A385480 Sum of the orders of every automorphism group of order n.

1, 1, 2, 8, 4, 8, 6, 212, 54, 24, 10, 64, 12, 48, 8, 20936, 16, 552, 18, 132, 54, 120, 22, 856, 500, 168, 11844, 216, 28, 192, 30, 10047248, 20, 288, 24, 2856, 36, 360, 180, 2128, 40, 468, 42, 520, 216, 528, 46, 61696, 2058, 13080, 32, 876, 52, 320400, 150, 3960

Offset: 1

Author

Miles Englezou, Jun 30 2025

Keywords

• nonn

Comments

Let G be a group of order n, let N = {1, 2, ..., n}, and let f: G -> N be a bijection whereby f(G) = I is an index set of G. An automorphism phi of G is a permutation of N via f(phi(G)). It is tempting to ask the question 'how many permutations of N obey the group laws?'. However this question is not well-defined since it would require there being a natural single choice of bijection for every group of order n, which in general does not exist. Enumerating permutations of N which are automorphisms for every isomorphism class G will therefore depend on the choice of bijection for G. a(n) is the upper bound for all such enumerations of permutations of size n since a(n) is either: the maximum enumeration when the choice of bijections ensures that all permutations are distinct; or a(n) is the enumeration including all multiplicities when the choice of bijections leads to permutations which are not distinct.

Examples

			a(3) = 2 since for the one group of order 3, C_3, |Aut(C_3)| = 2.
a(4) = 8 since for the two groups of order 4, C_4 and C_2 x C_2, |Aut(C_4)| + |Aut(C_2 x C_2)| = 2 + 6 = 8.
a(8) = 212 since for the five groups of order 8, the sum of the orders of their automorphism groups is 4 + 8 + 8 + 24 + 168 = 212.
		

Crossrefs

Cf. A137316.

Programs

GAP

a := function(n)
local T, k;
T := [];
for k in [1..NrSmallGroups(n)] do
    T := Concatenation(T, [Order(AutomorphismGroup(SmallGroup(n,k)))]);
od;
return Sum(T);
end;

Formula

a(n) is the sum of the k-th row in A137316.