2. M -solid varieties of groups

Vol. 34, No. 2, 2004, 127-139

Proc. Novi Sad Algebraic Conf. 2003 (eds. I. Dolinka, A. Tepavˇcevi´c)

HYPERSUBSTITUTIONS AND GROUPS

J¨org Koppitz¹

Abstract. We consider groups as algebras of type (2,1,0). A hypersub- stitution of type (2,1,0) is a mapping σ from the set of the operation symbols {·,⁻¹, e} into the set of terms of type (2,1,0) preserving the arity. For a monoid M of hypersubstitutions of type (2,1,0) a variety V is calledM-solid if for each group (G;·,⁻¹, e) ∈V the derived group (G;σ(·), σ(⁻¹), σ(e)) also belongs to V for all σ ∈M. The classS_M^Gr of allM-solid varieties of groups forms a complete sublattice of the lattice L(Gr) of all varieties of groups. In this way we get a tool for a better description of the whole lattice L(Gr) by characterization of complete sublatticesS_M^Gr.

AMS Mathematics Subject Classification (2000): 20M07, 08B15 Key words and phrases: hypersubstitution,M-solid variety, groups

1. Introduction

It is of some interest to know what the lattice of all varieties of some type τ looks like, but it has become clear that it is very complicated, even for such special case as the lattice of all varieties of semigroups. In [3] a new method to study these lattices was proposed, using complete sublattices consisting ofM- solid varieties, whereM is a monoid of hypersubstitutions. M-solid varieties of semigroups are considered in a range of papers (see for example [1], [2], and [7]).

Although groups can be considered as semigroups not every variety of groups correponds to a variety of semigroups. Considering groups as algebras of type (2,1,0) we can use the method of M-solid varieties for the description of the lattice of all varieties of groups.

In the next section we introduce the concept of a M-solid variety and collect some basic properties. In the third section we determine the set Hnt of all monoidsM of hypersubstitutions of type (2,1,0) such that there is a nontrivial M-solid variety of groups. It turns out that Hnt has infinitely many maximal and one minimal element, and Hnt consists of the submonoids of its maximal elements. The last section is devoted to the main result: For all maximal elementsHpofHntwe characterize the complete lattice of allHp-solid varieties

1University of Potsdam, Institute of Mathematics, Am Neuen Palais, 14415 Potsdam, Germany, e-mail:[email protected]

of groups. An open problem is the characterization of the lattice of all M- solid varieties of groups for arbitrary monoidsM. In the commutative case this problem is already solved (see [4]). The present paper will give the answer for another class of varieties of groups, namely for varieties of groups satisfying the identityx²y≈yx².

2. M -solid varieties of groups

Let W(X) be the set of all terms of type (2,1,0) over some fixed alpha- bet X := {x1, x2, x3, ...} where {f, g, e} denotes the set of operation symbols (f is binary, g is unary ande is 0-ary). Instead of x1, x2, x3, ... we write also x, y, z, .... Further,W(X2) (W(X1),W(∅)) denotes the set of all terms of type (2,1,0) overX2:={x1, x2}(X1:={x1},∅).

We recall that the identitiesg(f(y, x))≈f(g(x), g(y)),g(g(x))≈x,andg(e)≈ ehold in every variety of groups and usually one writesx⁻¹instead ofg(x) ([5]).

This allows us to write a termt∈W(X) as a semigroup word over the alphabet X^∗:=X∪ {w⁻¹|w∈X} ∪ {e}. For example, for t=f(f(g(x), x), g(f(x, e))) one can writet=x⁻¹xex⁻¹. (But if necessary we will write terms by using the operation symbolsf andg.)

For a variablew∈X^∗ and a termt∈W(X) we put:

w⁰:=e, w¹:=w, and w^m+1:=w^mwform≥1;

w^−m:= (w⁻¹)^m for anym≥2;

cw(t) - the number of occurrences ofw in the term tregarded as a semigroup word. For example, for t = f(f(g(x), x), g(f(x, e))) we have cx(t) = 1 and c_x⁻¹(t) = 2 since the semigroup wordx⁻¹xex⁻¹corresponds to this term.

