Subgroup structure of groups of order 128

Numerical information on counts of subgroups by order
We note the following:


 * Congruence condition on number of subgroups of given prime power order guarantees that the number of subgroups of a given order (among 1,2,4,8,16,32,64,128) is congruent to 1 mod 2 (i.e., it is odd). Since the non-normal subgroups occur in conjugacy classes of even sizes, we conclude the number of normal subgroups of a given order is also congruent to 1 mod 2.
 * Index two implies normal, so all the subgroups of order 64 are normal. Hence, the number of subgroups of order 64 equals the number of normal subgroups of order 64.
 * Formula for number of maximal subgroups of group of prime power order: The subgroups of order 64 are precisely the maximal subgroups, and hence they all contain the Frattini subgroup. Via the fourth isomorphism theorem, they correspond to maximal subgroups of the Frattini quotient, which is an elementary abelian group of order $$2^r$$ where $$r$$ is the minimum size of generating set for the original group. The number of such subgroups is $$(2^r - 1)/(2 - 1) = 2^r - 1$$, which is thus one of the numbers 1,3,7,15,31,63,127.
 * Formula for number of minimal normal subgroups of group of prime power order: The normal subgroups of order 2 are precisely the subgroups of order 2 contained in the socle, which, for a group of prime power order, is the first omega subgroup of the center, and is thus elementary abelian of order $$2^s$$ where $$s$$ is the rank of the center. The number of such normal subgroups is thus $$2^s - 1$$, and is thus one of the numbers 1,3,7,15,31,63,127. For a non-abelian group, it must be one of the numbers 1,3,7,15,31.
 * For an abelian group, all subgroups are normal, so the counts for number of subgroups and number of normal subgroups coincide for each order. Further, subgroup lattice and quotient lattice of finite abelian group are isomorphic, so we further have that (number of subgroups of order 2) = (number of subgroups of order 64), (number of subgroups of order 4) = (number of subgroups of order 32), and (number of subgroups of order 8) = (number of subgroups of order 16).

Counts of abelian subgroups and abelian normal subgroups

 * 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: The existence statement guarantees the existence of an abelian normal subgroup of order $$2^3 = 8$$ in a group of order $$2^7 = 128$$, because $$7 > 3(3 - 1)/2$$. The congruence condition then tells us that the number of such subgroups is odd.
 * Congruence condition on number of abelian subgroups of prime-fourth order and existence of abelian normal subgroups of small prime power order: The existence statement guarantees the existence of an abelian normal subgroup of order $$2^4 = 16$$ in a group of order $$2^7= 128$$, because $$7 > 4(4-1)/2$$. The congruence condition then tells us that the number of such subgroups is odd.
 * Abelian-to-normal replacement theorem for prime-square index guarantees that if there exists an abelian subgroup of index $$2^2$$ (or equivalently, order $$2^5 = 32$$) in a group of order $$2^7 = 128$$, then there exists an abelian normal subgroup of order $$2^5$$.
 * It also turns out experimentally that the number of abelian subgroups, if nonzero, is odd, so the congruence condition holds -- however, the general theorem is unclear.
 * Note that any subgroup of order 64 is normal, so the number of subgroups of order 64 equals the number of normal subgroups of order 64.
 * Congruence condition on number of abelian subgroups of prime index guarantees that if there exists an abelian subgroup of prime index, the number of such subgroups is odd. In fact, for a non-abelian group, the only possibilities for the number of abelian subgroups is 0, 1, or 3.