2 Preliminaries

An identity $\mathbf {u} \approx \mathbf {v}$ is directly deducible from an identity $\mathbf {s} \approx \mathbf {t}$ if there exist words $\mathbf {a},\mathbf {b} \in \mathcal {X}^\ast $ and a substitution $\varphi \colon \mathcal {X} \to \mathcal {X}^\ast $ such that $\{\mathbf {u}, \mathbf {v}\} = \{\mathbf {a}\varphi (\mathbf {s})\mathbf {b},\mathbf {a}\varphi (\mathbf {t})\mathbf {b}\}$ . A nontrivial identity $\mathbf {u} \approx \mathbf {v}$ is deducible from a set $\Sigma $ of identities if there exists some finite sequence $\mathbf {u} = \mathbf {w}_0,\ldots , \mathbf {w}_m = \mathbf {v}$ of words such that each identity $\mathbf {w}_i \approx \mathbf {w}_{i+1}$ is directly deducible from some identity in $\Sigma $ .

Given a variety $\mathbf {V}$ , a word $\mathbf {u}$ is called an isoterm for $\mathbf {V}$ if the only word $\mathbf {v}$ such that $\mathbf {V}$ satisfies the identity $\mathbf {u} \approx \mathbf {v}$ is the word $\mathbf {u}$ itself.

The alphabet of a word $\mathbf {w}$ , that is, the set of all letters occurring in $\mathbf {w}$ , is denoted by $\operatorname {\mathrm {alph}}(\mathbf {w})$ . For a word $\mathbf {w}$ and a letter x, let $\operatorname {\mathrm {occ}}_x(\mathbf {w})$ denote the number of occurrences of x in $\mathbf {w}$ . A letter x is called simple (respectively, multiple) in a word $\mathbf {w}$ if $\operatorname {\mathrm {occ}}_x(\mathbf {w})=1$ (respectively, $\operatorname {\mathrm {occ}}_x(\mathbf {w})>1$ ). The set of all simple (respectively, multiple) letters in a word $\mathbf {w}$ is denoted by $\operatorname {\mathrm {sim}}(\mathbf {w})$ (respectively, $\operatorname {\mathrm {mul}}(\mathbf {w})$ ).

The following statement was established in [Reference Gusev and Vernikov5, proof of Lemma 3.5].

If $\mathbf {w}$ is a word and $\mathcal Z\subseteq \operatorname {\mathrm {alph}}(\mathbf {w})$ , then we denote by $\mathbf {w}_{\mathcal Z}$ (respectively, $\mathbf {w}(\mathcal Z)$ ) the word obtained from $\mathbf {w}$ by removing all occurrences of letters from $\mathcal Z$ (respectively, $\operatorname {\mathrm {alph}}(\mathbf {w})\setminus \mathcal Z$ ). If $\mathcal Z=\{z\}$ , then we write $\mathbf {w}_z$ rather than $\mathbf {w}_{\{z\}}$ .

The expression $_{i\mathbf {w}}x$ means the ith occurrence of a letter x in a word $\mathbf {w}$ . If the ith occurrence of x precedes the jth occurrence of y in a word $\mathbf {w}$ , then we write $({_{i\mathbf {w}}x}) < ({_{j\mathbf {w}}y})$ .

3 Proof of Theorem 1.1

We will prove Theorem 1.1 following the scheme (i) $\Rightarrow $ (ii) $\Rightarrow $ (iii) $\Rightarrow $ (i).

(i) $\Rightarrow $ (ii). Let $\mathbf {V}$ be a proper monoid variety which is a modular element of the lattice $\mathbb {M}{\scriptstyle \mathbb {ON}}$ . Then, the variety $\mathbf {V}$ satisfies the identities

(3.1) $$ \begin{align} x^2\approx x^3,\ x^2y\approx yx^2 \end{align} $$

by [Reference Gusev and Lee3, Proposition 4.3].

If $M(xyx)\notin \mathbf {V}$ , then it follows from [Reference Gusev and Vernikov5, Lemma 3.3(i)] that $\mathbf {V}$ satisfies the identities $x^2y\approx xyx\approx yx^2$ which, evidently, imply all the identities listed in item (ii) of Theorem 1.1. So, we may further assume that $M(xyx)\in \mathbf {V}$ .

The rest of the proof proceeds in three steps.

Step 1: $\mathbf {V}$ satisfies the identities $xyzxty\approx yxzxty$ and $xzytxy\approx xzytyx$ .

Arguing by contradiction, suppose that $\mathbf {V}$ violates the identity $xyzxty\approx yxzxty$ . Then, $M(xyzxty)\in \mathbf {V}$ by Lemma 2.4. Let $\mathbf {X}^\prime $ denote the monoid variety defined by the identity

$$ \begin{align*} \mathbf{u}:=z_1t_1z_2t_2\,c^2z_1bz_2\,xcy\,b\,s_1xs_2y\approx z_1t_1z_2t_2\,c^2z_1bz_2\,ycx\,b\,s_1xs_2y=:\mathbf{v}. \end{align*} $$

We need the following auxiliary result.

