Introduction

Identifying a finite non-group semigroup (commutative or non-commutative) of order of a given positive integer n could be significant because of the interesting applications of finite semigroups in computer science, mathematics and finite machines. As a subclass of non-group semigroups, the non-commutative monoids of a given order were identified in 2014 by Ahmadi et al. [2] to show that for every positive integer \(n\ge 4,\) there exists a non-commutative monoid of order n. In this paper, we intend to identify two types of finite semigroups, commutative non-monoids of order \(p^\alpha q^\beta\) and commutative monoids of order of a given positive integer \(n\ge 4\). By giving the unique minimal generating set for the commutative non-monoids, we specify the number of all non-isomorphic semigroups of this type. Subsequently, we present a class of finitely presented non-isomorphic commutative monoids of a given positive integer n.

Our notation is merely standard, and we follow [3,4,5,6,7]. The detailed information on semigroup (or monoid) presentations may be found in [3, 4, 7]. These articles investigate the efficiency of finitely presented semigroups and the related monoids. We prefer to give a brief history on the finitely presented semigroups and monoids. A semigroup (or monoid) S is said to be presented by a semigroup (or monoid) presentation \(\langle A\mid R\rangle\) if \(S\cong A^{+}/\rho\) (or \(S\cong A^{*}/\rho\)) where A is an alphabet, \(A^{+}\) is the free semigroup on A, \(A^{*}=A\cup \{1\}\), \(\rho\) is a congruence on A (or \(A^{*}\)) generated by R, and \(R\subseteq A^{+}\times A^{+}\) (or \(R\subseteq A^{*}\times A^{*}\)). As usual, we will use the notation \(Sg(\pi )\) for the semigroup presented by the presentation \(\pi =\langle A\mid R\rangle\). Also, the deficiency of a finite semigroup presentation \(\pi =\langle A\mid R\rangle\) is defined to be \(\mid R\mid -\mid A\mid\) and is denoted by \(\hbox {def}(\pi )\). The semigroup deficiency \(\hbox {def}_s(S)\) of a finitely presented semigroup S is given by

$$\begin{aligned} \hbox {def}_s(S)=\min \{\hbox {def}(\pi ) \mid \pi \,\, \hbox {is}\,\, \hbox {a}\,\, \hbox {finite}\,\, \hbox {presentation}\,\, \hbox {for}\,\, S\} \end{aligned}$$

as well.

For every positive integers m, r and s, we define the presentations:

$$\begin{aligned} \begin{array}{ll} \pi _{m,r}=\langle a\mid a^{m+r}=a^r\rangle , &{} (m\ge 2, r\ge 1),\\ \pi '_{r,s}=\langle a,b\mid a^{r+1}=a^r, b^{s+1}=b^s, ab=ba=a\rangle , &{} (r,s\ge 2), \end{array} \end{aligned}$$

and let \(S_{m,r}=Sg(\pi _{m,r})\). Obviously, the semigroup \(S_{m,1}\) is the cyclic group of order \(m-1\) ; however, \(S_{m,r}\) is a non-monoid monogenic semigroup of order \(m+r-1\), for every positive integer \(r\ge 2\). These last monogenic semigroups will be used in construction of the commutative non-monoid semigroups. By using the well-known definition of direct product of semigroups [8], we may easily check that the direct product of two non-monoid monogenic semigroups is also a non-monoid semigroup. Note that, the direct product of semigroups \(S_{m,1}\) and \(S_{m',1}\), as studied in [1], is a 2-generated semigroup. However, the situation is completely different in the direct product of non-monoid monogenic semigroups as we will see in the following sections.

Main results

As a preliminary result, we have the following lemma on the monogenic semigroups.

Lemma 2.1

Let\(\alpha \ge 2\)be an integer, andpbe a prime. There are exactly\(p^\alpha -2\)number of non-isomorphic non-monoid monogenic semigroups of order\(p^\alpha\).

Proof

For every values of \(r\ge 2\), \(S_{m,r}=Sg(\pi _{m,r})\) is a non-monoid monogenic semigroup. Then, the equation \(m+r-1=p^\alpha\) gives us all of the possible values for m and r satisfying this equation. The number of partitions of \(p^\alpha +1\) as a summation of two positive integers m and r such that \(m,r\ge 2\), is equal to \(t=p^\alpha -2\). For two different partitions \(m+r=m'+r'=p^\alpha +1,\) consider the semigroups \(S_{m,r}\) and \(S_{m',r'}\). Evidently, \(m\ne m'\) and \(r\ne r'\). Hence \(S_{m,r}\) and \(S_{m',r'}\) are non-isomorphic semigroups because of their non-isomorphic maximal subgroups of orders m and \(m'\), respectively. Consequently, t is the number of non-isomorphic non-monoid monogenic semigroup of order \(p^\alpha\). \(\square\)

Theorem 2.2

There are exactly\((p^{\alpha }-2)(q^{\beta }-2)\)number of non-isomorphic semigroups of order\(n=p^\alpha q^\beta\)which are commutative and non-monoid. Moreover, each of these semigroups possesses the unique minimal generating set of the cardinality\(p^{\alpha }+q^{\beta }-1\).