A mapping σ : {f, g, e} −→ W(X2) with σ(g) ∈ W(X1) and σ(e) ∈ W(∅) is called a hypersubstitution of type (2,1,0) (for short hypersubstitution). Any hypersubstitutionσcan be uniquely extended to a mapσb:W(X)−→W(X), this is defined inductively by

(i)σ[w] :=b wfor anyw∈X∪ {e},

(ii)σ[fb (t1, t2)] :=σ(f)(bσ[t1],bσ[t2]), andσ[g(t)] :=b σ(g)(bσ[t]).

Hereσ(f) andσ(g) on the right-hand side of (ii) have to be interpreted as op- erations induced by the termσ(f) and σ(g), respectively, on the term algebra induced onW(X).

We denote byHypthe set of all hypersubstitutions. If we define a product ◦h

of hypersubstitutions byσ1◦hσ2:=bσ1◦σ2, where◦is the usual composition of functions, thenHyp= (Hyp;◦h, σid) is a monoid. Note thatσid is the identity hypersubstitution, defined byσid(f) =x1x2,σid(g) =x⁻¹₁ , and σid(e) =e.

Let M be a submonoid of Hyp. Further let V be a variety of type (2,1,0).

Then an identity s ≈ t of V is called an M-hyperidentity of V if for every σ∈M the equationσ[s]b ≈bσ[t] is an identity inV. If every identity inV is an M-hyperidentity thenV is calledM-solid. In the special case that M is all of Hyp, we speak of a hyperidentity and a solid variety. In order to show that any

identity is anM-hyperidentity inV we have not to check all σ∈M, we need only one representative of each equivalence class with respect to the following equivalence relation onHyp, established by J. PÃlonka ([6]):

σ1 ∼V σ2 iff σ1(µ) ≈ σ2(µ) is an identity in V for all operation symbols µ∈ {f, g, e}.

Ifσ₁ ∼_V σ₂ we say thatσ₁ and σ₂ areV-equivalent. In [6] was shown that if σ1 andσ2 areV-equivalent andσb1[s]≈bσ1[t] holds inV then alsoσb2[s]≈σb2[t]

holds inV.

By definition, to tell if a variety isM-solid, one has to test that application of any hypersubstitution σ to any identity of V results in an identity of V. De- necke and Reichel have developed a reduction in [3]. It suffices to show that every identity of the generating system ofV is anM-hyperidentity.

We denote byIdV the set of all identities inV and byL(V) we mean the subvari- ety lattice ofV.The setP(V) of all hypersubstitutionsσwithσ[s]b ≈bσ[t]∈IdV for all s ≈ t ∈ IdV forms a submonoid of Hyp [8]. An element of P(V) is called proper hypersubstitution ([6]). The varietyGr :=M od{f(f(x, y), z) ≈ f(x, f(y, z)), f(g(x), x)≈f(x, g(x))≈e, f(x, e)≈f(e, x)≈x} is the variety of all groups (considered as algebras of type (2,1,0)). For a set Σ of equations letGr(Σ) be the variety of groups satisfying Σ. By S_M^Gr we denote the class of all M-solid varieties of groups. S^Gr_M forms a complete sublattice of L(Gr).

Moreover, ifM1⊆M2 thenS_M^Gr₂ ⊆S_M^Gr₁ (see [3]).

3. Characterization of H

For each monoidM of hypersubstitutions the trivial varietyT R:=M od{x≈ y}belongs toS_M^Gr. This is clear, since the application of anyσ∈Hyptox≈y provides againx≈y, i.e. gives an identity ofT R. But there are monoids M such thatS_M^Grconsists only ofT R, for example in the caseM =Hyp. To make this clear we consider a hypersubstitutionσ∈Hyp withσ(f) andσ(e) =e. If we apply thisσto the group identity f(e, x)≈xwe gete≈xwhich holds only in the trivial variety. This shows thatT R is the only solid variety of groups.

Moreover, this example shows thatS_M^Gr={T R} for all monoidsM containing the previously defined σ. It raises the question: For which monoids M there are nontrivialM-solid varieties of groups. In this section we determine the set Hnt of all such submonoidsM ofHypfor whichS_M^Gr contains not onlyT R:

Hnt:={M |M ⊆Hyp, S_M^Gr6={T R}}.

Fora≥1 letV_a^c be the variety of all commutative groups of ordera:

V_a^c:=Gr({f(x, y)≈f(y, x), x^a ≈e}).

Note thatV₁^c=T R. Clearly,V_i^c6=V_j^c fori6=j.

