Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $p_0(2)=p_0(3)=2$ $p_0(4)=5$.)

share|improve this question
2  
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 $n$. –  Alex Becker Dec 22 '12 at 21:59

1 Answer 1

up vote 6 down vote accepted

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.