How to enumerate subgroups of each order of $S_4$ by hand I would like to count subgroups of each order
(2, 3, 4, 6, 8, 12) of $S_4$,
and, hopefully, convince others that I counted them correctly.  In
order to do this by hand in the term exam, I need a clever way to do
this because there can be as many subgroups of a group of order 24 as
$2^{23}$.
Do you know how to do this? 
(I would be most grateful if you could tell me what part of the answer to the old question answers my question before voting to close.)
 A: Well, the first step would be to write out $S_4$:
$$
S_4=\{(1), (12), (13), (14), (23), (24), (34), (123), (132), (142), (124), (134), (143), (234), (243), (1234), (1243), (1324), (1342), (1423), (1432), (12)(34), (13)(24), (14)(23)\}
$$
That's one subgroup down (29 to go).
Next, consider subgroups isomorphic to ${\mathbb Z_2}$. That's any element of order $2$ (from the list below) and the identity (9 total):
$$
(12), (13), (14), (23), (24), (34), (12)(34), (13)(24), (14)(23)
$$
Move on to those isomorphic to ${\mathbb Z_3}$. There are eight $3$-cycles in $S_4$ and each unique subgroup must contain two of them, so there are a total of 4.
Now, ${\mathbb Z_4}$. Consider $\langle (1234) \rangle$. We've got $\{(1),(1234),(13)(24),(1432) \}$ and two more like this.
From here, just keep going through the list. Think about what elements you need to form a particular isomorphism type and then figure out how many ways you can form each type. For instance, if you have a subgroup isomorphic to $D_8$, you'll need a cyclic subgroup of order $4$ (we have three of those) and 4 elements of order $2$.
Including what we've already come up with, there should be 4 subgroups isomorphic to the Klein 4-group (each contains 3 elements of order $2$), 4 to $S_3$ (think of this one as choosing 3 elements out of $\{1,2,3,4\}$ and considering the permutations on this new set of three symbols), 3 to $D_8$ (again, we're limited by the cyclic "subgroup of rotations", $A_4$ will be the only subgroup of order $12$, and don't forget $\{(1)\}$.
A: By Lagrange's theorem, the order of a subgroup divides 24, so we are looking for subgroups of orders 1, 2, 3, 4, 6, 8, 12, and 24. We go through the list, often using Sylow's theorems:


*

*The subgroups of order 1 and 24 are obviously unique.


Subgroups of order 2 are in 1–1 correspondance with elements of order 2, so you get 4-choose-2  = 6 transpositions $\langle (i,j) \rangle$ and 4-choose-2-over-2 = 3 double transpositions $\langle (i,j)(k,l) \rangle$.


*

*Nine subgroups of order 2, all cyclic, two conjugacy classes


By Sylow's theorem the subgroups of order 3 are all conjugate, so $\langle (1,2,3) \rangle$, $\langle (1,2,4) \rangle$, $\langle (1,3,4) \rangle$, and $\langle (2,3,4) \rangle$. 


*

*Four subgroups of order 3, all conjugate to the alternating group of degree 3


Size 4 is messy, so I delay it.
A subgroup of order 6 must have a normal Sylow 3-subgroup, so must live inside the normalizer (inside S4) of a Sylow 3-subgroup. The Sylow 3-subgroups are just the various alternating groups of degree 3, and their normalizers are various symmetric groups of degree 3, so are exactly the 4 subgroups of order 6.


*

*four subgroups of order 6, $\langle (i,j), (i,j,k) \rangle$ parameterized by sets $\{i,j,k\} \subset \{1,2,3,4\}$ of size 3.


All subgroups of order 8 are conjugate by Sylow's theorem, so we just have $\langle (i,k), (i,j,k,l) \rangle$ which is dihedral.


*

*three subgroups of order 8, all conjugate, all dihedral


A subgroup of order 4 is a subgroup of a Sylow 2-subgroup, so either cyclic $\langle (i,j,k,l) \rangle$ or one of the two kinds of Klein 4-subgroups $\langle (i,j), (k,l) \rangle$ (3 subgroups), or the true K4 $\langle (i,j)(k,l), (i,k)(j,l) \rangle$ (normal).


*

*seven subgroups of order 4, three conjugacy classes


A subgroup of order 12 either has a normal Sylow 2-subgroup (and the only subgroups of order 4 with normalizers having elements of order 3 are K4 with normalizer A4) or a normal Sylow 3-subgroup, but in the latter case the normalizer of a Sylow 3-subgroup is only size 6, not 12.


*

*one subgroup of order 12, the alternating group of degree 4


Those were all possible orders, and for each order we proved any subgroup of that order had a specific form, and then counted how many had that form.
