On automorphisms of strong semilattice of groups

1. Introduction

Let $(\mathrm{\Omega },\le )$ be a poset. Following [2] we say that $(\mathrm{\Omega },\le )$ is a lower (meet) semilattice if $a\wedge b\in \mathrm{\Omega }$ for all $a,b\in \mathrm{\Omega }$. Dually, one can define an upper (join) semilattice. In this paper by a semilattice, we shall always mean a lower semilattice. It is well known (see [2, Proposition 1.3.2]) that if $(\mathrm{\Omega },\le )$ is a semilattice, then a binary operation may be defined on $\mathrm{\Omega }$ as $\alpha \beta =\alpha \wedge \beta$ for all $\alpha ,\beta \in \mathrm{\Omega }$ under which $\mathrm{\Omega }$ becomes a commutative semigroup of idempotents and conversely if $(\mathrm{\Omega },\wedge )$ is a commutative semigroup of idempotents, then $(\mathrm{\Omega },\le )$, where $\alpha \le \beta$ if and only if $\alpha \beta =\alpha$ for all $\alpha ,\beta \in \mathrm{\Omega }$, is a semilattice.

Let $\mathrm{\Omega }$ be a semilattice and $\{S_{\alpha },\alpha \in \mathrm{\Omega }\}$ be the family of semigroups. A semigroup S is said to be the semilattice of semigroups $S_{\alpha }$, $\alpha \in \mathrm{\Omega }$, if $S={\bigcup }_{\alpha \in \mathrm{\Omega }}{S}_{\alpha }$ where $S_{\alpha }\cap S_{\beta }=\varnothing$ for $\alpha \ne \beta$ and $S_{\alpha }{S}_{\beta }\subseteq S_{\alpha \beta }$. If for each pair $\beta \le \alpha$ in $\mathrm{\Omega }$, ${\varphi }_{\alpha ,\beta }:S_{\alpha }\to S_{\beta }$ is a semigroup homomorphism such that

(i)	${\varphi }_{\alpha ,\alpha }={\mathrm{\text{id}}}_{S_{\alpha }}$ for any $\alpha \in \mathrm{\Omega }$.
(ii)	The homomorphisms are transitive: For any $\alpha ,\beta ,\gamma \in \mathrm{\Omega }$ with $\gamma \le \beta \le \alpha$ ${\varphi }_{\beta ,\gamma }\circ {\varphi }_{\alpha ,\beta }={\varphi }_{\alpha ,\gamma }$.

On $S={\bigcup }_{\alpha \in \mathrm{\Omega }}{S}_{\alpha }$ define a multiplication ∗ where for $s\in S_{\alpha }$ and $t\in S_{\beta }$,

Then S forms a semigroup, denoted $S=[\mathrm{\Omega };S_{\alpha },{\varphi }_{\alpha ,\beta }]$, known as a strong semilattice of semigroups. The homomorphisms ${\varphi }_{\alpha ,\beta }$ are called structure homomorphisms, $\mathrm{\Omega }$ is called the structure semilattice of S, and the semigroups $S_{\alpha }$ are called the components of S. If each component $S_{\alpha }$ of S is a group, then S forms a semigroup known as strong semilattice of groups. If S is strong semilattice of groups then unless stated otherwise the identity element of a component $S_{\alpha }$ will be denoted $e_{\alpha }$, and is the unique idempotent of $S_{\alpha }$, so that the set of idempotents $E_S$ of S is $\{e_{\alpha };\alpha \in \mathrm{\Omega }\}$.

An element a of a semigroup S is called $regular$ if there exists $x\in S$ such that $axa=a$ and S is called $regular$ if all its elements are regular. A regular semigroup S is Clifford if $E_S$ is central, that is, if idempotents commute with every element of S. The property of being Clifford has a number of alternate statements, to which we give only one.

Theorem 1.1 ([2, Theorem 4.2.1]). Let S be a semigroup. Then the following statements are equivalent :

(i)	S is a Clifford semigroup.
(ii)	S is a semilattice of groups.
(iii)	S is a strong semilattice of groups.
(iv)	S is regular and the idempotents of S are central.
(v)	S is regular and ${{𝒟}}^S\cap (E_S\times E_S)=1_{E_S}$.

In this paper, motivated by Theorem 1.1, we shall take the definition of a Clifford semigroup as a strong semilattice of groups. After looking for automorphisms of the strong semilattice of groups, we relate them to the isomorphisms and automorphisms of underlying groups, and provide a construction for non-trivial automorphisms of semilattices. We also illustrate our results with some particular examples.

2. Isomorphism of Strong Semilattice of Groups

In this section, first we mention a well-known result which provides the necessary and sufficient condition for two strong semilattices of groups to be isomorphic. For this, we first fix some notations without further mention. Let $S=[\mathrm{\Omega },{\{S_{\alpha }\}}_{\alpha \in \mathrm{\Omega }},{\{{\varphi }_{\alpha ,\beta }\}}_{\beta \le \alpha }]$ and $T=[\mathrm{\Lambda },{\{T_{\gamma }\}}_{\gamma \in \mathrm{\Lambda }},{\{{\psi }_{\gamma ,\delta }\}}_{\delta \le \gamma }]$ be two strong semilattices of groups. Now onwards by S or T, we mean the semigroups which are strong semilattice of groups.

A mapping ${\mathrm{\Theta }}_L$ of semilattice $\mathrm{\Omega }$ to a semilattice $\mathrm{\Lambda }$ is called a semilattice homomorphism if ${\mathrm{\Theta }}_L(\alpha \wedge \beta )={\mathrm{\Theta }}_L(\alpha )\wedge {\mathrm{\Theta }}_L(\beta )$ for all $\alpha ,\beta \in \mathrm{\Omega }$, and ${\mathrm{\Theta }}_L$ is an isomorphism if it is bijective. An isomorphism of semilattice $\mathrm{\Omega }$ onto itself is called an automorphism of $\mathrm{\Omega }$.