Proof. Take $\mathbf {w}\in \{\mathbf {u},\mathbf {v}\}$ and consider an arbitrary identity of the form $\mathbf {w}\approx \mathbf {w}^\prime $ that holds in $\mathbf {X}^\prime $ . We are going to verify that $\mathbf {w}^\prime \in \{\mathbf {u},\mathbf {v}\}$ . By Proposition 2.1 and evident induction, we may assume without any loss that the identity $\mathbf {w}\approx \mathbf {w}^\prime $ is directly deducible from the identity $\mathbf {u}\approx \mathbf {v}$ , that is, there exist words $\mathbf {a},\mathbf {b} \in \mathcal {X}^\ast $ and a substitution $\varphi \colon \mathcal {X} \to \mathcal {X}^\ast $ such that $(\mathbf {w}, \mathbf {w}^\prime ) = ( \mathbf {a}\varphi (\mathbf {u})\mathbf {b},\mathbf {a}\varphi (\mathbf {v})\mathbf {b} )$ . Notice that every subword of $\mathbf {w}$ of length $>1$ occurs in $\mathbf {w}$ exactly once and each letter occurs in $\mathbf {w}$ at most thrice. It follows that:

• (*) $\varphi (v)$ is either the empty word or a letter for any $v\in \operatorname {\mathrm {mul}}(\mathbf {u})=\operatorname {\mathrm {mul}}(\mathbf {v})$ .

Suppose first that $\varphi (c)=1$ . In this case, if $\varphi (x)=1$ or $\varphi (y)=1$ , then $\varphi (\mathbf {u})=\varphi (\mathbf {v})$ and so $\mathbf {w}^\prime \in \{\mathbf {u},\mathbf {v}\}$ , as required. So, we may assume that $\varphi (x)\ne 1$ and $\varphi (y)\ne 1$ . Then, $\varphi (x)$ and $\varphi (y)$ are letters by statement ( $\ast $ ). These letters must be distinct since no letter occurs in $\mathbf {w}$ more than thrice. However, the word $\mathbf {w}$ does not contain any subword consisting exclusively of nonlast occurrences of two distinct letters in $\mathbf {w}$ , while the first occurrences of the letters $\varphi (x)$ and $\varphi (y)$ are adjacent in $\mathbf {w}$ , which is a contradiction.

Suppose now that $\varphi (c)\ne 1$ . Then, $\varphi (c)=c$ because c is the only letter that occurs thrice in both $\mathbf {u}$ and $\mathbf {w}$ . Hence,

$$ \begin{align*} \varphi(z_1bz_2x)= \begin{cases} z_1bz_2x&\text{if } \mathbf{w}=\mathbf{u}, \\ z_1bz_2y&\text{if } \mathbf{w}=\mathbf{v}. \end{cases} \end{align*} $$

Further, it follows from statement ( $\ast )$ that $\varphi (b)=b$ , whence

$$ \begin{align*} (\varphi(x),\varphi(y))= \begin{cases} (x,y)&\text{if } \mathbf{w}=\mathbf{u}, \\ (y,x)&\text{if } \mathbf{w}=\mathbf{v}. \end{cases} \end{align*} $$

Since $({_{1\mathbf {u}}}x)<({_{1\mathbf {u}}}y)<({_{2\mathbf {u}}}x)<({_{2\mathbf {u}}}y)$ and $({_{1\mathbf {v}}}y)<({_{1\mathbf {v}}}x)<({_{2\mathbf {v}}}x)<({_{2\mathbf {v}}}y)$ , it follows that the case when $\mathbf {w}=\mathbf {v}$ is impossible. Therefore, $(\varphi (x),\varphi (y))=(x,y)$ in either case. This implies that $\mathbf {w}^\prime =\mathbf {v}$ and we are done.

Put $\mathbf {X}:=\mathbf M(\mathbf {u},\mathbf {v})\wedge \mathbf {X}^\prime $ . Consider an arbitrary identity of the form $\mathbf {u}\approx \mathbf {w}$ that is satisfied by the variety $\mathbf {X}\vee \mathbf {V}$ . In view of Claim 3.1, $\mathbf {w}\in \{\mathbf {u},\mathbf {v}\}$ . Since $M(xyzxty)\in \mathbf {V}$ , Lemma 2.2 implies that $\mathbf {w}(x,y,s_1,s_2)=\mathbf {u}(x,y,s_1,s_2)=xys_1xs_2y$ . Hence, $\mathbf {w} = \mathbf {u}$ . We see that $\mathbf {u}$ is an isoterm for the variety $\mathbf {X}\vee \mathbf {V}$ . By a similar argument, we can show that $\mathbf {v}$ is also an isoterm for $\mathbf {X}\vee \mathbf {V}$ . Then,

$$ \begin{align*} (\mathbf{X}\vee \mathbf{V})\wedge \mathbf M(\mathbf{u},\mathbf{v}) = \mathbf M(\mathbf{u},\mathbf{v}) \end{align*} $$

by Lemma 2.2. It is easy to see that the identity $xyzxty\approx yxzxty$ holds in $M(\mathbf {u},\mathbf {v})$ . Then, $\mathbf {V}\wedge \mathbf M(\mathbf {u},\mathbf {v})$ satisfies $\mathbf {u}\stackrel {\mathbf {V}}\approx \mathbf {u}_cc^2\stackrel {\mathbf M(\mathbf {u},\mathbf {v})}\approx \mathbf {v}_cc^2\stackrel {\mathbf {V}}\approx \mathbf {v}$ . By the very definition, the identity ${\mathbf {u}\approx \mathbf {v}}$ is satisfied by $\mathbf {X}$ as well. Since $\mathbf {X}\subset \mathbf M(\mathbf {u},\mathbf {v})$ , we have

$$ \begin{align*} \mathbf{X}\vee(\mathbf{V}\wedge\mathbf M(\mathbf{u},\mathbf{v}))\subset (\mathbf{X}\vee \mathbf{V})\wedge \mathbf M(\mathbf{u},\mathbf{v}) = \mathbf M(\mathbf{u},\mathbf{v}), \end{align*} $$

which contradicts the fact that $\mathbf {V}$ is a modular element of $\mathbb {M}{\scriptstyle \mathbb {ON}}$ . Therefore, $\mathbf {V}$ satisfies $xyzxty\approx yxzxty$ . By the dual argument, we can show that $xzytxy\approx xzytyx$ holds in $\mathbf {V}$ .

