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I've been looking at combinatorial group theory, but all the results seem to be about infinite groups. Are there any important results about the presentations finite groups specifically (or are useful for finite groups?). About how the minimum number of relations implies something about the structure of the group?

I'd prefer results that are applicable to all finite groups or to all finite simple or all simple groups.

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There's the Golod-Shafarevich inequality for finite p-groups, closely related to the still open conjecture of whether the minimal number of relators for a finite p-group is equal to the dimension of its second cohomology (mod p). – Steve D Mar 8 '12 at 4:43
@Steve D: As you gave the above comment, I would like to know if you have any comment on this question:… – Alireza Abdollahi Jun 18 '15 at 10:51

One such result that springs to mind is that, if the finite group $G$ has a presentation with $r$ generators and $s$ relations, then the Schur Multiplier $M(G) = H_2(G)$ of $G$ can be generated by at most $s-r$ elements. So in particular $s \ge r$. (Of course you can prove that more directly - a finitely presented group with $s<r$ has infinite abelianization.)

We can deduce for example that a finite abelian group $G$ of rank $r$ requires at least $r(r+1)/2$ relations to present it, because $M(G)$ has rank $r(r-1)/2$. In this case the converse holds - the obvious presentation of $G$ has $r$ generators and $r(r+1)/2$ relations.

The converse does not hold in general. Swan constructed examples of finite solvable groups with trivial multiplier and arbitrarily large minimum $s-r$. But I believe it is still an open problem for finite $p$-groups. i.e. does every $d$-generator finite $p$-group have a presentation with $d$ generators and $d + {\rm rk}(M(G))$ relations?

Finite groups of defect 0 - i.e. $r=s$ have also been much studied. There are lots of 2-generator examples known, a few 3-generator examples and none requiring 4 or more generators.

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I'd like to also mention that given a finite presentation for a finite group $G$, there is an algorithm for computing the Schur multiplier. If $G=\langle X\ |\ R\rangle$, then the subgroup $H=\langle R\rangle$ is central in $\langle X\ |\ [X,R]\rangle$. It is a finitely generated abelian group, and its torsion subgroup is $M(G)$. This is not hard to prove using Hopf's formula, and every step is something a computer can do (presentation of finite index subgroup, abelianization, etc.). – Steve D Mar 8 '12 at 4:45

Yes, of course, here are a few examples:

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The result by Guralnick, Kantor, Kassabov and Lubotzky (which is mentioned already in Mark Sapir's answer) is one of the most striking results in finite group theory of the last decade, in my opinion. It shows that there is a uniform bound on the length of the presentation for (probably all) nonabelian finite simple groups. This is much stronger than what anyone before this result even conjectured, so it's absolutely amazing. More precisely, they show the following:

All nonabelian finite simple groups of rank $n$ over a field of size $q$, with the possible exception of the Ree groups $^2G_2$, have presentations with at most 80 relations and bit-length $O(\log n + \log q)$.

(By the way, the assumption "nonabelian" is essential, the theorem is false for cyclic groups of prime order!) Their results appeared in a series of 3 papers:

  1. Presentation of finite simple groups: a quantitative approach
  2. Presentation of finite simple groups: cohomological and profinite approaches
  3. Presentation of finite simple groups: a computational approach

I would recommend you to have a look at the first of those 3 papers, it contains a wealth of information about presentations of finite (simple) groups, and it is written in a very readable and enjoyable style.

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It's not clear from what you write what exactly is false for abelian simple groups! Cyclic groups of order $n$ have a presentation of total length at most $O(\log n)$, but not with a fixed number of generators and relations. – Derek Holt Mar 8 '12 at 9:56

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