Identities of the plactic monoid

Abstract

It is shown that the plactic monoid M of rank $3$ satisfies the identity $wvvwvw=wvwvvw$ where $v=xyyx xy xyyx$ and $w= xyyx yx xyyx$. This is accomplished by first showing that certain simple monoids related to $M$ satisfy this identity. These simple monoids are natural generalizations of the bicyclic monoid $B$, which satisfies the identity $w=v$ by a result of Adjan.

1 Introduction

For an integer $n\ge 1$ we consider the finitely presented monoid $M_n=\left<a_1,\ldots ,a_n\right>$ defined by the relations

$$\begin{aligned} a_ia_ka_j=a_ka_ia_j\quad&\text {for }i\le j<k,\\ a_ja_ia_k=a_ja_ka_i\quad&\text {for }i<j\le k. \end{aligned}$$

It is called the plactic monoid of rank $n$ and was introduced in [14]. It is known that the elements of $M_n$ can be written in a canonical form, which implies that they are in a one-to-one correspondence with Young tableaux of certain type. In particular, $M_{n}$ has polynomial growth of degree $n(n+1)/2$. Because of its strong connections to Young tableaux, the plactic monoid has proved to be a very powerful tool in representation theory and in algebraic combinatorics, see [7, 13]. The combinatorics of $M_n$ has been extensively studied, recently including also the aspect of the Gröbner-Shirshov bases [2, 6, 11]. The algebraic structure and representations of the monoid algebra $K[M_n]$ of $M_n$ over a field $K$ have been investigated, [4, 10]. This paper is motivated by the following conjecture.

Conjecture

The plactic monoid of any rank $n\ge 1$ satisfies a nontrivial identity.

On one hand, a motivation comes from general problems concerning existence of identities in classes of finitely generated semigroups of polynomial growth. In particular, it is known (by results of Gromov and Grigorchuk) that a finitely generated cancellative semigroup of polynomial growth satisfies a nontrivial identity, see [15]. However, as shown by Shneerson in [17], this is no longer true without the cancellativity hypothesis. On the other hand, linear semigroups of polynomial growth satisfy a nontrivial identity, [15], Corollary 5.8, while there exist linear semigroups satisfying a nontrivial identity that have subexponential growth, see [15], Example on page 186, and [5] for further results in this direction. Further connections between polynomial growth and the theory of varieties of semigroups were explored in [18], motivated in particular by Sapir’s problem on relatively free semigroups of polynomial growth [19]. While simple monoids seem to be of special interest from the point of view of varieties of semigroups, it is also worth mentioning that the structure of $M_{n}$ heavily depends on certain simple monoids discovered in [10]. These monoids can be considered as generalizations of the bicyclic monoid $B$. Therefore, a result of Adjan [1], asserting that the bicyclic monoid $B=\langle p,q \mid qp=1 \rangle$ satisfies the identity

$$\begin{aligned} xyyx xy xyyx = xyyx yx xyyx. \end{aligned}$$

(1)

is the starting point for our approach.

On the other hand, the conjecture is motivated by the fact that a class of strongly related monoids, called Chinese monoids $C_{n}, n\ge 1$, satisfies identity (1). This class was introduced and studied in [3]. Actually, the monoids $C_{n}$ and $M_{n}$ are isomorphic if $n\le 2$. It was shown in [9] that $C_{n}$ embeds into the product $B^{i}\times \mathbb Z^{j}$, for some positive integers $i,j$ depending on $n$, where $\mathbb Z$ denotes the additive group of integers. This is a consequence of the description of the minimal prime spectrum of the semigroup algebra $K[C_{n}]$ over a field $K$. Therefore, since the bicyclic monoid satisfies identity (1), $C_{n}$ also satisfies this identity.

Applying some of the results of [10] we prove in this paper that $M=M_{3}$ satisfies the identity

$$\begin{aligned} wvvwvw = wvwvvw \end{aligned}$$

(2)

where $v=v(x,y)=xyyx xy xyyx$ and $w=w(x,y)= xyyx yx xyyx$. Our approach is based on homomorphic images of $M$ arising from natural maps $K[M]\rightarrow K[M]/P$, where $P$ are minimal prime ideals of $K[M]$ and on certain infinite dimensional linear representations of $M$. All such primes $P$ of $K[M]$ and all irreducible representations of $M$ have been determined [10], while in case of plactic monoids $M_{n}$ with $n\ge 4$ the corresponding problems remain a challenge. In particular, there is no obvious starting point for an attempt to extend our proof to the class of all plactic monoids. One might expect that for every $n\ge 4$ the monoid $M_{n}$ satisfies some identities, which however would be more complicated than those satisfied by $M_{n-1}$ (unlike the case of Chinese monoids mentioned above) and thus the proofs might be more involved, accordingly.

2 Identities of $M_{3}$

Adjan’s result was recently reproved in a completely different but quite complicated way in [8], starting from an embedding of $B$ into the semigroup of $2\times 2$ tropical matrices. We start with a simple conceptual proof of this result, and for a more detailed study of identities satisfied by the bicyclic monoid we refer to [16]. It is worth mentioning that one of the key ideas of the proof of our main result is based on a strategy similar to the argument given below.

