A note on isomorphisms between submonoids of $\mathbb N^k$ and numerical semigroups

TL;DR

This paper characterizes when submonoids of $\\mathbb{N}^k$ are isomorphic to submonoids of lower dimensions and proves the uniqueness of isomorphisms for numerical semigroups within $\\mathbb{N}$.

Contribution

It provides new characterizations of the minimal dimension for isomorphic embeddings of submonoids and establishes the uniqueness of isomorphisms for numerical semigroups.

Findings

Characterization of minimal $r$ such that $H$ is isomorphic to a submonoid of $\\mathbb{N}^r$

Proof that isomorphic numerical semigroups in $\\mathbb{N}$ are identical

Uniqueness of the isomorphism as the identity map for numerical semigroups

Abstract

Given a submonoid $H$ of $\mathbb N^k$, we give some characterizations of the minimum $r\in \mathbb N^+$ such that $H$ is isomorphic to a submonoid of $\mathbb N^r$. In the context of submonoids of $\mathbb N$, we prove that if two numerical semigroups are isomorphic submonoids of $\mathbb N$, then they are equal and the identity map is the unique isomorphism between them.

Equations

$$indH=\min \{r\in {\mathbb{N}}^{+}:H\cong K\text{ for some r-monoid }K\}.$$

$$freeH=max\{dimF:F\subseteq H\text{ and F is free}\}.$$

$$indH=|X|=freeH,$$

$$M^H=\left(\begin{array}{cccc}{h}_1(1)& h_2(1)& \cdots & h_r(1)\\ h_1(2)& h_2(2)& \cdots & h_r(2)\\ ⋮& ⋮& \ddots & ⋮\\ h_1(k)& h_2(k)& \cdots & h_r(k)\end{array}\right)$$

$$indH=rank{M}^H,$$

$$\sum _{i\in X}{c}_i{\alpha }_i^H(h)=\sum _{i\in X}{c}_i\left(\sum _{g\in \beta (H)}{d}_{g}g(i)\right)=\sum _{g\in \beta (H)}{d}_g\left(\sum _{i\in X}{c}_{i}g(i)\right)=0.$$

$$0={\varphi }^{\ast }(\sum _{j=1}^r{c}_j{\alpha }_j^{H|X})=\sum _{j=1}^r{c}_j{\varphi }^{\ast }({\alpha }_j^{H|X})=\sum _{j=1}^r{c}_j{\alpha }_{a_j}^H,$$

$${\alpha }_j^{F_n}={\alpha }_j^H{|}_{F_n}=\sum _{i\in X}{c}_i{\alpha }_i^H{|}_{F_n}=\sum _{i\in X}{c}_i{\alpha }_i^{F_n}.$$

$${\alpha }_j^{F_n}={\alpha }_i^{F_{n+1}}{|}_{F_n}=\sum _{i\in X_N}{c}_{i,n+1}{\alpha }_i^{F_{n+1}}{|}_{F_n}=\sum _{i\in X_N}{c}_{i,n+1}{\alpha }_i^{F_n},$$

$${\alpha }_j^H(h)={\alpha }_j^{F_n}(h)=\sum _{i\in X_N}{c}_i{\alpha }_i^{F_n}(h)=\sum _{i\in X_N}{c}_i{\alpha }_i^H(h).$$

$$indH=|X|=freeH.$$

$$indH=rank{M}^H.$$

$$n_1{h}_1+\cdots +n_r{h}_r+nh=m_1{h}_1+\cdots +m_r{h}_r+mh.$$

$$h_1\cdot k_1=h_1\phi (h_0)=\phi (h_1{h}_0)=h_0\phi (h_1)=h_0{k}_0,$$

$$h_0\cdot k=h_0\phi (h)=\phi (h_{0}h)=h\phi (h_0)=(h_0+s(r))\cdot m_K.$$

$$s(r)=\frac{h_0}{m_K}r$$

$${\phi }^{-1}(\{m_K,m_K+1,m_K+2,\dots \})=\{h_0+cr:r\ge 0\}.$$

Full text

1. Introduction

The symbol $\mathbb{N}$ stands for the set $\{0,1,2,\ldots,\}$. Elements in $\mathbb{N}^{k}$ will be regarded as functions $h:\{1,2,\ldots,k\}\to\mathbb{N}$. Sometimes it is more convenient to see $h\in\mathbb{N}^{k}$ as the $k$-tuple $(h(1),h(2),\ldots,h(k))$, and we also write $h=(h(1),h(2),\ldots,h(k))$ for simplicity.

The set $\mathbb{N}^{k}$ with the usual sum is a monoid with identity element $(0,0,\ldots,0)$ that we will denote by $0_{k}$.

For simplicity we will call a submonoid of $\mathbb{N}^{k}$ a $k$-monoid.

One interesting property of every $k$-monoid $H$ is that $H$ has a minimal generating set that we denote $\beta(H)$. This set $\beta(H)$ has the property that if $X\subseteq H$ generates $H$, then $\beta(H)\subseteq X$. Moreover, $\beta(H)=H^{*}\setminus(H^{*}+H^{*})$, where $H^{*}=H\setminus\{0_{k}\}$.

When $H$ is finitely generated, we can define the dimension of $H$, $dim\ H$, as the cardinality of the set $\beta(H)$, $dim\ H=|\beta(H)|$.

If $H$ is a $k$-monoid and $F$ is an $l$-monoid, then a function $\varphi:H\to F$ is additive if it satisfies $\varphi(h_{1}+h_{2})=\varphi(h_{1})+\varphi(h_{2})$ for all $h_{1},h_{2}\in H$. If $\varphi:H\to F$ is additive, then $\varphi(0_{k})=0_{l}$. An ismomorphism between $H$ and $F$ is a bijective additive function $\varphi:H\to F$.

