# Analogue of rank for finite abelian groups?

The rank of a finitely-generated abelian group is the size of the maximal $\mathbb{Z}$-linearly-independent subset; loosely this means the number of distinct copies of $\mathbb{Z}$ contained as direct summands. I have come across a situation where a notion somewhat like rank may be useful for finite abelian groups (which obviously all have rank zero). I'm looking at graphs embedded on surfaces and I have a procedure for associated certain finite abelian groups to a graph embedding. I've found that the groups I'm getting this way are all quotients of $\mathbb{Z}^{2g}$, where $g$ is the genus of the surface. So for example, I can realize $(\mathbb{Z}/2\mathbb{Z})^2$ as a group coming from a graph embedded on a torus, but I can't get $(\mathbb{Z}/2\mathbb{Z})^3$.

Here's my question: for a finite abelian group $G$, define the "finite rank" of $G$ to be the minimal rank of a free abelian group that surjects onto $G$. Is this a common notion? Although there is the possibility this could be an argumentative question, is it a useful notion (or conversely, can you make the case for it being useless)?

• You mean the maximal Z-linearly independent subset, I guess (I would not use this definition; it seems cleaner to say it is the dimension of the tensor product with Q). In any case, what you've defined is just the minimal number of generators. – Qiaochu Yuan Nov 19 '10 at 17:25
• To answer your last question: sure, the (minimal) number of generators of a finite abelian group is a very useful notion. It comes up all over the place. – Pete L. Clark Nov 19 '10 at 17:40

Unless I am mistaken, the "finite rank" is just the minimal size of a generating set for $G$. Indeed, if $X$ is a generating set for $G$, then the free abelian group on $X$ surjects onto $G$ by the map induced by the embedding of $X$ into $G$, so the "finite rank" is at most the smallest size of a generating set. And if a free abelian group $F$ of rank $k$ surjects onto $G$, then the image of a free generating set for $F$ maps to a generating set for $G$ (possibly not injectively), so $G$ has a generating set with at most $k$ elements.
This number, in an arbitrary group, is usually written $d(G)$; I don't know of any special name for it, and none of my books seems to have any name attached to it. In $p$-groups, it is the index of the Frattini subgroup, $[G:\Phi(G)]$.
• @NKS: I think that if you express a finitely generated abelian group using the invariant factor decomposition, then $d(G)$ is just the number of cyclic factors in that decomposition. – Arturo Magidin Nov 19 '10 at 17:39