Proposition 2.1

The bicyclic monoid $B$ satisfies the identity

$$\begin{aligned} w(x,y)=v(x,y). \end{aligned}$$

(3)

Proof

We apply the faithful representation $\phi :B\rightarrow {{\mathrm{End}}}(V)$, where $V$ is a vector space over a field $K$ with a basis $\{e_s:s\ge 0\}$ and

$$\begin{aligned} \phi (p)(e_{s})=e_{s+1},\quad \phi (q)(e_{s})= {\left\{ \begin{array}{ll} 0&{}\quad \text {if }s=0,\\ e_{s-1}&{}\quad \text {if }s>0, \end{array}\right. } \end{aligned}$$

see Exercise 11.9 in [12]. For simplicity we write $be_{s}= \phi (b)(e_{s})$ for $b\in B$. Notice that $p^{i}q^{j}(e_{s})=e_{s-j+i}$ if it is nonzero, which holds exactly when $j\le s$. In particular, if $be_{s}=0$ then $be_{t}=0$ for every $t\le s$.

Let $x=p^{i}q^{j}, y=p^k q^m\in B$. In order to show that $w(x,y)=v(x,y)$ we may assume that $i+ k - j - m \ge 0$, because otherwise we apply the involution on $B$ defined by $p\mapsto q$, $q\mapsto p$, which reduces the problem to this case. It is now enough to show that for every $s\ge 0$ we have $xy xyyx e_{s}= yx xyyx e_{s}$. We may assume that one of these elements is nonzero. Then $xyyx e_{s}\ne 0$. Write $yxe_{s}=e_{s'}$, $xyyxe_{s}=e_{s''}$. Then

$$\begin{aligned} s''= s' + (i+ k - j - m)\ge s'= s+ (i+ k - j - m) \ge s \end{aligned}$$

by the assumption.

Since $s''\ge s$ and $yx e_{s}\ne 0$, we must have $yx xyyx e_{s}=yx e_{s''}\ne 0$. Second, since $s''\ge s'$ and $xy e_{s'}=e_{s''}\ne 0$, we also have $xy xyyx e_{s}=xy e_{s''}\ne 0$.

Hence, both elements $we_{s}, ve_{s}$ are nonzero, and therefore they are both equal to $e_{s+3(i+ k -j - m)}$. The result follows. $\square$

It is known that $a_{n}a_{n-1} \cdots a_{2}a_{1}$ is a central and regular element of $K[M_{n}]$. In particular $M_{2}=\langle a,b \rangle \subseteq M_{2}\langle ba \rangle ^{-1}\cong B\times \mathbb {Z}$ and $M_{2}$ satisfies identity (1) by Proposition 2.1, as was already noticed in [9].

We recall from [10] basic information on certain homomorphic images $N_{1}$ and $N_{2}$ of $M=M_{3}$ that were used in the context of the classification of minimal prime ideals of the monoid algebra $K[M]$. Namely, $M$ is described by the presentation

$$\begin{aligned} M=\langle a,b,c \rangle , \end{aligned}$$

where

$$\begin{aligned} \begin{array}{l} aba =baa,\quad bab =bba,\quad aca =caa,\quad cac =cca,\\ cbb =bcb,\quad cbc =ccb,\quad bac =bca,\quad acb =cab \end{array} \end{aligned}$$

and we define $N_{1}=M/(ac=ca)$ and $N_{2}=M/(bacb=cbba)$. Here, for any elements $u_{1},u_{2}\in M$ by $M/(u_{1}=u_{2})$ we mean the factor semigroup $M/\tau$, where $\tau$ is the congruence on $M$ generated by the pair $(u_{1},u_{2})$. We will denote the generators of $M$ and their natural images in the considered homomorphic images of $M$ in the same way, if unambiguous.

The element $z=cba$ is a central and regular element of $K[M]$ and also of $K[N_{i}]$ for $i=1,2$, and one can consider the central localizations $M\langle z\rangle ^{-1}$and $N_{i}\langle z\rangle ^{-1}$. Moreover we have

$$\begin{aligned} M\langle z\rangle ^{-1} \cong M/(z=1)\times \langle z,z^{-1}\rangle \cong M/(z=1)\times \mathbb {Z}\end{aligned}$$

(4)

$$\begin{aligned} N_{i}\langle z\rangle ^{-1}\cong N_{i}/(z=1)\times \langle z,z^{-1}\rangle \cong N_{i}/(z=1)\times \mathbb Z \end{aligned}$$

(5)

and

$$\begin{aligned} N_{i}\langle z\rangle ^{-1}/(z=1)\cong N_{i}/(z=1). \end{aligned}$$

The map $f:M\rightarrow M$ defined by $a\mapsto cb$, $b\mapsto ca$, $c\mapsto ba$ is an antimonomorphism, which induces an involution $M/(z=1)\rightarrow M/(z=1)$. Moreover, $M$ is a subdirect product of $N_{1}$ and $N_{2}$ and $M/(z=1)$ is a subdirect product of $N_{1}/(z=1)$ and $N_{2}/(z=1)$. Notice that $f$ induces an antiisomorphim of the latter two monoids. It follows that in order to decide whether $M$ satisfies nontrivial identities it is sufficient to verify this for the monoid $N_{1}/(z=1)$. Namely, let us say that a pair of words $s=s(x,y)$, $t=t(x,y)$ in $x,y$ is reversive if one obtains the same two words (as a set) when reading $s(x,y)$, $t(x,y)$ backwards. Then we get

Lemma 2.2

The monoids $M, M/(cba =1)$ and $N_{1}/(cba=1)$ satisfy the same identities of the form $s(x,y)= t(x,y)$, where $s,t$ is a reversive pair of words.