Proof

Let \(T_1=S_{m,r}\) and \(T_2=S_{m',r'}\) be two non-monoids of orders \(p^\alpha\) and \(q^\beta\), respectively. So, \(m+r=p^\alpha +1\) and \(m'+r'=q^\beta +1\). Let \(S=T_1\times T_2\) be the direct product of \(T_1\) and \(T_2\). Every element of S is an ordered pair \((a^i,b^j),\) where \(1\le i\le p^\alpha\) and \(1\le j\le q^\beta\). Since \(T_1\) and \(T_2\) are commutative non-monoid semigroups, it may be easily checked that S is also a commutative non-monoid semigroup. Then, by the results of Lemma 2.1 and the definition of S as above, we conclude that there are exactly \((p^{\alpha }-2)(q^{\beta }-2)\) number of commutative non-monoid semigroups of order \(p^{\alpha }q^{\beta }\). To complete the proof, we have to give the minimal generating set for S. Without lose of generality, suppose that \(p^\alpha < q^\beta\). First of all by considering the elements

$$\begin{aligned} \begin{array}{ll} x=(a,b),&{}\\ A_i=(a^i,b),&{} (i=2,3,\dots , p^\alpha ),\\ B_j=(a,b^j),&{} (j=2,3,\dots , q^\beta ), \end{array} \end{aligned}$$

of S, we may easily check that every element \(c_{ij}=(a^i,b^j)\) of S may be presented as:

$$\begin{aligned} c_{ij}=\left\{ \begin{array}{ll} x^i,&{} {\text {if}}\;i=j,\\ x^{i-1}B_{j-i+1},&{} {\text {if}}\;i<j,\\ x^{j-1}A_{i-j+1},&{} {\text {if}}\;i>j, \end{array} \right. \end{aligned}$$

where \(2\le i\le p^\alpha\) and \(2\le j\le q^\beta\). Consequently, the subset

$$\begin{aligned} X=\{x\}\cup \{A_i\mid i=2,\dots , p^\alpha \}\cup \{B_j\mid j=2,\dots , q^\beta \} \end{aligned}$$

of cardinality \(p^\alpha +q^\beta -1\) is a generating set for S. To prove that X is the unique minimal generating set for S, it is sufficient to show that \(S\backslash S^2=X\), where \(S^2=\{yz\mid y,z\in S\}\). Remind the relators \(a^{p^{\alpha }+1}=a^r\) and \(b^{q^{\beta }+1}=b^s\) where \(r,s\ge 2,\) then, any product yz of the elements of X belongs to \(S\backslash X\). Also, any product yz of the elements of \(S\backslash X\) belongs to \(S\backslash X\). Consequently, \(S\backslash S^2=X\). \(\square\)

To identify the commutative monoids of a given integer \(n\ge 4,\) we show that the disjoint union of two finite cyclic groups \(H=\langle a\rangle\) and \(K=\langle b\rangle\) becomes a commutative non-group monoid if the multiplication is defined as \(ab=a=ba\).

Theorem 2.3

For a given integer\(n\ge 4\), the semigroup\(Sg(\pi '_{r,s})\)is a commutative monoid of ordern, for all values of the integers\(r, s\ge 2\), where\(n=r+s\). Moreover, there are exactly\(n-3\)non-isomorphic commutative monoids of ordern.

Proof

Let \(n=r+s\). Then, the set of elements of the semigroup \(Sg(\pi _{\{r,s\}})\) may be decomposed into the union of two groups as

$$\begin{aligned} \{a^i\mid i=1,2,\dots , r\}\cup \{b^j\mid j=1,2,\dots , s\}, \end{aligned}$$

because of the relators \(ab=ba=a\). So, \(Sg(\pi '_{r,s})\) is of order n. The element \(e=b^{s}\) is the identity element of \(Sg(\pi _{\{r,s\}})\) because of the relators

$$\begin{aligned} e.a^i=b^{s-1}.(ba).a^{i-1}=b^{s-1}.a.a^{i-1}=b^{s-1}.a^{i}=\dots =ba^i=(ba).a^{i-1}=a^i, \end{aligned}$$

for every \(i=1,2,\dots , r\). So, \(Sg(\pi '_{r,s})\) is a commutative monoid. Although \(Sg(\pi '_{r,s})\) is the disjoint union of two cyclic groups generated by a and b, it is not a group, because the element a has not the group inverse in \(Sg(\pi '_{r,s})\).