Step 2: $\mathbf {V}$ satisfies the identity $\mathbf {a}_{n,m}[\rho ]\approx \mathbf {a}_{n,m}^\prime [\rho ]$ for all $n,m\in \mathbb {N}$ and $\rho \in S_{n+m}$ .

Arguing by contradiction, suppose that $\mathbf {V}$ violates $\mathbf {a}_{p,q}[\pi ]\approx \mathbf {a}_{p,q}^\prime [\pi ]$ for some ${p, q\in \mathbb {N}}$ and $\pi \in S_{p+q}$ . Then, $M(\mathbf {a}_{n,m}[\rho ])\in \mathbf {V}$ for some $n,m\in \mathbb {N}$ and $\rho \in S_{n+m}$ by [Reference Gusev2, Lemma 4.8]. For brevity, put

$$ \begin{align*} &\mathbf{p}:= \begin{cases} z_1t_1\cdots z_nt_n &\text{if }1\le 1\rho\le n, \\ y_1s_1y_2s_2\,z_1t_1\cdots z_nt_n &\text{if }n< 1\rho\le n+m, \end{cases} \\ &\mathbf r:= \begin{cases} z_{n+1}t_{n+1}\cdots z_{n+m}t_{n+m}\,s_1y_1s_2y_2 &\text{if }1\le 1\rho\le n, \\ z_{n+1}t_{n+1}\cdots z_{n+m}t_{n+m} &\text{if }n< 1\rho\le n+m. \end{cases} \end{align*} $$

Let

$$ \begin{align*} \mathbf{X}:= \begin{cases} \mathbf M(y^2xtx,xy^2tx)&\text{if }1\le 1\rho\le n, \\ \mathbf M(xtxy^2,xty^2x)&\text{if }n< 1\rho\le n+m. \end{cases} \end{align*} $$

Consider an arbitrary identity of the form

$$ \begin{align*} \mathbf{u}:=\mathbf px\bigg(\prod_{i=1}^{n+m-1} z_{i\rho}\bigg)y_1z^2y_2z_{(n+m)\rho}x\mathbf r\approx \mathbf{v} \end{align*} $$

that is satisfied by the variety $\mathbf {X}\vee \mathbf {V}$ . In view of Lemma 2.2,

$$ \begin{align*} &\mathbf{v}_{\{y_1,y_2,s_1,s_2,z\}}=\mathbf{u}_{\{y_1,y_2,s_1,s_2,z\}}=\mathbf{a}_{n,m}[\rho],\\&\mathbf{v}_{\{x,z\}}=\mathbf{u}_{\{x,z\}}=\mathbf{p}\bigg(\prod_{i=1}^{n+m-1} z_{i\rho}\bigg)\,y_1y_2z_{(n+m)\rho}\mathbf r,\\&\mathbf{v}(y_1,s_1,z)=\mathbf{u}(y_1,s_1,z)= \begin{cases} y_1z^2s_1y_1&\text{if }1\le 1\rho\le n, \\y_1s_1y_1z^2&\text{if }n< 1\rho\le n+m, \end{cases} \\&\mathbf{v}(y_2,s_2,z)=\mathbf{u}(y_2,s_2,z)= \begin{cases} z^2y_2s_2y_2&\text{if }1\le 1\rho\le n, \\y_2s_2z^2y_2&\text{if }n< 1\rho\le n+m. \end{cases} \end{align*} $$

It follows that $\mathbf {v} = \mathbf {u}$ . We see that the word $\mathbf {u}$ is an isoterm for the variety $\mathbf {X}\vee \mathbf {V}$ . Then,

$$ \begin{align*} (\mathbf{X}\vee \mathbf{V})\wedge \mathbf M(\mathbf{u}) = \mathbf M(\mathbf{u}) \end{align*} $$

by Lemma 2.2.

Now, we are going to verify that $M(\mathbf {u})$ satisfies the identity

$$ \begin{align*} \mathbf{a}:=\mathbf{u}_z=\mathbf px\bigg(\prod_{i=1}^{n+m-1} z_{i\rho}\bigg)y_1y_2z_{(n+m)\rho}x\mathbf r \approx\mathbf pz_{1\rho}x\,\bigg(\prod_{i=2}^{n+m-1} z_{i\rho}\bigg)\,y_1y_2z_{(n+m)\rho}x\mathbf r=:\mathbf{a}^\prime. \end{align*} $$