Every element $u\in N_1$ can be uniquely written in the form

$$\begin{aligned} u=(cba)^{k_1}(ba)^{k_2}a^{k_3}(cb)^{k_4}b^{k_5}c^{k_6}, \end{aligned}$$

(6)

where $k_i\ge 0$, see [10], Lemma 2.3. Hence, in view of (5), every element $\overline{u}$ of the monoid $N_1/(z=1)$ can be uniquely written in the form

$$\begin{aligned} \overline{u}=(ba)^{k_2}a^{k_3}(cb)^{k_4}b^{k_5}c^{k_6}, \end{aligned}$$

(7)

where $k_i\ge 0$. And it follows that $N_1/(z=1)$ is a simple monoid, and hence so is $N_{2}/(z=1)$. Since $cb\cdot a= c\cdot ba$ is the unity of $N_1/(z=1)$, these monoids carry some flavor of the monoid $B$, see [10], Proposition 2.6.

Consider the natural homomorphisms

$$\begin{aligned} \phi _{1}:N_{1}\rightarrow N_{1}/(ab=ba) \end{aligned}$$

and

$$\begin{aligned} \phi _{2}:N_{1}\rightarrow N_{1}/(bc=cb). \end{aligned}$$

So, the former makes the image of $a$ central. It is easy to see that

$$\begin{aligned} N_{1}/(ab=ba)\cong \langle a \rangle \times \langle b,c \rangle \cong \mathbb N \times M_{2} \end{aligned}$$

(8)

and similarly

$$\begin{aligned} N_{1}/(bc=cb)\cong \langle c \rangle \times \langle a,b \rangle \cong \mathbb N \times M_{2}, \end{aligned}$$

(9)

with $\mathbb N$ denoting the additive monoid of non-negative integers.

Assume that two elements

$$\begin{aligned} \alpha&=(cba)^{k_1}(ba)^{k_2}a^{k_3}(cb)^{k_4}b^{k_5}c^{k_6},\\ \alpha '&=(cba)^{k'_1}(ba)^{k'_2}a^{k'_3}(cb)^{k'_4}b^{k'_5}c^{k'_6} \end{aligned}$$

of $N_{1}$ satisfy $\phi _{i}(\alpha )=\phi _{i}(\alpha ')$ for $i=1,2$. Using the canonical form of elements of $\mathbb N \times M_{2}$ it is easy to see that

$$\begin{aligned} \phi _{1} (\alpha ) = a^{k_{1}+k_{2}+k_{3}} (cb)^{k_{1}+k_{4}} b^{k_{2}+k_{5}}c^{k_{6}}, \quad \phi _{2} (\alpha ) = c^{k_{1}+k_{4}+k_{6}} (ba)^{k_{1}+k_{2}} a^{k_{3}} b^{k_{4}+k_{5}} \end{aligned}$$

and similar presentations hold for the elements $\phi _{1} (\alpha ' )$ and $\phi _{2} (\alpha ')$. Consequently, we get

$$\begin{aligned} k_{1}+k_{2}+k_{3}=k'_{1}+k'_{2}+k'_{3},\ k_{1}+k_{4}=k'_{1}+k'_{4},\ k_{2}+k_{5}=k'_{2}+k'_{5},\ k_{6}=k'_{6} \end{aligned}$$

and

$$\begin{aligned} k_{1}+k_{4}+k_{6}=k'_{1}+k'_{4}+k'_{6},\ k_{1}+k_{2}=k'_{1}+k'_{2},\ k_{4}+k_{5}=k'_{4}+k'_{5},\ k_{3}=k'_{3}. \end{aligned}$$

This is equivalent to the conditions

$$\begin{aligned} k_{1}+k_{2}=k'_{1}+k'_{2},\ k_{1}+k_{4}=k'_{1}+k'_{4},\ k_{2}+k_{5}=k'_{2}+k'_{5},\ k_{3}=k'_{3},\ k_{6}=k'_{6}. \end{aligned}$$

If $k_{1}=k_{1}'$ then clearly $k_{q}=k'_{q}$ for $q=1,2,3,4,5,6$, so that $\alpha =\alpha '$. Hence, assume for example that $k_{1}>k_{1}'$. Let $k_{1}'=i$ and $k_{1}=i+s$ for some $s\ge 1$. Then also $k_{2}'=k_{2}+s, k_{4}'=k_{4}+s, k_{5}=k_{5}'+s$ and $k_{2}'=k_{2}+s$. Put $j=k_{2}, k=k_{3}, l=k_{4}, m=k_{5}, r=k_{6}$. Then $\alpha , \alpha '$ are of the form

$$\begin{aligned} \alpha&= (cba)^{i+s}(ba)^{j}a^{k}(cb)^{l}b^{m+s}c^{r}= [(cba)^{i+s}(ba)^{j}a^{k}]b^{s}[(cb)^{l}b^{m}c^{r}], \end{aligned}$$