The number of different pairs (r, s) of the integers \(r, s\ge 2\) satisfying the equation \(n=r+s\), is equal to \(n-3\). Also, it is not very hard to see that the semigroups \(Sg(\pi '_{r,s})\) and \(Sg(\pi '_{r',s'})\) are non-isomorphic for any two different pairs (r, s) and \((r',s')\). Hence there are exactly \(n-3\) non-isomorphic commutative monoids of order n. \(\square\)

Non-isomorphic semigroups

Two special cases of non-group commutative semigroups of orders \(2p^\alpha\) and \(p^{2\alpha }\) are of interest to consider, for every odd prime p and every integer \(\alpha \ge 2\). Our results on the identification of semigroups are collected in the following theorems.

Theorem 3.1

There are exactly\(2p^\alpha\)non-isomorphic commutative non-monoid monogenic semigroups of order\(2p^\alpha\), for every odd prime p and every integer\(\alpha \ge 2\).

Proof

A similar method as in Lemma 2.1 may be used here to get the number of non-isomorphic semigroups. \(\square\)

Theorem 3.2

There are exactly\(p^\alpha -2\)non-isomorphic commutative non-monoid semigroups of order\(p^{2\alpha }\), for every integer\(\alpha \ge 2\)and every odd primep. Moreover, each semigroup possesses the unique minimal generating set of cardinality\(2p^\alpha -1\).

Proof

A commutative non-monoid semigroup of order \(p^{2\alpha }\) is indeed the direct product \(S_{m,r}\times S_{m,r},\) where \(m+r=p^\alpha +1\) and \(r\ge 2\). The first part may be verified in a similar method as in Theorem 2.3. Also, the subset

$$\begin{aligned} X=\{x,A_2,\dots ,A_{p^\alpha },B_2,\dots ,B_{p^\alpha }\} \end{aligned}$$

is the unique minimal generating set for each semigroup \(S_{m,r}\times S_{m,r}\) if \(m+r=p^\alpha +1\) and \(r=2,3,\dots ,p^\alpha -1\). Obviously, \(\mid X\mid =2p^\alpha -1\). \(\square\)

Remark 3.3

The proof of Lemma 2.1 may be used to identify the commutative non-monoid semigroups of order \(p^2\). Indeed, we deduce that there exists exactly \(p-2\) commutative non-monoid semigroups of order \(p^2\), for every odd prime p. This is a substantial difference between the groups and semigroups of order \(p^2\).

Conclusion

As an example we will give an efficient presentation for the commutative non-monoid semigroup of order \(3^2\). Before presenting this example, we have a look at the automorphism groups of the commutative non-monoid semigroups in the following theorem.

Theorem 4.1

For every integer\(\alpha \ge 2\)and every odd primep, the following statements hold:

(i):

The automorphism group of every non-monoid monogenic semigroup is the trivial group.

(ii):

The automorphism group of every commutative non-monoid semigroup of order\(p^{2\alpha }\)is a non-trivial group.

(iii):

For a given integer\(n\ge 4\)and the positive integersrandsas in Theorem 2.3, the automorphism group of the semigroup\(Sg(\pi '_{r,s})\)is a non-trivial group of order at least\(\phi (r)\phi (s)\), where\(\phi\)is the well-known Eulerian function.

Proof

To prove (i), observe that any automorphism \(\theta \in Aut(S_{m,r})\), where \(r\ge 2\), has to map a generator to a generator. Since \(S_{m,r}\) possesses a unique generating element, \(\theta\) is the identity map.

For (ii), let \(S=S_{m,r}\times S_{m,r}\) where \(m+r-1=p^\alpha\). We may define the involution automorphism \(\theta \in Aut(S)\) by \(\theta (a^i,b^j)=(a^j,b^i)\), where \(1\le i, j\le p^{\alpha }+1\). By considering the generating set

$$\begin{aligned} X=\{x\}\cup \{A_2,A_3,\dots , A_{p^\alpha }\}\cup \{B_2, B_3,\dots , B_{p^\alpha }\} \end{aligned},$$

we get \(\theta (x)=x\), \(\theta (A_i)=B_i\) and \(\theta (B_i)=A_i\), for every \(i=2,3,\dots ,p^\alpha\). Hence, we have to verify the relators:

$$\begin{aligned} \begin{aligned} \theta (A_iA_j)&=\theta (A_i)\theta (A_j),\\ \theta (B_iB_j)&=\theta (B_i)\theta (B_j),\\ \theta (A_iB_j)&=\theta (A_i)\theta (B_j),\\ \theta (xA_i)&=x\theta (A_i),\\ \theta (xB_i)&=x\theta (B_i). \end{aligned} \end{aligned}$$