We consider an arbitrary substitution $\psi \colon \mathscr X \to M(\mathbf {u})$ and show that $\psi (\mathbf {a})=\psi (\mathbf {a}^\prime )$ . We may suppose that at least one of the elements $\psi (\mathbf {a})$ or $\psi (\mathbf {a}^\prime )$ is nonzero and forms a subword of $\mathbf {u}$ . If $\psi (x)=1$ , then $\psi (\mathbf {a})=\psi (\mathbf {a}^\prime )$ and we are done. So, we may further assume that $\psi (x)\ne 1$ . Clearly, none of the letters $s_1,s_2,t_1,\ldots ,t_{n+m}$ belongs to $\operatorname {\mathrm {alph}}(\psi (x))$ because all these letters are simple in $\mathbf {u}$ , while $x\in \operatorname {\mathrm {mul}}(\mathbf {a})=\operatorname {\mathrm {mul}}(\mathbf {a}^\prime )$ . Further, none of the letters $y_1,y_2,z_1,\ldots ,z_{n+m}$ belongs to $\operatorname {\mathrm {alph}}(\psi (x))$ because there is a simple letter between the first and the second occurrences of any of these letters in $\mathbf {u}$ , while there are no simple letters between the first and the second occurrences of x in both $\mathbf {a}$ and $\mathbf {a}^\prime $ . Finally, $x\notin \operatorname {\mathrm {alph}}(\psi (x))$ because

$$ \begin{align*} \psi\bigg(\bigg(\prod_{i=1}^{n+m-1} z_{i\rho}\bigg)y_1y_2z_{(n+m)\rho}\bigg) \end{align*} $$

cannot contain $z^2$ as a subword. Therefore, $\psi (x)=z$ . Then, $\psi (y_1)=\psi (y_2)=\psi (z_{2\rho })=\cdots =\psi (z_{(n+m)\rho })=1$ . Assume that $\psi (z_{1\rho })\ne 1$ . Then, $\psi (\mathbf {a}^\prime )$ is a subword of $\mathbf {u}$ , and if $1\le 1\rho \le n$ (respectively, $n< 1\rho \le n+m$ ), the image of the second (respectively, first) occurrence of $z_{1\rho }$ in $\mathbf {a}^\prime $ under $\psi $ must contain the first (respectively, second) occurrence of $y_1$ in $\mathbf {u}$ , which is a contradiction. Therefore, $\psi (z_{1\rho })=1$ , whence $\psi (\mathbf {a})=\psi (\mathbf {a}^\prime )$ . Thus, we have proved that the identity $\mathbf {a}\approx \mathbf {a}^\prime $ holds in $M(\mathbf {u})$ .

Then, $\mathbf {V}\wedge \mathbf M(\mathbf {u})$ satisfies

$$ \begin{align*} \mathbf{u}\stackrel{\mathbf{V}}\approx \mathbf{u}_zz^2=\mathbf az^2\stackrel{\mathbf M(\mathbf{u})}\approx \mathbf{a}^\prime z^2\stackrel{\mathbf{V}}\approx \mathbf pz_{1\rho}x\bigg(\prod_{i=2}^{n+m-1} z_{i\rho}\bigg)y_1z^2y_2z_{(n+m)\rho}x\mathbf r=:\mathbf{u}^\prime. \end{align*} $$

It follows from the very definition of $\mathbf {X}$ that $\mathbf {X}\subset \mathbf M(\mathbf {u})$ and the identity $\mathbf {u}\approx \mathbf {u}^\prime $ is satisfied by $\mathbf {X}$ . Then,

$$ \begin{align*} \mathbf{X}\vee(\mathbf{V}\wedge\mathbf M(\mathbf{u}))\subset (\mathbf{X}\vee \mathbf{V})\wedge \mathbf M(\mathbf{u}) = \mathbf M(\mathbf{u}), \end{align*} $$

which contradicts the fact that $\mathbf {V}$ is a modular element of $\mathbb {M}{\scriptstyle \mathbb {ON}}$ . Therefore, $\mathbf {V}$ satisfies $\mathbf {a}_{n,m}[\rho ]\approx \mathbf {a}_{n,m}^\prime [\rho ]$ for all $n,m\in \mathbb {N}$ and $\rho \in S_{n+m}$ .

Step 3: $\mathbf {V}$ satisfies the identities $xzxtxysy\approx xzxtyxsy$ and $xzxytysy\approx xzyxtysy$ .

If $M(xyxzx)\notin \mathbf {V}$ , then, by Lemma 2.2, the variety $\mathbf {V}$ satisfies an identity $xyxzx\approx x^pyx^qzx^r$ with either $p>1$ or $q>1$ or $r>1$ . In any case, this identity together with the identities (3.1) imply the identity $xyxzx\approx x^2yz$ , and so the identities $xzxtxysy\approx xzxtyxsy$ and $xzxytysy\approx xzyxtysy$ . So, we may further assume that $M(xyxzx)\in \mathbf {V}$ .

Arguing by contradiction, suppose that $\mathbf {V}$ violates the identity $xzxtxysy\approx xzxtyxsy$ . Then, $M(xzxtxysy)\in \mathbf {V}$ by Lemma 2.4. Let $\mathbf {X}^\prime $ denote the monoid variety defined by the identity

$$ \begin{align*} \mathbf{u}:=xs_1xs_2z_1t_1z_2t_2\,cz_1bz_2\,xcy\,b\,s_3y\approx xs_1xs_2z_1t_1z_2t_2\,cz_1bz_2\,ycx\,b\,s_3y=:\mathbf{v}. \end{align*} $$

We need the following auxiliary result.

