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I'm wondering what total number of possible groups of order n would be if we are counting isomorphisms separately. I mean that if we were to write out the multiplication tables for elements $a_1, a_2, ... , a_n$, what is the maximum number of such tables? Can we make the bound tighter by noting that the table must satisfy the group axioms, or do we need to know something about n?

Matt Ross
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    https://oeis.org/wiki/Number_of_groups_of_order_n – angryavian Dec 08 '20 at 21:46
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    "Know something about $n$." Simplest is: if $n$ is prime, there is only one group of order $n$. For a tabulation, and lots of info, see http://oeis.org/A000001 – GEdgar Dec 08 '20 at 21:48
  • Do you mean two groups that are distinct as sets but isomorphic should be counted separately? If so, you're asking about labeled groups: A034383 – Jair Taylor Dec 08 '20 at 22:04
  • Yes I think labeled groups is what I was looking for but I was having a difficult time explaining it. Thanks! – Matt Ross Dec 09 '20 at 20:26

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This is A000001 on the OEIS. The answer is quite complicated and depends delicately on the prime factorization of $n$; only asymptotic results are known for general $n$. It's a result due to Higman and Sims that if $n = p^k$ then there are asymptotically $p^{\frac{2}{27} k^3 + O(k^{8/3})}$ groups of order $n$. This gives that if $n = \prod p_i^{k_i}$ then there are at least

$$\prod p_i^{\frac{2}{27} k_i^3 + O(k_i^{8/3})}$$

groups of order $n$ (this is a count of the nilpotent groups of order $n$, which are the products of their Sylow $p$-subgroups), and I'd guess this is an accurate asymptotic (up to that huge error in the exponent) but I don't know if this is known.

These results suggest, if you work with them for a bit, that almost all finite groups are finite $2$-groups (having order a power of $2$); this is an open problem to prove rigorously, though, as far as I know. As experimental evidence for this, it's known that over 99% of the first $50$ billion groups have order $1024 = 2^{10}$: there are known to be exactly

$$49487365422 \approx 4.9 \times 10^{10}$$

groups of this order.

A much easier special case to think about is the number of abelian groups of order $n$: by the structure theorem for finite abelian groups, if $n = \prod p_i^{k_i}$ then this is exactly $\prod p(k_i)$ where $p$ is the partition function.

In the opposite direction, there are a special set of positive integers $n$, including the primes, for which the only group of order $n$ is the cyclic group $C_n$; see this exposition by Keith Conrad for more, and also A003277 on the OEIS.

Qiaochu Yuan
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