Theorem 2.1 ([6, Proposition II.2.8]). Two strong semilattices of groups S and T are isomorphic if and only if there exists a semilattice isomorphism ${\mathrm{\Theta }}_L:\mathrm{\Omega }\to \mathrm{\Lambda }$ and a family of group isomorphisms $\{{\mathrm{\Theta }}_{\alpha }:\alpha \in \mathrm{\Omega }\}$, where ${\mathrm{\Theta }}_{\alpha }:S_{\alpha }\to T_{{\mathrm{\Theta }}_L(\alpha )}$ such that the diagram

commutes, i.e. ${\mathrm{\Theta }}_{\beta }{\varphi }_{\alpha ,\beta }={\psi }_{{\mathrm{\Theta }}_L(\alpha ),{\mathrm{\Theta }}_L(\beta )}{\mathrm{\Theta }}_{\alpha }$ for all $\beta \le \alpha$.

The isomorphism ${\mathrm{\Theta }}_L:\mathrm{\Omega }\to \mathrm{\Lambda }$ could be realized as: ${\mathrm{\Theta }}_L(\alpha )=\tau$, where if $\mathrm{\Theta }:S\to T$ is an isomorphism of strong semilattice of groups, then $\mathrm{\Theta }(e_{\alpha })=e_{\tau }$. Also the group isomorphism ${\mathrm{\Theta }}_{\alpha }$ could be realized as $\mathrm{\Theta }{|}_{S_{\alpha }}$, where $\alpha \in \mathrm{\Omega }$.

In the following result, we find the relationship between the images and kernels of the structure homomorphisms, respectively.

Corollary 2.2. Let $\mathrm{\Theta }:S\to T$ be an isomorphism of strong semilattices of groups. If $\alpha ,\beta \in \mathrm{\Omega }$ with $\beta \le \alpha$, then

(1)	${\mathrm{\Theta }}_{\beta }(\mathrm{\text{im}}{\varphi }_{\alpha ,\beta })=\mathrm{\text{im}}{\psi }_{{\mathrm{\Theta }}_L(\alpha ),{\mathrm{\Theta }}_L(\beta )};$
(2)	${\mathrm{\Theta }}_{\alpha }(\mathrm{\text{ker}}{\varphi }_{\alpha ,\beta })=\mathrm{\text{ker}}{\psi }_{{\mathrm{\Theta }}_L(\alpha ),{\mathrm{\Theta }}_L(\beta )}$.

Proof. (i) ${\mathrm{\Theta }}_{\beta }(\mathrm{\text{im}}{\varphi }_{\alpha ,\beta })\subseteq \mathrm{\text{im}}{\psi }_{{\mathrm{\Theta }}_L(\alpha ),{\mathrm{\Theta }}_L(\beta )}$ follows directly from diagram (1). For reverse inclusion, since $\mathrm{\Theta }$ is an isomorphism so by Theorem 2.1, ${\mathrm{\Theta }}_{\beta }^{-1}$ exists and by diagram (1), we have $\mathrm{\text{im}}{\psi }_{{\mathrm{\Theta }}_L(\alpha ),{\mathrm{\Theta }}_L(\beta )}\subseteq {\mathrm{\Theta }}_{\beta }(\mathrm{\text{im}}{\varphi }_{\alpha ,\beta })$. Hence, ${\mathrm{\Theta }}_{\beta }(\mathrm{\text{im}}{\varphi }_{\alpha ,\beta })=\mathrm{\text{im}}{\psi }_{{\mathrm{\Theta }}_L(\alpha ),{\mathrm{\Theta }}_L(\beta )}$.

(ii) Let $s_{\alpha }\in \mathrm{\text{ker}}{\varphi }_{\alpha ,\beta }$. Then ${\varphi }_{\alpha ,\beta }(s_{\alpha })=e_{\beta }$. Let us assume to the contrary that ${\mathrm{\Theta }}_{\alpha }(s_{\alpha })\notin \mathrm{\text{ker}}{\psi }_{{\mathrm{\Theta }}_L(\alpha ),{\mathrm{\Theta }}_L(\beta )}$. This implies that ${\psi }_{{\mathrm{\Theta }}_L(\alpha ),{\mathrm{\Theta }}_L(\beta )}({\mathrm{\Theta }}_{\alpha }(s_{\alpha }))\ne e_{{\mathrm{\Theta }}_L(\beta )}$. Therefore, by diagram (1), we have ${\mathrm{\Theta }}_{\beta }({\varphi }_{\alpha ,\beta }(s_{\alpha }))\ne e_{{\mathrm{\Theta }}_L(\beta )}$. This implies that ${\mathrm{\Theta }}_{\beta }(e_{\beta })\ne e_{{\mathrm{\Theta }}_L(\beta )}$, a contradiction as ${\mathrm{\Theta }}_{\beta }$ is a homomorphism. Thus, we have ${\mathrm{\Theta }}_{\alpha }(\mathrm{\text{ker}}{\varphi }_{\alpha ,\beta })\subseteq \mathrm{\text{ker}}{\psi }_{{\mathrm{\Theta }}_L(\alpha ),{\mathrm{\Theta }}_L(\beta )}$, as required. For reverse inclusion, since $\mathrm{\Theta }$ is an isomorphism so by Theorem 2.1, ${\mathrm{\Theta }}_{\alpha }^{-1}$ exists and so we have ${\mathrm{\Theta }}_{\alpha }^{-1}(\mathrm{\text{ker}}{\psi }_{{\mathrm{\Theta }}_L(\alpha ),{\mathrm{\Theta }}_L(\beta )})\subseteq \mathrm{\text{ker}}{\varphi }_{\alpha ,\beta }$, hence we have ${\mathrm{\Theta }}_{\alpha }(\mathrm{\text{ker}}{\varphi }_{\alpha ,\beta })=\mathrm{\text{ker}}{\psi }_{{\mathrm{\Theta }}_L(\alpha ),{\mathrm{\Theta }}_L(\beta )}$. □

It is well known that every automorphism of a finite chain (a finite semilattice in which every pair of elements is comparable) is trivial. So if the semilattice $\mathrm{\Omega }$ is a finite chain and $\mathrm{\Theta }\in \mathrm{\text{Aut}}(S)$, then ${\mathrm{\Theta }}_L$ is a trivial semilattice automorphism. A chain semilattice is a semilattice in which every pair of elements is comparable. The most natural example of a chain semilattice is $(\mathbb{N},\le )$, where $\le$ is natural order. Any finite subset of $\mathbb{N}$ with natural order is also a chain semilattice. It is well known that every automorphism of $(\mathbb{N},\le )$ or any of its finite subsemilattice is trivial. Therefore, if the semilattice $\mathrm{\Omega }$ has trivial automorphism, then we have the following corollary which immediately follows from Theorem 2.1.