(10)

$$\begin{aligned} \alpha '&= (cba)^{i}(ba)^{j+s}a^{k}(cb)^{l+s}b^{m}c^{r}= [(cba)^{i}(ba)^{j}a^{k}]((ba)^{s}(cb)^{s})[(cb)^{l}b^{m}c^{r}] \qquad \quad \end{aligned}$$

(11)

and the images of $\alpha , \alpha '$ in $N_{1}/(z=1)$ are of the form

$$\begin{aligned} (ba)^{j}a^{k}b^{s}(cb)^{l}b^{m}c^{r},\quad (ba)^{j}a^{k}((ba)^{s}(cb)^{s})(cb)^{l}b^{m}c^{r}, \end{aligned}$$

(12)

respectively. So $\alpha ,\alpha '$ are in the congruence $\rho$ on $N_{1}$ generated by all pairs $((cba)^{s}b^{s}, (ba)^{s}(cb)^{s})$, $s\ge 1.$ It follows that $\rho ={{\mathrm{Ker}}}\phi _1 \cap {{\mathrm{Ker}}}\phi _2$, the intersection of the congruences determined by $\phi _{1}$ and $\phi _{2}$ and $N_{1}/\rho$ embeds into the monoid $N_{1}/(ab=ba) \times N_{1}/(bc=cb)$.

We know that $\mathbb N \times M_{2}$ satisfies Adjan’s identity (1) because the plactic monoid of rank $2$ satisfies every identity holding in the bicyclic monoid. Hence, in view of (8) and (9), $N_{1}/\rho$ satisfies this identity.

In particular, for every $x,y\in N_{1}$ the elements $xy^{2}x xy xy^{2}x, xy^{2}x yx xy^{2}x$ must be of the form described in (10) and (11), while their images in $N_{1}/(z=1)$ are of the form (12). Moreover, if $k_{1}=k_{1}'$ (equivalently $s=0$) then $\alpha =\alpha '$, with the notation as above.

Hence we obtain the following consequence.

Lemma 2.3

Let $w_{1},w_{2}$ be two distinct words in the free monoid of rank two of the same total degree. Assume also that every evaluation of the words $w_{1}(v,w)$ and $w_{2}(v,w)$ in $N_{1}$ (with $v=v(x,y)$, $w=w(x,y)$ defined in (2)) leads to two elements of $N_{1}$ with equal exponents of $cba$ in the canonical form. Then $N_{1}$ satisfies a nontrivial identity, namely

$$\begin{aligned} w_{1}(xy^{2}xxyxy^{2}x,xy^{2}xyxxy^{2}x)= w_{2}(xy^{2}xxyxy^{2}x,xy^{2}xyxxy^{2}x). \end{aligned}$$

For our main result we will use an idea analogous to that used in the proof of Proposition 2.1. Consider the representation $\phi$ of $M$ coming from Proposition 3.6 in [10] (with $\beta =\gamma =1$). Namely, let $V$ be a vector space over a field $K$ with basis $\{e_{pq}:p,q\ge 0\}$. Let the action of $a,b,c\in M$ on $V$ be given by

$$\begin{aligned} ae_{pq}=e_{p,q+1},\quad be_{pq}= {\left\{ \begin{array}{ll}e_{pq}&{}\quad \text {if }q=0,\\ e_{p+1,q-1}&{}\quad \text {if }q>0,\end{array}\right. }\qquad ce_{pq}={\left\{ \begin{array}{ll}0&{}\quad \text {if }p=0,\\ e_{p-1,q}&{}\quad \text {if }p>0.\end{array}\right. } \end{aligned}$$

Then this action makes $V$ a (simple) left $K[M]$-module. Actually, this is also a $K[N_{1}]$-module and a $K[N_{1}/(z=1)]$-module.

We will use the fact that

$$\begin{aligned} b^{s}e_{pq}=e_{p+\min \{s,q\},q-\min \{s,q\}} \end{aligned}$$

and another easy consequence of the definition

$$\begin{aligned} (ba)^{s}(cb)^{s}e_{pq}={\left\{ \begin{array}{ll}0 &{}\quad \text {if }p+q<s,\\ e_{p+\min \{s,q\},q-\min \{s,q\}} &{}\quad \text {if}p+q\ge s.\end{array}\right. } \end{aligned}$$

In particular, $b^{s}e_{pq}=(ba)^{s}(cb)^{s}e_{pq}$ if $(ba)^{s}(cb)^{s}e_{pq}\ne 0$. Moreover, the latter is $0$ only if $p+q<s$ and then $b^{s}e_{pq}=e_{p+q,0}$.

Lemma 2.4

The representation $\phi$ is faithful on $N_{1}/(z=1)$.