A result of J. C. Rosales [2] establishes that a finitely generated commutative monoid is isomorphic to $k$-monoid if and only if it is cancellative, torsion free and has no other units than zero. There exist $k$-monoids that are not finitely generated. For instance, for $k>1$, the set $\{h\in\mathbb{N}^{k}:1\leq h(1)\leq h(2)\leq\cdots\leq h(k)\}\cup\{0_{k}\}$ is a $k$-monoid that is not finitely generated. If $k=1$, every 1-monoid is finitely generated, and furthermore every nontrivial 1-monoid is isomorphic to a numerical semigroup (see [1]).

To state our main results we make some comments and introduce terminology.

Let $H$ be a $k$-monoid. Then for every $r>k$ there exists an $r$-monoid that is isomorphic to $H$. In fact, for any $r\in\mathbb{N}^{+}$ consider the map $\iota_{r}:\mathbb{N}^{r}\to\mathbb{N}^{r+1}$ given by $\iota_{r}(h)=(0,h(1),\ldots,h(r))$ for all $h\in\mathbb{N}^{r}$. This map $\iota_{r}$ is a monomorphism. So, if $r>k$, then $H\cong\iota_{r-1}(\cdots\iota_{k}(H)\cdots)$. Thus, the question of interest is to determine when there exists an $r$-monoid isomorphic to $H$ if $r<k$.

We associate to $H$ the set \mathcal{I}(H):=\{r\in\mathbb{N}^{+}:H\cong K\text{ for some r-monoid }K\}. We define the index of $H$, $ind\ H$, as the minimum element of $\mathcal{I}(H)$:

[TABLE]

Given a nonempty subset $A$ of $\mathbb{N}^{k}$, let $\mathbb{Q}^{A}$ be the $\mathbb{Q}$-vector space of all functions $\alpha:A\to\mathbb{Q}$ with the usual sum and scalar product. For each $i\in I_{k}$ we define $\alpha_{i}^{A}:A\to\mathbb{Q}$ by $\alpha_{i}^{A}(f)=f(i)$ for all $f\in A$. We will denote by $vect\ A$ the subspace of $\mathbb{Q}^{A}$ spanned by the set $\{\alpha_{i}^{A}:i\in I_{k}\}$.

For any $k\in\mathbb{N}^{+}$, let $I_{k}=\{1,2,\ldots,k\}$. Let $A$ be a nonempty subset of $\mathbb{N}^{k}$. We say a nonempty subset $X$ of $I_{k}$ is independent over $A$ if $\{\alpha_{i}^{A}:i\in X\}$ is a linearly independent subset of $\mathbb{Q}^{A}$. If $\{\alpha_{i}^{A}:i\in X\}$ is linearly dependent in $\mathbb{Q}^{A}$, we say that $X$ is dependent over $A$.

We easily see that if $X\subseteq I_{k}$ is independent over $A$ and $\varnothing\neq Y\subseteq X$, then $Y$ is independent over $A$; also, if $X\neq\varnothing$ is dependent over $A$ and $X\subseteq Y\subseteq I_{k}$, then $Y$ is dependent over $A$. Besides, if $\varnothing\neq B\subseteq A\subseteq\mathbb{N}^{k}$ and $X$ is dependent over $A$, then $X$ is dependent over $B$, or equivalently, if $X$ is independent over $B$, then $X$ is independent over $A$.

We say that $X\subseteq I_{k}$ is maximal independent over $A$ if $X$ is independent over $A$ and there does not exist $Y\subseteq I_{k}$ independent over $A$ such that $X\subsetneq Y$. In other words, $X$ is maximal independent over $A$ if and only if $\{\alpha_{i}^{A}:i\in X\}$ is a basis of $vect\ A$. Note that if $A\neq\{0_{k}\}$ is nonempty, then there exists some nonempty set $X\subseteq I_{k}$ that is maximal independent over $A$, since there is some basis of $vect\ A$ contained in $\{\alpha_{i}^{A}:i\in I_{k}\}$.

For a $k$-monoid $H$ we define the index of free submonoids of $H$, denoted $free\ H$, as the maximal dimension of a free submonoid of $H$:

[TABLE]

One of our main results is that for any nontrivial $k$-monoid $H$, the equality

[TABLE]

holds, where $X\subseteq I_{k}$ is maximal independent over $H$.

When $H$ is a finitely generated $k$-monoid and $\beta(H)=\{h_{1},h_{2},\ldots,h_{r}\}$ where $r=dim\ H$, we associate to $H$ a $k\times r$ matrix $M^{H}$ that we will call the matrix of $H$ (that depends on the order of the elements in $\beta(H)$) given by

[TABLE]

In case $H$ is a finitely generated $k$-monoid, we also prove that

[TABLE]

where $rank\ M^{H}$ is the rank of the matrix $M^{H}$.

We will give some characterizations of $k$-monoids of index 1. Here, we come to the ambit of numerical semigroups. A numerical semigroup is a 1-monoid $H$ such that $\mathbb{N}\setminus H$ is finite. This condition is equivalent to say that $H$ is nontrivial and $\gcd H=1$. Our main result in this context is that if two numerical semigroups are isomorphic, then they are actually equal. This means that the relation of isomorphism on numerical semigroups is trivial.

2. Index of $k$-monoids

The following basic properties of $ind\ H$ are easy to prove.

Lemma 2.1**.**

The following statements are true.