Corollary 2.3. $\mathrm{\Theta }\in \mathrm{\text{Aut}}(S)$ if and only if ${\mathrm{\Theta }}_{\alpha }\in \mathrm{\text{Aut}}(S_{\alpha })$ with

for each $\alpha ,\beta \in \mathrm{\Omega }$.

Next we illustrate with the help of examples that while obtaining Eq. (2) in the above corollary the structure homomorphisms play a vital role, i.e. cardinality of Aut$(S)$ changes with choice of structure homomorphisms.

Example 2.1. Let $\mathrm{\Omega }={\{\alpha ,\beta \}}_{\beta \le \alpha }$ and $S=S_{\alpha }\cup S_{\beta }$, where $S_{\alpha }=\{0_{\alpha },1_{\alpha },2_{\alpha }\}$ is the cyclic group of order 3 and $S_{\beta }=\{e_{\beta },r_{\beta },r_{\beta }^2,s_{\beta },r_{\beta }{s}_{\beta },r_{\beta }^2{s}_{\beta }\}$ is the dihedral group of order 6. Let the structure homomorphism ${\varphi }_{\alpha ,\beta }$ be defined as ${\varphi }_{\alpha ,\beta }(s_{\alpha })=e_{\beta }$ for all $s_{\alpha }\in S_{\alpha }$. Table 1 illustrates the group $\mathrm{\text{Aut}}(S)$, where in each case ${\mathrm{\Theta }}_i{|}_{S_j}\in \mathrm{\text{Aut}}(S_j)$ for all $1\le i\le 12$ and $j\in \mathrm{\Omega }$.

From Table 1, we know that $|\mathrm{\text{Aut}}(S_{\beta })|=6$, $|\mathrm{\text{Aut}}(S_{\alpha })|=2$ and the structure homomorphism is defined as ${\varphi }_{\alpha ,\beta }(s_{\alpha })=e_{\beta }$ for all $s_{\alpha }\in S_{\alpha }$. Since, for any ${\mathrm{\Theta }}_{\beta }\in \mathrm{\text{Aut}}(S_{\beta })$ and ${\mathrm{\Theta }}_{\alpha }\in \mathrm{\text{Aut}}(S_{\alpha })$, the equation ${\mathrm{\Theta }}_{\beta }{\varphi }_{\alpha ,\beta }={\varphi }_{\alpha ,\beta }{\mathrm{\Theta }}_{\alpha }$ is satisfied, we have $|\mathrm{\text{Aut}}(S)|=12$.

Now if we take the structure homomorphism as

and the group automorphisms as then with this choice of ${\varphi }_{\alpha ,\beta }$ together with condition (A), ${\mathrm{\Theta }}_3,{\mathrm{\Theta }}_7$ and ${\mathrm{\Theta }}_{11}$ in Table 1 do not satisfy Eq. (2) so cannot be candidates for the group $\mathrm{\text{Aut}}(S)$.

Also with same ${\varphi }_{\alpha ,\beta }$ and the group automorphisms ${\mathrm{\Theta }}_{\alpha }$ and ${\mathrm{\Theta }}_{\beta }$ as

where $U_{\beta }=\{e_{\beta },r_{\beta },r_{\beta }^2\}$ is the image of ${\varphi }_{\alpha ,\beta }$ and is a cyclic subgroup of $S_{\beta }$. The automorphisms ${\mathrm{\Theta }}_2,{\mathrm{\Theta }}_6$ and ${\mathrm{\Theta }}_{10}$ in Table 1 again do not satisfy Eq. (2) so we exclude them from $\mathrm{\text{Aut}}(S)$. Rest, with ${\varphi }_{\alpha ,\beta }$ as above and any choice of ${\mathrm{\Theta }}_{\alpha }$ and ${\mathrm{\Theta }}_{\beta }$ in Table 1, Eq. (2) holds. So we conclude that with ${\varphi }_{\alpha ,\beta }$ chosen as above the cardinality of $\mathrm{\text{Aut}}(S)$ reduces to 6 as illustrated below.

Next we compute the cardinality of $\mathrm{\text{Aut}}(S)$ where S is the strong semilattice of finite and infinite cyclic groups, respectively.

Proposition 2.4. Let $S={\bigcup }_{i=1}^{\infty }{\mathbb{Z}}_{k_i}$ be a semilattice of groups ${\mathbb{Z}}_{k_i}$, where ${\mathbb{Z}}_{k_i}$ is the cyclic group of order $k_i$. If all the structure homomorphisms are identically zero, then $|\mathrm{\text{Aut}}(S)|={\prod }_{i=1}^{\infty }\phi (k_i)$, where $\phi (k_i)$ is Euler’s totient function.

Proof. Let $S={\bigcup }_{i=1}^{\infty }{\mathbb{Z}}_{k_i}$ be a semilattice of groups ${\mathbb{Z}}_{k_i}$, where ${\mathbb{Z}}_{k_i}$ is the cyclic group of order $k_i$. Let all the structure homomorphisms be identically zero. Then, for any ${\mathrm{\Theta }}_{k_i}\in \mathrm{\text{Aut}}({\mathbb{Z}}_{k_i})$ and ${\mathrm{\Theta }}_{k_j}\in \mathrm{\text{Aut}}({\mathbb{Z}}_{k_j})$, the equation ${\mathrm{\Theta }}_{k_j}{\varphi }_{k_i,k_j}={\varphi }_{k_i,k_j}{\mathrm{\Theta }}_{k_i}$ is satisfied. Therefore, by Corollary 2.6 and the fact that $|\mathrm{\text{Aut}}({\mathbb{Z}}_{k_i})|=\phi (k_i)$, we have $|\mathrm{\text{Aut}}(S)|={\prod }_{i=1}^{\infty }\phi (k_i)$. This completes the proof. □

Corollary 2.5. Let $S={\bigcup }_{i=1}^{\infty }{\mathbb{Z}}_{k_i}$ be a semilattice of groups ${\mathbb{Z}}_{k_i}$, where ${\mathbb{Z}}_{k_i}$ is a cyclic group of order $k_i$. If $\mathrm{\text{gcd}}(k_i,k_j)=1$ for all $1\le i,j<\infty$ with $i\ne j$, then $|\mathrm{\text{Aut}}(S)|={\prod }_{i=1}^{\infty }\phi (k_i)$ with $\phi (k_i)$ as the Euler’s totient function.