Proof. Take $\mathbf {w}\in \{\mathbf {u},\mathbf {v}\}$ and consider an arbitrary identity of the form $\mathbf {w}\approx \mathbf {w}^\prime $ that holds in $\mathbf {X}^\prime $ . We are going to verify that $\mathbf {w}^\prime \in \{\mathbf {u},\mathbf {v}\}$ . By Proposition 2.1 and evident induction, we may assume without any loss that the identity $\mathbf {w}\approx \mathbf {w}^\prime $ is directly deducible from the identity $\mathbf {u}\approx \mathbf {v}$ , that is, there exist words $\mathbf {a},\mathbf {b} \in \mathcal {X}^\ast $ and a substitution $\varphi \colon \mathcal {X} \to \mathcal {X}^\ast $ such that $(\mathbf {w}, \mathbf {w}^\prime ) = ( \mathbf {a}\varphi (\mathbf {u})\mathbf {b},\mathbf {a}\varphi (\mathbf {v})\mathbf {b} )$ . Notice that every subword of $\mathbf {w}$ of length $>1$ occurs in $\mathbf {w}$ exactly once. It follows that:

• (*) $\varphi (v)$ is either the empty word or a letter for any $v\in \operatorname {\mathrm {mul}}(\mathbf {u})$ .

Suppose first that $\varphi (c)=1$ . In this case, if $\varphi (x)=1$ or $\varphi (y)=1$ , then $\varphi (\mathbf {u})=\varphi (\mathbf {v})$ and so $\mathbf {w}^\prime \in \{\mathbf {u},\mathbf {v}\}$ , as required. So, we may assume that $\varphi (x)\ne 1$ and $\varphi (y)\ne 1$ . Then, $\varphi (x)$ and $\varphi (y)$ are letters by statement ( $\ast $ ). Since x is the only letter that occurs thrice in both $\mathbf {u}$ and $\mathbf {v}$ , we have $\varphi (x)=x$ . It follows that the image of ${_{1\mathbf {w}}}y$ under $\varphi $ must be either ${_{2\mathbf {w}}}c$ (if $\mathbf {w}=\mathbf {u}$ ) or ${_{2\mathbf {w}}}b$ (if $\mathbf {w}=\mathbf {v}$ ), which is a contradiction.

Suppose now that $\varphi (c)\ne 1$ . Then, $\varphi (c)\in \{b,c\}$ because b and c are the only multiple letters of $\mathbf {w}$ between the first and second occurrences of which there are no simple letters. If $\varphi (c)=b$ , then

$$ \begin{align*} \varphi(z_1bz_2x)= \begin{cases} z_2xcy&\text{if } \mathbf{w}=\mathbf{u}, \\ z_2ycx&\text{if } \mathbf{w}=\mathbf{v}, \end{cases} \end{align*} $$

and, by statement ( $\ast $ ), $\varphi (b)\in \{x,y\}$ . However, this contradicts the fact that there are simple letters between ${_{2\mathbf {w}}}x$ and ${_{3\mathbf {w}}}x$ as well as between ${_{1\mathbf {w}}}y$ and ${_{2\mathbf {w}}}y$ in $\mathbf {w}$ . Hence, $\varphi (c)=c$ and so

$$ \begin{align*} \varphi(z_1bz_2x)= \begin{cases} z_1bz_2x&\text{if } \mathbf{w}=\mathbf{u}, \\ z_1bz_2y&\text{if } \mathbf{w}=\mathbf{v}. \end{cases} \end{align*} $$

Further, it follows from statement ( $\ast )$ that $\varphi (b)=b$ , whence

$$ \begin{align*} (\varphi(x),\varphi(y))= \begin{cases} (x,y)&\text{if } \mathbf{w}=\mathbf{u}, \\ (y,x)&\text{if } \mathbf{w}=\mathbf{v}. \end{cases} \end{align*} $$

Since $({_{1\mathbf {u}}}x)<({_{2\mathbf {u}}}x)<({_{3\mathbf {u}}}x)<({_{1\mathbf {u}}}y)<({_{2\mathbf {u}}}y)$ and $({_{1\mathbf {v}}}x)<({_{2\mathbf {v}}}x)<({_{1\mathbf {v}}}y)<({_{3\mathbf {v}}}x)<({_{2\mathbf {v}}}y)$ , it follows that the case when $\mathbf {w}=\mathbf {v}$ is impossible. Therefore, $(\varphi (x),\varphi (y))=(x,y)$ in either case. This implies that $\mathbf {w}^\prime =\mathbf {v}$ and we are done.

Put $\mathbf {X}:=\mathbf M(\mathbf {u},\mathbf {v})\wedge \mathbf {X}^\prime $ . Consider an arbitrary identity of the form $\mathbf {u}\approx \mathbf {w}$ that is satisfied by the variety $\mathbf {X}\vee \mathbf {V}$ . In view of Claim 3.2, $\mathbf {w}\in \{\mathbf {u},\mathbf {v}\}$ . Since $M(xzxtxysy)\in \mathbf {V}$ , Lemma 2.2 implies that $\mathbf {w}(x,y,s_1,s_2,s_3)=\mathbf {u}(x,y,s_1,s_2,s_3)=xs_1xs_2xys_3y$ , then, ${\mathbf {w} = \mathbf {u}}$ . We see that $\mathbf {u}$ is an isoterm for the variety $\mathbf {X}\vee \mathbf {V}$ . By a similar argument, we can show that $\mathbf {v}$ is an isoterm for $\mathbf {X}\vee \mathbf {V}$ as well. Then,

