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Let $\mathcal{G}$ be the set of finite groups and for $G \in \mathcal{G}$, let $S(G)$ be the set of subgroups of $G$. I am interested in the ratio $R(G)=|S(G)|/|G|$. It is easy to show that by picking $G$ appropriately, $R(G)$ can be made arbitrarily large or arbitrarily close to zero. I am interested in some deeper properties of the set $R=(R(G) : G \in \mathcal{G})$, such as:

(1) For which $x \in \mathbb{R}$ do there exist sequences of finite groups $G_1, G_2, ...$ such that the sequence $R(G_1), R(G_2), ... $ converges to $x$?

(2) Does $R$ contain a (non-empty) interval $(a,b) \in \mathbb{Q}$?

(3) Which integers belong to $R$?

(4) How do these properties change when $\mathcal{G}$ is replaced by the set of finite abelian groups? Finite simple groups?

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For cyclic groups, it will be the number of divisors of n divided by n, which has an upper bound of 1, an easy upper bound of 2 sqrt(n) for most n, and for sufficiently large n will be O(log n). I suspect that for finite abelian groups, your spectrum set will be have only one limit point in the reals. Gerhard "Ask Me About System Design" Paseman, 2012.08.23 –  Gerhard Paseman Aug 23 '12 at 22:51
    
It seems to me that for groups of the form large number of copies of cyclic order 2 plus one cyclic order large prime you should be able to get the ratio arbitrarily close to any positive rational. –  Gerry Myerson Aug 23 '12 at 23:05
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"the set of finite groups" as well as "the set of finite abelian groups" are proper classes! –  tj_ Aug 23 '12 at 23:36
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For the finite simple groups I expect that the only accumulation point is $0$. This comes from the cyclic groups; one can of course ignore the sporadics, and then probably well-known lower bounds on the number of subgroups of the infinite families will do it (although I don't know them). It should more or less suffice to give such bounds for $A_n$ and $\text{PSL}_n(\mathbb{F}_q)$. –  Qiaochu Yuan Aug 24 '12 at 6:58
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I ran a quick check for Q3, iterating over all abelian groups of order at most $10^6$. The only integers I got were 1,2,4,6,7,14, and 21. –  Steve D Aug 24 '12 at 8:03

1 Answer 1

With regards to Q1 (and part of Q4), the numbers of the form $R(G)$ are dense in $\mathbb{R}_{\ge 0}$ even when $G$ is restricted to be abelian.

Some first results. $R(G \times H) = R(G) R(H)$ if $\gcd(|G|, |H|) = 1$. We also have $R(C_p) = \frac{2}{p}$ ($p$ is always a prime) and

$$R(C_p^4) = \frac{1}{p^4} \sum_{k=0}^4 {4 \choose k}_p = 1 + \frac{3}{p} + \frac{4}{p^2} + \frac{3}{p^3} + \frac{5}{p^4}$$

where ${n \choose k}_p$ is a Gaussian binomial coefficient.

Lemma: Let $a_1, a_2, ... $ be a sequence of positive reals such that $\lim_{n \to \infty} a_n = 0$ but such that $\sum a_n$ diverges. Then the set of sums of finite subsequences of the $a_i$ is dense in $\mathbb{R}_{\ge 0}$.

Proof. Let $r \in \mathbb{R}_{\ge 0}$ and fix $\epsilon > 0$. Choose $N$ such that $a_n < \epsilon$ for all $n \ge N$. Then the partial sums starting from $a_N$ diverge but begin less than $\epsilon$ and increase by at most $\epsilon$ at each step, so the conclusion follows. $\Box$

Applying the lemma to the sequence $\log R(C_p^4)$ (which satisfies the hypotheses of the lemma using the fact that $\sum \frac{1}{p}$ diverges), we conclude that the numbers of the form $R(G)$ where $G$ is a product of groups of the form $C_p^4$ for distinct primes $p$ are dense in $\mathbb{R}_{\ge 1}$, and if we allow in addition the groups of the form $C_p$, then the conclusion follows.

Q2 appears to be potentially very difficult and I have not thought about Q3.

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Curiously $4$ is the only dimension $d$ in which $R(C_p^d)$ approaches $1$ as $p \to \infty$, and the reason is something happening in the middle dimension. I wonder if this is related to the other reasons that $4$ is an exceptional dimension... –  Qiaochu Yuan Aug 24 '12 at 3:30
    
Isn't that just dependent on whether $n=\max_{k=0}^n k(n-k)$ - the max of the dimension of the various Grassmanians, since the number of points on an algebraic variety of dimension $d$ goes to $p^d$ as $p\to \infty$ by the Weil conjectures? It doesn't seem plausible to me that there are exotic $\mathbb R^4$s because $\operatorname{Gr}_2^2$ is $4$-dimensional, but I know nothing of the relevant topology and certainly stranger things have happened. –  Will Sawin Aug 24 '12 at 4:28
    
@Will: I had other other reasons in mind (for example the fact that $\Lambda^2(\mathbb{R}^4)$ is not an irreducible representation of $\text{SO}(4)$) but who knows, maybe they're all related. –  Qiaochu Yuan Aug 24 '12 at 4:49
    
It's not clear if "related" is even well-defined enough here to give that question a satisfying answer. Which mathematical statements are related to $2+2=2\times 2$? –  Will Sawin Aug 24 '12 at 18:07

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