Definition 1. Let a≥2 be a natural number. Let Ha be the set of all hyper-

substitutionsσ satisfying the following properties:

a) cx(σ(f)) − cx−1(σ(f)) ≡ 1(a);

b) cy(σ(f)) − cy⁻¹(σ(f)) ≡ 1(a);

c) c_x(σ(g)) − c_x−1(σ(g)) ≡ − 1(a).

Proposition 2. For all natural numbersa≥2 we haveHa=P(V_a^c).

Proof. Letσ∈P(V_a^c).We will show thatσsatisfies the properties a), b), and c).

Assume that a) does not hold. Then cx(σ(f))− c_x⁻¹(σ(f))≡ m(a) for some natural numberm with 1< m≤a. We applyσto f(x, e)≈x∈IdV_a^c and get x^{cx(σ(f))−cx−1(σ(f))} ≈ x∈IdV_a^c since σ is a proper hypersubstitution for V_a^c. Butx^{cx(σ(f))−cx−1(σ(f))} ≈x, x^a ≈e, andcx(σ(f))−c_x⁻¹(σ(f))≡m(a) imply x^m≈x, i.e. x^m−1≈ewith 1≤m−1< ais an identity inV_a^c, a contradiction.

Dually we get that b) is satisfied.

Assume that c) does not hold. Thencx(σ(g))−cx⁻¹(σ(g))≡m(a) for some nat- ural number m with 0 ≤ m < a − 1. Then σ(g)(x) ≈ x^{cx(σ(g))−cx−1(σ(g))} ≈ x^m ∈ IdV_a^c because of x^a ≈ e ∈ IdV_a^c. By a) and b) we have c_x(σ(f))− c_x−1(σ(f)) ≡ 1(a) and c_y(σ(f))− c_y−1(σ(f)) ≡ 1(a), re- spectively. Thusx^{m(cx(σ(f))−cx−1(σ(f)))+cy(σ(f))−cy−1(σ(f))} ≈x^m+1 because of x^a≈e∈IdV_a^c.

Further, there holds σ[fb (g(x), x)] = σ(f)(bσ[g(x)],bσ[x]) = σ(f) (σ(g)(x), x)≈σ(f)(x^m, x)≈x^{m(cx(σ(f))−cx−1(σ(f)))+cy(σ(f))−cy−1(σ(f))} ≈x^m+1. Sinceσis a proper hypersubstitution forV_a^c fromf(g(x), x)≈e∈IdV_a^cfollows b

σ[f(g(x), x)]≈σ[e]b ∈IdV_a^c, i.e. x^m+1≈ewith 1≤m+ 1< ais an identity in V_a^c, a contradiction.

Conversely, letσ∈Ha. We will show thatσis a proper hypersubstitution for Ha. For this we show thatσ isV_a^c-equivalent to the identity hypersubstitution σid. There are natural numbersk, l, m, nsuch thatcx(σ(f)) =k, c_x⁻¹(σ(f)) =l, cy(σ(f)) =m,andc_y⁻¹(σ(f)) =n. Thenσ(f)≈x^k−ly^m−nbecause of the com- mutative law. Because of a) and b) we have k−l ≡ 1(a) and m−n≡ 1(a), respectively. Thusσ(f)≈xy(because of x^a≈e).

Further, there are natural numbers i, j such that cx(σ(g)) = i, cx⁻¹(σ(g))

= j. Then σ(g) ≈ x^i−j because of the commutative law. Because of c) we havei−j ≡ −1(a). Thusσ(g)≈x⁻¹(because of x^a ≈e).

Obviously, we haveσ[e]b ≈e. 2

Notation 3 For a monoid M of hypersubstitutions of type (2,1,0) we define gcd(M) as be the greatest common divisor of the following integers:

cx(σ(f))−c_x⁻¹(σ(f))−1,cy(σ(f))−c_y⁻¹(σ(f))−1, andcx(σ(g))−c_x⁻¹(σ(g))+1 for allσ∈M.

Theorem 4. Let M be a monoid of hypersubstitutions of type (2,1,0). Then S_M^Gr 6={T R} iff there is a prime number pwithM ⊆Hp.

Proof. LetS^Gr_M 6= {T R}. Then there is an M-solid variety V of groups with V 6=T R.