Proof. Suppose that $S={\bigcup }_{i=1}^{\infty }{\mathbb{Z}}_{k_i}$ be a semilattice of groups ${\mathbb{Z}}_{k_i}$, where ${\mathbb{Z}}_{k_i}$ is a cyclic group of order $k_i$. Since we know that $|\mathrm{\text{Aut}}({\mathbb{Z}}_{k_i})|=\phi (k_i)$ and $\mathrm{\text{gcd}}(k_i,k_j)=1$ for all $1\le i,j<\infty$ with $i\ne j$, all structure morphisms are identically zero. Therefore, by Proposition 2.7, we have $|\mathrm{\text{Aut}}(S)|={\prod }_{i=1}^{\infty }\phi (k_i)$, as required. □

Proposition 2.6. Let $S={\bigcup }_{i=1}^{\infty }{𝔾}_i$ be a semilattice of groups ${𝔾}_i$, where each ${𝔾}_i$ is an infinite cyclic group. Then $|\mathrm{\text{Aut}}(S)|=2$.

Proof. Let ${𝔾}_i$ is an infinite cyclic group, then for each i, ${𝔾}_i$ behaves like $\mathbb{Z}$. Therefore, in this case, $S\cong \mathbb{Z}$ and, hence, $|\mathrm{\text{Aut}}(S)|=2=|\mathrm{\text{Aut}}(\mathbb{Z})|$. □

3. Constructing Automorphism from Structure Homomorphisms and Component Isomorphisms

Let Y be a non-empty subset of a partially ordered set $(\mathrm{\Omega },\le )$. An element a of Y is called minimum if $a\le y$, for all $y\in Y$. An element $a^{\prime }$ of Y is called maximum if $y\le a^{\prime }$, for all $y\in Y$. In this section, we consider $\mathrm{\text{Aut}}(S)$, the group of automorphisms of strong semilattices of groups S and take $\mathrm{\Omega }$ an arbitrary semilattice not necessarily a chain. The condition (1) thus becomes

for all $\alpha ,\beta \in \mathrm{\Omega },\beta \le \alpha$.

Now, by putting restrictions on the semilattice and the structure homomorphisms, we have

Theorem 3.1. Let the semilattice $\mathrm{\Omega }$ have the minimum element $\kappa$ and S be the strong semilattice of groups with all the structure homomorphisms one-to-one. Then every group isomorphism ${\mathrm{\Theta }}_{\alpha }$ associated with the automorphism $\mathrm{\Theta }$ of S is determined by ${\mathrm{\Theta }}_{\kappa }$ and is given

Proof. Let the semilattice $\mathrm{\Omega }$ have the minimum element $\kappa$. Then, by Eq. (3), we have ${\mathrm{\Theta }}_{\kappa }{\varphi }_{\alpha ,\kappa }={\varphi }_{{\mathrm{\Theta }}_L(\alpha ),{\mathrm{\Theta }}_L(\kappa )}{\mathrm{\Theta }}_{\alpha }$. As the structure homomorphism ${\varphi }_{{\mathrm{\Theta }}_L(\alpha ),{\mathrm{\Theta }}_L(\kappa )}$ is one-to-one, ${\varphi }_{{\mathrm{\Theta }}_L(\alpha ),{\mathrm{\Theta }}_L(\kappa )}$ restricted to $\mathrm{\text{im}}{\varphi }_{{\mathrm{\Theta }}_L(\alpha ),{\mathrm{\Theta }}_L(\kappa )}$ has an inverse. So, by Corollary 2.2, we have ${\mathrm{\Theta }}_{\alpha }={\varphi }_{{\mathrm{\Theta }}_L(\alpha ),{\mathrm{\Theta }}_L(\kappa )}^{-1}{\mathrm{\Theta }}_{\kappa }{\varphi }_{\alpha ,\kappa }$, as required. □

Next we have the lemma which helps us to prove the dual of Theorem 3.1 where all the structure homomorphisms are taken to be onto and the semilattice is assumed to have a maximum element.

Lemma 3.2. Let the semilattice $\mathrm{\Omega }$ have the maximum element $\eta$ and let all the structure homomorphisms be onto. Then

defines a map from $S_{\alpha }$ to $S_{{\mathrm{\Theta }}_L(\alpha )}$, where ${\varphi }_{\eta ,\alpha }^{-1}$ is the set of inverse images of the structure homomorphism ${\varphi }_{\eta ,\alpha }$.

Proof. Let the semilattice $\mathrm{\Omega }$ have the maximum element $\eta$. For each $g_{\alpha }\in S_{\alpha }$, and as the structure homomorphism ${\varphi }_{\eta ,\alpha }$ is onto, there exist $h\in S_{\eta }$ such that ${\varphi }_{\eta ,\alpha }(h)=g_{\alpha }$. Define ${\varphi }_{\eta ,\alpha }^{-1}(g_{\alpha })=h$. Now we have the following cases:

Case (i). If ${\varphi }_{\eta ,\alpha }$ is one-to-one, then we are done.

