Proof. If A is a submonoid S, not just a subset, condition (i) in the definition of invariant subset implies condition (ii) because $xy=xey\in A$ , and condition (ii) implies condition (iii) because $xy=xey\in (xAy)\cap A$ . Hence, conditions (i)–(iii) are equivalent if and only if condition (iii) implies condition (i).

Proof. Let S be a normal submonoid of G, and let us suppose that for some $x\in G$ and $s\in S$ , we have $xsx^{-1}\notin S$ . Using the equivalence between conditions (ii) and (iii) with $y=sx^{-1}$ , we conclude that

$$ \begin{align*} (xS(sx^{-1}))\cap S=\emptyset. \end{align*} $$

However, $e=xx^{-1}=xs^{-1}(sx^{-1})\in (xS(sx^{-1}))\cap S$ , contradicting our conclusion. Therefore, $xSx^{-1}\subseteq S$ for every $x\in G$ . Conversely, let S be a normal subgroup of G. We have to see that S satisfies Equation (2-1). Indeed, let us suppose that $(xSy)\cap S\neq \emptyset $ for some $x,y\in G$ . This means that there exist $s_1,s_2\in S$ such that $xs_1y=s_2$ . Then for each $s\in S$ , we have $xsy=\tilde {s}s_2$ with $\tilde {s}=xss^{-1}x^{-1}$ . Now, $\tilde {s}\in S$ because S is a normal subgroup of G. Hence, $xsy\in S$ and $xSy\subseteq S$ .

Proof. Clearly, $f^{-1}(1)$ is always a groupal submonoid of M. Suppose that for some $x,y\in M$ , there exists $z_1,z_2\in f^{-1}(1)$ such that $xz_1y=z_2$ . Then $1_N=f(z_2)=f(xz_1y)=f(x)f(y)$ and hence, for every $z\in f^{-1}(1)$ , we have $f(xzy)=f(x)f(y)=1_N$ , that is, $x\,f^{-1}(1)\,y\subseteq f^{-1}(1)$ . Thus, $f^{-1}(1)$ is also invariant.

Proof. The submonoid $\langle n\rangle $ is clearly groupal. Moreover, if $(k,l)\in X_{\langle n\rangle }$ , there exist $q,q'\geq 0$ such that $k+nq+l=nq'$ . Hence, $q'\geq q$ , and $k+l\in \langle n\rangle $ so that $k+\langle n\rangle +l\subseteq \langle n\rangle $ , that is, $\langle n\rangle $ is invariant. Let us now prove that these are the only normal submonoids of $\mathbb {N}_+$ . Since all submonoids of $\mathbb {N}_+$ are groupal, we need to show that these are the only invariant submonoids. To prove this, let S be any submonoid of $\mathbb {N}_+$ and $n:=\min (S\setminus \{0\})$ , so that $\langle n\rangle \subseteq S$ . If $S\neq \langle n\rangle $ , let $n':=\min (S\setminus \langle n\rangle )$ . Then we have $n'=nq'+r$ for some $0<r<n$ and $q'\geq 1$ ( $q' \neq 0$ because n is the least nonzero element of S). It follows that $n'\in (0+S+r)\cap S$ , that is, $(0,r)\in X_S$ . However, because $r\in 0+S+r$ while $r\notin S$ .

Proof. Let $\{S_\lambda \}_{\lambda \in \Lambda }$ be a family of normal submonoids of M, and let

$$ \begin{align*} S:=\bigcap_{\lambda\in\Lambda}S_\lambda. \end{align*} $$

Clearly, S is a groupal submonoid. Moreover, since $S\subseteq S_\lambda $ , we have $X_{S}\subseteq X_{S_\lambda }$ for each $\lambda \in \Lambda $ . Thus, for each $(x,y)\in X_{S}$ ,

$$ \begin{align*} xSy\subseteq xS_\lambda y\subseteq S_\lambda \end{align*} $$

because each $S_\lambda $ is invariant. Hence, $xSy\subseteq S$ , that is, S is also invariant.

Proof. Every normal submonoid of M containing A must contain $\mathrm {ncl}_M(A)$ . Hence, it is enough to see that $\mathrm {ncl}_M(A)$ is a normal submonoid. Let $z,z'\in \mathrm {ncl}_M(A)$ . Then $z\in A_n$ and $z'\in A_{n'}$ for some $n,n'\geq 0$ . Since $A_n,A_{n'}\subseteq A_{\max \{n,n'\}}$ , it follows that $z,z'\in A_{\max \{n,n'\}}$ and hence, $zz'\in A_{\max \{n,n'\}}\subseteq \mathrm {ncl}_M(A)$ because $A_{\max \{n,n'\}}$ is a submonoid. Moreover, if $z\in U(M)$ , we also have $z^{-1}\in A_{\max \{n,n'\}}$ because it is a groupal submonoid. This proves that $\mathrm {ncl}_M(A)$ is a groupal submonoid of M. To prove that it is invariant, let us first observe that

(2-5) $$ \begin{align} X_{\mathrm{ncl}_M(A)}=\bigcup_{n\geq 0}X_{A_n}. \end{align} $$