Proof

We use the canonical form (7) of elements in $N_{1}/(z=1)$. Suppose that $\phi (t) =\phi (t')$ for some

$$\begin{aligned}t=(ba)^{k_2}a^{k_3}(cb)^{k_4}b^{k_5}c^{k_6},\quad t' =(ba)^{k'_2}a^{k'_3}(cb)^{k'_4}b^{k'_5}c^{k'_6} \in N_{1}/(z=1),\end{aligned}$$

where $k_i\ge 0$. If $k'_{6}>k_{6}$ then $t'e_{k_{6},k_{4}}=0$ and $te_{k_{6},k_{4}}\ne 0$. So we may assume $k'_{6}=k_{6}$. If $k'_{4}>k_{4}$ then $t'e_{k_{4}+k_{6,0}}= 0$ and $te_{k_{4}+k_{6,0}}\ne 0$. So $k_{4}=k'_{4}$. Now $te_{pq}= e_{p-k_{6}+k_{5}+k_{2},q-k_{5}-k_{4}+k_{3}}$ if $p\ge k_{6}$ and $q\ge k_{4}+k_{5}$. Hence $k_{5}+k_{2}=k'_{5}+k'_{2}$ and $-k_{5}+k_{3}=-k'_{5}+k'_{3}$. If $k'_{5}\ge k_{5}$ then $t'e_{k_{6} , k_{4}+k'_{5}}=e_{k'_{5}+k_{2},k_{3}}$ and $te_{k_{6} , k_{4}+k'_{5}}= e_{k_{5}+k_{2},k_{3}}$. Hence $k'_{5}=k_{5}$ and the assertion follows easily. $\square$

We will often use the following standard fact [10, 14].

Lemma 2.5

If $t\in M$ then $t= cubu'a$ for some $u,u'\in M$ if and only if $t=cba t'$ for some $t'\in M$.

We will also use the involution $\epsilon :M\rightarrow M$ determined by $\epsilon (a)=c$, $\epsilon (b)=b$, $\epsilon (c)=a$. It also leads to an involution of $N_{1}$, also denoted by $\epsilon$ for simplicity.

Theorem 2.6

The monoid $M=M_{3}$ satisfies the identity $wvvwvw=wvwvvw$, where $v=xyyx xy xyyx$ and $w= xyyx yx xyyx$ represent the left-hand and respectively the right-hand side of identity (1).

Proof

The main difficulty of the proof is to show that this identity is satisfied on $N_{1}$. Our approach will be based on Lemma 2.4 and on Lemma 2.3. For $x,y\in N_{1}$ and $w=w(x,y), v=v(x,y)$ we have

$$\begin{aligned} \{ w, v \}= \{ (cba)^{i+s}(ba)^{j}a^{k}b^{m}\ b^{s}(cb)^{l}c^{r},\ (cba)^{i}(ba)^{j}a^{k}b^{m}\ ((ba)^{s}(cb)^{s})(cb)^{l}c^{r}\} \end{aligned}$$

for some $s$, see (10),(11). Hence, in order to decide whether $\alpha (w(x,y), v(x,y)) = \beta (w(x,y), v(x,y))$ in $N_{1}$ for some $x,y\in N_{1}$ and some words $\alpha , \beta$ of the same degree in each of the two variables, it is enough to show that $\alpha (w(x,y), v(x,y))$ and $\beta (w(x,y), v(x,y))$ have the same kernel on the basis $\{ e_{pq}: p,q\ge 0\}$ of the module $V$ introduced before Lemma 2.4. (Indeed, in this case their images must be equal in $N_{1}/(z=1)$ by the observation made before Lemma 2.4, so that a degree argument shows that they are equal in $N_{1}$.) We will use this observation several times without further comment.

First, we consider the case where

$$\begin{aligned} \begin{aligned} w&=(cba)^{i+s}(ba)^{j}a^{k}b^{m}\ b^{s}(cb)^{l}c^{r},\\ v&= (cba)^{i}(ba)^{j}a^{k}b^{m}\ ((ba)^{s}(cb)^{s})(cb)^{l}c^{r}. \end{aligned} \end{aligned}$$

(13)

We will show that in $N_{1}$ either $wvwv=vwwv$ and $vwvw=wvvw$, or $wvwv=wvvw$ and $vwvw=vwwv$. The case where the roles of $w$ and $v$ are switched then also follows because of the symmetry of these equalities.

We write $u=(ba)^{j}a^{k}$, $t = (cb)^{l}b^{m}c^{r}$, so that

$$\begin{aligned} w=(cba)^{i+s}u b^{s}t,\quad v= (cba)^{i}u((ba)^{s}(cb)^{s})t \end{aligned}$$

(14)

and consider the following four cases:

• 1. $r\le j$ and $l\le k$,

• 2. $r\le j$ and $l> k$,

• 3. $r>j$ and $l\le k$,

• 4. $r> j$ and $l> k$.

It is easy to see that the product $tu$ is divisible by $(cba)^{r+l}, (cba)^{r+k+ \min \{ l-k, j-r\}}$, $(cba)^{j+l},(cba)^{k+j}$ in these cases, respectively. So, the central regular element $cba$ appears in this way the same number of times in each of the elements $wvwv, vwwv, vwvw$ and $wvvw$. Moreover, the initial segment $u$ of each of these elements and the terminal segment $t$ do not affect the exponent of $cba$ in the canonical form of these elements in $N_{1}$. Therefore, deleting from $u$ and from $t$ the appropriate factors (those that produce the maximal power of $cba$ in the product $tu$) we may assume that $u,t$ have one of the following forms (corresponding to the cases 1–4 listed above):

• 1. $u=(ba)^{j}(a)^{k}$ and $t=(b)^{m}$,

• 2. $u =(ba)^{j}$ and $t=(cb)^{l}(b)^{m}$,

• 3. $u=(a)^{k}$ and $t=(b)^{m}(c)^{r}$,

• 4. $u=1$ and $t=(cb)^{l}(b)^{m}(c)^{r}$.