1. (1)

If $H$ is $k$-monoid, then $ind\ H\leq k$. 2. (2)

$ind\ \mathbb{N}^{k}=k$. 3. (3)

$\mathcal{I}(H)=\{r\in\mathbb{Z}^{+}:r\geq ind\ H\}$. 4. (4)

If $H$ and $F$ are $k$-monoids and $F\subseteq H$, then $ind\ F\leq ind\ H$. 5. (5)

For any $f\in\mathbb{N}^{k}$, $ind\ \langle f\rangle=1$. 6. (6)

If $H\cong F$, then $ind\ H=ind\ F$.

Property (3) in Lemma 2.1 means that if $H$ is a $k$-monoid and $r\geq ind\ H$, then we can find an $r$-monoid $F$ isomorphic to $H$, but if $r<ind\ H$, there does not exist such $r$-monoid $F$. Thus, we can find $r$-monoids isomorphic to $H$ for some $r<k$ only if $ind\ H<k$. We will see how under certain conditions on $H$ and $1\leq r<k$ we can construct an $r$-monoid isomorphic to $H$ (Proposition 2.3 below).

Proposition 2.2**.**

If $H$ is a free and nontrivial $k$-monoid, then $ind\ H=dim\ H$.

Proof.

If $r=dim\ H$, then $H\cong\mathbb{N}^{r}$ and therefore $ind\ H=ind\ \mathbb{N}^{r}=r=dim\ H$. ∎

Let $I_{k}=\{1,2,\ldots,k\}$. For a subset $X=\{a_{1},a_{2},\ldots,a_{r}\}$ of $I_{k}$, where $a_{1}<a_{2}<\cdots<a_{r}$, let $\eta_{X}:I_{r}\to I_{k}$ be defined by $\eta_{X}(j)=a_{j}$, for $j\in I_{r}$. For $h\in\mathbb{N}^{k}$ we define $h|X:=h\circ\eta_{X}\in\mathbb{N}^{r}$; for $A\subseteq\mathbb{N}^{k}$ we define $A|X:=\{h|X:h\in A\}\subseteq\mathbb{N}^{r}$.

Proposition 2.3**.**

Let $H$ be a submonoid of $\mathbb{N}^{k}$ and $X\subseteq I_{k}$ nonempty. Assume that for each $j\in I_{k}\setminus X$, $\alpha_{j}^{H}\in span\ \{\alpha_{i}^{H}:i\in X\}$. Then the map $\phi:H\to H|X$ given by $\phi(h)=h|X$ for all $h\in H$ is an isomorphism.

Proof.

It is easy to see that $\phi$ is additive and surjective. By hypothesis, if $j\in I_{k}\setminus X$, then for each $i\in X$ there exists $c_{i,j}\in\mathbb{Q}$ such that $h(j)=\sum_{i\in X}c_{i,j}h(i)$ for all $h\in H$. If $g|X=h|X$ for $g,h\in H$, then $g(i)=h(i)$ for all $i\in X$ and then, for all $j\in I_{k}\setminus X$, $g(j)=\sum_{i\in X}c_{i,j}g(i)=\sum_{i\in X}c_{i,j}h(i)=h(j)$. Thus $g=h$. This shows that $\phi$ is injective and therefore $\phi$ is an isomorphism. ∎

Proposition 2.4**.**

Let $H$ be a nontrivial $k$-monoid and $X\subseteq I_{k}$ nonempty. Then $X$ is independent over $H$ if and only if $X$ is independent over $\beta(H)$.

Proof.

Since $\beta(H)\subseteq H$, if $X$ is independent over $\beta(H)$, then $X$ is also independent over $H$. Now, if $X$ is dependent over $\beta(H)$, then there are rationals numbers $c_{i}$, $i\in X$, not all zero, such that $\sum_{i\in X}c_{i}\alpha_{i}^{\beta(H)}=0$; this means that $\sum_{i\in X}c_{i}g(i)=0$ for all $g\in\beta(H)$. If $h\in H$, then $h=\sum_{g\in\beta(H)}d_{g}g$ for some nonnegative integers $d_{g}$, $g\in\beta(H)$, so

[TABLE]

Thus, $X$ is dependent over $H$. ∎

Corollary 2.5**.**

If $H$ is a nontrivial $k$-monoid, then $X\subseteq I_{k}$ is maximal independent over $H$ if and only if $X$ is maximal independent over $\beta(H)$.

If $H$ is a $k$-monoid, $F$ is an $l$-monoid and $\phi:H\to F$ is an additive map, then we can define a $\mathbb{Q}$-linear map $\phi^{\ast}:\mathbb{Q}^{F}\to\mathbb{Q}^{H}$ given by $\phi^{\ast}(\alpha)=\alpha\circ\phi$ for $\alpha\in\mathbb{Q}^{F}$.

Proposition 2.6**.**

Let $H$ be a nontrivial $k$-monoid and $X\subseteq I_{k}$ independent over $H$. If $r=|X|$, then $I_{r}$ is independent over $H|X$.

Proof.

Write $X=\{a_{1},a_{2},\ldots,a_{r}\}$ where $a_{1}<a_{2}<\cdots<a_{r}$. If $h\in H$, then $(h|X)(j)=h(a_{j})$, that is, $\alpha_{j}^{F|X}(h|X)=\alpha_{a_{j}}^{H}(h)$ for all $1\leq j\leq r$. If $\phi:H\to H|X$ is the function defined by $\phi(h)=h|X$ for $h\in H$, then we have that $\phi^{*}(\alpha_{j}^{H|X})=\alpha_{j}^{H|X}\circ\phi=\alpha_{a_{j}}^{H}$, $j=1,\ldots,r$.