The proofs are straightforward because of the commutativity of S and the key relators \(A_iA_j=xA_{i+j-1}\), \(B_iB_j=xB_{i+j-1}\) and

$$\begin{aligned} A_iB_j=\left\{ \begin{array}{ll} x^{j}A_{i-j+1},&{}{\text {if}}\;i>j,\\ x^{i}B_{j-i+1},&{}{\text {if}}\;i<j,\\ x^{i+1},&{}{\text {if}}\;i=j. \end{array}\right. \end{aligned}$$

Finally, consider the subgroups \(H=\{a, a^2,\dots , a^r\}\) and \(K=\{b, b^2,\dots , b^s\}\) of the semigroup \(Sg(\pi '_{\{r,s\}})\). For any \(\theta \in Aut(Sg(\pi '_{r,s}))\), the restrictions \(\theta \mid _{H}\) and \(\theta \mid _{K}\) are automorphisms of H and K, respectively. As a well-known result of the cyclic groups, there are exactly \(\phi (r)\) number of generating elements for H, and then, \(\mid Aut(H)\mid =\phi (r)\). Similarly, \(\mid Aut(K)\mid =\phi (s)\). Since the cross-product of two automorphisms is an automorphism, (iii) follows at once. \(\square\)

Obtaining a finite presentation for the non-group commutative semigroups of order \(n=p^\alpha q^\beta\) requires too long and tedious hand calculation. As an example, we give here a finite presentation for the non-group commutative non-monoid semigroup of order \(p^2\) where \(p=3\). Let \(T=Sg(S_{2,2}\times S_{2,2})\). Then, T is of order 9 and possesses an efficient presentation as in the following example.

Example 4.2

The semigroup T may be presented as \(\langle X\mid R\rangle\) where \(X=\{x,\alpha _1, \alpha _2, \alpha _3, \alpha _4\}\) and R is the set of 29 relators:

$$\begin{aligned} \begin{array}{ll} x^4=x^2,\,\,,\alpha _i^4=\alpha _i^2=x\alpha _i&{} (1\le i\le 4),\\ x\alpha _i=\alpha _ix,\,\,\alpha _i\alpha _j=\alpha _j\alpha _i&{}(1\le i<j\le 4),\\ x\alpha _1=\alpha _1\alpha _2,\,\,x\alpha _3=\alpha _3\alpha _4,\,\,x\alpha _2=x^2=x\alpha _4,&{}\\ \alpha _1\alpha _3=x^3,\,\,\alpha _2\alpha _4=x^2,&{}\\ x\alpha _1\alpha _4=\alpha _2\alpha _3=x\alpha _1\alpha _2,&{}\\ x\alpha _2\alpha _3=\alpha _1\alpha _4= x\alpha _3\alpha _4.&{} \end{array} \end{aligned}$$

To verify these relators, one may use the original definition of the minimal generating set X. Indeed, by letting \(\alpha _1=A_2=(a^2,b)\), \(\alpha _2=A_3=(a^3,b)\), \(\alpha _3=B_2=(a,b^2)\), \(\alpha _4=B_3=(a,b^3)\), all of the products of two elements of X could be calculated by using Lemma 2.1. Then, the redundant relators have to be deleted. By using Gap code [9], we now believe that this presentation is an efficient semigroup presentation for T.

This example along with our computations results the following conjecture on the non-group commutative non-monoid semigroups of order \(p^2\).

Conjecture 4.3

For every odd primep, there are\(p-2\)number of non-group commutative non-monoid semigroups of order\(p^2\). Each semigroup of this type has an efficient presentation of semigroup deficiency\(4p(p-1)\).

The first part of this conjecture may be verified in a similar method as in Theorem 3.2. For \(p=3,\) the above example verifies the conjecture and shows that \(\hbox {def}(T)=29-5=24=4\times 3\times 2\). Also, for \(p=5,\) we examined the semigroup \(T_1=Sg(S_{4,2}\times S_{4,2})\) of order \(p^2\) with the generating set \(X=\{x,\alpha _1, \alpha _2, \alpha _3, \alpha _4, \alpha _5, \alpha _6, \alpha _7, \alpha _8\}\). As well as in the above example, we managed to get an efficient presentation with 89 relators. This verifies the conjecture showing that \(\hbox {def}(T_1)=89-9=80=4\times 5\times 4\).