In the second case, if the exponents of $ba$ and $cb$ are nonzero, the situation is also reduced to case 1. or case 4., due to the possible cancellation of $cba$ coming from the factor $cb \ ba$ arising from $tu$.

So, cancelling the appropriate exponents of $cba$ in $w$ and $v$, we may assume that the elements $w,v$ are of one of the following forms (corresponding to cases 1, 4, and 3, respectively):

• (i) $w=(ba)^{j}a^{k}b^{m+s}$ and $v=a^{k}(ba)^{j+s}(cb)^{s}b^{m}$,

• (ii) $w=b^{m+s}(cb)^{l}c^{r}$ and $v=(ba)^{s}(cb)^{l+s}b^{m}c^{r}$,

• (iii) $w=a^{k}b^{m+s}c^{r}$ and $v=a^{k}(ba)^{s}(cb)^{s}b^{m}c^{r}$.

In the following computations we often apply Lemma 2.5 and the relations defining $N_{1}$ without further comment.

We consider the three cases indicated above.

Case (i). The proof of this case will be based on Lemma 2.4. First, we will show that $vwvw=wvvw$. As observed in the first paragraph of the proof, to establish the latter equality it is enough to check that

$$\begin{aligned} vwvwe_{pq}\ne 0\text { iff }wvvwe_{pq} \ne 0\text { for all }p,q\ge 0. \end{aligned}$$

(15)

Before we proceed to the proof of (15) let us first note that $we_{pq}\ne 0$ for all $p,q\ge 0$. Moreover $ve_{pq}=0$ if and only if $(cb)^sb^me_{pq}=0$. Since we have

$$\begin{aligned} (cb)^sb^me_{pq}={\left\{ \begin{array}{ll}0, &{}\quad \text {if }p+q<s,\\ e_{p-s+\min \{q,m+s\},q-\min \{q,m+s\}}, &{}\quad \text {if }p+q\ge s,\end{array}\right. } \end{aligned}$$

it is easy to see that $ve_{pq}=0$ if and only if $p+q<s$. Next, let us note that elements $ba$ and $a$ acting on $e_{pq}$ increase $p+q$ exactly by $1$. Similarly, $cb$ annihilates $e_{pq}$ or decreases $p+q$ exactly by $1$, while $b$ does not change $p+q$. Hence we conclude that if $e_{p'q'}=we_{pq}$ or $e_{p'q'}=ve_{pq}\ne 0$, then necessarily $p+q\le p'+q'$.

In order to prove (15) we may assume that one of the elements $vwvwe_{pq}$, $wvvwe_{pq}$ is nonzero. In this situation, we have $vwe_{pq}\ne 0$. Therefore, $e_{p_1q_1}=we_{pq}\ne 0$, $e_{p_2q_2}=ve_{p_1q_1}\ne 0$ and $e_{p_3q_3}=we_{p_2q_2}\ne 0$, for some natural numbers $p_{1},p_{2},p_{3},q_{1},q_{2},q_{3}$. As observed above, in this case $p_1+q_1\le p_2+q_2\le p_3+q_3$. If $ve_{p_2q_2}=0$, then using $p_1+q_1\le p_2+q_2<s$ we would get $ve_{p_1q_1}=0$, a contradiction. Hence $e_{p_3'q_3'}=ve_{p_2q_2}\ne 0$ for some $p_{3}',q_{3}'$. Finally, from $p_2+q_2\le p_3+q_3$ we conclude that $ve_{p_3q_3}\ne 0$ and consequently we get

$$\begin{aligned} vwvwe_{pq}=vwve_{p_1q_1}=vwe_{p_2q_2}=ve_{p_3q_3}\ne 0,\\ wvvwe_{pq}=wvve_{p_1q_1}=wve_{p_2q_2}=we_{p_3'q_3'}\ne 0. \end{aligned}$$

Hence, by (15), in Case (i) the elements $vwvw$ and $wvvw$ are equal in $N_{1}$. In a similar way one shows that $wvwv e_{pq}\ne 0$ if and only if $vwwve_{pq}\ne 0$ for all $p,q\ge 0$ and hence $wvwv=vwwv$.

Case (ii). In this case we apply the involution $\epsilon$ of $N_{1}$ and we get

$$\begin{aligned} \epsilon (w)&= \epsilon (b^{m+s}(cb)^{l}c^{r}) =a^{r}(ba)^{l}b^{m+s},\\ \epsilon (v)&= \epsilon ((ba)^{s}(cb)^{l+s}b^{m}c^{r}) = a^{r}b^{m}(ba)^{l+s}(cb)^{s}. \end{aligned}$$

This leads to Case (i). Therefore, we get relations: $vwvw=vwwv$ and $wvwv=wvvw$ in $N_{1}$.

Case (iii). First, assume that $r\ge k$. We consider three subcases:

• (I) $s\le k\le r$,

• (II) $k\le s \le r$,

• (III) $k\le r \le s$.

