A338757 Number of splitting-simple groups of order n; number of nontrivial groups of order n that are not semidirect products of proper subgroups.

0, 1, 1, 1, 1, 0, 1, 2, 1, 0, 1, 0, 1, 0, 0, 2, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 5, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 19, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 2, 0, 1, 0, 0, 0, 0, 0, 1

Offset: 1

Author

Jianing Song, Nov 07 2020

Keywords

• nonn
• hard

Comments

The other names for groups of this kind include "semidirectly indecomposable groups" or "inseparable groups". Note that the following are equivalent definitions for a nontrivial group to be a splitting-simple group:

- It is not the (internal) semidirect product of proper subgroups;

- It is not isomorphic to the (external) semidirect product of nontrivial groups;

- It has no proper nontrivial normal subgroups with a permutable complement.

- It is the non-split extension of every proper nontrivial normal subgroup by the corresponding quotient group.

Also note that being simple is a stronger condition than being splitting-simple, while being directly indecomposable (see A090751) is weaker.

a(p^e) >= 1 since C_p^e cannot be written as the semidirect product of proper subgroups. For e >= 3, a(2^e) >= 2 by the existence of the generalized quaternion group of order 2^e, which is the only non-split extension of C_2^(e-1) by C_2 other than C_2^e.

The smallest numbers here with a(n) > 0 that are not prime powers are 48, 60, 120, 144, 168, 192, 240, 320, 336, 360 and so on. Are there any odd numbers n that are not prime powers satisfying a(n) > 0 ?

Conjecture: a(n) = 0 for squarefree n which is not a prime.

The conjecture that a(n) = 0 for nonprime squarefree n is true. Proof: It is known that every group G of squarefree order is supersolvable; hence G contains a normal series with prime cyclic factors. Since every Sylow subgroup of G is prime cyclic, these cyclic factors are isomorphic to the Sylow subgroups of G. Let P be one such factor; then for an appropriate M in G, P = G/M, where |G| = |P|*|M|. By the Schur-Zassenhaus theorem, G is a semidirect product of M and P, and a(n) = 0 when n is squarefree. - Miles Englezou, Oct 24 2024

Examples

			a(48) = 1 because the binary octahedral group, which is of order 48, cannot be written as the semidirect product of proper subgroups.
a(16) = 2, and the corresponding groups are C_16 and Q_16 (generalized quaternion group of order 16).
a(81) = 2, and the corresponding groups are C_81 and SmallGroup(81,10).
a(64) = 19, and the corresponding groups are SmallGroup(64,i) for i = 1, 11, 13, 14, 19, 22, 37, 43, 45, 49, 54, 79, 81, 82, 160, 168, 172, 180 and 245.
For n = 60 or 168, the unique simple group is the only group of order n that cannot be written as the semidirect product of proper subgroups, hence a(60) = a(168) = 1. [The unique simple groups are respectively Alt(5) and PSL(2,7). - _Bernard Schott_, Nov 08 2020]
For n = 12, we have C_12 = C_3 X C_4, C_6 X C_2 = C_6 X C_2, D_6 = C_6 : C_2, Dic_12 = C_3 : C_4 and A_4 = (C_2 X C_2) : C_3, all of which can be written as the semidirect product of nontrivial groups. So a(12) = 0.
		

Links

• Jianing Song, Table of n, a(n) for n = 1..255
• Tim Dokchitser, Group extensions
• The Group Properties Wiki, Splitting-simple group
• The Group Properties Wiki, Permutable complements
• Jianing Song, List of splitting-simple groups of order up to 255 by ID in GAP's SmallGroup library
• Index entries for sequences related to groups

Crossrefs

Cf. A000001, A090751 (number of directly indecomposable groups of order n), A001034, A120944.

Programs

GAP

IsSplittingSimple := function(G)
  local c, l, i;
  c := NormalSubgroups(G);
  l := Length(c);
  if l > 1 then
    for i in [2..l-1] do
    if Length(ComplementClassesRepresentatives(G,c[i])) > 0 then
      return false;
    fi;
    od;
    return true;
  else
    return false;
  fi;
end;
A338757 := n -> Length(AllSmallGroups( n, IsSplittingSimple ));

Formula

For primes p != q:

a(p) = a(p^2) = 1; a(p^3) = 2 for p = 2, 1 otherwise;

a(p^4) = 2 for p = 2 or 3, 1 otherwise;

a(pq) = 0;

a(4p) = a(8p) = 0, p > 2.

a(n) <= A090751(n) for all n, and the equality holds if n = 1, p, p^2 for primes p or n = pq for primes p < q and p does not divide q-1.

a(A001034(k)) >= 1, since A001034 lists the orders of (non-Abelian) simple groups.

a(A120944(n)) = 0. - Miles Englezou, Oct 24 2024