Subgroup structure of groups of order 60

Numerical information on counts of subgroups by order
We have:

$$\! 60 = 2^2 \cdot 3 \cdot 5 = 4 \cdot 3 \cdot 5$$

Note that, apart from alternating group:A5, all the other groups of order 60 are finite solvable groups. $$A_5$$ is a simple non-abelian group. For more, see A5 is the simple non-abelian group of smallest 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 2,4,3,6,12,5,10,20,15,30.

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). So is the number of subgroups of order 4. Similarly, the number of subgroups of order 3 is congruent to 1 mod 3, and 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 any Sylow subgroup:
 * The number of subgroups of order 4 (i.e., 2-Sylow subgroups) divides $$60/4 = 15$$. Combining with the congruence condition, we obtain that this number is 1, 3, 5, or 15.
 * The number of subgroups of order 3 (i.e., 3-Sylow subgroups) divides $$60/3 = 20$$. Combining with the congruence condition, we obtain that this number is either 1 or 4.
 * The number of subgroups of order 5 (i.e., 5-Sylow subgroups) divides $$60/5 = 12$$. Combining with the congruence condition, we obtain that this number is either 1 or 6.
 * In the case of a finite nilpotent group (which in this case coincides with finite abelian 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 4) = (number of subgroups of order 3) = (number of subgroups of order 5) = (number of subgroups of order 12) = (number of subgroups of order 15) = (number of subgroups of order 20) = 1.
 * (number of subgroups of order 2) = (number of subgroups of order 6) = (number of subgroups of order 10) = (number of subgroups of order 30), and this number is either 1 or 3, depending on whether the 2-Sylow subgroup is cyclic group:Z4 or Klein four-group.
 * Nilpotent of cube-free order implies abelian
 * Finite supersolvable implies subgroups of all orders dividing the group order
 * 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. Thus, for all the solvable groups of order 60:
 * There exists a (2,3)-Hall subgroup (of order 12), and the number of subgroups of order 12 divides $$60/12 = 5$$.
 * There exists a (2,5)-Hall subgroup (of order 20), and the number of subgroups of order 20 divides $$60/20 = 3$$.
 * There exists a (3,5)-Hall subgroup (of order 15), and the number of subgroups of order 15 divides $$60/15 = 4$$.

The only non-solvable group of order 60 is alternating group:A5. This has a subgroup of order 12 but does not have subgroups of order 15 or 20.