A376349 Number of isomorphism classes k of groups G of order p*2^n when G contains a unique Sylow p subgroup and the maximal 2^m dividing p-1 is such that 2^m >= 2^n.

1, 2, 5, 15, 54, 247, 1684, 21820, 1118964

Offset: 0

Author

Miles Englezou, Sep 19 2024

Keywords

• nonn
• more
• hard

Comments

A Sylow p subgroup is a subgroup of order p^r that necessarily exists when r is a maximal power of p. It is not necessarily unique, but when it is unique it is normal in G.

The condition that G of order p*2^n contains a unique Sylow p subgroup places an upper bound on the number of isomorphism classes of G; it is equivalent to stating that the minimal 2^r such that 2^r == 1 (mod p) be such that 2^r > 2^n. The condition that the maximal 2^m dividing p-1, i.e. for p == 1 (mod 2^m), is such that 2^m >= 2^n ensures a lower bound which is equal to the upper bound. See the Miles Englezou link for a proof.

If we relax the two conditions and just consider an arbitrary odd prime p and the number of isomorphism classes for |G| = p*2^n, it is likely that the set of such numbers is unique to p. Since every odd prime has a minimal 2^r such that 2^r == 1 (mod p) (a consequence of Fermat's little theorem), when 2^r = 2^n for |G| = p*2^n, the number of isomorphism classes will differ from a(n) due to the existence of groups where the Sylow p subgroup is not unique.

Examples

			a(2) = 5 since D_(p*2^2), C_(p*2^2), C_(p*2^1) x C_2, and two semidirect products C_p : C_4 are all the groups of order p*2^2 for p satisfying the two conditions.
Table showing minimal 2^r and maximal 2^m (as defined in the Comments) for some primes:
---------------------------------------------------------------------------
p |      Minimal 2^r == 1 (mod p)       |   Maximal 2^m, p == 1 (mod 2^m)  |
---------------------------------------------------------------------------
2 |             2^0  = 1                |              2^0 = 1             |
3 |             2^2  = 4                |              2^1 = 2             |
5 |             2^4  = 16               |              2^2 = 4             |
7 |             2^3  = 8                |              2^1 = 2             |
11|             2^10 = 1024             |              2^1 = 2             |
13|             2^12 = 4096             |              2^2 = 4             |
17|             2^8  = 256              |              2^4 = 16            |
19|             2^18 = 262144           |              2^1 = 2             |
23|             2^11 = 2048             |              2^1 = 2             |
29|             2^28 = 268435456        |              2^2 = 4             |
31|             2^5  = 32               |              2^1 = 2             |
37|             2^36 = 68719476736      |              2^2 = 4             |
---------------------------------------------------------------------------
Table of primes satisfying 2^r > 2^n, and 2^m >= 2^n:
-------------------------------------------------------------------------------
   2^n   |                          primes                           |   a(n)  |
-------------------------------------------------------------------------------
2^0 = 1  |  all primes                                    = A000040  | 1       |
2^1 = 2  |  all primes > 2                                = A065091  | 2       |
2^2 = 4  |  5, 13, 17, 29, 37, 41, 53, ...                = A002144  | 5       |
2^3 = 8  |  17, 41, 73, 89, 97, 113, 137, ...             = A007519  | 15      |
2^4 = 16 |  17, 97, 113, 193, 241, 257, 337 ...           = A094407  | 54      |
2^5 = 32 |  97, 193, 257, 353, 449, 577, 641, ...         = A133870  | 247     |
2^6 = 64 |  193, 257, 449, 577, 641, 769, 1153, ...       = A142925  | 1684    |
2^7 = 128|  257, 641, 769, 1153, 1409, 2689, 3329, ...    = A208177  | 21820   |
2^8 = 256|  257, 769, 3329, 7937, 9473, 14081, 14593 ...  = A105131  | 1118964 |
-------------------------------------------------------------------------------
		

Links

• Miles Englezou, Proof

Crossrefs

Cf. A000001, A000040, A065091, A002144, A007519, A094407, A133870, A142925, A208177, A105131, A139669, A014664, A023506, A058384.

Programs

GAP

S:=[];
for i in [0..8] do
    n:=7681*2^i; # 7681 is an appropriate prime for reproducing up to a(8)
    S:=Concatenation(S,[NrSmallGroups(n)]);
od;
Print(S);

Formula

a(n) = A000001(p*2^(n)) for every p satisfying the two conditions mentioned in Comments.