If $I_{r}$ were dependent over $H|X$, then $\sum_{j=1}^{r}c_{j}\alpha_{i}^{H|X}=0$, where $c_{1},\ldots,c_{r}\in\mathbb{Q}$ are not all zero. Hence

[TABLE]

that is, $X$ would be dependent over $H$. ∎

Proposition 2.7**.**

Let $H$ be a nontrivial $k$-monoid and $X\subseteq I_{k}$ maximal independent over $H$. Then, there exists a finitely generated $k$-monoid $F$ such that $F\subseteq H$ and $X$ is maximal independent over $F$.

Proof.

If $H$ is finitely generated, we take $F=H$. Suppose that $H$ is not finitely generated and that $\beta(H)=\{h_{1},h_{2},\ldots,h_{n},\ldots\}$. For each $n\in\mathbb{N}^{+}$ let $F_{n}=\langle h_{1},h_{2},\ldots,h_{n}\rangle\subseteq H$.

For each $n\in\mathbb{N}^{+}$, the set $\{\alpha_{i}^{F_{n}}:i\in X\}$ spans $vect\ F_{n}$. In fact, given that $\{\alpha_{i}^{H}:i\in X\}$ is a basis of $vect\ H$, if $j\in I_{k}\setminus X$, then $\alpha_{j}^{H}=\sum_{i\in X}c_{i}\alpha_{i}^{H}$ for some $c_{i}\in\mathbb{Q}$, $i\in X$. Now, since the relation $\alpha_{r}^{H}|_{F_{n}}=\alpha_{r}^{F_{n}}$ holds for $1\leq r\leq k$, it results that

[TABLE]

This shows that $\{\alpha_{i}^{F_{n}}:i\in X\}$ spans $vect\ F_{n}$. Now, $\{\alpha_{i}^{F_{1}}:i\in X\}$ spans $vect\ F_{1}$ and so there exists $X_{1}\subseteq X$ such that $\{\alpha_{i}^{F_{1}}:i\in X_{1}\}$ is a basis of $vect\ F_{1}$, that is, $X_{1}$ is maximal independent over $F_{1}$. Since $F_{1}\subseteq F_{2}$, $X_{1}$ is independent over $F_{2}$, and this means that $\{\alpha_{i}^{F_{2}}:i\in X_{1}\}$ is linearly independent in $vect\ F_{2}$. Now, $\{\alpha_{i}^{F_{2}}:i\in X\}$ spans $vect\ F_{2}$ and so there exists $X_{2}\subseteq X$ that contains $X_{1}$ and is maximal independent over $F_{2}$. By continuing this way we construct an increasing sequence of subsets of $X$, $X_{1}\subseteq X_{2}\subseteq\cdots\subseteq X_{n}\subseteq\cdots\subseteq X$ in such a way that $X_{n}$ is maximal independent over $F_{n}$ for all $n\in\mathbb{N}^{+}$. Since $X$ is finite, the sequence $\{X_{n}\}$ stabilizes and so there exists $N\in\mathbb{N}^{+}$ such that $X_{n}=X_{N}$ for all $n\geq N$.

We claim that $X_{N}=X$. Assume, on the contrary that $X_{N}\neq X$. Let $n\geq N$. Since $\{\alpha_{i}^{F_{n}}:i\in X_{N}\}$ is a basis of $vect\ F_{n}$, there exist unique $c_{i,n}\in\mathbb{Q}$, $i\in X_{N}$, such that $\alpha_{j}^{F_{n}}=\sum_{i\in X_{N}}c_{i,n}\alpha_{i}^{F_{n}}$.

Now, for $n\geq N$, the relations $\alpha_{r}^{F_{n+1}}|_{F_{n}}=\alpha_{r}^{F_{n}}$, $1\leq r\leq k$, implies that

[TABLE]

so that by uniqueness of the $c_{i,n}$, it follows that $c_{i,n}=c_{i,n+1}$ for all $i\in X_{N}$. In particular, $c_{i,n}=c_{i,N}$ for all $i\in X_{N}$ and all $n\geq N$.

Let us set $c_{i}=c_{i,N}$ for each $i\in X_{N}$. Since $X\neq X_{N}$, there is some $j\in X\setminus X_{N}$. For $n\geq N$ we have $\alpha_{j}^{F_{n}}=\sum_{i\in X_{N}}c_{i}\alpha_{i}^{F_{n}}$. Let $h\in H$. Then $h=a_{1}g_{1}+\cdots a_{t}g_{t}$ where $a_{1},\ldots,a_{t}\in\mathbb{N}$ and $g_{1},\ldots,g_{t}\in\beta(H)$. Choose $n\geq N$ big enough so that $\{g_{1},\ldots,g_{t}\}\subseteq\{h_{1},\ldots,h_{n}\}\subseteq F_{n}$. Then $h\in F_{n}$ and

[TABLE]

This shows that $\alpha_{j}^{H}=\sum_{i\in X_{N}}c_{i}\alpha_{i}^{H}$, that contradicts the fact that $X$ is independent over $H$.

Let $F=F_{N}$. Then $F\subseteq H$ is finitely generated and $X$ is independent over $F$. That $X$ is maximal independent over $F$ follows from the fact that if $X\subseteq Y\subseteq I_{k}$ and $Y$ is independent over $F$, then, since $F\subseteq H$, $Y$ is also independent over $H$, and thus $Y=X$. ∎

Now we are ready to prove our main result.

Theorem 2.8**.**

Let $H$ be a nontrivial $k$-monoid and $X\subseteq I_{k}$ maximal independent over $H$. Then

[TABLE]

Moreover, if $H$ is finitely generated, then

[TABLE]

Proof.

By Proposition 2.3, $H\cong H|X$ and by Lemma 2.1, since $H|X$ is a $|X|$-monoid, $ind\ H=ind\ (H|X)\leq|X|$. Besides, if $F\subseteq H$ is a free $k$-monoid, then by Proposition 2.2 and Lemma 2.1, $dim\ L=ind\ L\leq ind\ H$, so $free\ H\leq ind\ H$.