If (I) holds then

$$\begin{aligned} wv= [a^{k}b^{m+s}c^{r}][a^{k}(ba)^{s}(cb)^{s}b^{m}c^{r}]=(cba)^{s} u \end{aligned}$$

for some $u\notin cba M$ and

$$\begin{aligned} vw=[a^{k}(ba)^{s}(cb)^{s}b^{m}c^{r}][a^{k}b^{m+s}c^{r}]=(cba)^{s}u' \end{aligned}$$

for some $u'\notin cba M$. By Lemma 2.3 we get $wv=vw$ in $N_{1}$.

If (II) holds then

$$\begin{aligned} wv&= [a^{k}b^{m+s}c^{r}][a^{k}(ba)^{s}(cb)^{s}b^{m}c^{r}]\\&=a^{k}b^{m+s} c^{r-s} a^{k}(cba)^{s} (cb)^{s} b^{m}c^{r}\\&= (cba)^{s} a^{k} (ba)^{k} b^{m+s-k} c^{r-s} (cb)^{s}b^{m}c^{r}\\&=(cba)^{s} a^{k} (ba)^{k} t \end{aligned}$$

where $t\in \langle b,c\rangle$. Then

$$\begin{aligned} v(wv)= u (cb)^{s}c^{r} (cba)^{s} a^{k} (ba)^{k} t = u (cba)^{s+k+k} t_{1} \end{aligned}$$

where $u\in \langle a,b\rangle , t_{1}\in \langle b,c\rangle$. So $vwv \notin (cba)^{s+2k+1}M$. Moreover

$$\begin{aligned} vw&=[a^{k}(ba)^{s}(cb)^{s}b^{m}c^{r}][a^{k}b^{m+s}c^{r}]\\&=a^{k}(ba)^{s} (cba)^{k} b^{m}(cb)^{s-k} c^{r} b^{m+s} c^{r}\\&=(cba)^{k} a^{k}(ba)^{s} t' \end{aligned}$$

where $t'\in \langle b,c\rangle$. Hence

$$\begin{aligned} v(vw)= u (cb)^{s}c^{r} (cba)^{k} a^{k}(ba)^{s} t' = u (cba)^{s+k+k} t_{2} \end{aligned}$$

where $t_{2}\in \langle b,c\rangle$. So $vvw \notin (cba)^{s+2k+1}M$.

By Lemma 2.3 we get $vwv=vvw$ in $N_{1}$.

Similarly

$$\begin{aligned} w(wv)= [a^{k}b^{m+s}c^{r}][(cba)^{s} a^{k} (ba)^{k} t]\in (cba)^{s+k} M\setminus (cba)^{s+k+1}M \end{aligned}$$

and

$$\begin{aligned} w(vw)= [a^{k}b^{m+s}c^{r}][(cba)^{k} a^{k}(ba)^{s} t']\in (cba)^{s+k}M\setminus (cba)^{s+k+1}M. \end{aligned}$$

Hence we also have $wwv=wvw$. So in Case (II) we also get $wvwv=wvvw$ and $vwwv=vwvw$ in $N_{1}$.

If (III) holds then

$$\begin{aligned} wv&=[a^{k}b^{m+s}c^{r}][a^{k}(ba)^{s}(cb)^{s}b^{m}c^{r}]\\&=a^{k}b^{m+s} (cba)^{r}(ba)^{s-r} a^{k} (cb)^{s} b^{m} c^{r} \\&=(cba)^{r} a^{k}b^{2m+s-k} (ba)^{s-r+k} (cb)^{s} c^{r}\\ vw&=[a^{k}(ba)^{s}(cb)^{s}b^{m}c^{r}][a^{k}b^{m+s}c^{r}]\\&=a^{k}(ba)^{s}b^{m} (cba)^{k} (cb)^{s-k} c^{r}b^{m+s}c^{r} \\&=(cba)^{k} a^{k}(ba)^{s}b^{m} (cb)^{s-k+r} b^{m+s-r}c^{r}\\&=(cba)^{k} a^{k}(ba)^{s}b^{2m+s-r} (cb)^{s-k+r} c^{r}. \end{aligned}$$

Then

$$\begin{aligned} wvvw&=[(cba)^{r} a^{k} b^{2m+s-k} (ba)^{s-r+k}(cb)^{s} c^{r}][(cba)^{k} a^{k}(ba)^{s}b^{2m+s-r} (cb)^{s-k+r} c^{r}]\\&=(cba)^{k+r} u (cb)^{s} c^{r} a^{k}(ba)^{s} t\\&=(cba)^{k+r} u (cba)^{k+r}(cb)^{s-k} (ba)^{s-r}t\\&=(cba)^{2k+s+r}u't' \end{aligned}$$

where $u,u'\in \langle a,b\rangle$, $t,t'\in \langle b,c\rangle$ and

$$\begin{aligned} wvwv&=[(cba)^{r} a^{k} b^{2m+s-k} (ba)^{s-r+k} (cb)^{s} c^{r}] [(cba)^{r} a^{k} b^{2m+s-k} (ba)^{s-r+k} (cb)^{s} c^{r}] \\&=(cba)^{r+r} u' (cb)^{s} c^{r} a^{k}(ba)^{s-r+k} t''\\&=(cba)^{r+r} u' (cba)^{k} (cb)^{s-k}c^{r} (ba)^{s-r+k} t''\\&=(cba)^{s+r+2k} u't''' \end{aligned}$$

where $u'\in \langle a,b\rangle$, $t'',t'''\in \langle b,c\rangle$. By Lemma 2.3 we get $wvwv=wvvw$ in $N_{1}$.