Assume that M * Hp for all prime numbers p. Then for each prime num- ber p there is a σ ∈ M with cx(σ(f))−cx⁻¹(σ(f)) 6≡ 1(p) or cy(σ(f))−

cy⁻¹(σ(f))6≡1(p) orcx(σ(g))−cx⁻¹(σ(g))6≡ −1(p). This means gcd(M) = 1.

On the other hand we have

{bσ[f(x, e)] ≈ σ[x]b | σ ∈ M} ∪ {bσ[f(e, x)] ≈ bσ[x] | σ ∈ M}∪

{bσ[f(g(x), x)] ≈ bσ[e] | σ ∈ M} ⊆ IdV. This provides {x^{cx(σ(f))−cx−1(σ(f))}

≈ x | σ ∈ M} ∪ {x^{cy(σ(f))−cy−1(σ(f))} ≈ x | σ ∈ M}∪

{x^{[cx(σ(f))−cx−1(σ(f))][cx(σ(g))−cx−1(σ(g))]+cy(σ(f))−cy−1(σ(f))} ≈ e | σ ∈ M} ⊆ IdV. For σ ∈M, using x^{cx(σ(f))−cx−1(σ(f))} ≈xand x^{cy(σ(f))−cy−1(σ(f))} ≈x from x^{[cx(σ(f))−cx−1(σ(f))][cx(σ(g))−cx−1(σ(g))]+cy(σ(f))−cy−1(σ(f))} ≈ e it follows x^{cx(σ(g))−cx−1(σ(g))+1} ≈ e and thus x^{cx(σ(g))−cx−1(σ(g))+2} ≈ x. This shows that {x^{cx(σ(f))−cx−1(σ(f))} ≈ x | σ ∈ M} ∪ {x^{cy(σ(f))−cy−1(σ(f))} ≈ x | σ ∈ M} ∪ {x^{cx(σ(g))−cx−1(σ(g))+2}≈x|σ∈M} ⊆IdV. From these identities we can derive x^gcd(M)+1 ≈ x∈IdV. Since gcd(M) = 1, we have x² ≈ x∈IdV, i.e.

x≈e∈IdV andx≈y∈IdV. ThusV =T R, a contradiction.

Conversely, let M ⊆ H_p for some prime number p. Then S_H^Gr_p ⊆ S_M^Gr. Since P(V_p^c) =Hp (Proposition 2) we haveV_p^c∈S_H^Gr_p ⊆S_M^Gr and thus S_M^Gr 6={T R}.

2

Remark 5. The previous theorem shows that the monoidsHp are maximal el- ements in Hnt, where for two different prime numbers p1 andp2 the monoids Hp1 andHp2 are different.

Moreover, it is easy to check that

M1:={σid} forms a monoid. M₁ is the least element in H_nt.

The following set D of hypersubstitutions of type (2,1,0) is the set of all proper hypersubstitutions of the variety of all commutative groups ([4]).

Definition 6. LetD be the set of all hypersubstitutionsσsatisfying the follow- ing properties:

a) c_x(σ(f)) − c_x−1(σ(f)) = b) cy(σ(f)) − c_y−1(σ(f)) = c) cx(σ(g)) − c_x⁻¹(σ(g)) = −

1;

1;

1.

Obviously, we haveD ⊆Hn for all natural numbersn≥2. We will deter- mine such monoidsM withM ⊆Hn for all natural numbersn≥2.

Definition 7. For any submonoidM ⊆Hypwe denote byL(M)the submonoid lattice ofM.

Proposition 8. There holds T

2≤i∈N

L(Hi) =L(D).

Proof. ” ⊇” : Clearly, for 2≤i∈N we haveD ⊆Hi, i.e. D ∈ L(Hi). Thus L(D)⊆ L(Hi) for 2≤i∈N,i.e. L(D)⊆ T

2≤i∈N

L(Hi).

”⊆” : Let M ∈ T

2≤i∈N

L(Hi) and letσ ∈M. Then there is a natural number n≥1 with cx(σ(f))−c_x⁻¹(σ(f)) =n. Assume that n 6= 1. Then n 6≡1(n), i.e. σ /∈ Hn and M /∈ L(Hn), contradicts M ∈ T

2≤i∈N

L(Hi). Thuscx(σ(f))− c_x−1(σ(f)) = 1. Similarly, one can show that cy(σ(f))−c_y−1(σ(f)) = 1 and cx(σ(g))−c_x−1(σ(g)) =−1.