To complete the proof, we construct a free $k$-monoid $F\subseteq H$ such that $dim\ F=|X|$.

By Proposition 2.7, there exists a $k$-monoid $L\subseteq H$ such that $X$ is maximal independent over $L$. Consider the matrix $M^{L}$, whose columns are basically the elements in $\beta(L)$. By Corollary 2.5, $X$ is maximal independent over $\beta(L)$. This means that the matrix $M^{L}$ has rank $|X|$. Since the rank of $M^{L}$ equals the maximal number of linearly independent columns of $M^{L}$, we find that there are $|X|$ elements of $\beta(L)$, say $h_{1},h_{2},\ldots,h_{|X|}$, that are linearly independent in $\mathbb{Q}^{k}$.

Let $F=\langle h_{1},h_{2},\ldots,h_{|X|}\rangle$. Then $F$ is a free $k$-monoid, $F\subseteq L\subseteq H$ and $dim\ F=|X|$.

Note that in case $H$ is finitely generated, we can take $L=M$ and have therefore $ind\ H=rank\ M^{H}$. ∎

Corollary 2.9**.**

Let $H$ be a nontrivial $k$-monoid and $X\subseteq I_{k}$ independent over $H$. Then $ind\ (H|X)=|X|$.

Proof.

By Proposition 2.6, if $r=|X|$ then $I_{r}$ is independent over $H|X$, and therefore, $I_{r}$ is maximal independent over $H|X$. By Theorem 2.8, $ind\ (H|X)=|I_{r}|=r=|X|$. ∎

Proposition 2.10**.**

Let $H$ be a nontrivial $k$-monoid. Then $r=ind\ H$ if and only if there exists $B\subseteq\beta(H)$ such that

1. (1)

$|B|=r$, 2. (2)

$\langle B\rangle$* is free, and* 3. (3)

for any $h\in\beta(H)\setminus B$ there exist $c\in\mathbb{N}^{+}$ and $f,g\in\langle B\rangle$ such that $f+ch=g$.

Proof.

If $r=ind\ H$, then, as in the last part of the proof of Theorem 2.8, we can take $B=\{h_{1},h_{2},\ldots,h_{r}\}$ (we know that $r=|X|$ where $X$ is maximal independent over $H$). Then $|B|=r$ and $\langle B\rangle$ is free. Let $h\in\beta(H)\setminus B$. Since $\langle B\cup\{h\}\rangle$ cannot be free (otherwise $ind\ H\geq r+1$), there are $n,n_{1},\cdots,n_{r},m,m_{1},\ldots,m_{r},\in\mathbb{N}$ such that $\{n,n_{1},\ldots,n_{r}\}\neq\{m,m_{1},\ldots,m_{r}\}$ and

[TABLE]

If $n=m$, then $n_{1}h_{1}+\cdots+n_{r}h_{r}=m_{1}h_{1}+\cdots+m_{r}h_{r}$ and $n_{1}=m_{1},\ldots,n_{r}=m_{r}$ (since $\langle B\rangle$ is free), that is, $\{n,n_{1},\ldots,n_{r}\}=\{m,m_{1},\ldots,m_{r}\}$, a contradiction. Thus, $n\neq m$, say $m<n$. If $c=n-m$, $f=n_{1}h_{1}+\cdots+n_{r}h_{r}$ and $g=m_{1}h_{1}+\cdots+m_{r}h_{r}$, then $c>0$ and $f+ch=g$.

Conversely, if there is some $B\subseteq\beta(H)$ that satisfy the three conditions, then by $(1)$, $(2)$ and Theorem 2.8, $r\leq ind\ H$. If $ind\ H>r$, then, using the last part of the proof of Theorem 2.8, we can find a subset $C\subseteq\beta(H)$ such that $|C|=ind\ H$. We can form a $k\times(r+ind\ H)$ matrix $M$ whose first $r$ columns are the elements of $B$ and its last $ind\ H$ columns are those elements in $C$. Condition $(3)$ assures that the rank of the matrix $M$ is $r$. Also, there are $ind\ H$ linearly independent columns coming from $C$, so that the rank of $M$ is at least $ind\ H>r$. This is a contradiction. Therefore, $ind\ H=r$. ∎

For instance, if $H_{k}:=\{h\in\mathbb{N}^{k}:1\leq h(1)\leq h(2)\leq\cdots\leq h(k)\}\cup\{0_{k}\}$ then $H_{k}$ is a $k$-monoid that is not finitely generated. We show that $ind\ H_{k}=k$ by finding a free submonoid of $H_{k}$ of dimension $k$. In fact, the $k$ elements $(1,1,1,\ldots,1),(1,2,2,\ldots,2),(1,2,3\ldots,3),\ldots,(1,2,3,\ldots,k)\in H_{k}$ are linearly independent in $\mathbb{Q}^{k}$, so they generate a free submonoid of $H_{k}$ of dimension $k$. Hence $k\leq H_{k}$, but really $ind\ H_{k}=k$ because $H_{k}$ is a $k$-monoid. In particular, $H_{k}\not\cong H_{r}$ if $k\neq r$.

3. $k$-monoids of index 1

In this section we give characterizations of $k$-monoids of index 1. We start with the following lemma.

Lemma 3.1**.**

Let $f,g\in\mathbb{N}^{k}$. If $cf=dg$ for some $c,d\in\mathbb{N}^{+}$, then there exist $h\in\mathbb{N}^{k}$, $r,s\in\mathbb{N}^{+}$ such that $f=rh$ y $g=sh$. In particular, $f,g\in\langle h\rangle$.

Proof.

