Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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)?

share|improve this question
    
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
2  
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

1 Answer 1

up vote 5 down vote accepted

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, $[G:\Phi(G)]$.

share|improve this answer
    
Ah of course... I guess I got hung up on trying to describe this notion in more intrinsic terms. Forest-for-trees issue. –  NKS Nov 19 '10 at 17:30
    
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.