Idle question:

Let $g(n)$ be the sum, over all isomorphism classes of groups of order $n$, of $\frac{1}{|Aut(G)|}$ where $G$ is a group in the class. Thus $g(n)n!$ is the number of group laws on a fixed set of size $n$. Is anything known about the asymptotic behavior of this quantity? I could easily believe that abelian groups account for most of it. If we only count abelian groups, calling the analogous number $a(n)$, then the function $a$ is clearly multiplicative in the sense that $a(mn)=a(m)a(n)$ when $m$ an $n$ are relatively prime, and I believe that the function $a(p^k)$ can be written as an explicit function of $k\ge 0$ and the prime $p$: $a(p)=\frac{1}{p-1}$, $a(p^2)=\frac{p}{(p-1)(p^2-1)}$, $a(p^3)=\frac{p^3}{(p-1)(p^2-1)(p^3-1)}$.

ADDED: So it looks like $a(p^k)=p^{\frac{(k(k-1)}{2}}\prod_{1\le j\le k}(p^j-1)^{-1}$. (I checked it up to $k=4$.) If you let $m\ge k$ and use the fact that every abelian group of order $p^k$ is isomorphic to a subgroup of $(\mathbb Z/p^m)^k$ and the fact that every isomorphism between two such subgroups is induced by an automorphism of $(\mathbb Z/p^m)^k$, you can interpret this as saying that the sum, over all automorphisms $g$ of $(\mathbb Z/p^m)^k$, of the number of subgroups of order $p^k$ in the fixed set of $g$, is a certain power of $p$. But I can't think of a reason why that should be true.

`$\sum_{k \geq 0} p^{k(k-1)/2} \prod_{1 \leq j \leq k} (p^j-1)^{-1} = \prod(1-p^{-n})^{-1}$`

. Proof: The left hand summand can be rewritten as $p^{-k} / \prod_{1 \leq j \leq k} (1-p^{-j})$, which is well known to be the generating function for partitions with precisely $k$ parts. The right hand side, of course, is the generating function for all partitions. $\endgroup$ – David E Speyer Jul 7 '10 at 15:06`$\prod_{n=1}^{\infty} (1-p^{-n}) |\mathrm{Aut}(G)|^{-1}$`

. Your conjecture is that the probability that the determinant is $p^k(\mathrm{unit})$ is`$p^{-k} \prod_{n > k} (1-p^{-n})$`

. Equivalently, the probability that the determinant is not $0$ modulo $p^k$ is`$\prod_{n \geq k} (1-p^{-n})$`

. Don't know how to prove this, but it sounds simpler. $\endgroup$ – David E Speyer Aug 24 '10 at 23:08