We have that $cf(i)=dg(i)$ for $1\leq i\leq k$. By cancelling common factors we can assume that $c$ and $d$ are relatively prime. Then, for all $1\leq i\leq k$, $f(i)=df^{\prime}(i)$ and $g(i)=cg^{\prime}(i)$ for some $f^{\prime}(i),g^{\prime}(i)\in\mathbb{N}$. Then, $cdf^{\prime}(i)=cf(i)=dg(i)=dcg^{\prime}(i)$ for all $1\leq i\leq k$, from where $f^{\prime}(i)=g^{\prime}(i)$ for all $i$. If $h\in\mathbb{N}^{k}$ is given by $h(i)=f^{\prime}(i)$ for all $i$, then $f=dh$ and $g=ch$. ∎

Proposition 3.2**.**

A $k$-monoid $H$ has index 1 if and only if there exists $f\in\mathbb{N}^{k}$ such that $H\subseteq\langle f\rangle$.

Proof.

If $H\leq\langle f\rangle$, then $1\leq ind\ H\leq ind\ \langle f\rangle=1$, so that $ind\ H=1$. Conversely, suppose that $ind\ H=1$. If $H=\{0_{k}\}$, then we can take $f=0_{k}$.

Assume $H$ is nontrivial. Since $ind\ H=1$, there exists a nontrivial 1-monoid $F$ such that $H\cong F$, but every 1-monoid is finitely generated (see [1]), so $H$ is finitely generated. Let $\beta(H)=\{h_{1},h_{2},\ldots,h_{r}\}$ where $r=dim\ H$. The matrix $M^{H}$ associated to $H$ has rank 1 by Theorem 2.8. Then, taking $h_{1}$ and $h_{2}$, there is a rational $c_{1}/d_{1}$ where $c_{1},d_{1}\in\mathbb{N}^{+}$ such that $h_{2}=(c_{1}/d_{1})h_{1}$, or the same, $c_{1}h_{1}=d_{1}h_{2}$. By Lemma 3.1, there is some $f_{1}\in\mathbb{N}^{k}$ such that $h_{1},h_{2}\in\langle f_{1}\rangle$. Now, take $h_{2}$ and $h_{3}$. Again, there are $c_{2},d_{2}\in\mathbb{N}^{+}$ such that $c_{2}h_{2}=d_{2}h_{3}$; also, there is some $e\in\mathbb{N}^{+}$ such that $h_{2}=ef_{1}$. Then, $(c_{2}e)f_{1}=d_{2}h_{3}$. By Lemma 3.1, there is some $f_{2}\in\mathbb{N}^{k}$ such that $f_{1},h_{3}\in\langle f_{2}\rangle$ and therefore, $h_{1},h_{2},h_{3}\in\langle f_{2}\rangle$. We continue in this way until we find $f=f_{r-1}\in\mathbb{N}^{k}$ such that $h_{1},h_{2},\ldots,h_{r}\in\langle f\rangle$ and hence $H\subseteq\langle f\rangle$. ∎

The element $f$ such that $H\subseteq\langle f\rangle$ in Proposition 3.2 is not unique in general. We describe a method to determine all such $f$.

Let us call an element $h\in\mathbb{N}^{k}\setminus\{0_{k}\}$ primitive if $\gcd h(I_{k})=1$.

If $h\in\mathbb{N}^{k}\setminus\{0_{k}\}$, then there are unique $c_{h}\in\mathbb{N}^{+}$ and $g_{h}\in\mathbb{N}^{k}$ such that $h=c_{h}g_{h}$ and $g_{h}$ is primitive. In fact, $c_{h}=\gcd h(I_{k})$ and $g_{h}:I_{k}\to\mathbb{N}$ is given by $g_{h}(i)=h(i)/c_{h}$ for $i\in I_{k}$. Let us call $g_{h}$ the primitive part of $h$.

Proposition 3.3**.**

Let $H$ be a $k$-monoid generated by $h_{1},h_{2},\ldots,h_{r}$ (all nonzero). Then $ind\ H=1$ if and only if $h_{1},h_{2},\ldots,h_{r}$ have the same primitive part.

Proof.

If $ind\ H=1$, then by Proposition 3.2, $H\subseteq\langle f\rangle$ for some $f\in\mathbb{N}^{k}$. Of course $f\neq 0_{k}$. For each $1\leq j\leq r$, $h_{j}=d_{j}f$ for some $d_{j}\in\mathbb{N}^{+}$. Then $h_{j}=(d_{j}c_{f})g_{f}$ and the primitive part of $h_{j}$ is $g_{f}$. Thus, all the $h_{j}$ have the same primitive part $g_{f}$.

Conversely, if all $h_{j}$ have the same primitive part, say it is $f\in\mathbb{N}^{k}$, then $H\subseteq\langle f\rangle$ and so $ind\ H=1$ by Propostition 3.2. ∎

Proposition 3.4**.**

Let $H$ be a nontrivial $k$-monoid such that $ind\ H=1$. Then there is a unique primitive $f\in\mathbb{N}^{k}$ such that $H\subseteq\langle f\rangle$. If $g\in\mathbb{N}^{k}$ is such that $H\subseteq\langle g\rangle$, then $g=cf$ for some $c\in\mathbb{N}^{+}$.

Proof.

Assume $H\subseteq\langle f_{1}\rangle$ and $H\subseteq\langle f_{2}\rangle$ where $f_{1},f_{2}\in\mathbb{N}^{k}$ are primitive. Let $h\in H\setminus\{0_{k}\}$. Then there are $c_{1},c_{2}\in\mathbb{N}^{+}$ such that $h=c_{1}f_{1}$ and $h=c_{2}f_{2}$. Thus $c_{1}f_{1}=c_{2}f_{2}$ and by Lemma 3.1 there are $r,s\in\mathbb{N}^{+}$ and $g\in\mathbb{N}^{k}$ such that $f_{1}=rg$ and $f_{2}=sg$. Since $f_{1}$ and $f_{2}$ are primitive, $r=s=1$ and thus $f_{1}=g=f_{2}$. This proves uniqueness.