Case (ii). If ${\varphi }_{\eta ,\alpha }$ is not one-to-one, then there may exist $h_1,h_2\in S_{\eta }$ such that ${\varphi }_{\eta ,\alpha }(h_1)=g_{\alpha }={\varphi }_{\eta ,\alpha }(h_2)$. Let us suppose that ${\varphi }_{{\mathrm{\Theta }}_L(\eta ),{\mathrm{\Theta }}_L(\alpha )}{\mathrm{\Theta }}_{\eta }{\varphi }_{\eta ,\alpha }^{-1}(g_{\alpha })={\varphi }_{{\mathrm{\Theta }}_L(\eta ),{\mathrm{\Theta }}_L(\alpha )}{\mathrm{\Theta }}_{\eta }(h_i)$ for $i=1,2$. Since $h_1\in S_{\eta }$ and $S_{\eta }$ is a group, so $h_1^{-1}$ in $S_{\eta }$ exists. Therefore, ${\varphi }_{\eta ,\alpha }(h_1{h}_1^{-1})={\varphi }_{\eta ,\alpha }(h_2{h}_1^{-1})$ which implies that ${\varphi }_{\eta ,\alpha }(e_{\eta })={\varphi }_{\eta ,\alpha }(h_2{h}_1^{-1})$; that is, ${\varphi }_{\eta ,\alpha }(h_2{h}_1^{-1})=e_{\alpha }$. Thus, we have $h_2{h}_1^{-1}\in \mathrm{\text{ker}}{\varphi }_{\eta ,\alpha }$ and, so ${\mathrm{\Theta }}_{\eta }(h_2{h}_1^{-1})\in {\mathrm{\Theta }}_{\eta }(\mathrm{\text{ker}}{\varphi }_{\eta ,\alpha })$. As, by Corollary 2.2, ${\mathrm{\Theta }}_{\eta }(\mathrm{\text{ker}}{\varphi }_{\eta ,\alpha })\subseteq \mathrm{\text{ker}}{\varphi }_{{\mathrm{\Theta }}_L(\eta ),{\mathrm{\Theta }}_L(\alpha )}$, we have

Therefore, we have $e_{{\mathrm{\Theta }}_L(\alpha )}={\varphi }_{{\mathrm{\Theta }}_L(\eta ),{\mathrm{\Theta }}_L(\alpha )}{\mathrm{\Theta }}_{\eta }(h_2){({\varphi }_{{\mathrm{\Theta }}_L(\eta ),{\mathrm{\Theta }}_L(\alpha )}{\mathrm{\Theta }}_{\eta }(h_1))}^{-1}$; that is Hence, ${\varphi }_{{\mathrm{\Theta }}_L(\eta ),{\mathrm{\Theta }}_L(\alpha )}{\mathrm{\Theta }}_{\eta }{\varphi }_{\eta ,\alpha }^{-1}$ is a well-defined map. □

The following theorem is the dual of Theorem 3.1.

Theorem 3.3. Let the semilattice $\mathrm{\Omega }$ have the maximum element $\eta$ and S be the strong semilattice of groups with all the structure homomorphisms onto. Then every group isomorphism ${\mathrm{\Theta }}_{\alpha }$ associated with the automorphism $\mathrm{\Theta }$ of S is determined by ${\mathrm{\Theta }}_{\eta }$ and is given by

Proof. Let the semilattice $\mathrm{\Omega }$ have the maximum element $\eta$. Then, by Eq. (3), we have ${\mathrm{\Theta }}_{\alpha }{\varphi }_{\eta ,\alpha }={\varphi }_{{\mathrm{\Theta }}_L(\eta ),{\mathrm{\Theta }}_L(\alpha )}{\mathrm{\Theta }}_{\eta }$. By Lemma 3.2, we have ${\varphi }_{{\mathrm{\Theta }}_L(\eta ),{\mathrm{\Theta }}_L(\alpha )}{\mathrm{\Theta }}_{\eta }{\varphi }_{\eta ,\alpha }^{-1}$ is a well-defined map. Since the structure homomorphism ${\varphi }_{\eta ,\alpha }$ is surjective thus for any $g_{\alpha }\in S_{\alpha }$ the set ${\varphi }_{\eta ,\alpha }^{-1}(g_{\alpha })=\{h\in S_{\eta }:{\varphi }_{\eta ,\alpha }(h)=g_{\alpha }\}$ is non-empty. Now let $g_{\alpha }\in S_{\alpha }$, then we have

Therefore, we get the required equality as ${\mathrm{\Theta }}_{\alpha }={\varphi }_{{\mathrm{\Theta }}_L(\eta ),{\mathrm{\Theta }}_L(\alpha )}{\mathrm{\Theta }}_{\eta }{\varphi }_{\eta ,\alpha }^{-1}$. □

In the following theorem, we take the structure homomorphisms as isomorphisms.

Theorem 3.4. Let S be the strong semilattice of groups with all the structure homomorphisms bijective. Then every group isomorphism ${\mathrm{\Theta }}_{\alpha }$ associated with the automorphism $\mathrm{\Theta }$ of S is given by

Proof. By Eq. (3), we have ${\mathrm{\Theta }}_{\kappa }{\varphi }_{\alpha ,\kappa }={\varphi }_{{\mathrm{\Theta }}_L(\alpha ),{\mathrm{\Theta }}_L(\kappa )}{\mathrm{\Theta }}_{\alpha }$ and ${\mathrm{\Theta }}_{\kappa }{\varphi }_{\beta ,\kappa }={\varphi }_{{\mathrm{\Theta }}_L(\beta ),{\mathrm{\Theta }}_L(\kappa )}{\mathrm{\Theta }}_{\beta }$. Since all the structure homomorphisms are bijective, we have

and ${\mathrm{\Theta }}_{\kappa }={\varphi }_{{\mathrm{\Theta }}_L(\beta ),{\mathrm{\Theta }}_L(\kappa )}{\mathrm{\Theta }}_{\beta }{\varphi }_{\beta ,\kappa }^{-1}$. Now substituting the value of ${\mathrm{\Theta }}_{\kappa }$ in Eq. (5), we have ${\mathrm{\Theta }}_{\alpha }={\varphi }_{{\mathrm{\Theta }}_L(\alpha ),{\mathrm{\Theta }}_L(\kappa )}^{-1}{\varphi }_{{\mathrm{\Theta }}_L(\beta ),{\mathrm{\Theta }}_L(\kappa )}{\mathrm{\Theta }}_{\beta }{\varphi }_{\beta ,\kappa }^{-1}{\varphi }_{\alpha ,\kappa }$, as required. □

Now we provide a construction for the automorphisms of S from the automorphisms of underlying groups $S_{\alpha }$:

Theorem 3.5. Suppose $\mathrm{\Omega }={\{\alpha ,\beta \}}_{\alpha \le \beta }$. Consider S as the strong semilattice of $S_{\alpha }$ and $S_{\beta }$ with all the structure homomorphisms bijective. Then each automorphism of $S_{\alpha }$ $(respectively,S_{\beta })$ provides an automorphism of S.