Indeed, since $A_n\subseteq \mathrm {ncl}_M(A)$ , we have $X_{A_n}\subseteq X_{\mathrm {ncl}_M(A)}$ for each $n\geq 0$ and hence, $\bigcup _{n\geq 0}X_{A_n}\subseteq X_{\mathrm {ncl}_M(A)}$ . Conversely, let $(x,y)\in X_{\mathrm {ncl}_M(A)}$ . This means that there exist $t^*_1,t^*_2\in \mathrm {ncl}_M(A)$ such that $x t^*_1y=t^*_2$ and hence, some $n_1,n_2\geq 0$ such that $t^*_1\in A_{n_1}$ , $t^*_2\in A_{n_2}$ , and $xt^*_1y=t^*_2$ . Since $A_{n_1},A_{n_2}\subseteq A_{\max \{n_1,n_2\}}$ , it follows that $(x,y)\in A_{\max \{n_1,n_2\}}\subseteq \bigcup _{n\geq 0}X_{A_n}$ . Let now $(x,y)\in X_{\mathrm {ncl}_M(A)}$ , and let us prove that $x\mathrm {ncl}_M(A)y\subseteq \mathrm {ncl}_M(A)$ . Indeed, it follows from Equation (2-5) that $(x,y)\in X_{A_{n_0}}$ for some $n_0\geq 0$ and hence, $(x,y)\in X_{A_n}$ for each $n\geq n_0$ . Then for any element $t^*\in \mathrm {ncl}_M(A)$ , we have $t^*\in A_{m_0}$ for some $m_0\geq 0$ and hence, $t^*\in A_m$ for each $m\geq m_0$ . If $m_0\leq n_0$ , we have $t^*\in A_{n_0}$ and $(x,y)\in X_{A_{n_0}}$ , from which it follows that $xt^*y\in A_{n_0+1}\subseteq \mathrm {ncl}_M(A)$ . Similarly, if $m_0>n_0$ , we have $t^*\in A_{m_0}$ and $(x,y)\in X_{A_{m_0}}$ , from which it follows that $xt^*y\in A_{m_0+1}\subseteq \mathrm {ncl}_M(A)$ .

Proof. (i) Let M be a finite monoid that is not a group, that is, such that $M\setminus U(M)\neq \emptyset $ . Then the stability set of $U(M)$ is

$$ \begin{align*} X_{U(M)}=U(M)\times U(M), \end{align*} $$

from which the statement follows readily. The inclusion $U(M)\times U(M)\subseteq X_{U(M)}$ is obvious. Regarding the reverse inclusion, it is ultimately a consequence of the so-called stability property of finite monoids, according to which for a finite monoid M, and for every $x,y\in M$ , the two following equivalences hold:

1. (1) $MxM=MxyM$ $\Longleftrightarrow xM = xyM$ ;

2. (2) $MxM=MyxM$ $\Longleftrightarrow Mx = Myx$ .

Using either of these equivalences, it is not difficult to see that

$$ \begin{align*} U(M)=\{x\in M:MxM=M\}, \end{align*} $$

and that its complement $M\setminus U(M)$ , nonempty by hypothesis, is a (two-sided) ideal of M; see, for example, [Reference Steinberg12, Section 1.3] for the details. Now, if $M\setminus U(M)$ is an ideal and for some $x,y\in M$ , there exists some $u\in U(M)$ such that $xuy\in U(M)$ , we necessarily have $x,y\in U(M)$ . Hence, $X_{U(M)}=U(M)\times U(M)$ .

(ii) If M is commutative and $xuy\in U(M)$ for some $u\in U(M)$ , it follows that $xy\in U(M)$ and hence, $xu'y=xyu'\in U(M)$ for each $u'\in U(M)$ .

Proof. Every subgroup H of $U(M)$ is trivially a groupal submonoid of M. Let us assume that H is a normal subgroup of $U(M)$ and that for some $x,y\in M$ , we have $(xHy)\cap H\neq \emptyset $ . Since $H\subseteq U(M)$ , we also have $(xU(M)y)\cap U(M)\neq \emptyset $ and hence, $xU(M)y\subseteq U(M)$ because $U(M)$ is a normal submonoid of M. Then for every $h\in H$ , we have $xhy\in U(M)$ .

Proof. Here, $T_n$ is a normal monoid because it is finite. To prove that it is normally simple it is enough to see that for each $f\in T_n$ , the normal closure of $\{f\}$ in $T_n$ is given by

(2-8) $$ \begin{align} \mathrm{ncl}_{T_n}(f)= \begin{cases} \mathrm{ncl}_{S_n}(f) & \mbox{if } f\in S_n, \\ T_n & \mbox{if } f\notin S_n, \end{cases} \end{align} $$

where $\mathrm {ncl}_{S_n}(f)$ stands for the normal closure of $\{f\}$ in the symmetric group $S_n$ . Indeed, let this be true and let S be a normal submonoid of $T_n$ . If $S\setminus S_n\neq \emptyset $ and $f\in S\cap S_n$ , it follows from Equation (2-8) that $T_n=\mathrm {ncl}_{T_n}(f)\subseteq S$ and hence, $S=T_n$ . Otherwise, we have $S\subseteq S_n$ and, being a normal submonoid of $T_n$ , S is also a normal submonoid of $S_n$ and hence, a normal subgroup of $S_n$ .

Let us prove Equation (2-8). We need to compute the sequence of groupal submonoids $\{f\}_0,\{f\}_1,\ldots .$ Let us first consider the case where f is a permutation. By definition,

$$ \begin{align*} \{f\}_0=\langle f\rangle'=\{f^p : p\in\mathbb{Z}\}. \end{align*} $$

Then the stability set of $\{f\}_0$ consists of all pairs $(g,h)\in T_n\times T_n$ such that $gf^ph=f^q$ for some $p,q\in \mathbb {Z}$ . Since f is a bijection, h must be injective and g surjective. Being endomorphisms of $\mathbf {n}$ , it follows that $g,h\in S_n$ . Hence, we can take as g any permutation in $S_n$ , and $h=f^qg^{-1}f^{-p}$ , that is,

$$ \begin{align*} X_{\{f\}_0}=\{(\tau,f^q\tau^{-1}f^{-p}) : \tau\in S_n,\,p,q\in\mathbb{Z}\}. \end{align*} $$

In particular, $(\tau ,\tau ^{-1})\in X_{\{f\}_0}$ for each $\tau \in S_n$ . Thus,

$$ \begin{align*} \mathrm{ncl}_{T_n}(f)\supseteq\{f\}_1=\bigg\langle\bigcup_{(g,h)\in X_{\{f\}_0}}g\{f\}_0h\ \bigg\rangle'\supseteq\bigg\langle \bigcup_{\tau\in S_n}\tau\{f\}_0\tau^{-1}\,\bigg\rangle'\supseteq\mathrm{ncl}_{S_n}(f). \end{align*} $$

However, $\mathrm {ncl}_{S_n}(f)$ is already a normal submonoid of $T_n$ containing f because of Proposition 2.20. Therefore, $\mathrm {ncl}_{T_n}(f)=\mathrm {ncl}_{S_n}(f)$ when f is invertible.