Now, if $H\subseteq\langle f\rangle$ and $H\subseteq\langle g\rangle$ where $f,g\in\mathbb{N}^{k}$ and $f$ is primitive, then by taking $h\in H$, $h\neq 0_{k}$, we have that $h=c_{1}f$ and $h=c_{2}g$ for some $c_{1},c_{2}\in\mathbb{N}^{+}$, so that $c_{1}f=c_{2}g$ and by Lemma 3.1, there are $c,d\in\mathbb{N}^{+}$ and $h_{0}\in\mathbb{N}^{k}$ such that $g=ch_{0}$ and $f=dh_{0}$. Since $f$ is primitive, $d=1$, so $f=h_{0}$ and $g=cf$. ∎

If $H$ is a $k$-monoid generated by $h_{1},h_{2},\ldots,h_{r}$ and all the $h_{j}$ have the same primitive part $f$, then $f$ is the unique such that $H\subseteq\langle f\rangle$ with $f$ primitive. For $1\leq j\leq r$ we have $h_{j}=d_{j}f$ for some $d_{j}\in\mathbb{N}^{+}$. The function $\{h_{1},h_{2},\ldots,h_{r}\}\to\mathbb{N}$ that sends $h_{j}$ to $d_{j}$ extends additively to a monomorphism $\varphi:H\to\mathbb{N}$ that gives an isomorphism between $H$ and $\langle d_{1},d_{2},\ldots,d_{r}\rangle$. We can take $d=\gcd(d_{1},d_{2},\ldots,d_{r})$ and $H$ is isomorphic to $\langle d_{1}/d,d_{2}/d,\ldots,d_{r}/d\rangle$ (a numerical semigroup). Finally, if $H\subseteq\langle g\rangle$ for some $g\in\mathbb{N}^{k}$, then $g=cf$ where $c$ is some factor of $d$. This determines all $f$ in Proposition 3.2.

4. Isomorphic numerical semigroups are equal

Let $H$ be a numerical semigroup. If $H\neq\mathbb{N}$, then there exists $m_{H}\in\mathbb{N}^{+}$ such that $H$ contains every natural number $n\geq m_{H}$ and $m_{H}-1\notin H$. This number $m_{H}-1$ is known as the Frobenius number of $H$. If $H=\mathbb{N}$ we define $m_{H}=1$.

If $H$ is a numerical semigroup, then the multiplicity of $H$ is the number $h_{0}=min\ (H\setminus\{0\})$. The set $H\setminus\{h_{0}\}$ is a numerical semigroup, for $h_{0}$ cannot be written as the sum of two nonzero elements of $H$.

If $H$ and $K$ are numerical semigroups and $\varphi:H\to K$ is an additive map, then, for $n\in H$ and $m\in\mathbb{N}$, $\varphi(mn)=m\varphi(n)$, and if $n,m\in H$, then $\varphi(nm)=n\varphi(m)=m\varphi(n)$.

The main result of this section is that if $H$ and $K$ are numerical semigroups and they are isomorphic, then $H=K$ and there is only one automorphism of $H$, the identity map.

Lemma 4.1**.**

If $H$ and $K$ are isomomorphic numerical semigroups, then there is only one isomorphism $\varphi:H\to K$, that is an increasing function.

Proof.

Let $\varphi:H\to K$ be an isomorphism. Then $\varphi(min\ H\setminus\{0\})=min\ K\setminus\{0\}$. In fact, if $h_{0}=min\ H\setminus\{0\}$, $k_{0}=min\ K\setminus\{0\}$, $k_{1}=\varphi(h_{0})$ and we assume that $k_{1}>k_{0}$, then there is $h_{1}\in H$ with $h_{1}>h_{0}$ such that $\varphi(h_{1})=k_{0}$. Then

[TABLE]

and since $h_{1}>h_{0}$ and $k_{1}>k_{0}$, also $h_{1}k_{1}>h_{0}k_{0}$, a contradiction. Therefore $k_{1}=k_{0}$, that is, $\varphi(h_{0})=k_{0}$.

If we write $H=\{0<h_{0}<h_{1}<h_{2}<\cdots\}$ and $K=\{0<k_{0}<k_{1}<k_{2}<\cdots\}$, then $\varphi(h_{0})=k_{0}$. Now, $\varphi|_{H\setminus\{h_{0}\}}:H\setminus\{h_{0}\}\to K\setminus\{k_{0}\}$ is an ismomorphism of numerical semigroups, thus, by a similar argument as above, $\varphi(h_{1})=k_{1}$. And by induction on $n$ we obtain that $\varphi(h_{n})=k_{n}$ for all $n\in\mathbb{N}$. This shows that $\varphi$ is unique and is an increasing function. ∎

Lemma 4.2**.**

Suppose that $\varphi:H\to K$ is an isomorphism between numerical semigroups. If $L\subseteq H$ is a numerical semigroup, then $\varphi(L)\subseteq K$ is a numerical semigroup.

Proof.

Since $L$ is a numerical semigroup, $\mathbb{N}\setminus L$ is finite, and therefore $H\setminus L$ is finite. Then, $\varphi(H\setminus L)=K\setminus\varphi(L)$ is finite and $\mathbb{N}\setminus\varphi(L)=(\mathbb{N}\setminus K)\cup(K\setminus\varphi(L))$ is finite. Thus, $\varphi(L)$ is a numerical semigroup. ∎

Theorem 4.3**.**

If $H$ and $K$ are isomorphic numerical semigroups, then $H=K$.

