Known bounds for the number of groups of a given order.

The number of nonisomorphic groups of order $n$ is usually called $\nu(n)$. I found a very good survey about the values. $\nu(n)$ is completely known absolutely up to $n=2047$, and for many other values of $n$ too (for squarefree n, there is a formula).

In general, however, $\nu$ is very hard to calculate.

So, I would like to at least have easy to calculate lower and upper bounds for $\nu$. Are such bounds known?

• Well, there's always $n^{n^2}$. You probably want a good upper bound though. – Chris Eagle Jun 16 '13 at 18:59
• May be relevant: math.stackexchange.com/questions/415891/… – Seirios Jun 16 '13 at 19:02
• @ChrisEagle Actually, if I'm counting correctly, you can get $(n-1)^{n-1} + (n-2)^{2T(n-2)} = (n-1)^{n-1} + (n-2)^{(n-2)(n-1)}$ (for $n\geq 3$). Still a very elementary bound, though. – Stahl Jun 16 '13 at 19:26
• Martin Isaacs writes in his Finite Group Theory (after reciting the number of groups of order $2^7,2^8, 2^9$ and $2^{10}$) : "These numbers were computed by a remarkable algorithm for counting $p$-groups that was developed by E. O'Brien." Maybe that's a start? (Possibly the reference is the same as the one by Derek Holt in the other topic.) – Myself Jun 16 '13 at 19:26

Geoff's answer is exactly correct, but I wanted to give the specifics.

If you only want bounds that are easy to compute without being able to prove them yourself, then this answer should be just fine. The bounds are easy and reasonably tight. If you want to understand the proofs (which are not too bad, and involve many fun areas of finite groups), then read the book: Blackburn–Neumann–Venkataraman (2007).

Pyber (1993) showed that for $f(n)$ the number of isomorphism classes of groups of order $n$:

$$f(n) \leq n^{(2/27)\mu(n)^2+O(\mu(n)^{3/2})}$$

where $\mu(n) \leq \log_2(n)$ is the highest power of any prime dividing $n$. When $\mu(n)=1$ you have the square-free case you mentioned, and when $n=p^k$, then $k=\mu(n)$ and the bound is asymptotically sharp. For $p$-groups, pretty decent but slightly weaker bounds were first proven in Higman (1960), and improved in Sims (1965).

The best lower bounds that aren't ridiculously complex to compute and that I can think of just follow from $f(k) \leq f(n)$ if $k$ divides $n$. In other words, count the nilpotent groups of that order as a lower bound. For reference an explicit form of Higman's lower bound is:

$$f(p^k) \geq p^{\tfrac{2}{27} k^2(k-6)}$$

All of these results and more are explained very nicely in the book Blackburn–Neumann–Venkataraman (2007). I recommend it highly.

• Blackburn, Simon R.; Neumann, Peter M.; Venkataraman, Geetha. “Enumeration of finite groups.” Cambridge Tracts in Mathematics, 173. Cambridge University Press, Cambridge, 2007. xii+281 pp. ISBN: 978-0-521-88217-0 MR2382539 DOI:10.1017/CBO9780511542756
• Pyber, L. “Enumerating finite groups of given order.” Ann. of Math. (2) 137 (1993), no. 1, 203–220. MR1200081 DOI:10.2307/2946623
• Sims, Charles C. “Enumerating p-groups.” Proc. London Math. Soc. (3) 15 (1965) 151–166. MR169921 DOI:10.1112/plms/s3-15.1.151
• Higman, Graham. “Enumerating p-groups. I. Inequalities.” Proc. London Math. Soc. (3) 10 (1960) 24–30. MR113948 DOI:10.1112/plms/s3-10.1.24

The number of $p$-groups of order $p^{n}$ is (asymptotically) around $p^{ \frac{2 n^{3}}{27}}$. This suggests that one can't expect to do better than $n^{c\log(n)^{2}}$ for some constant $c$, for the number of isomorphism types of groups of order $n.$ I don't know whether such a bound has been conjectured ( or shown to be wrong), but this is the sort of bound I would expect, and a proof (or otherwise) might be accessible, given CFSG. If I was looking seriously into this, I would check the papers of L. Pyber ( some work was done on this sort of thing several years ago by P.M. Neumann).

We can obtain an even better lower bound by looking at the groups of exponent $p$ and nilpotency class $2$.

Apparently this tells us that there are at least $p^{\frac{2}{27}m^3-\frac{2}{3}m^2}$ groups of order $p^m$.

This is found in the same book mentioned in the other solution.

• But $p^{\frac{2}{27}m^2(m - 6)} > p^{\frac{2}{27}m^3 - \frac{2}{3}m^2}$. – Deedlit Jun 29 '17 at 14:06