Article 3Z3S6 Number of groups of prime power order

Number of groups of prime power order

by
John
from John D. Cook on (#3Z3S6)

John Baez left a comment on my post on group statistics saying

It's known that the number of groups of order p^n for prime p is p^{2n^3/27+O(n^(8/3))}. It might be fun to compare this to what Mathematica says.

Here goes. First let's let p = 2. Mathematica's FiniteGroupCount function tops out at n = 10. And for the range of values Mathematica does support, 2^{2n^3/27} grossly overestimates the number of groups.

 Table[{FiniteGroupCount[2^n], 2^((2./27) n^3)}, {n, 1, 11}] {{1, 1.05269}, {2, 1.50795}, {5, 4.}, {14, 26.7365}, {51, 612.794}, {267, 65536.}, {2328, 4.45033*10^7}, {56092, 2.61121*10^11}, {10494213, 1.80144*10^16}, {49487365422, 1.98847*10^22}, {FiniteGroupCount[2048], 4.7789*10^29}}

OEIS has entries for the sizes of various groups, and the entry for powers of 2 confirms the asymptotic formula above. Maybe it's not a good approximation until n is very large. (Update: Here's a reference for the asymptotic expression for all p.)

The OEIS entry for number of groups of order powers of 5 has an interesting result:

For a prime p >= 5, the number of groups of order p^n begins 1, 1, 2, 5, 15,
61 + 2*p + 2*gcd (p - 1, 3) + gcd (p - 1, 4),
3*p^2 + 39*p + 344 + 24*gcd(p - 1, 3) + 11*gcd(p - 1, 4) + 2*gcd(p - 1, 5), "

We can duplicate this with Mathematica.

 Table[FiniteGroupCount[5^n], {n, 0, 6}] {1, 1, 2, 5, 15, 77, 684}

and the last two numbers match the calculations given in OEIS.

There's something interesting going on with Mathematica. It doesn't seem to know, or agree with, the formula above for groups of order p4. For example,

 Table[FiniteGroupCount[7^n], {n, 0, 6}] {1, 1, 2, 5, FiniteGroupCount[2401], 83, 860}

I get similar results when I use larger primes: it can't handle the fourth power.

 Table[FiniteGroupCount[389^n], {n, 0, 6}] {1, 1, 2, 5, FiniteGroupCount[22898045041], 845, 469548}

The results for n = 5 and 6 agree with OEIS.

Is OEIS wrong about the number of groups of order p4 or should Mathematica simply return 15 but there's a gap in the software?

Also, does anybody know why the agreement with the asymptotic formula above is so bad? It's just as bad or worse for other primes that I tried.

Related posts4GJiYchuggo
External Content
Source RSS or Atom Feed
Feed Location http://feeds.feedburner.com/TheEndeavour?format=xml
Feed Title John D. Cook
Feed Link https://www.johndcook.com/blog
Reply 0 comments