Subgroup structure of groups of order 120

Numerical information on counts of subgroups by order
We have

$$\! 120 = 2^3 \cdot 3 \cdot 5 = 8 \cdot 3 \cdot 5$$

Note that, apart from symmetric group:S5 (an almost simple group), special linear group:SL(2,5) (a quasisimple group), and direct product of A5 and Z2, all groups of order 120 are solvable groups, and in particular finite solvable groups.

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 2,4,8,3,6,12,24,5,10,20,40,15,30,60.

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


 * Congruence condition on number of subgroups of given prime power order: For any prime power $$p^r$$ 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). Similarly, the number of subgroups of order 4 is odd, and the number of subgroups of order 8 is odd.
 * 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 5 is congruent to 1 mod 5.
 * 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 the Sylow subgroup.
 * Case $$p = 2$$: The number of subgroups of order 8 divides $$120/8=15$$. Combining with the congruence condition, we get that the number of subgroups of order 8 is 1, 3, 5, or 15.
 * Case $$p = 3$$: The number of subgroups of order 3 divides $$120/3 = 40$$. Combining with the congruence condition, we get that the number of subgroups of order 3 is 1 or 4.
 * Case $$p = 5$$: The number of subgroups of order 5 divides $$120/5 = 24$$. Combining with the congruence condition, we get that the number of subgroups of order 5 is 1 or 6.
 * 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 5) = (number of subgroups of order 24) = (number of subgroups of order 15) = (number of subgroups of order 40) = 1, and also, (number of subgroups of order 2) = (number of subgroups of order 6) = (number of subgroups of order 10) = (number of subgroups of order 30). Separately, we have (number of subgroups of order 4) = (number of subgroups of order 12) = (number of subgroups of order 20) = (number of subgroups of order 60).
 * Subgroup lattice and quotient lattice of finite abelian group are isomorphic: This states that the number of subgroups of order $$d$$ equals the number of subgroups of order $$120/d$$. In particular, we get (number of subgroups of order 8) = (number of subgroups of order 3) = (number of subgroups of order 5) = (number of subgroups of order 24) = (number of subgroups of order 15) = (number of subgroups of order 40) = 1. Similarly, we have (number of subgroups of order 2) = (number of subgroups of order 4) = (number of subgroups of order 6) = (number of subgroups of order 12) = (number of subgroups of order 10) = (number of subgroups of order 20) = (number of subgroups of order 30) = (number of subgroups of order 60).
 * Hall subgroups exist in finite solvable (the converse is also true, see Hall's theorem): For the solvable groups, we have that there exist Hall subgroups of all possible orders, i.e., there exists a (2,3)-Hall subgroup (order 24), a (2,5)-Hall subgroup (order 40), and a (3,5)-Hall subgroup (order 15). Further, Hall implies order-dominating in finite solvable, so we get:
 * The number of (2,3)-Hall subgroups, of order 24, divides $$120/24 = 5$$, hence it is either 1 or 5.
 * The number of (2,5)-Hall subgroups, of order 40, divides $$120/40 = 3$$, hence it is either 1 or 3.
 * The number of (3,5)-Hall subgroups, of order 15, divides $$120/15 = 8$$, hence it is one of the numbers 1,2,4,8.
 * Finite supersolvable implies subgroups of all orders dividing the group order: For supersolvable groups, there are subgroups of every order dividing the group order.