Consequently,σ∈D and thusM ⊆D, i.e. M ∈ L(D). 2

4. All H

-solid varieties of groups

The monoidsHp,for prime numbersp,are the maximal elements inHnt. In particular, for anyM ∈ Hntthere is a prime numberpwithM ⊆Hp, i.e. S^Gr_Hp⊆ S_M^Gr. If we have a characterization of the lattice S_H^Gr_p for all prime numbers p then we have some knowledge about a complete sublattice of S_M^Gr for any monoidM ∈ Hnt. The main theorem of the present paper, the characterization ofS_H^Gr_p for all prime numbersp,is the topic of this section. We start with some properties ofHp-solid varieties of groups.

Lemma 9. Let n≥2 be a natural number. Then in each Hn-solid variety V of groups there holdsxyx⁻¹zxy⁻¹x⁻¹≈yzy⁻¹.

Proof. We consider the following hypersubstitutionσ:

σ(f) :=x²yx⁻¹ σ(g) :=x⁻¹ σ(e) :=e.

We have cx(σ(f))−cx⁻¹(σ(f)) = 2−1 = 1 ≡1(n), cy(σ(f))−cy⁻¹(σ(f)) = 1−0 = 1≡1(n), andc_x(σ(g))−c_x−1(σ(g)) = 0−1 =−1≡ −1(n), i.e. σ∈H_n. Since V is Hn-solid, the application of σ to the associative law provides the identities x²yx⁻¹x²yx⁻¹z(x²yx⁻¹)⁻¹ ≈ x²y²zy⁻¹x⁻¹, x²yxyx⁻¹zxy⁻¹x⁻² ≈ x²y²zy⁻¹x⁻¹, xyx⁻¹zxy⁻¹x⁻¹≈yzy⁻¹in V. 2 For a groupA= (A;·,⁻¹, e), byC(A) :={a∈A|xa=axfor allx∈A}we denote the centre ofA. In particular,C(A) forms a subgroup ofA(see [5]). For a, b∈ A let [a, b] :=aba⁻¹b⁻¹be the commutator of aandb.The commutator group ofA, i.e. the group generated by the set{[a, b]|a, b∈ A}, is denoted by [A,A].

Proposition 10. Let n≥2 be a natural number, V be an Hn-solid variety of groups andA ∈V. Then

[A,A]⊆ C(A), i.e. the commutator group is a subgroup of the centre.

Proof. We will show that for any a, b∈A the commutator [a, b] belongs to the centre ofA, i.e. {[a, b]|a, b∈ A} ⊆ C(A).

Let a, b ∈A. Then for any x ∈ A holds ba⁻¹b⁻¹xbab⁻¹ ≈ a⁻¹xa by Lemma 9. This impliesaba⁻¹b⁻¹xbab⁻¹ba⁻¹b⁻¹ ≈aa⁻¹xaba⁻¹b⁻¹, i.e. aba⁻¹b⁻¹x≈ xaba⁻¹b⁻¹ and thus the commutator [a, b] = aba⁻¹b⁻¹ belongs to the centre ofA. Since {[a, b] |a, b∈ A} ⊆ C(A) andC(A) is a subgroup of A the group generated by the set{[a, b]|a, b∈ A}, i.e. the commutator group [A,A], is a

subgroup ofC(A). 2

Lemma 11. Let n≥2 be a natural number. Then in each Hn-solid varietyV of groups there holdsxⁿ ≈e.

Proof. We consider the following hypersubstitutionσ:

σ(f) :=xⁿ⁺¹y σ(g) :=x⁻¹ σ(e) :=e.

We havecx(σ(f))−c_x⁻¹(σ(f)) =n+1−0 =n+1≡1(n),cy(σ(f))−c_y⁻¹(σ(f)) = 1−0 = 1≡1(n), andcx(σ(g))−c_x⁻¹(σ(g)) = 0−1 =−1≡ −1(n), i.e. σ∈Hn. Since V is Hn-solid, the application of σ to the group identity f(x, e) ≈ x provides an identity inV, namelyxⁿ⁺¹≈x, i.e.xⁿ ≈e. 2 By Proposition 10 and Lemma 11, respectively, it becomes clear that an Hn-solid variety of groups consists of solvable groups.

