A051532 The abelian orders (or abelian numbers): numbers m such that every group of order m is abelian.

1, 2, 3, 4, 5, 7, 9, 11, 13, 15, 17, 19, 23, 25, 29, 31, 33, 35, 37, 41, 43, 45, 47, 49, 51, 53, 59, 61, 65, 67, 69, 71, 73, 77, 79, 83, 85, 87, 89, 91, 95, 97, 99, 101, 103, 107, 109, 113, 115, 119, 121, 123, 127, 131, 133, 137, 139, 141, 143, 145, 149, 151, 153, 157, 159, 161

Offset: 1

Author

Des MacHale, Dec 11 1999

Keywords

• nonn
• nice
• easy

Comments

Except for a(2)=2 and a(4)=4, all of the terms in the sequence are odd. This is because of the existence of a non-abelian dihedral group of order 2m for each m > 2.

Cubefree terms of A056867; A212793(a(n)) = 1. - Reinhard Zumkeller, Jun 28 2013

See similar comment with "squarefree terms" in A003277 (Donald J. McCarthy link). - Bernard Schott, Feb 20 2023

Examples

			a(4) = 4 because every group of order 4 is abelian.
These two abelian groups of order 4 are the cyclic group C_4 and the Klein four-group = C_2 X C_2, the smallest non-cyclic abelian group. - _Bernard Schott_, Feb 21 2023
		

References

• W. R. Scott, Group Theory, Dover, 1987, page 217.

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..10000 (terms 1..1000 from T. D. Noe)
• Donald J. McCarthy, A survey of partial converses to Lagrange's theorem on finite groups, Transactions of the New York Academy of Sciences, 1971, page 592.
• J. Pakianathan and K. Shankar, Nilpotent Numbers, Amer. Math. Monthly, 107, August-September 2000, pp. 631-634.
• Index entries for sequences related to groups

Crossrefs

Subsequence of A056867 and supersequence of A003277.

Complement of A060652.

Intersection of A004709 and A056867.

Cf. A064899, A212793.

Programs

Haskell

a051532 n = a051532_list !! (n-1)
a051532_list = filter ((== 1) . a212793) a056867_list
-- Reinhard Zumkeller, Jun 28 2013

Mathematica

okQ[n_] := Module[{f, lf, p, e, v}, f = FactorInteger[n]; lf = Length[f]; p = f[[All, 1]]; e = f[[All, 2]]; If[AnyTrue[e, # > 2&], Return[False]]; v = p^e; For[i = 1, i <= lf, i++, For[j = i+1, j <= lf, j++, If[Mod[v[[i]], p[[j]]]==1 || Mod[v[[j]], p[[i]]]==1, Return[False]]]]; Return[True]];
Select[Range[200], okQ] (* Jean-François Alcover, May 03 2012, after PARI, updated Jan 10 2020 *)

PARI

is(n)=my(f=factor(n),v=vector(#f[,1])); for(i=1,#v, if(f[i,2]>2, return(0), v[i]=f[i,1]^f[i,2])); for(i=1,#v, for(j=i+1,#v, if(v[i]%f[j,1]==1 || v[j]%f[i,1]==1, return(0)))); 1 \\ Charles R Greathouse IV, Feb 13 2011

Python

from sympy import factorint
def ok(n):
    if n == 1: return True
    f = factorint(n)
    p, e = f.keys(), f.values()
    if max(e) >= 3: return False
    return all((pi**k)%pj!=1 for pi in p for pj in p if pj!=pi for k in range(1, f[pi]+1))
print([k for k in range(1, 162) if ok(k)]) # Michael S. Branicky, Feb 20 2023

Formula

m must be cubefree and its prime divisors must satisfy certain congruences.

Let the prime factorization of m be p1^e1 * ... * pr^er. Then m is in this sequence if ei < 3 for all i and pi^k is not congruent to 1 (mod pj) for all i and j and 1 <= k <= ei. - T. D. Noe, Mar 25 2007