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Integers k such that there exists a group of order k with exactly k subgroups.
3

%I #47 Feb 18 2024 11:47:03

%S 1,2,6,8,28,40,48,54,72,96,100,104,128,132,144,160,176,180,192,216,

%T 240,252,260,288,324,336,368,384,416,456,480,496

%N Integers k such that there exists a group of order k with exactly k subgroups.

%C Powers of 4 cannot appear in this sequence. This is because for a group of order p^n, the number of subgroups of order p^k is congruent to 1 mod p, for 0 <= k <= n. It follows from p=2 and Lagrange's theorem that the number of subgroups of order 2^n for n even is congruent to 1 mod 2, i.e. not equal to 2^n. - _Robin Jones_, Feb 17 2024

%C a(33) >= 512. The smallest term strictly larger than 512 is 560. -_Robin Jones_, Feb 18 2024

%e 1 is a term since the trivial group (order 1) has exactly 1 subgroup.

%e 2 is a term since the cyclic group C_2 has exactly 2 subgroups.

%e 6 is a term since the symmetric group S_3 has exactly 6 subgroups.

%o (Magma, to get the terms up to 100)

%o i:=1;

%o while i lt 100 do // terms up to 100

%o for G in SmallGroups(i) do

%o if #AllSubgroups(G) eq i then

%o i; break;

%o end if;

%o ; end for;

%o i:=i+1;

%o end while;

%Y Cf. A018216, A061034.

%K nonn,more

%O 1,2

%A _Robin Jones_, Dec 29 2023