$$ \begin{align*} (\mathbf{X}\vee \mathbf{V})\wedge \mathbf M(\mathbf{u},\mathbf{v}) = \mathbf M(\mathbf{u},\mathbf{v}) \end{align*} $$

by Lemma 2.2. It is easy to see that $xzxtxysy\approx xzxtyxsy$ holds in $M(\mathbf {u},\mathbf {v})$ . Recall that $\mathbf {V}$ satisfies the identities (3.1) and $\mathbf {a}_{n,m}[\rho ]\approx \mathbf {a}_{n,m}^\prime [\rho ]$ for all $n,m\in \mathbb {N}$ and $\rho \in S_{n+m}$ . Then, $\mathbf {V}\wedge \mathbf M(\mathbf {u},\mathbf {v})$ satisfies $\mathbf {u}\stackrel {\mathbf {V}}\approx \mathbf {u}_cc^2\stackrel {\mathbf M(\mathbf {u},\mathbf {v})}\approx \mathbf {v}_cc^2\stackrel {\mathbf {V}}\approx \mathbf {v}$ . By the very definition, the identity ${\mathbf {u}\approx \mathbf {v}}$ is satisfied by $\mathbf {X}$ as well. Since $\mathbf {X}\subset \mathbf M(\mathbf {u},\mathbf {v})$ , we have

$$ \begin{align*} \mathbf{X}\vee(\mathbf{V}\wedge\mathbf M(\mathbf{u},\mathbf{v}))\subset (\mathbf{X}\vee \mathbf{V})\wedge \mathbf M(\mathbf{u},\mathbf{v}) = \mathbf M(\mathbf{u},\mathbf{v}), \end{align*} $$

which contradicts the fact that $\mathbf {V}$ is a modular element of $\mathbb {M}{\scriptstyle \mathbb {ON}}$ . Therefore, $\mathbf {V}$ satisfies $xzxtxysy\approx xzxtyxsy$ . By the dual argument, we can show that $xzxytysy\approx xzyxtysy$ holds in $\mathbf {V}$ .

This proves the implication (i) $\Rightarrow $ (ii).

(ii) $\Rightarrow $ (iii). If $M(xyx)\notin \mathbf {V}$ , then it follows from [Reference Gusev and Vernikov5, Lemma 3.3(i)] that $\mathbf {V}$ coincides with one of the varieties $\mathbf M(\emptyset )$ , $\mathbf M(1)$ , $\mathbf M(x)$ or $\mathbf M(xy)$ and we are done. Assume now that $M(xyx)\in \mathbf {V}$ . Denote by $\mathscr W$ the set of all words in $\{xzxyty,\,xt_1x\cdots t_nx\mid n\in \mathbb {N}\}$ that are isoterms for $\mathbf {V}$ . Notice that, in this case, the variety $\mathbf M(\mathscr W)$ coincides with one of the varieties

$$ \begin{align*} &\mathbf M(xt_1x),\,\ldots,\, \mathbf M(xt_1x\cdots t_nx),\,\ldots,\,\mathbf M(\{xt_1x\cdots t_nx\mid n\in\mathbb{N}\}),\\ &\mathbf M(xzxyty,xt_1x),\,\ldots,\, \mathbf M(xzxyty,xt_1x\cdots t_nx),\,\ldots,\,\mathbf M(\{xzxyty,xt_1x\cdots t_nx\mid n\in\mathbb{N}\}). \end{align*} $$

According to Lemma 2.2, $\mathbf M(\mathscr W)\subseteq \mathbf {V}$ . It remains to verify that $\mathbf {V}\subseteq \mathbf M(\mathscr W)$ . Arguing by contradiction, suppose that $\mathbf M(\mathscr W)\ne \mathbf {V}$ . Then, there is an identity $\sigma $ which holds in $\mathbf M(\mathscr W)$ but does not hold in $\mathbf {V}$ . Now, [Reference Gusev and Vernikov5, Proposition 3.15] and its proof allow us to assume that $\sigma $ coincides with either the identity

(3.2) $$ \begin{align} x\bigg(\prod_{i=1}^nt_ix\bigg)\approx x^2\bigg(\prod_{i=1}^nt_i\bigg), \end{align} $$

where $n\in \mathbb {N}$ , or the identity

(3.3) $$ \begin{align} \bigg(\prod_{i=1}^k a_it_i\bigg) xy \bigg(\prod_{i=k+1}^{k+\ell} t_i a_i\bigg)\approx\bigg(\prod_{i=1}^k a_it_i\bigg) yx \bigg(\prod_{i=k+1}^{k+\ell} t_ia_i\bigg), \end{align} $$

where $k,\ell \ge 0$ and $\{a_1,\ldots ,a_{k+\ell }\}=\{x,y\}$ . Notice that an identity of the form (3.3) does not hold in $\mathbf {V}$ if and only if it coincides (up to renaming of letters) with the identity $xzxyty\approx xzyxty$ . If $\sigma $ equals (3.2), the word $xt_1x\cdots t_nx$ is an isoterm for $\mathbf {V}$ by [Reference Gusev and Vernikov5, Lemma 3.3(ii)]. This word must belong to the set $\mathscr W$ , which contradicts our assumption that $\sigma $ is satisfied by $\mathbf M(\mathscr W)$ . If $\sigma $ coincides with the identity $xzxyty\approx xzyxty$ , then the word $xzxyty$ is an isoterm for $\mathbf {V}$ by Lemma 2.4. However, this contradicts our assumption that $\sigma $ holds in $\mathbf M(\mathscr W)$ . Therefore, $\mathbf {V}= \mathbf M(\mathscr W)$ and we are done.

