Let $S$ be a numerical semigroup. Let $\mathbb N$ denote the monoid of non-negative integers under addition. Let $F(S)=\max (\mathbb N \setminus S)$ be the Frobenius number of $S$; let $g(S)=|\mathbb N \setminus S|$ be the genus of $S$; and let $m(S)=\min (S \setminus \{0\})$ be the multiplicity of $S$. If $m(S)=3$, then is it true that $S$ is the submonoid of $\mathbb N$ generated by $3, 3g(S)- F(S)$ and $3+F(S)$?
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2 Answers
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Yes, this is true and it follows immediately from
[1, Corollary 4] Two numerical semigroups with multiplicity three are equal if and only if they have the same Froebenius number and the same gender.
together with
[1, Lemma 6] Let $S$ be a numerical semigroup with multiplicity three, Frobenius number $F(S)$ and gender $G(S)$. Then $\frac{F(S) + 1}{2} \le G(S) < \frac{2F(S) + 3}{3}$.
and
[1, Theorem 7] Let $F$ be a positive integer greater than or equal to four that is not a multiple of three. Let $G$ be a positive integer such that $\frac{F + 1}{2} \le G < \frac{2F + 3}{3}$. Then $S = \langle 3, 3G − F, F + 3 \rangle$ is a numerical semigroup with multiplicity three, Frobenius number $F$ and gender $G$.
[1] J. C. Rosales, "Numerical semigroups with multiplicity three and four", 2005.
answered Jun 21, 2018 at 17:03
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Some way to do this is to say $S=\langle a,b,c\rangle=\langle 3,3k+1,3j+2\rangle$ then it is easy to see that $g(S)=k+j$. Now take a generator of $S$ it can not be less than $F(S)$ as in any case modulo $3$, $F(S)$ is equal to one of the generators, so one of them is $>$ $F(S)$ and by the definition of minimal generator and $F(S)$ we get say $a=3$, $c=3+F(S)$. Lastly $b+c=3(j+k)+3$ and we are done.