Let now f be a noninvertible endomorphism. To prove that $\mathrm {ncl}(f)=T_n$ , we proceed by induction on the rank $k<n$ of f. If $k=1$ , there exists $i\in \mathbf {n}$ such that $f=c_i$ , the constant function mapping all of $\mathbf {n}$ to i. Now,

$$ \begin{align*} \{c_i\}_0=\{\mathrm{id}_{\mathbf{n}},c_i\}, \end{align*} $$

and for every $h\in T_n$ , we have $c_ih=c_i$ , that is, $(\mathrm {id}_{\mathbf {n}},h)\in X_{\{c_i\}_0}$ . It follows that for every $h\in T_n$ ,

$$ \begin{align*} h\in id_{\mathbf{n}}\{c_i\}_0h\subseteq\{c_i\}_1\subseteq\mathrm{ncl}_{T_n}(f), \end{align*} $$

that is, $\mathrm {ncl}_{T_n}(f)=T_n$ . Let us now assume that $ncl_{T_n}(\phi )=T_n$ for every endomorphism $\phi \in T_n$ of rank $l\in \{1,\ldots ,k-1\}$ , with $1<k<n$ . Since f is not invertible,

$$ \begin{align*} \{f\}_0=\{f^p : p\geq 0\}. \end{align*} $$

Let $\mathrm {Im}f=\{i_1<i_2<\cdots <i_k\}\subset \mathbf {n}$ and $\{A_1,\ldots ,A_k\}$ the partition of $\mathbf {n}$ such that $f^{-1}(i_j)=A_j$ . Thus, $f=\varepsilon \eta $ , with $\eta :\mathbf {n}\to \{i_1,\ldots ,i_k\}$ given by $A_j\mapsto i_j$ , and $\varepsilon :\{i_1,\ldots ,i_k\}\to \mathbf {n}$ the canonical inclusion. Then the powers of f can be described in terms of the map $\alpha :\{i_1,\ldots ,i_k\}\to \{i_1,\ldots ,i_k\}$ defined by

$$ \begin{align*} \alpha(i_j)=i_{j'}, \end{align*} $$

where $j'$ is the unique element in $\{1,\ldots ,k\}$ such that $i_j\in A_{j'}$ . Thus, $f^2$ is given by $A_j\overset {f}{\mapsto } i_j\overset {f}{\mapsto } i_{j'}$ , that is, we have $f^2=\varepsilon \alpha \eta $ . More generally, for each $p\geq 1$ ,

$$ \begin{align*} f^p=\varepsilon\alpha^{p-1}\eta. \end{align*} $$

The map $\alpha $ can be bijective or not. If it is not bijective, it follows that the powers of f all have rank less than k. However, if $\{f\}_0$ contains a map $\phi $ of rank $l<k$ , it follows by the induction hypothesis that $T_n=\mathrm {ncl}_{T_n}(\phi )$ , and hence $\mathrm {ncl}_{T_n}(f)=T_n$ because $\mathrm {ncl}_{T_n}(\phi )\subseteq \mathrm {ncl}_{T_n}(f)$ . It remains to prove that the same is true when $\alpha $ is not bijective. In this case, all powers of f are of rank k. However, we claim that when $\alpha $ is not bijective, $\{f\}_1$ contains maps of rank less than k, so that the same argument can be applied. Indeed, since f is not a bijection, there is some $j\in \{1,\ldots ,k\}$ such that $A_j$ contains more than one element. Without loss of generality, let us assume that $|A_1|>2$ . Moreover, since $\alpha $ is a bijection, each $i_j$ belongs to a different $A_{j'}$ and hence, there is some $a_1\in A_1$ such that $a_1\notin \{i_1,\ldots ,i_k\}$ . Then let us consider the pair $(g,h)\in T_n\times T_n$ given as follows:

1. (i) h maps all of $A_1$ to $a_1$ , and each $A_j$ for $j\in \{2,\ldots ,k\}$ to any $a_j\in A_j$ ;

2. (ii) g acts as the identity on $\{i_1,\ldots ,i_k\}$ , and collapses $\mathbf {n}\setminus \{i_1,\ldots ,i_k\}$ to $i_2$ (in particular, we have $g(a_1)=i_2$ ).

Then we have $gfh=f$ , that is, $(g,h)\in X_{\{f\}_0}$ and hence, $gh\in \{f\}_1$ . However, $gh$ is of rank at most $k-1$ because even if $\{a_2,\ldots ,a_k\}=\{i_2,\ldots ,i_k\}$ , we have that $a_1$ is also mapped to $i_2$ .

Proof. Since M is assumed to be cancellative, $\widehat {M}$ is given (up to isomorphism) by the quotient of the product monoid $M\times M$ modulo the congruence relation

$$ \begin{align*} (m,n)\sim(m',n')\ \Leftrightarrow\ m+n'=m'+n. \end{align*} $$

We denote by $[m,n]$ the equivalence class of $(m,n)$ in $\widehat {M}$ . Then for every normal submonoid $S\lhd M$ , let $\widehat {S}$ be the subgroup of $\widehat {M}$ given by

$$ \begin{align*} \widehat{S}:=\{u\in\widehat{M} : u=[s,s'] \mbox{ for some } (s,s')\in S\times S\}. \end{align*} $$

The notation is justified by the fact that this subgroup is indeed isomorphic to the Grothendieck group of S. This is again a consequence of the cancellative character of M, which ensures that $(s,s'),(t,t')\in S\times S$ are equivalent in the Grothendieck group of S if and only if they are equivalent in $\widehat {M}$ . Then let $f:\mathsf {NorSub}(M)\to \mathsf {Sub}(\widehat {M})$ be the map defined by $S\mapsto \widehat {S}$ .