Definition 12. We define a hypersubstitution σd by σd(f) :=yx

σd(g) :=x⁻¹ σd(e) :=e.

A varietyV of groups is called self-dual if the application ofσd to any identity ofV gives again an identity inV:

{bσdu]≈bσd[v]|u≈v∈IdV} ⊆IdV.

Lemma 13. Let n≥2 be a natural number. AnyHn-solid varietyV of groups is self-dual.

Proof. We have cx(σd(f))−cx⁻¹(σd(f)) = cy(σd(f))−cy⁻¹(σd(f)) = 1−0 = 1 ≡1(n), and cx(σd(g))−cx⁻¹(σd(g)) = 0−1 =−1 ≡ −1(n), i.e. σd ∈Hn. Since V is Hn-solid, the application of σd to an identity of V gives again an

identity ofV. 2

Lemma 14. Let V be a variety of groups satisfying xyx⁻¹zxy⁻¹x⁻¹ ≈ yzy⁻¹. Then for any integera there holds

xyx⁻¹y^a≈y^axyx⁻¹∈IdV.

Proof. All is clear fora= 0. Leta6= 0 be an integer. Then we havexyx⁻¹y^a≈ y^axy^−ayy^ax⁻¹y^−ay^a ≈y^axyx⁻¹ (usingxyx⁻¹zxy⁻¹x⁻¹≈yzy⁻¹). 2 Lemma 15. Let V be a variety of groups satisfying xyx⁻¹zxy⁻¹x⁻¹ ≈ yzy⁻¹. Then for integers r, s, t, u6= 0 the following identities (i)-(iv) are satis- fied inV:

(i) x^ry^sx^−ry^tx^u≈y^tx^ry^sx^u−r (ii) x^ry^sx^−ty^ux^t≈x^r−ty^ux^ty^s (iii) x^ry^sx^ry^tx^u≈y^−tx^ry^s+2tx^r+u (iv) x^ry^sx^ty^ux^t≈x^r+ty^u+2sx^ty^−s.

Proof. The identities (i) and (ii) are immediate consequences of Lemma 14.

We show (iii). The identity (iv) can be checked dually. Using Lemma 14 we havex^ry^sx^ry^tx^u≈x^ry^sx^ry^tx^−rx^u+r

≈x^rx^ry^tx^−ry^sx^u+r

≈x^2ry^tx^−ry^−ty^s+tx^u+r

≈y^tx^−ry^−tx^2ry^s+tx^u+r

≈y^tx^−ry^−tx^2ry^s+tx^−2rx^u+3r

≈y^tx^−rx^2ry^s+tx^−2ry^−tx^u+3r

≈y^tx^ry^s+tx^−rx^−ry^−tx^u+3r

≈x^ry^s+tx^−ry^tx^−ry^−tx^u+3r

≈x^ry^s+ty^tx^−ry^−tx^−rx^u+3r

≈x^ry^s+2tx^−ry^−tx^u+2r

≈y^−tx^ry^s+2ty^tx^−rx^u+2r

≈y^−tx^ry^s+2ty^tx^u+r. 2

Theorem 16. Let r ≥ 2 be a natural number and let V be a variety of groups. V is Hr-solid iff V is self-dual and satisfies both identities x^r ≈eand xyx⁻¹zxy⁻¹x⁻¹≈yzy⁻¹.

Proof. Suppose that V is Hr-solid. Then V is self-dual by Lemma 13, satisfies x^r ≈ e (i.e. it is a variety of r-group) by Lemma 11 and satisfies xyx⁻¹zxy⁻¹x⁻¹≈yzy⁻¹by Lemma 9.

Suppose now thatV is a self-dual variety ofr-groups satisfying xyx⁻¹zxy⁻¹x⁻¹≈yzy⁻¹(i).

Letσ∈Hr. We will show thatσ(f)≈x^ay^bx^cy^d or σ(f)≈y^dx^cy^bx^a for some natural numbersa, b, c, d witha+c≡b+d≡1(r).

For this we check that for natural numbersa, n2, n3, n4, n5 we have

x^an3yⁿ²xⁿ³yⁿ⁴xⁿ⁵ ≈y^{(−a+1)n2−an4}xⁿ³y^an2+(a+1)n4x^n5+an3∈IdV (ii).

