Subgroup structure of groups of order 72

This article gives specific information, namely, subgroup structure, about a family of groups, namely: groups of order 72.
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Numerical information on counts of subgroups by order

FACTS TO CHECK AGAINST FOR SUBGROUP STRUCTURE: (finite solvable group)
Lagrange's theorem (order of subgroup times index of subgroup equals order of whole group, so both divide it), |order of quotient group divides order of group (and equals index of corresponding normal subgroup)
Sylow subgroups exist, Sylow implies order-dominating, congruence condition on Sylow numbers|congruence condition on number of subgroups of given prime power order
Hall subgroups exist in finite solvable|Hall implies order-dominating in finite solvable| normal Hall implies permutably complemented, Hall retract implies order-conjugate
MINIMAL, MAXIMAL: minimal normal implies elementary abelian in finite solvable | maximal subgroup has prime power index in finite solvable group

The prime factorization is as follows:

$\!72=2^{3}\cdot 3^{2}=8\cdot 9$

There are only two prime factors of this number. Order has only two prime factors implies solvable (by Burnside's $p^{a}q^{b}$-theorem) and hence all groups of this order are solvable groups (specifically, finite solvable groups). Another way of putting this is that the order is a solvability-forcing number. In particular, there is no simple non-abelian group of this order.

Note that, by Lagrange's theorem, the order of any subgroup must divide the order of the group. Thus, the order of any proper nontrivial subgroup is one of the numbers 2,4,8,3,6,12,24,9,18,36.

Here are some observations on the number of subgroups of each order:

• Congruence condition on number of subgroups of given prime power order: The number of subgroups of order 2 is congruent to 1 mod 2 (i.e., it is odd). The same is true for the number of subgroups of order 4, as well as for the number of subgroups of order 8. The number of subgroups of order 3 is congruent to 1 mod 3. The same is true for the number of subgroups of order 9.
• By the fact that Sylow implies order-conjugate, we obtain that Sylow number equals index of Sylow normalizer, and in particular, divides the index of the Sylow subgroup. Thus, the number of subgroups of order 8 (i.e., 2-Sylow subgroups) divides 9, so combining with the congruence condition, we obtain that this number must be 1, 3, or 9. Similarly, the number of subgroups of order 9 (i.e., 3-Sylow subgroups) must divide 8. Combining with the congruence condition, we obtain that this number must be 1 or 4.
• In the case of a finite nilpotent group, the number of subgroups of a given order is the product of the number of subgroups of order equal to each of its maximal prime power divisors, in the corresponding Sylow subgroup.
• In the case of a finite abelian group, we further have that the number of subgroups of a particular order equals the number of subgroups whose index equals that order, because subgroup lattice and quotient lattice of finite abelian group are isomorphic.