We claim that f is a lattice isomorphism. It is clearly order-preserving. Moreover, it is injective. Indeed, let $S_1,S_2\lhd M$ be such that $\widehat {S}_1=\widehat {S}_2$ , and let $x\in S_1$ . Then $[x,0]\in \widehat {S}_1$ and hence, we also have $[x,0]\in \widehat {S}_2$ . However, this means that there exists a pair $(s_2,s^{\prime }_2)\in S_2\times S_2$ such that $[x,0]=[s_2,s^{\prime }_2]$ , that is, such that $x+s^{\prime }_2=s_2$ . It follows that x belongs to the left stability set of $S_2$ and consequently, $x=x+0\in S_2$ because $S_2$ is invariant. This proves that $S_1\subseteq S_2$ . A similar argument proves that $S_2\subseteq S_1$ and hence, $S_1=S_2$ . To prove it is surjective, let

$$ \begin{align*} n(H):=\{x\in M:[x,0]\in H\} \end{align*} $$

for every (normal) subgroup H of $\widehat {M}$ . We claim that $n(H)$ is a normal submonoid of M such that

(2-9) $$ \begin{align} \widehat{n(H)}=H. \end{align} $$

Indeed, it is a groupal submonoid of M because H is a subgroup of $\widehat {M}$ . Moreover, let $z\in LX_{n(H)}$ so that $z+x\in n(H)$ for some $x\in n(H)$ . This means that $[z+x,0],[x,0]\in H$ and hence,

$$ \begin{align*} [z,0]=[z+x,x]=[z+x,0]+[0,x]=[z+x,0]-[x,0] \end{align*} $$

is also in H. Therefore, $z\in n(H)$ and consequently, $z+n(H)\subseteq n(H)$ , that is, $n(H)$ is invariant. Let us now prove Equation (2-9). The inclusion $\widehat {n(H)}\subseteq H$ holds even if M does not satisfy (*). Thus, if $u\in \widehat {n(H)}$ , then $u=[x,x']$ for some $x,x'\in n(H)$ and hence, such that $[x,0],[x',0]\in H$ . Then,

$$ \begin{align*} u=[x,0]+[0,x']=[x,0]-[x',0]\in H \end{align*} $$

because H is a subgroup. It is to prove the reverse inclusion that condition (*) is needed. Thus, let $h\in H$ , that is, $h=[y,y']$ for some $y,y'\in M$ . By condition (*), we have either $y=y'+z$ or $y'=y+z$ for some $z\in M$ . In the first case,

$$ \begin{align*} h=[y'+z,y']=[z,0]\in H \end{align*} $$

and hence, $z\in n(H)$ and $h=[z,0]\in \widehat {n(H)}$ . Similarly, in the second case,

$$ \begin{align*} h=[y,y+z]=[0,z]=-[z,0]\in H \end{align*} $$

and hence, also $[z,0]\in H$ because H is a subgroup. Consequently, $z\in n(H)$ and again $h\in \widehat {n(H)}$ . This proves that f is a bijective order-preserving map. Since the inverse map $H\mapsto n(H)$ is also order-preserving, it follows that f is a lattice isomorphism. The last assertion then follows from the modularity of the lattice of normal subgroups of a group.

Proof. We have to see that the above injective map $f:\mathsf {NorSub}(M)\to \mathsf {Sub}(\widehat {M})$ is join-preserving. Let $S_1,S_2\lhd M$ . The inclusion $\widehat {S}_1\vee \widehat {S}_2\subseteq \widehat {S_1\vee S}_2$ follows because f is order-preserving, and $\widehat {S}_1\vee \widehat {S}_2$ is the smallest subgroup containing both $\widehat {S}_1$ and $\widehat {S}_2$ . To prove the reverse inclusion, let $x\in \widehat {S_1\vee S}_2$ . This means that $x=[t,t']$ for some $t,t'\in S_1\vee S_2$ . Since $\widehat {S}_1\vee \widehat {S}_2=\widehat {S}_1+\widehat {S}_2$ , we have to see that

(2-10) $$ \begin{align} (t,t')\sim(s_1+s_2,s^{\prime}_1+s^{\prime}_2) \end{align} $$

for some pairs $(s_1,s^{\prime }_1)\in S_1\times S_1$ and $(s_2,s^{\prime }_2)\in S_2\times S_2$ . To prove this, recall that

$$ \begin{align*} S_1\vee S_2=\mathrm{ncl}_M(S_1\cup S_2)=\bigcup_{n\geq 0}(S_1\cup S_2)_n, \end{align*} $$

with $(S_1\cup S_2)_n$ the sequence of groupal submonoids recursively defined by

$$ \begin{align*} (S_1\cup S_2)_0&=\langle S_1\cup S_2\rangle', \\ (S_1\cup S_2)_n&=\bigg\langle\bigcup_{z\in LX_{(S_1\cup S_2)_{n-1}}}(z+(S_1\cup S_2)_{n-1})\bigg\rangle',\quad n\geq 1 \end{align*} $$

(see Section 2.2). Since the sequence $(S_1\cup S_2)_n$ is a nondecreasing chain, $t,t'\in S_1\vee S_2$ implies that $t,t'\in (S_1\cup S_2)_n$ for some $n\geq 0$ . Then we prove Equation (2-10) by induction on $n\geq 0$ . If $n=0$ , we have $t,t'\in \langle S_1\cup S_2\rangle $ because both $S_1,S_2$ are groupal submonoids. Hence, both are finite sums of elements in $S_1$ and/or $S_2$ . By the commutativity of M, it follows that $(t,t')$ is in fact equal to a pair as in the right-hand side of Equation (2-10). Let us now assume that for some $n\geq 1$ , every pair $(u,u')$ with $u,u'\in (S_1\cup S_2)_{n-1}$ is equivalent to a pair as in the right-hand side of Equation (2-10), and let $t,t'\in (S_1\cup S_2)_n$ . This means that there exist $z_1,\ldots ,z_{k+l},z^{\prime }_1,\ldots ,z^{\prime }_{k'+l'}\in LX_{(S_1\cup S_2)_{n-1}}$ and $u_1,\ldots ,u_{k+l},u^{\prime }_1,\ldots ,u^{\prime }_{k'+l'}\in (S_1\cup S_2)_{n-1}$ such that $z_{i}+u_{i},z^{\prime }_{i'}+u^{\prime }_{i'}$ are invertible for each $i\in \{k+1,\ldots ,k+l\}$ and $i'\in \{k'+1,\ldots ,k'+l'\}$ , and $(t,t')=(A+B,A'+B')$ with

