Subgroup structure of groups of order 96

Numerical information on counts of subgroups by order
There are two distinct prime factors of this number, as follows:

$$\! 96 = 2^5 \cdot 3 = 32 \cdot 3$$

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 2,4,8,16,32,3,6,12,24,48.

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, the number of subgroups of order 16, and the number of subgroups of order 32.
 * The number of subgroups of order 3 is congruent to 1 mod 3.
 * By the fact that Sylow implies order-conjugate, we obtain that Sylow number equals index of Sylow normalizer, and in particular, divides the index of the Sylow subgroup. Combined with the congruence condition, we get the following: the number of 2-Sylow subgroups is either 1 or 3, and the number of 3-Sylow subgroups is one of the number 1, 4, and 16.
 * In the case of a finite nilpotent group, the number of subgroups of a given order is the product of the number of subgroups of order equal to 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), (number of subgroups of order 48) = (number of subgroups of order 16), and (number of subgroups of order 32) = 1.
 * In the special case of a finite abelian group, we have (number of subgroups of order 3) = (number of subgroups of order 32) = 1. We also have (number of subgroups of order 2) = (number of subgroups of order 16) = (number of subgroups of order 6) = (number of subgroups of order 48). Separately, we have (number of subgroups of order 4) = (number of subgroups of order 8) = (number of subgroups of order 12) = (number of subgroups of order 24).