(iii) $\Rightarrow $ (i). It follows from [Reference Gusev and Lee3, Theorem 1.1] that the varieties $\mathbf M(\emptyset )$ , $\mathbf M(1)$ , $\mathbf M(x)$ and $\mathbf M(xy)$ are modular elements of $\mathbb {M}{\scriptstyle \mathbb {ON}}$ . So, we may further assume that $\mathbf {V}=\mathbf M(\mathscr W)$ for some $\mathscr W\subseteq \{xzxyty,xt_1x\cdots t_nx\mid n\in \mathbb {N}\}$ . In particular, $M(xyx)\in \mathbf {V}$ .

Arguing by contradiction, suppose that $\mathbf {V}$ is not a modular element of the lattice $\mathbb {M}{\scriptstyle \mathbb {ON}}$ . This means that there are monoid varieties $\mathbf {X}$ and $\mathbf {Y}$ such that $\mathbf {X}\vee \mathbf {V}=\mathbf {Y}\vee \mathbf {V}$ , $\mathbf {X}\wedge \mathbf {V}=\mathbf {Y}\wedge \mathbf {V}$ but $\mathbf {X}\subset \mathbf {Y}$ .

We need the following auxiliary result.

Proof. If the identity $\mathbf {u}\approx \mathbf {v}$ holds in $\mathbf {V}$ , then it also holds in $\mathbf {X}\vee \mathbf {V}$ and so in $\mathbf {Y}$ because $\mathbf {X}\vee \mathbf {V}=\mathbf {Y}\vee \mathbf {V}$ . So, we may further assume that $\mathbf {V}$ violates $\mathbf {u}\approx \mathbf {v}$ . This is only possible when $M(xzxyty)$ violates $\mathbf {u}\approx \mathbf {v}$ and $M(xzxyty)\in \mathbf {V}$ , because ${\mathbf {V}\subseteq \mathbf M(\{xzxyty,xt_1x\cdots t_nx\mid n\in \mathbb {N}\})}$ and $\mathbf {u}\approx \mathbf {v}$ is satisfied by $M(\{xt_1x\cdots t_nx\mid n\in \mathbb {N}\})$ . Then, $M(xzxyty)\notin \mathbf {Y}$ because $M(xzxyty)\in \mathbf {X}$ otherwise, which contradicts the fact that $M(xzxyty)$ violates $\mathbf {u}\approx \mathbf {v}$ .

If $M(xyx)\in \mathbf {Y}$ , then $\mathbf {Y}$ satisfies $xzxyty\approx xzyxty$ by Lemma 2.4. Now, let $M(xyx)\notin \mathbf {Y}$ . Then, Lemma 2.2 implies that the variety $\mathbf {Y}$ satisfies an identity $xyx\approx x^pyx^q$ with either $p>1$ or $q>1$ . Further, it follows from Lemma 2.4 that $\mathbf {X}\vee \mathbf M(xyx)$ satisfies $xzxyty\approx xzyxty$ . Then, $\mathbf {X}$ also satisfies the identity $x^pzx^qyty\approx x^pzyx^qty$ . Evidently, the last identity holds in the variety $\mathbf {V}$ as well. Therefore, it is satisfied by $\mathbf {Y}$ because $\mathbf {Y}\subset \mathbf {X}\vee \mathbf {V}$ . Thus, we see that the identity $xzxyty\approx xzyxty$ holds in the variety $\mathbf {Y}$ in any case.

Let $\mathbf {u}=\mathbf {u}_0t_1\mathbf {u}_1\cdots t_m\mathbf {u}_m$ , where $t_1,\ldots ,t_m$ are all the simple letters of the word $\mathbf {u}$ and $\mathbf {u}_0,\ldots ,\mathbf {u}_m\in \mathscr X^\ast $ . Then, by Lemma 2.3, $\operatorname {\mathrm {sim}}(\mathbf {v})=\{t_1,\ldots ,t_m\}$ and $\mathbf {v}=\mathbf {v}_0t_1\mathbf {v}_1\cdots t_m\mathbf {v}_m$ for some $\mathbf {v}_0,\ldots ,\mathbf {v}_m\in \mathscr X^\ast $ .

It is easy to see that the identity $xzxyty\approx xzyxty$ implies the identities

$$ \begin{align*} \mathbf{u}\approx \mathbf{p}_0\mathbf{q}_0t_1 \mathbf{p}_1\mathbf{q}_1\cdots t_m \mathbf{p}_m\mathbf{q}_m=:\mathbf{u}^\prime,\ \ \mathbf{v}\approx \mathbf{p}_0^\prime\mathbf{q}_0^\prime t_1 \mathbf{p}_1^\prime\mathbf{q}_1^\prime\cdots t_m \mathbf{p}_m^\prime\mathbf{q}_m^\prime=:\mathbf{v}^\prime, \end{align*} $$

