It's well-known that for each prime number $p$ there are exactly two groups of order $p^2$, five of order $p^3$, and fifteen of order $p^4$ (at least when $p>3$).

I know that the classification of $p$-groups gets much harder for higher exponents. But I wonder if $p$-groups with higher dimensions are still structured enough that they would at least theoretically allow a classification.

In particular: Is there, for every natural $n$, a prime $p_0(n)$ and a number $m(n)$ such that for every $p \ge p_0(n)$ the number of groups of order $p^n$ equals $m(n)$?

(For example, $m(2)=2,m(3)=5,m(4)=15$ for $p_0(2)=p_0(3)=2$ $p_0(4)=5$.)

  • 2
    $\begingroup$ The wikipedia page for p-groups has asymptotics, but it seems no formula is known. I would be surprised if the existence of a formula is known but not the formula itself. Edit: In fact, since the asymptotics depend on $p$ I think such a $p_0$ cannot exist for large $n$. $\endgroup$ – Alex Becker Dec 22 '12 at 21:59

The answer is no. For $n \ge 5$ the number of isomorphism classes of groups of order $p^n$ increases with $p$.

The Higman PORC conjecture (polynomial on residue classes) is that, for large enough $p$, this number is a polynomial function of $p$, and of the value of $p$ modulo $n_p$ for some finite number of constants $n_p$. For example, for $n=5$ and $p \ge 5$, the number of groups is

$2p + 61 + 2 \gcd(p-1, 3) + \gcd(p-1, 4)$.

The PORC conjecture has been proved for $n=5,6,7$, but is open for $n=8$.

Recent work of Vaughan-Lee and du Sautoy suggests that the conjecture is likely to be false for large $n$. See users.ox.ac.uk/~vlee/PORC/porcsurvey.pdf

  • $\begingroup$ It appears that newer results by Vaughan-Lee settle the case $n=8$. $\endgroup$ – Hagen von Eitzen Jan 16 '16 at 15:32
  • $\begingroup$ Isn't that just for groups of exponent $p$? $\endgroup$ – Derek Holt Jan 16 '16 at 15:37

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