Subgroup structure of groups of order 48

Numerical information on counts of subgroups by order
Note that, by Lagrange's theorem, the order of any subgroup must divide the order of the group. Thus, the order of any proper nontrivial subgroup is one of the numbers 2,4,8,16,3,6,12,24.

Here are some observations on the number of subgroups of each order:


 * Congruence condition on number of subgroups of given prime power order: The number of subgroups of order 2 is congruent to 1 mod 2 (i.e., it is odd). The same is true for the number of subgroups of order 4, the number of subgroups of order 8, and the number of subgroups of order 16. Also, the number of subgroups of order 3 is congruent to 1 mod 3.
 * Sylow implies order-conjugate, and hence Sylow number equals index of Sylow normalizer. In particular, it divides the index of the Sylow subgroup. Combining with the congruence condition, we obtain that the number of 2-Sylow subgroups (i.e., subgroups of order 16) is either 1 or 3, and the number of 3-Sylow subgroups (i.e., subgroups of order 3) is either 1, 4, or 16.
 * In the case of a finite nilpotent group, the number of subgroups of each order equals the product of the number of subgroups of order each of its maximal prime power divisors, in the corresponding Sylow subgroup. In particular, we get (number of subgroups of order 3) = 1, (number of subgroups of order 6) = (number of subgroups of order 2), (number of subgroups of order 12) = (number of subgroups of order 4), (number of subgroups of order 24) = (number of subgroups of order 8), and (number of subgroups of order 16) = 1.
 * In the special case of a finite abelian group, we have (number of subgroups of order 3) = (number of subgroups of order 16) = 1, (number of subgroups of order 2) = (number of subgroups of order 8) = (number of subgroups of order 6) = (number of subgroups of order 24), and (number of subgroups of order 4) = (number of subgroups of order 12).