Subgroup structure of groups of order 72

Numerical information on counts of subgroups by order
The prime factorization is as follows:

$$\! 72 = 2^3 \cdot 3^2 = 8 \cdot 9$$

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 2,4,8,3,6,12,24,9,18,36.

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, as well as for the number of subgroups of order 8. The number of subgroups of order 3 is congruent to 1 mod 3. The same is true for the number of subgroups of order 9.
 * 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. Thus, the number of subgroups of order 8 (i.e., 2-Sylow subgroups) divides 9, so combining with the congruence condition, we obtain that this number must be 1, 3, or 9. Similarly, the number of subgroups of order 9 (i.e., 3-Sylow subgroups) must divide 8. Combining with the congruence condition, we obtain that this number must be 1 or 4.
 * 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 the case of a finite abelian group, we further have that the number of subgroups of a particular order equals the number of subgroups whose index equals that order, because subgroup lattice and quotient lattice of finite abelian group are isomorphic.