$$ \begin{align*} A&=(z_1+u_1)+\cdots+(z_k+u_k), \\ A'&=(z^{\prime}_1+u^{\prime}_1)+\cdots+(z^{\prime}_{k'}+u^{\prime}_{k'}), \\ B&=[-(z_{k+1}+u_{k+1})]+\cdots+[-(z_{k+l}+u_{k+l})], \\ B'&=[-(z^{\prime}_{k'+1}+u^{\prime}_{k'+1})]+\cdots+[-(z^{\prime}_{k'+l'}+u^{\prime}_{k'+l'})] \end{align*} $$

(here we are using the commutativity of M to write first all ‘positive terms’ in the expressions of both $t,t'$ ). However,

$$ \begin{align*} (t,t')&=(A+B,A'+B') \\ &\sim(A+B,A'+B')+((-B)+(-B'),(-B)+(-B')) \\ &=(A+(-B'),A'+(-B)). \end{align*} $$

Moreover, since M is commutative, and both $LX_{(S_1\cup S_2)_{n-1}}$ and $(S_1\cup S_2)_{n-1}$ are submonoids, reordering terms,

$$ \begin{align*} A+(-B')&=y+v, \\ A'+(-B)&=y'+v', \end{align*} $$

with $y\in LX_{(S_1\cup S_2)_{n-1}}$ , $v\in (S_1\cup S_2)_{n-1}$ given by

$$ \begin{align*} y&=z_1+\cdots+z_k+z^{\prime}_{k'+1}+\cdots+z^{\prime}_{k'+l'}, \\ v&=u_1+\cdots+u_k+u^{\prime}_{k'+1}+\cdots+u^{\prime}_{k'+l'}, \end{align*} $$

and similarly $y',v'$ . Now, since $y,y'\in LX_{(S_1\cup S_2)_{n-1}}$ , there exists $w,w'\in (S_1\cup S_2)_{n-1}$ such that $y+w,y'+w'\in (S_1\cup S_2)_{n-1}$ . Therefore,

$$ \begin{align*} (t,t')\sim(y+v,y'+v')\sim(y+w+v+w',y'+w'+v'+w)\in (S_1\cup S_2)_{n-1}. \end{align*} $$

By the induction hypothesis, it follows that $(t,t')$ is also equivalent to a pair as in the right-hand side of Equation (2-10). Hence,

$$ \begin{align*} x=[t,t']=[s_1,s^{\prime}_1]+[s_2,s^{\prime}_2]\in\widehat{S}_1+\widehat{S}_2, \end{align*} $$

and $\widehat {S}_1\vee \widehat {S}_2\supseteq \widehat {S_1\vee S}_2$ .

Proof. If M is commutative, we clearly have that $v\in M$ is an elementary A-deformation of $u\in M$ if and only if $v=u+a$ for some $a\in A$ . Hence, if $x+a=y+a'$ for some $a,a'\in A$ , then we have and $(x,y)\in R_A$ . Conversely, let $(x,y)\in R_A$ . Then there exists a sequence $z_0,\ldots ,z_{2k}$ as before. If $k=1$ , we have $x+a=y+a'$ for some $a,a'\in R$ . The cases $k\geq 2$ follow then by induction on k together with the fact that A is closed under sums. Thus, let us assume that $z_0+a_0=z_{2k-2}+a_1$ for some $a_0,a_1$ . Then from the case $k=1$ , it follows that we also have $z_{2k-2}+a^{\prime }_1=z_{2k}+a_2$ for some $a^{\prime }_1,a_2\in A$ . Hence,

$$ \begin{align*} z_0+a_0+a^{\prime}_1=z_{2k-2}+a_1+a^{\prime}_1=z_{2k}+a_1+a_2 \end{align*} $$

with $a_0+a^{\prime }_1,a_1+a_2\in A$ .

Proof. Clearly, $\mathsf {Cong}(M,A)\supseteq \mathsf {Cong}(M,\langle A\rangle ')$ because $A\subseteq \langle A\rangle '$ . To prove the reverse inclusion, let $R\in \mathsf {Cong}(M)$ be such that $A\subseteq [1]_R$ . Then for every $a_1,a_2\in A$ , we have $(1,a_1),(1,a_2)\in R$ and hence,

$$ \begin{align*} (1,a_1a_2)=(1,a_1)(1,a_2)\in R. \end{align*} $$

Moreover, if $a\in A$ is invertible, we have $(a^{-1},a^{-1}),(1,a)\in R$ and hence,

$$ \begin{align*} (a^{-1},1)=(a^{-1},a^{-1})(1,a)\in R. \end{align*} $$

Since every element in $\langle A\rangle '$ is a finite product of elements of A and/or their inverses (when they exist), it follows that $[1]_R$ also contains $\langle A\rangle '$ . This proves item (a). As for item (b), the inclusion $\mathsf {Cong}(M,A)\supseteq \mathsf {Cong}(M,xAy)$ is again obvious. To prove the reverse inclusion, let $R\in \mathsf {Cong}(M)$ be such that $[1]_R$ contains A, and let $(x,y)\in X_A$ . Then there exists $a_1,a_2\in A$ such that $xa_1y=a_2$ , and $(a_1,1),(1,a_2)\in R$ by hypothesis. Then for every $a\in A$ , we have $(1,a)\in R$ and hence, $(a_1,a)=(a_1,1)(1,a)\in R$ . Therefore,

$$ \begin{align*} (a_2,xay)=(x,x)(a_1,a)(y,y)\in R. \end{align*} $$

By transitivity, we conclude that $(1,xay)\in R$ . Since this is true for each $a\in A$ and for each $(x,y)\in X_A$ , it follows that $[1]_R$ also contains every generalized conjugate of A.

