|
%I #30 Sep 28 2023 15:23:39
%S 1,1,1,2,1,2,1,2,1,3,1,2,1,2,2,2,1,4,1,4,1,2,2,2,1,4,1,2,1,3,1,2,1,2,
%T 2,2,1,2,1,4,1,4,1,2,2,3,1,4,1,3,2,2,1,6,1,2,2,2,1,4,1,4,1,6,1,4,1,6,
%U 1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,2,2,1,4,1,2,2,2,1,4,1,2,3,2,1,4,1,2,2,2,1,6
%N Irregular table read by rows: T(m,n) is the number of non-isomorphic groups G such that G is the semidirect product of C_m and C_n, where C_m is a normal subgroup of G and C_n is a subgroup of G, 1 <= n <= A002322(m).
%C The semidirect product of C_m and C_n has group representation G = <x, y|x^m = y^n = 1, yxy^(-1) = x^r>, where r is any number such that r^n == 1 (mod m). Two groups G = <x, y|x^m = y^n = 1, yxy^(-1) = x^r> and G' = <x, y|x^m = y^n = 1, yxy^(-1) = x^s> are isomorphic if and only if there exists some k, gcd(k,n) = 1 such that r^k == s (mod m), in which case f(x^i*y^j) = x^i*y^(k*j) is an isomorphic mapping from G to G'.
%C Given m, T(m,n) only depends on the value of gcd(n,psi(m)), psi = A002322 (Carmichael lambda). So each row of A327924 is periodic with period psi(m), so we have this for an alternative version.
%C Every number k occurs in the table. By Dirichlet's theorem on arithmetic progressions, there exists a prime p such that p == 1 (mod 2^(k-1)), then T(p,2^(k-1)) = d(gcd(2^(k-1),p-1)) = k (see the formula below). For example, T(5,4) = 3, T(17,8) = 4, T(17,16) = 5, T(97,32) = 6, T(193,64) = 7, ...
%C Row m and Row m' are the same if and only if (Z/mZ)* = (Z/m'Z)*, where (Z/mZ)* is the multiplicative group of integers modulo m. The if part is clear; for the only if part, note that the two sequences {(number of x in (Z/mZ)* such that x^n = 1)}_{n>=1} and {T(m,n)}_{n>=1} determine each other, and the structure of a finite abelian group G is uniquely determined by the sequence {(number of x in G such that x^n = 1)}_{n>=1}. - _Jianing Song_, May 16 2022
%H Jianing Song, <a href="/A327925/b327925.txt">Table of n, a(n) for n = 1..8346</a> (the first 200 rows)
%H Math Overflow, <a href="https://mathoverflow.net/q/455396/494268">When are two semidirect products of two cyclic groups isomorphic</a>
%F T(m,n) = Sum_{d|n} (number of elements x such that ord(x,m) = d)/phi(d), where ord(x,m) is the multiplicative order of x modulo m, phi = A000010.
%F Equivalently, T(m,n) = Sum_{d|gcd(n,psi(m))} (number of elements x such that ord(x,m) = d)/phi(d). - _Jianing Song_, May 16 2022
%F For odd primes p, T(p^e,n) = d(gcd(n,(p-1)*p^(e-1))) = A051194((p-1)*p^(e-1),n), d = A000005; for e >= 3, T(2^e,n) = 2*(v2(n)+1) for even n and 1 for odd n, where v2 is the 2-adic valuation.
%e Table starts
%e m = 1: 1;
%e m = 2: 1;
%e m = 3: 1, 2;
%e m = 4: 1, 2;
%e m = 5: 1, 2, 1, 3;
%e m = 6: 1, 2;
%e m = 7: 1, 2, 2, 2, 1, 4;
%e m = 8: 1, 4;
%e m = 9: 1, 2, 2, 2, 1, 4;
%e m = 10: 1, 2, 1, 3;
%e m = 11: 1, 2, 1, 2, 2, 2, 1, 2, 1, 4;
%e m = 12: 1, 4;
%e m = 13: 1, 2, 2, 3, 1, 4, 1, 3, 2, 2, 1, 6;
%e m = 14: 1, 2, 2, 2, 1, 4;
%e m = 15: 1, 4, 1, 6;
%e m = 16: 1, 4, 1, 6;
%e m = 17: 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5;
%e m = 18: 1, 2, 2, 2, 1, 4;
%e m = 19: 1, 2, 2, 2, 1, 4, 1, 2, 3, 2, 1, 4, 1, 2, 2, 2, 1, 6;
%e m = 20: 1, 4, 1, 6;
%e Example shows that T(21,6) = 6: The semidirect product of C_21 and C_6 has group representation G = <x, y|x^21 = y^6 = 1, yxy^(-1) = x^r>, where r = 1, 2, 4, 5, 8, 10, 11, 13, 16, 17, 19, 20. Since 2^5 == 11 (mod 21), 4^5 == 16 (mod 21), 5^5 == 17 (mod 21), 10^5 == 19 (mod 21), there are actually four pairs of isomorphic groups, giving a total of 8 non-isomorphic groups.
%o (PARI) numord(n,q) = my(v=divisors(q),r=znstar(n)[2]); sum(i=1,#v,prod(j=1,#r,gcd(v[i],r[j]))*moebius(q/v[i]))
%o T(m,n) = my(u=divisors(n)); sum(i=1,#u,numord(m,u[i])/eulerphi(u[i]))
%o Row(m) = my(l=if(m>2,znstar(m)[2][1],1), R=vector(l,n,T(m,n))); R
%Y Cf. A327925, A002322, A000010, A000005.
%Y Cf. also A060594, A060839, A073103, A319099, A319100, A319101, A051194.
%K nonn,tabf
%O 1,4
%A _Jianing Song_, Sep 30 2019
|