Proof. Suppose all the structure homomorphisms are bijective. Then $S_{\alpha }\cong S_{\beta }\cong 𝔾$ (say). Let $\psi \in \mathrm{\text{Aut}}(𝔾)$ be any arbitrary automorphism of $𝔾$. Since $S_{\alpha }\cong S_{\beta }\cong 𝔾$, therefore we have isomorphisms ${\mathrm{\Theta }}_{\alpha }:𝔾\to S_{\alpha }$ and ${\mathrm{\Theta }}_{\beta }:𝔾\to S_{\beta }$.

Since ${\mathrm{\Theta }}_{\beta }:𝔾\to S_{\beta }$ and ${\varphi }_{\beta ,\alpha }:S_{\beta }\to S_{\alpha }$ are isomorphisms, the morphism ${\varphi }_{\beta ,\alpha }{\mathrm{\Theta }}_{\beta }:G\to S_{\beta }$ is an isomorphism. Therefore, we may let ${\mathrm{\Theta }}_{\alpha }={\varphi }_{\beta ,\alpha }{\mathrm{\Theta }}_{\beta }$. Let ${\psi }^{\mathrm{\Theta }}:S\to S$ be the map defined by

We show that ${\psi }^{\mathrm{\Theta }}\in \mathrm{\text{Aut}}(S)$. For this, we first show that for all $s\in S_{\beta }$ Therefore, we have Now, for any $s\in S_{\beta }$, we have Hence, for all $s\in S_{\beta }$, we have ${\psi }^{\mathrm{\Theta }}{\varphi }_{\beta ,\alpha }(s)={\varphi }_{\beta ,\alpha }{\psi }^{\mathrm{\Theta }}(s)$.

It is clear that ${\psi }^{\mathrm{\Theta }}$ is bijective. Now we show that ${\psi }^{\mathrm{\Theta }}$ is a homomorphism. For this, we consider the following cases:

Case (i). If $s,t\in S_{\alpha }$ $(s,t\in S_{\beta })$, then

Case (ii). If $s\in S_{\alpha }$ and $t\in S_{\beta }$, then Hence, ${\psi }^{\mathrm{\Theta }}\in \mathrm{\text{Aut}}(S)$ and every automorphism of S can be constructed in this way. □

Next, we prove the lemma which helps us to prove Theorem 3.5 for arbitrary semilattices.

Lemma 3.6. Let $\mathrm{\Omega }$ be any semilattice and S be a strong semilattice of groups with all the structure homomorphisms being bijective. Then, for any $\alpha \in \mathrm{\Omega }$, we have $S\cong \mathrm{\Omega }\times S_{\alpha }(\cong \mathrm{\Omega }\times 𝔾)$.

Proof. Fix $\alpha \in \mathrm{\Omega }$. Then, for each $\lambda \in \mathrm{\Omega }$, we have an isomorphism

Now define a map $\chi :S\to \mathrm{\Omega }\times S_{\alpha }$ by $\chi (s)=(\lambda ,{\sigma }_{\lambda }(s))$ if $s\in S_{\lambda }$. We show that $\chi$ is an isomorphism. To show this, take any $s_1,s_2\in S$. If $s_1=s_2$, then $s_1,s_2\in S_{\lambda }$ for some $\lambda \in \mathrm{\Omega }$. Since ${\sigma }_{\lambda }:S_{\lambda }\to S_{\alpha }$ is an isomorphism, we have That is, $\chi$ is well-defined and injective. Next take any $(\lambda ,t)\in \mathrm{\Omega }\times S_{\alpha }$. As ${\sigma }_{\lambda }$ is surjective, $t={\sigma }_{\lambda }(s^{\prime })$ for some $s^{\prime }\in S_{\lambda }$. Therefore $(\lambda ,t)=(\lambda ,{\sigma }_{\lambda }(s^{\prime }))=\chi (s^{\prime })$; that is, $\chi$ is surjective.

Finally to show that $\chi$ is a homomorphism and, hence, an isomorphism, take any $s,t\in S$. Then $s\in S_{\lambda }$ and $t\in S_{\delta }$ for some $\lambda ,\delta \in \mathrm{\Omega }$. If $\lambda =\delta$, then there is nothing to prove. So we suppose that $\lambda \ne \delta$. Now

Therefore, $\chi$ is an isomorphism. □

The proof of the following corollary follows from Theorem 3.5 and Lemma 3.6.

Corollary 3.7. Let $\mathrm{\Omega }$ be any semilattice and S be a strong semilattice of groups with all the structure homomorphisms are bijective. Then each automorphism of $S_{\alpha }$ $(\alpha \in \mathrm{\Omega })$ gives rise to an automorphism of S.

4. Constructing Non-Trivial Automorphisms of Semilattices

Let $\mathrm{\Omega }$ be a semilattice and $F_{\alpha }$ denote a set of non-comparable elements of $\mathrm{\Omega }$ such that each pair of elements of $F_{\alpha }$ has to meet $\alpha$. Let ${ℱ}={\{F_{\alpha }\}}_{\alpha \in \mathrm{\Omega }}$ and $F_{\tau }\in {ℱ}$ satisfy the following two conditions:

(i)	for any $\alpha \notin F_{\tau }$ if $\eta \le \alpha$ or ($\alpha \le \eta$) for some $\eta \in F_{\tau }$, then $\delta \le \alpha$ or ($\alpha \le \delta$) for all $\delta \in F_{\tau }$;
(ii)	for any $\alpha \notin F_{\tau }$, if $\alpha \wedge \beta =\gamma$ for some $\beta \in F_{\tau }$, then $\alpha \wedge \delta =\gamma$ for all $\delta \in F_{\tau }$.

Now we have the theorem which provides a construction for non-trivial automorphisms of semilattices.

Theorem 4.1. Let $\mathrm{\Omega }$, ${ℱ}={\{F_{\alpha }\}}_{\alpha \in \mathrm{\Omega }}$ be as above and $F_{\tau }\in {ℱ}$ satisfies conditions (i) and (ii). Any mapping $\xi$ on $\mathrm{\Omega }$, which fixes every element of $\mathrm{\Omega }\setminus F_{\tau }$ and permutes bijectively the elements of $F_{\tau }$ is a semilattice automorphism of $\mathrm{\Omega }$.