Proof. The inclusion $\mathsf {Cong}(M,\mathrm {ncl}(A))\subseteq \mathsf {Cong}(M,A)$ is obvious. To prove the reverse inclusion, let us assume that $R\in \mathsf {Cong}(M,A)$ . Then an easy induction on $n\geq 0$ using Lemma 3.3 shows that $R\in \mathsf {Cong}(M,A_n)$ for each $n\geq 0$ , where $A_n$ is the sequence of groupal submonoids of M defined by Equations (2-3)–(2-4). Since $\mathrm {ncl}(A)$ is the union of all $A_n$ ’s (compare with Proposition 2.15 above), it follows that $R\in \mathsf {Cong}(M,\mathrm {ncl}(A))$ . The last assertion follows from the fact that $R_A$ is the minimum of the sublattice $\mathsf {Cong}(M,A)$ , and $R_{\mathrm {ncl}(A)}$ is the minimum of $\mathsf {Cong}(M,\mathrm {ncl}(A))$ .

Proof. Clearly, $[1]_R$ is a groupal submonoid of M because of the compatibility of R. To see that it is invariant, let $(x,y)\in X_{[1]_R}$ . This means that there exist $z,z'\in [1]_R$ such that $xzy=z'$ . Then by the compatibility of R,

$$ \begin{align*} [1]_R=[z']_R=[xzy]_R=[xy]_R, \end{align*} $$

that is, $xy\in [1]_R$ . Therefore, for each $z_1\in [1]_R$ , we have $[xz_1y]_R=[xy]_R=[1]_R$ , that is, $x[1]_Ry\subseteq [1]_R$ .

Proof. The map $\Phi _M$ is order-preserving. Thus, if $S\subseteq S'$ , then $R_{S'}$ is a congruence containing $\{1\}\times S'$ and consequently, also $\{1\}\times S$ so that $R_S\subseteq R_{S'}$ by definition of $R_S$ . Since $S,S'\subseteq S\vee S'$ , it follows that $R_S,R_{S'}\subseteq R_{S\vee S'}$ and hence, $R_S\vee R_{S'}\subseteq R_{S\vee S'}$ by definition of the join $R_S\vee R_{S'}$ . To prove the reverse inclusion, it is enough to observe that for all normal submonoids $S,S'$ , the equivalence class of the identity element modulo $R_S\vee R_{S'}$ is a normal submonoid that contains both S and $S'$ and hence, also $S\vee S'$ . Therefore, $R_{S\vee S'}\subseteq R_S\vee R_{S'}$ by definition of $R_{S\vee S'}$ . This proves that $\Phi _M$ is join-preserving. As for $\Psi _M$ , it is clearly order-preserving, and for all congruences $R,R'$ , and every $x\in M$ , we have $[x]_{R\cap R'}=[x]_R\cap [x]_{R'}$ . In particular, this is true when $x=1$ , that is, $\Psi _M$ is meet-preserving. This proves item (a).

To prove item (b), we have to see that $[1]_{R_S}=S$ for every normal submonoid S of M. The inclusion $S\subseteq \,[1]_{R_S}$ follows from the definition of $R_S$ . To prove the reverse inclusion, let us consider an element $z\in [1]_{R_S}$ . By the previous description of $R_S$ , this means that there exists a finite sequence $x_0,\ldots ,x_k$ of elements in M such that $x_0=1$ , $x_k=z$ , and either $x_i\underset {S}{\rightsquigarrow} x_{i+1}$ or $x_{i+1}\underset {S}{\rightsquigarrow} x_i$ for each $i\in \{0,1,\ldots ,k-1\}$ . Then to see that $z\in S$ , we prove the following facts:

1. (i) every elementary S-deformation of an element in S is again an element in S; and

2. (ii) an element in S can be an elementary S-deformation of an element $v\in M$ only if $v\in S$ .

Clearly, if both of these facts are true, every element in the above sequence $x_0,\ldots ,x_k$ is in S because $x_0=1\in S$ . In particular, we have $z=x_k\in S$ and hence, $[1]_{R_S}\subseteq S$ . Let us prove items (i) and (ii).

Proof of item (i). Let $w\in M$ be an S-deformation of some element $s\in S$ . This means that there exist $x,y\in M$ , and $s'\in S$ such that $s=xy$ , and $w=xs'y$ . However, $s=xy\in S$ implies that $w=xs'y\in xSy\subseteq S$ because of the invariance of S.

Proof of item (ii). Let $s\in S$ be an elementary S-deformation of $v\in M$ . This means that there exist $v_1,v_2\in M$ and $s'\in S$ such that $v=v_1v_2$ and $s=v_1s'v_2$ . In particular, we have $s\in (v_1Sv_2)\cap S$ and hence, $v=v_1v_2\in S$ because of the invariance of S.

Proof. Both items are immediate consequences of Proposition 3.4 and Theorem 3.6.

Proof. It follows from Theorem 3.6 that S is the preimage of the identity element by the projection of M onto its quotient $M/R_S$ .

Proof. It is enough to see that $\Phi _M$ is also meet-preserving when M is as in the statement. For all normal submonoids $S,S'$ of any monoid M, we always have $R_{S\cap S'}\subseteq R_S\cap R_{S'}$ because $R_S\cap R_{S'}$ is a congruence containing $\{1\}\times (S\cap S')$ . Conversely, let us assume that $(x,y)\in R_S\cap R_{S'}$ . Since M is commutative, this means that there exist $s_1,s_2\in S$ and $s^{\prime }_1,s^{\prime }_2\in S'$ such that $x+s_1=y+s_2$ and $x+s^{\prime }_1=y+s^{\prime }_2$ . Hence,

$$ \begin{align*} x+s_1+s^{\prime}_2=y+s_2+s^{\prime}_2=x+s^{\prime}_1+s_2 \end{align*} $$

