This question sounds like it should be very well known, but for some reason I failed to find a decent answer anywhere. Let $G\subset\mathbb{Z}^n$ be a subgroup, and $G_+=G\cap\mathbb{Z}_{\ge0}^n$ be a cone of elements in $G$ whose coordinates are all nonnegative. The semigroup $G_+$ is finitely generated; it can be proved in a couple of ways. My question is, is there an effective upper bound on the number of generators in terms of $n$, or, even better, the rank of $G$?
1 Answer
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No, there is no upper bound on the number of generators in terms of $n$.
Let $k$ be arbitrary positive integer. Consider the subgroup $$G=\{(x,y)\in\mathbb Z^2\mid x+y\equiv 0\pmod k\}.$$ Any set of generators of $G_+$ contains all the elements $(x,y)$ such that $x,y\ge 0$ and $x+y=k$. Therefore the number of generators of $G_+$ is $k+1$.
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$\begingroup$ OK, that's discouraging but convincing enough. I am definitely accepting it, and will post another question if I will figure out a way to put a restrictions on $G$'s that I had in mind. Thanks! $\endgroup$ Commented Jul 14, 2011 at 20:52