Subgroup structure of groups of order 16

Number of subgroups per isomorphism type
The number in each column is the number of subgroups in the given group of that isomorphism type:

Number of normal subgroups per isomorphism type
The number in each column is the number of normal subgroups in the given group of that isomorphism type.

Note that for groups in the same Hall-Senior genus, the total number of normal subgroups is equal, and in fact, the number of subgroups of each order, and even of each Hall-Senior family within an order, is equal. However, the number of subgroups of each specific isomorphism type may not be equal.

Number of characteristic subgroups per isomorphism type
The number in each column is the number of characteristic subgroups in the given group of that isomorphism type.

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 odd. 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 2) = (number of normal subgroups of order 2) = (number of subgroups of order 8) = (number of normal subgroups of order 8), and separately, (number of subgroups of order 4) = (number of normal subgroups of order 4).
 * Since index two implies normal, we have, in all cases, that the number of subgroups of order 8 = number of normal subgroups of order 8. See also equivalence of definitions of maximal subgroup of group of prime power order.
 * Formula for number of maximal subgroups of group of prime power order: The subgroups of order 8 (all of which are normal) correspond to the maximal subgroups of the Frattini quotient, which is an elementary abelian group of order $$2^r, 1 \le r \le 4$$. The number of such subgroups is $$(2^r - 1)/(2 - 1) = 2^r - 1$$, and hence must be one of the numbers 1,3,7,15.
 * Formula for number of minimal normal subgroups of group of prime power order: The normal subgroups of order 2 correspond precisely to the subgroups of order 2 in the socle, which is an elementary abelian subgroup defined as $$\Omega_1$$ of the center. If the socle has order $$2^s, 1 \le s \le 4$$, the number of normal subgroups of order 2 is $$(2^s - 1)/(2 - 1) = 2^s - 1$$. Thus, it must be 1, 3, 7, or 15. Moroever, for a non-abelian group, the socle can have order either 2 or 4 (cannot have order 8 or 16) so the number of normal subgroups is either 1 or 3.
 * For groups of the same Hall-Senior genus, the number of normal subgroups of a given order are equal. Thus, groups with Hall-Senior symbols $$16\Gamma_3a_1$$ and $$16\Gamma_3a_2$$ have the same number of normal subgroups of order 4, for instance.

Subgroups of order 2
The table below provides information on the counts of subgroups of order 2. Note the following:

Central series
We have nilpotency class and order determine group up to commutator map-equivalence for up to prime-fourth order. Further/in particular, the nilpotency class and order determines the equivalence class up to isoclinism.

In particular, the isomorphism classes of the successive quotients for the lower central series as well as the isomorphism classes for the successive quotients for the upper central series are determined by the nilpotency class.

With the exception of the whole group, the isomorphism classes of the lower central series members and upper central series members are determined by the Hall-Senior genus.

Below are some of the details that are determined via the Hall-Senior family and genus about the lower and upper central series. Note the following special thing about the order 16 (and more generally, order $$p^k$$ where $$p$$ is prime and $$k \le 4$$) -- that the nilpotency class determines the equivalence class up to isoclinism.

Abelian groups
We first consider the abelian groups, i.e., the groups of nilpotency class exactly one. There are five such groups: cyclic group:Z16 (ID: (16,1)), direct product of Z4 and Z4 (ID: (16,2)), direct product of Z8 and Z2 (ID: (16,5)), direct product of Z4 and V4 (ID: (16,10)), and elementary abelian group:E16 (ID: (16,14)). In all these cases, the derived subgroup is trivial (so the lower central series reaches the trivial subgroup in its second step) and the center is the whole group (so the upper central series reaches the whole group at once). Thus, the abelianization is the whole group and the inner automorphism group is trivial.

Groups of class two
These all form a single Hall-Senior family $$\Gamma_2$$. Note that the third member of the lower central series is trivial, and so is the second member of the upper central series. See below:

Groups of class three
They all form a single Hall-Senior family $$\Gamma_3$$ and a single Hall-Senior genus $$16\Gamma_3a$$. The groups are dihedral group:D16 (ID: (16,7)), semidihedral group:SD16 (ID: (16,8)), generalized quaternion group:Q16 (ID: (16,9)).

All the groups of order 16 and class three are UL-equivalent groups, i.e., the lower and upper central series coincide. In other words, the center is the third member of the lower central series, and the derived subgroup is the second member of the upper central series.

Counts of abelian subgroups and abelian normal subgroups
Below is a table with information on the counts of abelian subgroups and abelian normal subgroups in groups of order 16.

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).
 * Index two implies normal, so all the abelian subgroups of order 8 are normal. Thus the count for abelian subgroups of order 8 is the same as the count for abelian normal subgroups of order 8.
 * 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 8 = number of abelian normal subgroups of order 8. Separately, we have number of abelian subgroups of order 4 = number of abelian normal subgroups of order 4.
 * 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 $$16\Gamma_2a_1$$ and $$16\Gamma_2a_2$$ both belong to the Hall-Senior genus $$16\Gamma_2a$$.

Counts of abelian characteristic subgroups
Below is information on the counts of abelian characteristic subgroups in groups of order 16.