Subgroup structure of groups of order 64

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 2 (i.e., it is odd). Since the non-normal subgroups occur in conjugacy classes whose size is a nontrivial power of 2, the number of normal subgroups is congruent to 1 mod 2. In particular, for orders 2 and 4, 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 2, 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 8 is congruent to 1 mod 2 (i.e., it is odd). Hence, the number of abelian normal subgroups of order 8 is also congruent to 1 mod 2 (i.e., it is odd).
 * Congruence condition on number of abelian subgroups of prime-fourth order and existence of abelian normal subgroups of small prime power order: This gives us that the number of abelian subgroups of order 16 is congruent to 1 mod 2 (i.e., it is odd). Hence, the number of abelian normal subgroups of order 16 is also congruent to 1 mod 2 (i.e., it is odd).
 * Congruence condition on number of abelian subgroups of prime index: This gives us that the number of abelian subgroups of order 32 is congruent to 1 mod 2 (i.e., it is odd) or equal to 0. Hence, the number of abelian normal subgroups of order 32 is also congruent to 1 mod 2 (i.e., it is odd) or equal to 0.
 * Index two implies normal, so all the abelian subgroups of order 32 are normal. Thus the count for abelian subgroups of order 8 is the same as the count for abelian normal subgroups of order 32.
 * 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 2 = number of abelian normal subgroups of order 2 = number of abelian subgroups of order 32 = number of abelian normal subgroups of order 32. Separately, we have number of abelian subgroups of order 4 = number of abelian normal subgroups of order 4 = number of abelian subgroups of order 16 = number of abelian normal subgroups of order 16. Finally, we also have number of abelian subgroups of order 8 = number of abelian normal subgroups of order 8.
 * The "number of abelian normal subgroups" columns depend only on the Hall-Senior genus, i.e., two groups with the same Hall-Senior genus have the same "number of abelian normal subgroups" of each order. The Hall-Senior genus is the part of the Hall-Senior symbol excluding the very final subscript, so for instance $$64\Gamma_2a_1$$ and $$64\Gamma_2a_2$$ both belong to the Hall-Senior genus $$64\Gamma_2a$$.