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Before my question, some background:

Definition 1: A numerical semigroup is a subsemigroup $N$ of the additive semigroup $\mathbb N$ of the non-negative integers such that $\mathbb N-N$ is finite. A numerical semigroup has genus $g$ if $L=\mathbb N-N$ has $g$ elements. The elements of $L$, $l_1\lt l_2\lt \ldots\lt l_g$, is called gaps of $N$, while the elements of $N$, $n_1\lt n_2\lt \ldots$ are called non-gaps of $N$.

Theorem 1: In a numerical semigroup of genus $g$, the value of a gap can't be greater than $2g-1$.

Definition 2: A numerical semigroup of genus $g$ is called symmetric when its greater gap is $2g-1$.

My doubt is in the following comment:

I couldn't prove if $n_0\lt n_1\lt\ldots\lt n_g\lt \ldots$ are the non-gaps of $N$, then $l_g-n_0,\ldots, l_g-n_{g-1}$ are the $g$ gaps of $N$.

Thanks in advance

user74141
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1 Answers1

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Suppose that we have $g$ gaps among $0,1,2,\dots,\,2g-1$ with the largest gap $l_g=2g-1$.

We have $g$ pairs of the form $(a,\ l_g-a)$, $\,a=0...(g-1)$, and as their sum $l_g$ is a gap, it cannot be that both $a$ and $l_g-a$ are nongaps. So, each such pair must contain at least one gap.

But, as we have exactly $\,g$ gaps, it must mean that each pair contains exactly one gap.

Update: For the other direction, we have $g$ gaps among the first $2g$ numbers by the theorem, so the other $g$ numbers are the nongaps: $\{n_0,n_1,\dots,n_{g-1},\,l_1,l_2,\dots,l_g\}=\{0,1,2,\dots,\,2g-1\}$.
Now, the hypothesis says that $l_g-n_0,\dots,\ l_g-n_{g-1}$ are the gaps in $N$, which statement implies that each of them are nonnegative (by the definition of gaps).

So, if the biggest gap, $l_g$ was less than $2g-1$, then $2g-1$ is a nongap, moreover it must be the $g$th nongap, ie. $n_{g-1}=2g-1$, but then $l_g-(2g-1)$ would be negative, so it cannot be a gap.

Berci
  • 90,745
  • I've already tried this argument. Counterexample: $N=0,4,5,6,\ldots$, in this case $l_g=3$. Take the pair $(3,3-4)$, it doesn't work. Maybe you should use the fact $l_g=2g-1$ – user74141 Aug 16 '14 at 04:21
  • Yes, $l_g=2g-1$ is meant to be used. However, the other direction seems not true (neither in your counterexample): if $N={2,3,\dots}$ then $g=2$, and the gaps and nongaps are also symmetric to $3=2g-1$. – Berci Aug 16 '14 at 22:33