7

While bounds on the number of isomorphism classes of groups of order $p^n$ where $p$ is prime have been known for quite a while (such as the work of Higman$^{[1]}$ and Sims$^{[2]}$) which give us the bounds for $f(n,p)$ (which returns the number of isomorphism classes of groups of order $p^n$) as follows:

$f(n,p) = p^{An^3} \operatorname{where} A(n,p) = \frac{2}{27} + O(n^{-1/3})$

However, while this tells us that for a fixed $p$ the value of $f(n,p)$ grows very rapidly with n, when I've been reading through the papers I don't think I have enough knowledge to understand why exactly they should grow this rapidly - for example, the number of abelian groups of order $p^n$ is simply the number of partitions of $n$, and while this number grows quite rapidly with n, this number is tiny in comparinson to the number of non-abelian groups of the same exponent.

Therefore, would it be possible for someone to try and explain to me some reasons for why there are so many groups of order $p^n$ for $n>2$ (as it is fairly straightforward to categorize such groups for $n=1$ and $2$), and also why there are so many such $p$-groups when compared to groups of similar sized order? (I suspect these two questions are closely linked in some respects, which is why I ask them both, although answers on one or the other are equally appreciated)

For those who would like them, the references to the papers of Higman and Sims are below:

$[1]$ - Proc. London Math. Soc. (1960) s3-10 (1): 24-30. doi: 10.1112/plms/s3-10.1.24

$[2]$ - Proc. London Math. Soc. (1965) s3-15 (1): 151-166. doi: 10.1112/plms/s3-15.1.151

Andrew D
  • 2,370
  • I am not sure that this is a plausible explanation, but there may be some correlation with the fact that the Lagrange Theorem has an inverse for $p$-groups. – Andrea Mori Jul 31 '13 at 14:12
  • 1
    One possible reason is that the automorphism group of a $p$-group will generally have order divisible by a fairly large power of $p$, which means there is a large number of possible semidirect products to be made when increasing the order. And then there are the possibilities of non-split extensions of course. – Tobias Kildetoft Jul 31 '13 at 14:13
  • I think one reason would be that the number of generators of a cyclic group of order $p^r$ is $p^{r-1}(p-1)$ which is divisible by $p^{r-1}$ - so the automorphism group will give plenty of scope for nonabelian group extensions (eg semidirect products). – Mark Bennet Jul 31 '13 at 14:16
  • @AndreaMori Are you referring to the Sylow theorems? – Andrew D Jul 31 '13 at 14:17
  • No, the converse of Lagrange for $p$-groups is not related to the Sylow theorems (the Sylow theorems are trivial for $p$-groups). – Tobias Kildetoft Jul 31 '13 at 14:22
  • 9
    As to why there are so many $p$-groups, just look at the paper of Higman and see how they are constructed. It is harder to explain why $p$-grpoups predominate over all groups, and it has not been proved to be true, so it may not be! – Derek Holt Jul 31 '13 at 14:25
  • This other SE discussion seems relevant: http://math.stackexchange.com/questions/241369/more-than-99-of-groups-of-order-less-than-2000-are-of-order-1024?rq=1 – Andrea Mori Jul 31 '13 at 14:33
  • @TobiasKildetoft Ah, so this is just the literal converse of Lagrange's theorem which happens to hold for p-groups (as a consequence of Cauchy's theorem, I presume)? – Andrew D Jul 31 '13 at 14:36
  • @AndrewD : To prove that a $p$-group $G$ has subgroups for every divisor of its order, show first that the center $Z(G)$ is non-trivial and then use the quotient $G/Z(G)$ to set up an induction step. – Andrea Mori Jul 31 '13 at 14:39
  • @AndreaMori I fail to see how I can use the quotient in an induction step, seeing as I don't believe it is necessarily a subgroup of G, unless I'm completely misunderstanding what to do here. – Andrew D Jul 31 '13 at 15:00
  • @AndrewD: it is a standard technique. If the center is non trivial, the quotient is strictly smaller. then you can construct subgroups of $G$ by pulling back subgroups of the quotients (which exist of every order, by inductive hypothesis) along the quotient map. This doesn't work quite right if $|Z(G)|\geq p^2$ (why?) but it is easy to fix it. – Andrea Mori Jul 31 '13 at 15:35
  • @AndreaMori Unfortunately, I'm still fairly confused here - what exactly do you mean by your sentence starting "then you can construct subgroups of $G$..."? I feel like there may be a lack of knowledge on my behalf here - while I know of the quotient map, I don't know anyway of relating it to constructing subgroups (again, I feel quite confused here, apologies). – Andrew D Jul 31 '13 at 16:29
  • @AndrewD : we are way off topic here and I suggest that you acquire more expertise on basic group theory by studying some tetbook. The point is that if $\phi:G\rightarrow H$ is a surjective homomorphism of (finite) groups, and $H^\prime<H$ then $\phi^{-1}(H^\prime)$ is a subgroup of $G$ with $|H^\prime|\cdot|\ker(\phi)|$ elements. – Andrea Mori Jul 31 '13 at 17:07

1 Answers1

2

I don't think that a precise answer to your question exists (up to this moment). More or less your question is Question 2 in Mann's paper Some Questions about $p$-Groups (you can get it freely on the net), I find it better to refer you to Sections 1 and 2 of that paper, than any word I can say.

user1729
  • 31,015
Yassine Guerboussa
  • 1,544