and consequently, $s_1+s^{\prime }_2=s^{\prime }_1+s_2$ because M is cancellative. Let us now observe that if M satisfies condition (*), then every normal submonoid of M also satisfies condition (*). Indeed, if $s_1, s_2\in S$ , with $S\lhd M$ , are such that $s_2=s_1+z$ for some $z\in M$ , then $z\in LX_S$ and hence, $z=z+0\in S$ . Then we may assume that $s^{\prime }_2=s^{\prime }_1+s'$ for some $s'\in S'$ . Then,

$$ \begin{align*} s^{\prime}_1+s_2=s_1+s^{\prime}_2=s_1+s^{\prime}_1+s' \end{align*} $$

and hence, $s_2=s_1+s'$ , that is, $s'\in LX_S$ . Since S is invariant, it follows that $s'\in S$ and consequently, $s'\in S\cap S'$ . Coming back to the initial hypothesis that $x+s_1=y+s_2$ , we conclude that

$$ \begin{align*} x+s_1=y+s_1+s' \end{align*} $$

and hence $x=y+s'$ , that is, $(x,y)\in R_{S\cap S'}$ .

Proof. It follows from Theorem 3.6 that the corestriction of $\Phi _M$ to $\mathsf {NorCon}(M)$ is an order-preserving bijection from $\mathsf {NorSub}(M)$ to $\mathsf {NorCon}(M)$ whose inverse map is given by $R\mapsto [1]_R$ and hence, also an order-preserving map.

Proof. Let us first observe that every congruence $R\in \mathsf {Cong}_S(M)$ induces a unital congruence on the quotient $M/S$ given by

$$ \begin{align*} \tilde{R}:=\{([x]_{R_S},[y]_{R_S})\in M/S\times M/S : (x,y)\in R\}. \end{align*} $$

Indeed, if $(x,y)\in R$ , and $x',y'\in M$ are such that $(x',x)\in R_S$ , $(y,y')\in R_S$ , then we also have $(x,x')\in R$ and $(y,y')\in R$ because $R_S$ is the minimum of $\mathsf {Cong}_S(M)$ . By transitivity, it follows that $(x',y')\in R$ and hence, $\tilde {R}$ is a well-defined binary relation on $M/S$ . Moreover, it is immediate to check that $\tilde {R}$ is a congruence because this is true for R. Finally, $[x]_{R_S}\tilde {R}\,[1]_{R_S}$ means that $x\,R\,1$ and hence, $x\in [1]_{R}=S=[1]_{R_S}$ , that is, $\tilde {R}$ is unital. In particular, when R is $R_S$ , then $\tilde {R}$ is the trivial congruence on $M/S$ . Conversely, every unital congruence T on $M/S$ induces a congruence $T^*$ on M given by

$$ \begin{align*} T^*:=\{(x,y)\in M\times M : ([x]_{R_S},[y]_{R_S})\in T\}, \end{align*} $$

and $T^*$ is such that $[1]_{T^*}=S$ . Indeed, $[[1]_{R_S}]_T=\{[1]_{R_S}\}$ because T is unital. Hence, for each $x\in M$ ,

$$ \begin{align*} (1,x)\in T^*\ \Leftrightarrow\ ([1]_{R_S},[x]_{R_S})\in T \ \Leftrightarrow [x]_{R_S}=[1]_{R_S}\ \Leftrightarrow\ (1,x)\in R_S\ \Leftrightarrow\ x\in S. \end{align*} $$

Then it is easy to check that the maps $t_S:\mathsf {Cong}_S(M)\to \mathsf {Cong}_1(M/S)$ and $t^*_S:\mathsf {Cong}_1(M/S)\to \mathsf {Cong}_S(M)$ respectively given by $R\mapsto \tilde {R}$ , and $T\mapsto T^*$ are inverses of each other. Moreover, both are order-preserving and hence, they define a meet semilattice isomorphism.

Proof. Let R be unital, so that [1]_R=\{1\} , and let $x,y\in M$ such $x\in U(M)$ , and $(x,y)\in R$ . Then $(1,x^{-1}y),(1,yx^{-1})\in R$ and hence, $x^{-1}y=yx^{-1}=1$ , that is $x=y$ . This proves that (i) implies (ii). The remaining implications (ii) $\Rightarrow $ (iii), and (iii) $\Rightarrow $ (i) are immediate.

Proof. It follows from Example 2.24 that the only nontrivial, proper normal quotients of $T_n$ for $n\geq 5$ are $T_n/S_n$ and $T_n/A_n$ . Hence, it is enough to prove that:

1. (a) $T_n/S_n\cong (\{0,1\},\cdot ,1)$ ;

2. (b) $T_n/A_n\cong (\{-1,0,1\},\cdot ,1)$ .

Proof of item (a). Recall that $T_n/S_n$ means the quotient of $T_n$ modulo the smallest congruence $R=R_{S_n}$ on $T_n$ such that $[1]_{R}=S_n$ . In particular, for each $u\in S_n$ , we have $[u]_R=S_n$ . It turns out that $[u]_R=T_n\setminus S_n$ if $u\notin S_n$ . To be precise, if $c_1:\mathbf {n}\to \mathbf {n}$ denotes the constant map defined by $c_1(k)=1$ for each $k\in \mathbf {n}$ , let us prove that

(3-2) $$ \begin{align} (u,c_1)\in R \ \Leftrightarrow\ u\notin S_n. \end{align} $$

The implication to the right is an immediate consequence of the fact that $[1]_R=S_n$ . Thus, if $(u,c_1)\in R$ for some $u\in S_n$ , then we also have $(1,c_1)\in R$ and hence, $c_1\in [1]_R=S_n$ which is clearly false. To prove the converse, we proceed by induction on the rank $r\in \{1,\ldots ,n-1\}$ of u (that is, on the cardinality of its image). If $r=1$ , then u is the constant map $c_i$ sending every $k\in \mathbf {n}$ to i for some $i\in \mathbf {n}$ , and $c_1$ is clearly an elementary $S_n$ -deformation of $c_i$ . Indeed, $c_1$ is the composite of $c_i$ with the transposition $(1i)\in S_n$ . Hence, $(u,c_1)\in R$ . Let us now assume that for some $r\in \{1,\ldots ,n-2\}$ , every map of rank r is R-related to $c_1$ , and let u be of rank $r+1$ . Since $r<n-1$ , we still have $r+1<n$ and hence, there exists some $i\in \mathbf {n}$ not in the image of u. Let j be any element in the image of u and l such that $u(l)\neq j$ . Such an l exists because u is of rank $r+1\geq 2$ . Then let us consider the map $u':\mathbf {n}\to \mathbf {n}$ defined by

$$ \begin{align*} u'(k)=\begin{cases} k & \mbox{if } k\neq i, \\ j & \mbox{if } k=i. \end{cases} \end{align*} $$

