}} → G^{S} be a nontrivial homomorphism. Then there exists s_{∗} ∈ S_{0} such that h(c_{s_{∗}}) is nontrivial (clearly for every finite sequence s starting with s_{∗} also h(c_{s}) is nontrivial). We let c_{s} = c_{s}^{S_{0}} and d_{t} = c_{t}^{S} and F_{m} = F_{m}^{S}.

We have n such that h(c_{s_{∗}}) ∈ F_{n}. Since F_{n} is free, Im(h) is not included in F_{n} and hence there exists s_{0} ∈ S_{0} starting with s_{∗} and such that h(c_{s_{0}}) ∈ F_{n}, but h(c_{s_{0}0}) ∉ F_{n} or h(c_{s_{0}1}) ∉ F_{n}. Then by Lemma 3.9 we have d_{t_{0}} ∈ E_{n} such that h(c_{s_{0}}) is conjugate to d_{t_{0}} or d_{t_{0}}^{−} and t_{0} is binary branched.

Moreover, Lemma 2.5 implies that neither h(c_{s_{0}0}) nor h(c_{s_{0}1}) belongs to F_{n}. We show the following by induction on k ∈ ℕ:

(1) For u ∈ Seq(\underline{2}) with lh(u) = k

(a) h(c_{s_{0}u}) is conjugate to d_{t_{0}v} or d_{t_{0}v}^{−} in F_{n+k} and t_{0}v is binary branched for some v ∈ Seq(\underline{2}) with lh(v) = k;

(b) Neither h(c_{s_{0}u0}) nor h(c_{s_{0}u1}) belongs to F_{n+k};

(2) For every v ∈ Seq(\underline{2}) with lh(v) = k there exists u ∈ Seq(\underline{2}) such that lh(u) = k and h(c_{s_{0}u}) is conjugate to d_{t_{0}v} or d_{t_{0}v}^{−} in F_{n+k}.

We have shown that this holds when k = 0.

Suppose that (1) and (2) hold for k. Let lh(u) = k and h(c_{s_{0}u}) is conjugate to d_{t_{0}v} or d_{t_{0}v}^{−} etc. Then [h(c_{s_{0}u0}), h(c_{s_{0}u1})] is conjugate to [d_{t_{0}v0}, d_{t_{0}v1}] or [d_{t_{0}v1}, d_{t_{0}v0}] in F_{n+k+1}. We claim h(c_{s_{0}u0}) ∈ F_{n+k+1}. To show this by contradiction, suppose that h(c_{s_{0}u0}) ∉ F_{n+k+1}. Apply Lemma 3.9 to F_{n+k+1}, then we have [h(c_{s_{0}u0}), h(c_{s_{0}u1})] is a conjugate to d_{t} or d_{t}^{−} with lh(t) = n + k + 1 in F_{n+k+1}, which is impossible since [h(c_{s_{0}u0}), h(c_{s_{0}u1})] ∈ [F_{n+k+1}, F_{n+k+1}]. Similarly we have h(c_{s_{0}u1}) ∈ F_{n+k+1}.

On the other hand, neither h(c_{s_{0}u0}) nor h(c_{s_{0}u1}) belongs to [F_{n+k+1}, F_{n+k+1}] by Lemma 3.10. Hence at least one of h(c_{s_{0}u00}) and h(c_{s_{0}u01}) does not belong to F_{n+k+1} and consequently neither h(c_{s_{0}u00}) nor h(c_{s_{0}u01}) belongs to F_{n+k+1} by Lemma 2.5.

Hence h(c_{s_{0}u0}) is conjugate to d_{t} or d_{t}^{−} with lh(t) = n + k + 1 by Lemma 3.9. Similarly, h(c_{s_{0}u1}) is conjugate to d_{t′} or d_{t′}^{−} with lh(t′) = n+ k+1. Since [h(c_{s_{0}u0}), h(c_{s_{0}u1})] is conjugate to [d_{t_{0}v0}, d_{t_{0}v1}] or [d_{t_{0}v0}, d_{t_{0}v1}] in F_{n+k+1}, h(c_{s_{0}u0}) and h(c_{s_{0}u1}) are conjugate to d_{t_{0}vj} or d_{t_{0}vj}^{−} for some j ∈ \underline{2} and for each j ∈ \underline{2} the element d_{t0vj} is conjugate to exactly one of h(c_{s_{0}u0}), h(c_{s_{0}u1}), h(c_{s_{0}u0})^{−} or h(c_{s_{0}u1})^{−} by Lemma 3.11. Hence (1) and (2) hold for k + 1. Now we have shown the induction step and finished the proof. □

Remark 3.12. Though the conclusion of Theorem 3.1 is rather simple, embeddings from G^{S_{0}} into G^{S} may be complicated. In particular automorphisms on G^{S_{0}} may be complicated, since the following hold:

[dc^{−}d^{−}, dcd^{−}c^{−}d^{−}] = dc^{−}d^{−}dcd^{−}c^{−}d^{−}dcd^{−}dcdc^{−}d^{−} = cdc^{−}d^{−} = [c, d].

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IMFM, University of Ljubljana, Jadranska 19, Ljubljana, SLOVE-NIA

E-mail address: matija.cencelj@guest.arnes.si

School of Science and Engineering, Waseda University, Tokyo 169-8555, JAPAN

E-mail address: eda@waseda.jp

IMFM, University of Ljubljana, Jadranska 19, Ljubljana, SLOVE-NIA

E-mail address: ales.vavpetic@fmf.uni-lj.si