Subgroup structure of groups of order 3.2^n

Numerical information on counts of subgroups by order
By Lagrange's theorem, the only possible orders of subgroups are of the form $$2^r, 0 \le r \le n$$, and $$3 \cdot 2^r, 0 \le r \le n$$.

Here is some information on the existence and counts of subgroups of various orders:


 * Congruence condition on number of subgroups of given prime power order: If $$p$$ is a prime and $$p^r$$ divides the order of the group, the number of subgroups of order $$p^r$$ is congruent to 1 mod $$p$$.
 * Case $$p = 2$$: 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 $$2^r$$ is odd for each fixed $$r$$ satisfying $$0 \le r \le n$$.
 * Case $$p = 3$$: 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.
 * Case $$p = 2$$: The number of 2-Sylow subgroups (in this case, subgroups of order $$2^n$$ forming a single conjugacy class) divides 3, hence it is either 1 or 3. If it is 1, the 2-Sylow subgroup is a normal Sylow subgroup and the whole group is an internal semidirect product of that by the action of the 3-Sylow subgroup. If the action is trivial, we get a finite nilpotent group, otherwise we get a non-nilpotent group.
 * Case $$p = 3$$: The number of 3-Sylow subgroups (in this case, all isomorphic to cyclic group:Z3 and all forming a single conjugacy class) is one of the numbers $$1,2,4,8,\dots,2^n$$. Combining with the congruence condition, we can eliminate odd powers of 2, so we get $$1,4,16,\dots,2^l$$ where $$l$$ is $$n$$ or $$n - 1$$ depending on whether $$n$$ is even or odd.
 * 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) = (number of subgroups of order $$2^n$$) = 1. Also, we get that, for each $$1 \le r \le n - 1$$, (number of subgroups of order $$2^r$$) = (number of subgroups of order $$3 \cdot 2^r$$).
 * In the special case of a finite abelian group, we have subgroup lattice and quotient lattice of finite abelian group are isomorphic. Thus, we get that (number of subgroups of order $$2^r$$) = (number of subgroups of order $$2^{n-r}$$) = (number of subgroups of order $$3 \cdot 2^r$$) = (number of subgroups of order $$3 \cdot 2^{n-r}$$), for each $$1 \le r \le n - 1$$. Also, we get (number of subgroups of order 3) = (number of subgroups of order $$2^n$$) = 1.
 * Finite supersolvable implies subgroups of all orders dividing the group order: For any finite supersolvable group, there are subgroups of every possible order, i.e., there are subgroups of order $$2^r$$ and $$3 \cdot 2^r$$ for $$0 \le r \le n$$.