Subgroup structure of groups of order 168

Numerical information on counts of subgroups by order
We have:

$$\! 168 = 2^3 \cdot 3 \cdot 7 = 8 \cdot 3 \cdot 7$$

Note that projective special linear group:PSL(3,2) is a simple non-abelian group of order 168, and all other groups have order 168.

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 2,4,8,3,6,12,24,7,14,28,56,21,42,84.

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


 * Congruence condition on number of subgroups of given prime power order: If $$p^r$$ is a prime power dividing the order of the group, the number of subgroups of order $$p^r$$ is congruent to 1 mod $$p$$.
 * Case $$p = 2$$: The number of subgroups of order 2 is congruent to 1 mod 2 (i.e., it is odd). The number of subgroups of order 4 is odd, and so is the number of subgroups of order 8.
 * Case $$p = 3$$: The number of subgroups of order 3 is congruent to 1 mod 3.
 * Case $$p = 5$$: The number of subgroups of order 7 is congruent to 1 mod 7.
 * Sylow implies order-conjugate, which yields that Sylow number equals index of Sylow normalizer, and in particular, the number of Sylow subgroups divides the index of any Sylow subgroup:
 * Case $$p = 2$$: The number of subgroups of order 8 divides $$168/8 = 21$$. Thus, it is one of the numbers 1, 3, 7, and 21.
 * Case $$p = 3$$: The number of subgroups of order 3 divides $$168/3 = 56$$. Combining with the congruence condition, we obtain that it is one of the numbers 1, 4, 7, and 28.
 * Case $$p = 7$$: The number of subgroups of order 7 divides $$168/7 = 24$$. Combining with the congruence condition, we obtain that it is one of the numbers 1 and 8.
 * 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 particular, we get:
 * (number of subgroups of order 8) = (number of subgroups of order 3) = (number of subgroups of order 7) = (number of subgroups of order 21) = (number of subgroups of order 24) = (number of subgroups of order 56) = 1.
 * (number of subgroups of order 2) = (number of subgroups of order 6) = (number of subgroups of order 14) = (number of subgroups of order 42).
 * (number of subgroups of order 4) = (number of subgroups of order 12) = (number of subgroups of order 28) = (number of subgroups of order 84).
 * Hall subgroups exist in finite solvable and Hall implies order-dominating in finite solvable, and the converse is also true -- if all possible Hall subgroups exist, then the group is solvable, by Hall's theorem. In particular, for all the solvable groups of order 168, we have the following:
 * There exists a (2,3)-Hall subgroup, i.e., a subgroup of order 24, and the number of such subgroups divides $$168/24 = 7$$, so it is either 1 or 7.
 * There exists a (2,7)-Hall subgroup, i.e., a subgroup of order 56, and the number of such subgroups divides $$168/56 = 3$$, so it is either 1 or 3.
 * There exists a (3,7)-Hall subgroup, i.e., a subgroup of order 21, and the number of such subgroups divides $$168/21 = 8$$, so it is one of the numbers 1, 2, 4, and 8.
 * Finite supersolvable implies subgroups of all orders dividing the group order