Proof. We only need to show that $\xi$ is a semilattice homomorphism. Take any ${\alpha }_1,{\alpha }_2\in \mathrm{\Omega }$, the following cases arise:

Case (i). ${\alpha }_1,{\alpha }_2\notin F_{\tau }$. Then we claim that ${\alpha }_1{\alpha }_2\notin F_{\tau }$. For if ${\alpha }_1{\alpha }_2\in F_{\tau }$, then ${\alpha }_1{\alpha }_2=\lambda$ for some $\lambda \in F_{\tau }$; that is, $\lambda \le {\alpha }_1$ and $\lambda \le {\alpha }_2$. Now by condition (i), we have $x\le {\alpha }_1$ and $x\le {\alpha }_2$ for all $x\in F_{\tau }$. As ${\alpha }_1{\alpha }_2=\lambda$, we have $x\le \lambda$ for all $x\in F_{\tau }$. This is a contradiction to the choice of $F_{\tau }$, hence ${\alpha }_1{\alpha }_2\notin F_{\tau }$. Now

as required.

Case (ii). ${\alpha }_1,{\alpha }_2\in F_{\tau }$. By the definition of $\xi$, we have $\xi ({\alpha }_1),\xi ({\alpha }_2)\in F_{\tau }$ and by the choice of $F_{\tau }$, we have ${\alpha }_1{\alpha }_2=\tau \notin F_{\tau }$. Thus, $\xi ({\alpha }_1{\alpha }_2)=\xi (\tau )=\tau$ and therefore $\xi ({\alpha }_1)\xi ({\alpha }_2)=\tau =\xi ({\alpha }_1{\alpha }_2)$. Hence, $\xi$ is a homomorphism.

Case (iii). One of ${\alpha }_1$ or ${\alpha }_2$ is in $F_{\tau }$. Without loss assume that ${\alpha }_1\notin F_{\tau }$ and ${\alpha }_2\in F_{\tau }$. Then, we have the following two subcases:

(a)	either ${\alpha }_1\le {\alpha }_2$ or ${\alpha }_2\le {\alpha }_1$; and
(b)	${\alpha }_1$ and ${\alpha }_2$ are not comparable and ${\alpha }_1\wedge {\alpha }_2=\eta (\eta \ne \tau )$.

Subcase (a). If ${\alpha }_1\le {\alpha }_2$, then ${\alpha }_1\wedge {\alpha }_2={\alpha }_1$. So $\xi ({\alpha }_1\wedge {\alpha }_2)=\xi ({\alpha }_1)={\alpha }_1$. Also $\xi ({\alpha }_1)\wedge \xi ({\alpha }_2)={\alpha }_1\wedge y$, where $y=\xi ({\alpha }_2)$ and $y\in F_{\tau }$. Therefore, by Condition (ii), we have ${\alpha }_1\wedge y={\alpha }_1$. Thus $\xi ({\alpha }_1\wedge {\alpha }_2)=\xi ({\alpha }_1)={\alpha }_1={\alpha }_1\wedge y=\xi ({\alpha }_1)\wedge \xi ({\alpha }_2)$, as required.

Again, when ${\alpha }_2\le {\alpha }_1$, we have $\xi ({\alpha }_1\wedge {\alpha }_2)=\xi ({\alpha }_2)=y$, where $\xi ({\alpha }_2)=y$ and $y\in F_{\tau }$. Again, as ${\alpha }_1\wedge {\alpha }_2={\alpha }_2$, we have ${\alpha }_2\le {\alpha }_1$ whence, by Condition (i), $y\le {\alpha }_1$. So ${\alpha }_1\wedge y=y$. Thus $\xi ({\alpha }_1)\wedge \xi ({\alpha }_2)={\alpha }_1\wedge y=y$. Now $\xi ({\alpha }_1\wedge {\alpha }_2)=\xi ({\alpha }_2)=y=\xi ({\alpha }_1)\wedge \xi ({\alpha }_2)$, as required.

Subcase (b). First we show that $\eta \notin F_{\tau }$. Suppose to the contrary that $\eta \in F_{\tau }$. As ${\alpha }_1\wedge {\alpha }_2=\eta$, by Condition (ii), we have ${\alpha }_1\wedge x=\eta$ for all $x\in F_{\tau }$. This implies $\eta \le x$ for all $x\in F_{\tau }$, a contradiction to the fact that no pair of members of $F_{\tau }$ is comparable. Thus, $\eta \notin F_{\tau }$. Now $\xi ({\alpha }_1\wedge {\alpha }_2)=\xi (\eta )=\eta$. Also $\xi ({\alpha }_1)\wedge \xi ({\alpha }_2)={\alpha }_1\wedge y$, where $\xi ({\alpha }_2)=y\in F_{\tau }$. As ${\alpha }_1\wedge {\alpha }_2=\eta$, by Condition (ii), we have ${\alpha }_1\wedge y=\eta$. Thus, $\xi ({\alpha }_1\wedge {\alpha }_2)=\xi (\eta )=\eta ={\alpha }_1\wedge y=\xi ({\alpha }_1)\wedge \xi ({\alpha }_2)$, as required.

Hence, in all cases, $\xi$ is a homomorphism and therefore $\xi \in \mathrm{\text{Aut}}(\mathrm{\Omega })$. □

We hasten to note that if ${ℱ}$ has only one element satisfying conditions (i) and (ii), then we have the following corollary which provides the number of such automorphisms of a semilattice.

Corollary 4.2. Let $\mathrm{\Omega }$, ${ℱ}={\{F_{\alpha }\}}_{\alpha \in \mathrm{\Omega }}$ be as in the above theorem, if there exists unique $F_{\tau }\in {ℱ}$ satisfying conditions (i) and (ii) with $|F_{\tau }|=n$, then $|\mathrm{\text{Aut}}(\mathrm{\Omega })|=n!$.

Proof. Since any $\xi \in \mathrm{\text{Aut}}(\mathrm{\Omega })$ fixes every element of $\mathrm{\Omega }\setminus F_{\tau }$ and permutes bijectively the elements of $F_{\tau }$, so $\xi$ has $n!$ possible choices. Therefore, we have $|\mathrm{\text{Aut}}(\mathrm{\Omega })|=n!$, as required. □

Next we illustrate the above corollary with some particular examples.

