Subgroup structure of groups of order 81

Number of subgroups per order
Due to congruence condition on number of subgroups of given prime power order, all the counts of subgroups, as well as of normal subgroups, are congruent to 1 modulo 3. Further, note the following:


 * For an abelian group, the number of subgroups of a given order equals the number of normal subgroups. Moreover, because subgroup lattice and quotient lattice of finite abelian group are isomorphic, we get that (number of subgroups of order 3) = (number of normal subgroups of order 3) = (number of subgroups of order 27) = (number of normal subgroups of order 27), and separately, (number of subgroups of order 9) = (number of normal subgroups of order 9).
 * Since prime power order implies nilpotent, and nilpotent implies every maximal subgroup is normal, we have, in all cases, that the number of subgroups of order 27 = number of normal subgroups of order 27.
 * The subgroups of order 27 (all of which are normal) correspond to the maximal subgroups of the Frattini quotient, which is an elementary abelian group of order $$3^r, 1 \le r \le 4$$ where $$r$$ is the minimum size of generating set for the group. The number of such subgroups is $$(3^r - 1)/(3 - 1) = 3^{r-1} + 3^{r-2} + \dots + 3 + 1$$, and hence must be one of the numbers 1,4,13,40.
 * The normal subgroups of order 3 correspond precisely to the subgroups of order 3 in the socle, which is an elementary abelian subgroup defined as $$\Omega_1$$ of the center. If the socle has order $$3^s, 1 \le s \le 4$$, the number of normal subgroups of order 3 is $$(3^s - 1)/(3 - 1) = 3^{s-1} + 3^{s-2} + \dots + 3 + 1$$. See minimal normal implies central in nilpotent. Thus, the count of normal subgroups of order 3 must be 1, 4, 13, or 40. Moroever, for a non-abelian group, the socle can have order either 3 or 9 (cannot have order 27 or 81) so the number of normal subgroups is either 1 or 3.

Counts of abelian subgroups and abelian normal subgroups
Note the following:

The upshot is that all counts in the table below are odd.
 * Congruence condition on number of subgroups of given prime power order tells us that for any fixed order, the number of subgroups is congruent to 1 mod 3. Since the non-normal subgroups occur in conjugacy classes whose size is a nontrivial power of 3, the number of normal subgroups is congruent to 1 mod32. In particular, for orders 3 and 9, since every subgroup of that order is abelian anyway, the congruence condition tells us that the number of abelian subgroups is congruent to 1 mod 3, and so is the number of abelian normal subgroups.
 * Congruence condition on number of abelian subgroups of prime-cube order and existence of abelian normal subgroups of small prime power order: This gives us that the number of abelian subgroups of order 27 is congruent to 1 mod 3 (i.e., it is odd). Hence, the number of abelian normal subgroups of order 8 is also congruent to 1 mod 3 (i.e., it is odd).
 * Prime power order implies nilpotent and nilpotent implies every maximal subgroup is normal, so all the abelian subgroups of order 27 are normal. Thus the count for abelian subgroups of order 8 is the same as the count for abelian normal subgroups of order 27.
 * For the abelian groups: note that abelian implies every subgroup is normal and also that subgroup lattice and quotient lattice of finite abelian group are isomorphic. Thus, when the whole group is abelian, we have: number of abelian subgroups of order 3 = number of abelian normal subgroups of order 3 = number of abelian subgroups of order 27 = number of abelian normal subgroups of order 27. Separately, we have number of abelian subgroups of order 9 = number of abelian normal subgroups of order 9.