Clearly, we have $u'u=u$ . Hence, the composite $\hat {u}=u'\tau u$ , with $\tau $ the transposition $(iu(l))\in S_n$ is an elementary $S_n$ -deformation of u and consequently, $(u,\hat {u})\in R$ . However, $\hat {u}$ is of rank r because all elements in $u^{-1}(u(l))$ are now mapped not to $u(l)\neq j$ but to j, while $\hat {u}(k)=u(k)$ for each $k\notin u^{-1}(u(l))$ . By the induction hypothesis, $(\hat {u},c_1)\in R$ and hence, $(u,c_1)\in R$ by transitivity. This proves Equation (3-2).

It follows that the map $T_n\to \{0,1\}$ sending each $u\notin S_n$ to $0$ and each $u\in S_n$ to $1$ is a monoid epimorphism that factors through the quotient $T_n/S_n$ , and the induced map $T_n/S_n\to \{0,1\}$ is a monoid isomorphism.

Proof of item (b). Here, $T_n/A_n$ means the quotient of $T_n$ modulo the smallest congruence $R=R_{A_n}$ on $T_n$ such that $[1]_{R}=A_n$ . In particular, for each $u\in A_n$ , we have $[u]_R=A_n$ . It turns out now that $[u]_R=S_n\setminus A_n$ if $u\in S_n\setminus A_n$ and $[u]_R=T_n\setminus S_n$ if $u\notin S_n$ . To be precise, let us prove that

(3-3) $$ \begin{align} (u,c_1)\in R \ &\Leftrightarrow\ u\notin S_n, \end{align} $$

(3-4) $$ \begin{align} (u,(12))\in R \ &\Leftrightarrow\ u\in S_n\setminus A_n. \end{align} $$

The implication to the right in Equation (3-3) follows from the fact that $[u]_R\subseteq S_n$ for each $u\in S_n$ . To see this, recall that $(u,v)\in R$ if and only if there exists a finite sequence $w_0,\ldots ,w_k$ of elements in $T_n$ such that $u=w_0$ , $v=w_k$ , and for each $i\in \{0,1,\ldots ,k-1\}$ , either $w_i$ or $w_{i+1}$ is an elementary $A_n$ -deformation of the other. Actually, since $A_n$ is a group, both conditions are equivalent. For instance, if $w_{i+1}=z_1\tau z_2$ for some $z_1,z_2\in T_n$ such that $w_i=z_1z_2$ and some $\tau \in A_n$ , then $w_i=(z_1\tau )\tau ^{-1} z_2$ . Therefore, it is enough to see that every elementary $A_n$ -deformation of an element in $S_n$ is still in $S_n$ , and this is clearly true because every factorization of an element in $S_n$ is necessarily as a composite of elements in $S_n$ . To prove the converse, we proceed again by induction on the rank r of u. If $r=1$ , then u is the constant map $c_i$ for some $i\in \mathbf {n}$ , and when $i\neq 1$ , the map $c_1$ is the composite of $c_i$ and the permutation $(1ji)\in A_n$ for some $j\neq 1,i$ (here we use that $n\geq 3$ ). Hence, $(u,c_1)\in R$ . Let us now assume that for some $r\in \{1,\ldots ,n-2\}$ , every map of rank r is R-related to $c_1$ , and let u be of rank $r+1$ . Let i be not in the image of u, j any element in the image of u, and l such that $u(l)\neq j$ (we are using again that $n\geq 3$ ), and let $u':\mathbf {n}\to \mathbf {n}$ be the same map as in the proof of the previous proposition. In particular, we have $u'u=u$ . Then the composite $\tilde {u}=u'\sigma u$ with $\sigma $ the permutation $(iu(l)j)\in A_n$ is an elementary $A_n$ -deformation of u and consequently, $(u,\tilde {u})\in R$ . However, $\tilde {u}$ is of rank r because all elements in $u^{-1}(u(l))$ are now mapped to j, while $\tilde {u}(k)=u(k)$ for each $k\notin u^{-1}(u(l))$ . Indeed, $\sigma u$ maps all of $u^{-1}(j)$ to i, but this is next mapped again to j by $u'$ . By the induction hypothesis, $(\tilde {u},c_1)\in R$ and hence, $(u,c_1)\in R$ . This proves Equation (3-3).

Let us now prove Equation (3-4). The implication to the right follows now from the fact that for every odd permutation (for instance, $(12)$ ), $[\sigma ]_R\subseteq S_n\setminus A_n$ . Indeed, every factorization of an odd permutation is necessarily as a composite of an odd and an even permutation, and hence, every elementary $A_n$ -deformation of it is also an odd permutation. The converse follows from the obvious fact that every odd permutation u can be connected to $(12)$ by a finite number of elementary $A_n$ -deformations. It is enough to write u as the composite of an odd number of transpositions including $(12)$ (two consecutive times if necessary), and then proceed to eliminate consecutive pairs of transpositions by performing elementary $A_n$ -deformations until we are left only with the transposition $(12)$ . This proves Equation (3-4).

By an easy case-by-case check, it follows that the map $T_n\to \{-1,0,1\}$ sending each $u\notin S_n$ to $0$ , each $u\in A_n$ to $1$ , and each $u\in S_n\setminus A_n$ to $-1$ preserves products and hence, is a monoid epimorphism that factors through the quotient $T_n/A_n$ , and whose induced map $T_n/A_n\to \{-1,0,1\}$ is a monoid isomorphism.