We show by induction on k that x^kn3yⁿ²xⁿ³yⁿ⁴xⁿ⁵ ≈ y^{(−k+1)n2−kn4}xⁿ³y^kn2+(k+1)n4x^n5+kn3∈IdV.

For k = 1 we have x¹ⁿ³yⁿ²xⁿ³yⁿ⁴xⁿ⁵ ≈y^{(−1+1)n2−1n4}xⁿ³y^1n2+(1+1)n4xⁿ⁵⁺¹ⁿ³

∈IdV by Lemma 15(iii).

Suppose now that the statement is true for k = m, i.e. x^mn3yⁿ²xⁿ³yⁿ⁴xⁿ⁵

≈y^{(−m+1)n2−mn4}xⁿ³y^mn2+(m+1)n4x^n5+mn3 ∈IdV (hypothesis).

Then fork=m+ 1 holdsx^(m+1)n3yⁿ²xⁿ³yⁿ⁴xⁿ⁵

≈xⁿ³x^mn3yⁿ²xⁿ³yⁿ⁴xⁿ⁵

≈xⁿ³y^{(−m+1)n2−mn4}xⁿ³y^mn2+(m+1)n4x^n5+mn3 (by hypothesis)

≈y^{−mn2−(m+1)n4}xⁿ³y^{(−m+1)n2−mn4+2mn2+2(m+1)n4}x^n5+mn3+n3(by Lemma 15(iii))

≈y(−(m+1)+1)n2−(m+1)n4xⁿ³y^(m+1)n2+((m+1)+1)n4x^n5+(m+1)n3. This shows that (ii) holds.

We show now that the following statement (iii) holds:

For any natural numbers n1, n2, n3, n4, n5 there are natural numbers a, b, c, d such thatxⁿ¹yⁿ²xⁿ³yⁿ⁴xⁿ⁵≈y^ax^by^cx^d,n1+n3+n5≡b+d(r), andn2+n4≡ a+c(r).

Leta1, b1, c1, d1, e1be natural numbers. Then there are natural numbersk1and r1withr1< c1 such thata1=k1c1+r1. Then we havex^a1y^b1x^c1y^d1x^e1

≈x^r1x^k1c1y^b1x^c1y^d1x^e1

≈x^r1y^{(−k1+1)b1−k1d1}x^c1y^{k1b1+(k1+1)d1}x^e1+k1c1 (by (ii))

≈x^r1y^{(−k1+1)b1−k1d1}x^c1y^{(k1−1)b1+k1d1}y^b1+d1x^e1+k1c1

≈y^{(−k1+1)b1−k1d1}x^c1y^{(k1−1)b1+k1d1}x^r1y^b1+d1x^e1+k1c1 (by Lemma 14)

≈y^f2x^a2y^b2x^c2y^d2x^e2 with a2 :=c1, b2 := (k1−1)b1+k1d1, c2 := r1, d2 :=

b₁+d₁,e₂:=e₁+k₁c₁andf₂:= (−k₁+ 1)b₁−k₁d₁whereb₂+d₂+f₂=b₁+d₁ and a2 +c2+e2 = a1+c1+e1. In n ≥ 1 such steps we can derive from x^a1y^b1x^c1y^d1x^e1 a term

y^f2... y^fn+1x^an+1y^bn+1x^cn+1y^dn+1x^en+1with integersan+1, bn+1, cn+1, dn+1,en+1, f2, ..., fn+1 such that cn+1 = 0 and bn+1 +dn+1+ Pⁿ

i=1

fi+1 = b1+d1 and an+1+cn+1+en+1=a1+c1+e1. Because ofx^r≈ethere are natural num- bers a, b, c, d such that Pⁿ

i=1

fi+1 ≡a(r), an+1 ≡ b(r), bn+1+ dn+1 ≡c(r) and en+1 ≡ d(r), i.e., y^f2... y^fn+1x^an+1y^bn+1x^cn+1y^dn+1x^en+1 ≈ y^ax^by^cx^d ∈ IdV and altogether we have xⁿ¹yⁿ²xⁿ³yⁿ⁴xⁿ⁵ ≈y^ax^by^cx^d ∈IdV. This shows the

statement (iii).

