Subgroup structure of groups of order 36

Numerical information on counts of subgroup by order
The number 36 has only two prime factors: 2 and 3:

$$\! 36 = 2^2 \cdot 3^2 = 4 \cdot 9$$

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 2,4,3,6,9,18.

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


 * Congruence condition on number of subgroups of given prime power order: This states that if $$p^r$$ divides the order of the group, then the number of subgroups of order $$p^r$$ is congruent to 1 mod $$p$$.
 * For $$p = 2$$: We obtain that the number of subgroups of order 2 is congruent to 1 mod 2 (i.e., it is odd). Similarly, the number of subgroups of order 4 is congruent to 1 mod 2 (i.e., it is odd).
 * For $$p = 3$$: We obtain that the number of subgroups of order 3 is congruent to 1 mod 3. Similarly, the number of subgroups of order 9 is congruent to 1 mod 3.
 * Sylow implies order-conjugate yields that Sylow number equals index of Sylow normalizer: In particular, the Sylow number divides the index of the Sylow subgroup. We get:
 * For $$p = 2$$: We obtain that the number of subgroups of order 4 divides 9. It must thus be one of the numbers 1, 3, and 9.
 * For $$p = 3$$: We obtain that the number of subgroups of order 9 divides 4. It must thus be one of the numbers 1 and 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 particular, we get (number of subgroups of order 4) = (number of subgroups of order 9) = 1. Separately, (number of subgroups of order 2) = (number of subgroups of order 18) and (number of subgroups of order 3) = (number of subgroups of order 12).
 * Nilpotent of cube-free order implies abelian: Since the order is cube-free, all nilpotent groups of the order are abelian.