A137315 a(n) is the least number m such that any finite group of order at least m has at least n automorphisms.

1, 3, 7, 7, 13, 13, 19, 19, 31, 31, 31, 31, 43, 43, 43, 43, 61, 61, 61, 61, 67, 67, 67, 67, 91, 91, 91, 91, 91, 91, 91, 91, 121, 121, 121, 121, 127, 127, 127, 127, 151, 151, 151, 151, 151, 151, 151, 151

Offset

1,2

Comments

a(n) <= (n-1)^(n + (n-2)[log_2(n-1)]) for n > 4 [Ledermann, Neumann, Thm. 6.6].

a(n) is odd [MacHale, Sheehy, Thm. 15].

a(2n-1) = a(2n) for 1 < n < 204 [ibid.].

The case of cyclic groups shows that a(n)>=A139795(n). This inequality can be strict: if M denotes the Mathieu group M_{22} of order 2^7.3^2.5.7.11, then Aut(12.M) = M.2, so that a(2^8.3^2.5.7.11 + 1) > 2^9.3^3.5.7.11, but A139795(2^8.3^2.5.7.11 + 1) = 2.3.5.7^2.11.13.23 + 1 < 2^9.3^3.5.7.11.

Links

Table of n, a(n) for n=1..48.

John N. Bray and Robert A. Wilson, On the orders of automorphism groups of finite groups, Bull. London Math. Soc. 37 (2005) 381--385.

W. Ledermann, B. H. Neumann, On the order of the automorphism group of a finite group. I, Proc. Roy. Soc. Lon., 233A(1195) (1956), 494-506

D. MacHale and R. Sheehy, Finite groups with few automorphisms, Math. Proc. Roy. Irish Acad., 104A(2) (2004), 231--238.

Benjamin Sambale, On a theorem of Ledermann and Neumann, arXiv:1909.13220 [math.GR], 2019.

Example

a(3) = a(4) = 7 because every finite group with at least 7 elements has at least 4 automorphisms while the cyclic group of order 6 has only phi(6)=2 automorphisms.

Crossrefs

Different from A139795 (see Comments).

See also A340521.

Sequence in context: A024612 A227025 A073881 * A139795 A064829 A290649

Adjacent sequences: A137312 A137313 A137314 * A137316 A137317 A137318

Keyword

nonn,hard,more

Author

Benoit Jubin, Apr 06 2008, May 26 2008

Status

approved