On the other hand there are natural numbers n≥ 1 and a1, ..., a2n such that σ(f) ≈ x^a1y^a2...x^a2n−1y^a2n ∈ IdV with ⁿ⁻¹P

i=0

a2i+1 ≡ Pⁿ

i=1

a2i ≡ 1(r). Using (iii) we get σ(f) ≈ x^ay^bx^cy^d ∈ IdV or σ(f) ≈ y^dx^cy^bx^a ∈ IdV for some natural numbers a, b, c, d with a+c ≡ⁿ⁻¹P

i=0

a2i+1 and Pⁿ

i=1

a2i ≡ b +d, i.e.

a+c≡b+d≡1(r).

Now we check that the application of σ to the group identities gives again identities in V. We note that σ(e) ≈ e and σ(g) ≈ x⁻¹ since cx(σ(g))− c_x⁻¹(σ(g))≡ −1(r) andx^r≈e∈IdV. Thus we have

b

σ[f(x, e)]≈x^a+c≈x=σ[x] andb b

σ[f(x, g(x))]≈x^a+c(x⁻¹)^b+d ≈xx⁻¹=e=bσ[e] sincea+c≡b+d≡1(r) and x^r≈e∈IdV. Dually we getσ[f(e, x)]b ≈σ[x]b ∈IdV andbσ[f(g(x), x)]≈bσ[e]∈ IdV.

Now we show that the application ofσ to the associative law gives an identity inV. For this we check by induction onk that

(x^ay^bx^cy^d)^kz(y^−dx^−cy^−bx^−a)^k ≈x^ky^kzy^−kx^−k∈IdV (iv) Fork= 1 we havex^ay^bx^cy^dzy^−dx^−cy^−bx^−a

≈x^ax^cy^bx^−cx^cy^dzy^−dx^−cx^cy^−bx^−cx^−a (by (i))

≈x^a+cy^b+dzy^−(b+d)x^−(a+c)

≈xyzy⁻¹x⁻¹ sincea+c≡b+d≡1(r) andx^r≈e∈IdV.

Suppose now that (iv) is true for k = m, i.e. it holds (x^ay^bx^cy^d)^mz (y^−dx^−cy^−bx^−a)^m≈x^my^mzy^−mx^−m∈IdV (hypothesis).

Then fork=m+ 1 we have (x^ay^bx^cy^d)^m+1z(y^−dx^−cy^−bx^−a)^m+1

≈(x^ay^bx^cy^d)x^my^mzy^−mx^−m(y^−dx^−cy^−bx^−a) (by hypothesis)

≈x^ax^cy^bx^−cx^cy^dx^my^mzy^−mx^−my^−dx^−cx^cy^−bx^−cx^−a (by (i))

≈x^a+cy^b+dx^my^mzy^−mx^−my^−(b+d)x^−(a+c)

≈x^a+cx^my^b+dx^−mx^my^mzy^−mx^−mx^my^−(b+d)x^−mx^−(a+c) (by (i))

≈x^a+c+my^b+d+mzy^−(b+d+m)x^−(a+c+m)

≈x^1+my^1+mzy^−(1+m)x^−(1+m)sincea+c≡b+d≡1(r) andx^r≈e∈IdV. Now we havebσ[f(f(x, y), z)]≈(x^ay^bx^cy^d)^az^b(x^ay^bx^cy^d)^cz^d

≈(x^ay^bx^cy^d)^az^b(x^ay^bx^cy^d)^−a(x^ay^bx^cy^d)z^d (since a+c ≡ 1(r) and x^r ≈ e∈ IdV)

≈(x^ay^bx^cy^d)^az^b(y^−dx^−cy^−bx^−a)^a(x^ay^bx^cy^d)z^d

≈x^ay^az^by^−ax^−a(x^ay^bx^cy^d)z^d (by (iv))

≈x^ay^az^by^−a+bx^cy^dz^d

≈x^ay^az^by^−a+bx^cy^a+c−bz^d (sincea+c≡b+d≡1(r) andx^r≈e∈IdV)

≈x^ay^by^a−bz^by^−a+bx^cy^a−bz^−by^b−ay^−by^az^by^cz^d

≈x^ay^bz^bx^cz^−by^−by^az^by^cz^d (by (i))

≈x^a(y^az^by^cz^d)^bx^c(z^−dy^−cz^−by^−a)^b(y^az^by^cz^d) (by (iv))