Example 4.1. Let $\mathrm{\Omega }=\{\alpha ,\lambda ,𝜖,\delta ,\rho ,\chi ,\beta ,o\}$ be a semilattice defined in Fig. 1.

We have that $F_{\beta }=\{\lambda ,𝜖,\delta \}$ and $F_o=\{\rho ,\chi \}$ are the elements of the family ${ℱ}$. Also the element $F_{\beta }$ of ${ℱ}$ satisfies the conditions of Theorem 4.1, and so we have the map $\xi :\mathrm{\Omega }\to \mathrm{\Omega }$, defined by $\xi (\alpha )=\alpha$ for $\alpha \notin F_{\beta }$ and $\xi {|}_{F_{\beta }}$ is a permutation on $F_{\beta }$, is an automorphism of $\mathrm{\Omega }$ and by Corollary 4.2, $|\mathrm{\text{Aut}}(\mathrm{\Omega })|=6$.

Now we provide an example to validate the Theorems 3.1, 3.3 and 3.4.

Example 4.2. Let $\mathrm{\Omega }=\{0,1,2,3,4\}$ be the semilattice defined in Fig. 2.

Then $\{1,2,3\}$ is the subset of $\mathrm{\Omega }$ in which elements are not comparable to each other with meet 0. Let $S=[\mathrm{\Omega },{\{S_{\alpha }\}}_{\alpha \in \mathrm{\Omega }},{\{{\varphi }_{\alpha ,\beta }\}}_{\alpha \ge \beta }]$ be a strong semilattice of groups with all structure morphisms being bijective. Then, by Corollary 4.2, we have $|\mathrm{\text{Aut}}(\mathrm{\Omega })|=3!=6$. Suppose that $\mathrm{\text{Aut}}(\mathrm{\Omega })=\{{\mathrm{\Theta }}_{L_1},{\mathrm{\Theta }}_{L_2},\dots ,{\mathrm{\Theta }}_{L_6}\}$. Thus for every $\mathrm{\Theta }\in \mathrm{\text{Aut}}(S)$, we may consider the following correspondence:

Now consider the product of elements in $\mathrm{\text{Aut}}(S)$. Let $\lambda ,\mu \in \mathrm{\text{Aut}}(S)$ and $s\in S$. Then $s\in S_{\alpha }$ for some $\alpha \in \mathrm{\Omega }$. For instance, if $\alpha =2$, then $s\in S_2$. Now consider ${\mathrm{\Theta }}_{L_4}\in \mathrm{\text{Aut}}(\mathrm{\Omega })$ associated with $\lambda$ such that ${\mathrm{\Theta }}_{L_4}(2)=3$ and ${\mathrm{\Theta }}_{L_4}(3)=1$, say. Then $\mu \lambda (s)=\mu ({\lambda }_2(s))={\mu }_{{\mathrm{\Theta }}_{L_4}(2)}({\lambda }_2(s))={\mu }_3{\lambda }_2(s)$. Now consider a group isomorphism ${\mathrm{\Theta }}_2$, say, where ${\mathrm{\Theta }}_2:S_2\to S_{{\mathrm{\Theta }}_{L_4}(2)}$. Then ${\mathrm{\Theta }}_2={\varphi }_{1,3}{\mathrm{\Theta }}_3{\varphi }_{3,2}^{-1}$. Thus ${\mathrm{\Theta }}_2$ is determined by ${\mathrm{\Theta }}_3$. In fact, by Theorem 3.1, any group isomorphism of $S_{\alpha }$ $(\alpha \in \mathrm{\Omega })$ can be determined by ${\mathrm{\Theta }}_0$, where 0 is the least element in $\mathrm{\Omega }$. For instance, let us choose any group isomorphism ${\mathrm{\Theta }}_1$, say, associated with an automorphism $\mathrm{\Theta }$ of S and ${\mathrm{\Theta }}_{L_j}$ for some $j\in \{1,2,\dots ,6\}$. Suppose ${\mathrm{\Theta }}_{L_j}(1)=3$ and ${\mathrm{\Theta }}_{L_j}(0)=0$. Then ${\mathrm{\Theta }}_1$ can be determined by ${\mathrm{\Theta }}_0$ as Therefore, ${\mathrm{\Theta }}_0$ determines any group isomorphism associated with the automorphism $\mathrm{\Theta }$ of S.

Now we validate Theorem 3.3 by showing that any group isomorphism of $S_{\alpha }$ $(\alpha \in \mathrm{\Omega })$ can be determined by ${\mathrm{\Theta }}_4$, where 4 is the maximum element in $\mathrm{\Omega }$. For instance, let us choose any group isomorphism ${\mathrm{\Theta }}_3$, say, associated with an automorphism $\mathrm{\Theta }$ of S and ${\mathrm{\Theta }}_{L_k}$ for some $k\in \{1,2,\dots ,6\}$. Suppose ${\mathrm{\Theta }}_{L_k}(3)=2$ and ${\mathrm{\Theta }}_{L_k}(4)=4$. Then ${\mathrm{\Theta }}_3$ can be determined by ${\mathrm{\Theta }}_4$ as

Therefore, ${\mathrm{\Theta }}_4$ determines any group isomorphism associated with the automorphism $\mathrm{\Theta }$ of S.

Again, by Theorem 3.4, any group isomorphism of $S_{\alpha }$ $(\alpha \in \mathrm{\Omega })$ can be determined by ${\mathrm{\Theta }}_{\beta }$, where $\alpha ,\beta \in \mathrm{\Omega }$. For instance, let us choose $\alpha =2$ and $\beta =3$. Then $2\wedge 3=0$. Let ${\mathrm{\Theta }}_{\alpha }$ is associated with an automorphism $\mathrm{\Theta }$ of S and ${\mathrm{\Theta }}_{L_p}$ for some $p\in \{1,2,\dots ,6\}$. Suppose ${\mathrm{\Theta }}_{L_p}(2)=1$, ${\mathrm{\Theta }}_{L_p}(3)=2$ and ${\mathrm{\Theta }}_{L_p}(0)=0$. Then, ${\mathrm{\Theta }}_2$ can be determined by ${\mathrm{\Theta }}_3$ as

Therefore, ${\mathrm{\Theta }}_3$ determines any group isomorphism associated with the automorphism $\mathrm{\Theta }$ of S.