Moreover

$$\begin{aligned} vwwv&=[(cba)^{k} a^{k}(ba)^{s}b^{2m+s-r} (cb)^{s-k+r} c^{r} ][ (cba)^{r} a^{k} b^{2m+s-k} (ba)^{s-r+k} (cb)^{s} c^{r}]\\&=u (cba)^{k+r} (cb)^{s-k+r} c^{r} a^{k} (ba)^{s-r+k} t\\&=u (cba)^{k+r +s-r +k} c^{r} b^{s-r+k} (cb)^{2r-2k} a^{k}t \\&=u (cba)^{2k+s+k}t' \end{aligned}$$

where $u\in \langle a,b\rangle$, $t,t'\in \langle b,c\rangle$ (because the degree of $b$ in $(cb)^{2r-2k}b^{s-r+k}$ is $s+r-k\ge k$). Also

$$\begin{aligned} vwvw&=[(cba)^{k} a^{k}(ba)^{s}b^{2m+s-r} (cb)^{s-k+r} c^{r} ][ (cba)^{k} a^{k}(ba)^{s}b^{2m+s-r} (cb)^{s-k+r} c^{r}]\\&=u' (cba)^{k+k} (cb)^{s-k+r} c^{r} a^{k} (ba)^{s} t''\\&=u' (cba)^{k+k} (cb)^{s-k+r} a^{k} (cba)^{r} (ba)^{s-r} t''\\&=u' (cba)^{2k+s+k} t''' \end{aligned}$$

where $u'\in \langle a,b\rangle$, $t'',t'''\in \langle b,c\rangle$ (because the degree of $a$ in $a^{k} (ba)^{s-r}$ is $s+k-r\le s-k+r$).

By Lemma 2.3 we get $vwwv=vwvw$ in $N_{1}$.

It remains to consider the situation where $r<k$ in Case (iii). Then applying the involution $\epsilon$ on $N_{1}$ we get

$$\begin{aligned} \epsilon (v)=(a)^{r}b^{m}(ba)^{s}(cb)^{s}c^{k},\quad \epsilon (w)=a^{r}b^{m+s}c^{k}. \end{aligned}$$

Therefore, this reduces the situation to the case where $r>k$, considered above. This means that $vwwv=wvwv$ and $wvvw=vwvw$ in $N_{1}$.

Conclusion: if $w=w(x,y)$ and $v =v(x,y)$ for some $x,y\in N_{1}$ then either $wvwv=vwwv$ and $vwvw=wvvw$ or $wvwv=wvvw$ and $vwvw=vwwv$. Hence in the former case we have

$$\begin{aligned} wv (vwvw) = wv (wvvw) \end{aligned}$$

while in the latter

$$\begin{aligned} (wvvw) vw = (wvwv) vw. \end{aligned}$$

Thus, the identity $wvvwvw=wvwvvw$ is satisfied in $N_{1}$, as claimed at the beginning of the proof.

Recall that $v$ (viewed as a word in $x,y$) is obtained by reading backwards the word $w$. Therefore, an argument as that leading to Lemma 2.2 shows that also in $N_{2}$ we get that either $wvwv=vwwv$ and $vwvw=wvvw$, or $wvwv=wvvw$ and $vwvw=vwwv$. So $N_{2}$ also satisfies the identity $wvvwvw=wvwvvw$. Since $M$ is a subdirect product of $N_{1}$ and $N_{2}$, this identity is also satisfied in $M$. $\square$

One can ask whether $M$ satisfies a simpler identity. We conclude by showing that $M$ does not satisfy Adjan’s identity $w(x,y)=v(x,y)$. Let $x=a^{3}c$ and $y=(ba)b(cb)^{2}$. Then

$$\begin{aligned} xy=(cba)a^3 b (cb)^2,\qquad yx=(cba)^{2}(ba)^{2}c \end{aligned}$$

and

$$\begin{aligned} xyyx= (cba)a^3 b (cb)^2\ (cba)^{2}(ba)^{2}c = (cba)^{5}a^{3}b^{3}c, \end{aligned}$$

so that it is easy to see that

$$\begin{aligned} w (x,y)&= (cba)^{5}a^{3}b^{3}c \ (cba)a^3 b (cb)^2 \ (cba)^{5}a^{3}b^{3}c\\&= (cba)^{11} a^{3}b^{3}a^{3}(cb)^{3} a^{3}b^{3}c \\&= (cba)^{14} a^{3} (ba)^{3}b^{3}c,\\ v (x,y)&= (cba)^{5}a^{3}b^{3}c\ (cba)^{2}(ba)^{2}c \ (cba)^{5}a^{3}b^{3}c \\&= (cba)^{12} a^{3}b^{3}c (ba)^{2}c a^{3}b^{3}c \\&= (cba)^{13} a^{3} (ba) b^{3}c a^{3}b^{3}c \\&= (cba)^{13} a^{3} (ba)^{4}b^{2} (cb) c \end{aligned}$$

are written in the canonical forms in $M$, whence they are different. Notice that $vw$ and $wv$ again have a different exponent of $cba$ in their canonical forms, namely $29$ and $28$, respectively; whence they are different. But we get $wvvw=vwvw$ and $vwwv=wvwv$.