Subgroup structure of groups of prime-cube order

Note that the tables here work correctly only for groups of order $$p^3$$ where $$p$$ is an odd prime. The case $$p = 2$$ behaves differently. For that case, see subgroup structure of groups of order 8.

Number of subgroups per isomorphism type
The number in each column is the number of subgroups in the given group (by row) of the isomorphism type of the (column) group:

Number of conjugacy classes of subgroups per isomorphism type
The number in each column is the number of conjugacy classes of subgroups in the given group (by row) of the isomorphism type of the (column) group:

Number of automorphism classes of subgroups per isomorphism type
The number in each column is the number of automorphism classes of subgroups in the given group (by row) of the isomorphism type of the (column) group:

Number of subgroups per order
We have the following:


 * Congruence condition on number of subgroups of given prime power order tells us that the number of subgroups of order $$p$$ is congruent to 1 mod $$p$$, and so is the number of subgroups of order $$p^2$$. Since the non-normal subgroups form conjugacy classes of size equal to a nontrivial power of $$p$$, the number of normal subgroups of each order is also congruent to 1 mod $$p$$.
 * All the subgroups of order $$p^2$$ are normal, because prime power order implies nilpotent and nilpotent implies every maximal subgroup is normal. In particular, we obtain that (number of subgroups of order $$p^2$$) = (number of normal subgroups of order $$p^2$$).
 * The subgroups of order $$p^2$$ all contain the Frattini subgroup, and thus correspond, via the fourth isomorphism theorem, to the maximal subgroups of the Frattini quotient, which is an elementary abelian group of order $$p^r, 1 \le r \le 3$$, where $$r$$ is the minimum size of generating set. The number of subgroups is $$(p^r - 1)/(p - 1)$$, which could be one of the numbers $$1, p + 1, p^2 + p + 1$$.
 * The normal subgroups of order $$p$$ are precisely the subgroups of order $$p$$ inside the socle, which is the first omega subgroup of the center. In particular, it is an elementary abelian group of order $$p^s, 1 \le s \le 3$$, and the number of subgroups of order $$p$$ in it is $$(p^s - 1)/(p - 1) = p^{s-1} + p^{s-2} + \dots + p + 1$$. Thus, the number of normal subgroups of order $$p$$ is $$1, p + 1, p^2 + p + 1$$. For a non-abelian group, it must be 1.
 * For abelian groups, every subgroup is normal. Moreover, we have subgroup lattice and quotient lattice of finite abelian group are isomorphic, so we get (number of subgroups of order $$p$$) = (number of normal subgroups of order $$p$$) = (number of subgroups of order $$p^2$$) = (number of normal subgroups of order $$p^2$$).