Proof.

Let $H$ and $K$ be numerical semigroups and let $\varphi:H\to K$ be the unique isomorphism, that by Lemma 4.1 is increasing.

First we prove that $\varphi(m_{H})=m_{K}$. Let $h_{0}\in H$ such that $\varphi(h_{0})=m_{K}$. If $k\in K$ is such that $k\geq m_{K}$, then $k=m_{K}+r$ for some $r\geq 0$ and there exists $h\in H$ such that $\varphi(h)=k$. Since $\varphi$ is increasing, we have that $h=h_{0}+s(r)$ for some $s(r)\geq 0$. Then we have

[TABLE]

By replacing $k=m_{K}+r$ we get $h_{0}(m_{K}+r)=(h_{0}+s(r))m_{K}$, that yields to $rh_{0}=s(r)m_{K}$ and therefore

[TABLE]

for all $r\geq 0$. Let $c=h_{0}/m_{K}$. Since $cr=s(r)$ is a natural number for all $r\geq 0$, $c$ is a natural number. Hence $\varphi(h_{0}+cr)=m_{K}+r$ for all $r\geq 0$, and so $\varphi^{-1}(m_{K}+r)=h_{0}+cr$, which means that

[TABLE]

Now we apply Lemma 4.2 to the isomorphism $\varphi^{-1}:K\to H$ and the numerical semigroup $L=\{0,m_{K},m_{K}+1,m_{K}+2,\ldots\}\subseteq K$ to obtain that $\varphi^{-1}(L)=\{0\}\cup\{h_{0}+cr:\ r\geq 0\}$ is a numerical semigroup. Then, for some $r$ big enough, $h_{0}+rc$ and $h_{0}+c(r+1)$ must be consecutive natural numbers, that is, $h_{0}+c(r+1)=h_{0}+cr+1$, that yields to $c=1$.

Therefore $\varphi^{-1}(\{m_{K},m_{K}+1,m_{K}+2,\ldots\})=\{h_{0}+r:\ r\geq 0\}$. In particular, $\{h_{0}+r:\ r\geq 0\}\subseteq H$ and so $m_{H}\leq h_{0}$. Then, $\varphi(m_{H})\leq\varphi(h_{0})=m_{K}$.

By a symmetric argument applied to the isomorphism $\varphi^{-1}:K\to H$ we obtain that $\varphi^{-1}(m_{K})\leq m_{H}$, so that $m_{K}\leq\varphi(m_{H})$. Thus, we have prover that $\varphi(m_{H})=m_{K}$.

Now we show that $m_{H}=m_{K}$. In fact, the restriction of $\varphi$ to $M=\{0,m_{H},m_{H}+1,m_{H}+2,\ldots\}$ gives an isomorphism between $M$ and $L=\{0,m_{K},m_{K}+1,m_{K}+2,\ldots\}$, so these two numerical semigroups have the same dimension, but $M$ has dimension $m_{H}$ and $L$ has dimension $m_{K}$, so that $m_{H}=m_{K}$.

By Lemma 4.1 there is only one isomorphism from $M$ to $L=M$, that is indeed the identity map. The restriction $\varphi|_{M}:M\to M$ is thus the identity of $M$, so for all $h\geq m_{H}$, $\varphi(h)=h$.

Now we can show that $H=K$. If $h\in H\setminus\{0\}$, then for some $m\in\mathbb{N}^{+}$, $mh>m_{H}$. Then, $m\varphi(h)=\varphi(mh)=mh$, and since $m>0$ it results that $h=\varphi(h)\in K$. This shows that $H\subseteq K$. The analog argument applied to $\varphi^{-1}:K\to H$ shows the inclusion $K\subseteq H$. Thus, $H=K$. ∎

Note that by Lemma 4.1, if $H$ is a numerical semigroup, then the identity map $I:H\to H$ is the only isomorphism that exists.

Corollary 4.4**.**

Every nontrivial $k$ monoid of index 1 is isomorphic to a unique numerical semigroup. If $H=\langle h_{1},h_{2},\ldots,h_{r}\rangle$ is a numerical semigroup, then a $k$-monoid $F$ is isomorphic to $H$ if and only if there exists $f\in\mathbb{N}^{k}\setminus\{0_{k}\}$ such that $F=\langle h_{1}f,h_{2}f,\ldots,h_{r}f\rangle$.

Proof.

The first part of the corollary follows directly from Theorem 4.3. Let $H=\langle h_{1},h_{2},\ldots,h_{r}\rangle$ be a numerical semigroup. Let us assume without loss of generality that $\beta(H)=\{h_{1},h_{2},\ldots,h_{r}\}$ and $0<h_{1}<h_{2}<\cdots<h_{r}$. If $F$ is a $k$-monoid isomorphic to $H$, then there is a unique primitive $g\in\mathbb{N}^{k}$ and $d_{1},d_{2},\ldots,d_{r}\in\mathbb{N}^{+}$ such that $d_{1}<d_{2}<\cdots<d_{r}$ and $\beta(F)=\{d_{1}g,d_{2}g,\ldots,d_{r}g\}$. If $d=\gcd\{d_{1},d_{2},\ldots,d_{r}\}$, then $F$ is isomomorphic to the numerical semigroup $\langle d_{1}/d,d_{2}/d,\ldots,d_{r}/d\rangle$. By Theorem 4.3, $h_{j}=d_{j}/d$, $j=1,2,\ldots,r$. Thus, $F=\langle(dh_{1})g,(dh_{2})g,\ldots,(dh_{r})g\rangle$ and we can take $f=dg$. The converse is obvious. ∎