where the word $\mathbf {p}_i$ (respectively, $\mathbf {q}_i$ ) is obtained from the word $\mathbf {u}_i$ by retaining only the first (respectively, nonfirst) occurrences of letters in $\mathbf {u}$ , while the word $\mathbf {p}_i^\prime $ (respectively, $\mathbf {q}_i^\prime $ ) is obtained from the word $\mathbf {v}_i$ by retaining only the first (respectively, nonfirst) occurrences of letters in $\mathbf {v}$ . By the very construction of the words $\mathbf {u}^\prime $ and $\mathbf {v}^\prime $ , the identity $\mathbf {u}^\prime \approx \mathbf {v}^\prime $ holds in the monoid $M(xzxyty)$ and so in the variety $\mathbf {V}$ . Since $\mathbf {X}$ satisfies $xzxyty\approx xzyxty$ and $\mathbf {u}\approx \mathbf {v}$ , the identity $\mathbf {u}^\prime \approx \mathbf {v}^\prime $ also holds in $\mathbf {X}$ . Then, $\mathbf {Y}$ satisfies $\mathbf {u}^\prime \approx \mathbf {v}^\prime $ as $\mathbf {Y}\subseteq \mathbf {X}\vee \mathbf {V}$ . It follows that $\mathbf {u}\approx \mathbf {v}$ holds in $\mathbf {Y}$ because the identities $\mathbf {u}\approx \mathbf {u}^\prime $ and $\mathbf {v}\approx \mathbf {v}^\prime $ are consequences of $xzxyty\approx xzyxty$ .

It follows from Lemma 2.2 and Claim 3.3 that there is a least $k\in \mathbb {N}$ such that the word $xt_1x\cdots t_kx$ is not an isoterm for $\mathbf {X}$ . Then, $\mathbf {X}$ satisfies a nontrivial identity $xt_1x\cdots t_kx\approx \mathbf {w}$ for some $\mathbf {w}\in \mathscr X^\ast $ . It follows from Lemma 2.3 that $\mathbf {w}=x^{e_0}t_1x^{e_1}\cdots t_kx^{e_k}$ for some $e_0,\ldots ,e_k\ge 0$ . Since k is the least natural number for which the word $xt_1x\cdots t_kx$ is not an isoterm for $\mathbf {X}$ , at least one of the numbers $e_0,\ldots ,e_k$ must exceed  $1$ .

Now, consider an arbitrary identity $\mathbf {a}\approx \mathbf {b}$ of $\mathbf {X}$ . Applying the identity $xt_1x\cdots t_kx\approx \mathbf {w}$ to the words $\mathbf {a}$ and $\mathbf {b}$ , we can replace their subwords of the form $x\mathbf f_1x\cdots \mathbf f_kx$ by $x^{e_0}\mathbf f_1x^{e_1}\cdots \mathbf f_kx^{e_k}$ . In other words, the identity $xt_1x\cdots t_kx\approx \mathbf {w}$ can be used to convert the words $\mathbf {a}$ and $\mathbf {b}$ into some words $\mathbf {a}^\prime $ and $\mathbf {b}^\prime $ , respectively, so that the identity $\mathbf {a}^\prime \approx \mathbf {b}^\prime $ holds in $M(\{xt_1x\cdots t_nx\mid n\in \mathbb {N}\})$ (because the word $xt_1x\cdots t_{k-1}x$ is an isoterm for $\mathbf {X}$ and at least one of the numbers $e_0,\ldots ,e_k$ exceeds $1$ ). Therefore, the variety $\mathbf {X}$ can be defined by $\{xt_1x\cdots t_kx\approx \mathbf {w}\}\cup \Sigma $ for some set $\Sigma $ of identities holding in $M(\{xt_1x\cdots t_nx\mid n\in \mathbb {N}\})$ .

Further, if $M(xt_1x\cdots t_kx)\in \mathbf {V}$ , then $M(xt_1x\cdots t_kx)\notin \mathbf {Y}$ because $\mathbf {X}\wedge \mathbf {V}=\mathbf {Y}\wedge \mathbf {V}$ . In this case, it is easy to deduce from Lemmas 2.2 and 2.3 that $\mathbf {Y}$ satisfies a nontrivial identity $xt_1x\cdots t_kx\approx x^{f_0}t_1x^{f_1}\cdots t_kx^{f_k}=:\mathbf {w}^\prime $ , where at least one of the numbers $f_0,\ldots ,f_k$ exceeds $1$ . Since $\mathbf {X}\subset \mathbf {Y}$ , the identity $\mathbf {w}\approx \mathbf {w}^\prime $ holds in $\mathbf {X}$ . Clearly, this identity is satisfied by $\mathbf {V}$ as well. Hence, $\mathbf {X}\vee \mathbf {V}=\mathbf {Y}\vee \mathbf {V}$ satisfies $\mathbf {w}\approx \mathbf {w}^\prime $ and, therefore, $\mathbf {Y}$ satisfies $xt_1x\cdots t_kx\approx \mathbf {w}$ .

Suppose now that $M(xt_1x\cdots t_kx)\notin \mathbf {V}$ . In this case, it is easy to deduce from Lemmas 2.2 and 2.3 that $\mathbf {V}$ satisfies the identity $xt_1x\cdots t_kx\approx \mathbf {w}$ . Since $\mathbf {X}\vee \mathbf {V}=\mathbf {Y}\vee \mathbf {V}$ , we see that this identity holds in $\mathbf {Y}$ .

According to Claim 3.3, all the identities in $\Sigma $ are satisfied by $\mathbf {Y}$ , which contradicts $\mathbf {X}\subset \mathbf {Y}$ . Therefore, $\mathbf {V}$ is a modular element of the lattice $\mathbb {M}{\scriptstyle \mathbb {ON}}$ . Theorem 1.1 is thus proved.