Finding all subgroups of S4

$|S_4|=24$. So, the possible orders of subgroups of $S_4$ are 12, 8, 6, 4, 3, 2, 1.

$S_4$ is composed of the following elements:

$$\begin{array}{rl}{S}_4=& \{\\ & e,(01)(23),(02)(13),(03)(12),\\ & (01),(12),(23),(30),(02),(13),\\ & (012),(123),(230),(301),(021),(132),(203),(310),\\ & (0123),(0132),(0231),(0213),(0321),(0312),\\ & \}\end{array}$$

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Subgroups of order 12

We know that the alternating group $A_4$ is the kernel of the sign homomorphism $\sigma :S_n\to {\mathbb{Z}}_2$, and hence is a subgroup of $S_n$. The members of $A_4$ are the members of $S_4$ with a positive sign:

$$\begin{array}{rl}{A}_4=& \{\\ & e,(01)(23),(02)(13),(03)(12),\\ & (012),(123),(230),(301),(021),(132),(203),(310),\\ & \}\end{array}$$

Notice that $A_4$ contains all 3-cycles of $S_4$.

Let $H$ be a subgroup of $G$ with order 12 such that it does not contain some 3-cycle $(abc)$. Note that $(bac)\notin H$, since $(bac)\in H⟹(bac{)}^2=(abc)\in H$. Now, consider the cosets $(abc)H$ and $(bac)H$. Since $(abc),(bac)\notin H$ and $[G:H]=2$, It must be that $(abc)H=(bac)H$. But, $(bac{)}^{-1}(abc)=(abc)(abc)=(bac)\notin H$, which is a contradiction. Thus, any subgroup of $G$ of order 12 must have all 3-cycles.

Now, let $H$ be a subgroup of $G$ with order 12 containing some 2-cycle $(ab)$. By composing $(ab)$ with the four 3-cycles that do not contain $a$ or $b$, we get four 4-cycles $(ab)(bcd)=(bcda)$, $(ab)(bdc)=(bdca)$, $(ab)(cda)=(cdab)$, $(ab)(cad)=(cbad)$. $H$ does not have the space to have these elements. Thus, any subgroup of $G$ of order 12 must not have any 2-cycles.

If these is a subgroup of $G$ of order 12 that contains some 4-cycle, we can compose that four cycle with some 3-cycle to obtain a 2-cycle, which we know cannot be in $H$: $(abcd)(adc)=(bc)$. Thus, any subgroup of $G$ of order 12 must not have any 4-cycles.

Thus, $A_4$ is the only subgroup of $S_4$ of order 12.

A shorter way to prove the above statement is to notice that any subgroup of index 2 must be normal, and so must be made of whole conjugacy classes. The only way to do this is $A_4$.

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Subgroups of order 8

Let $H$ be a subgroup of order 8. The elements of $H$ can have order 2 or 4. Say all the elements of $H$ have order 2, i.e, a 2-cycle, or a double 2-cycle. $H$ must have some 2-cycle $(a,b)$. Notice that if $H$ has any other 2-cycle where $a$ or $b$ is not fixed, like $(b,c)$, then the two 2-cycles can be composed to get a three cycle. Thus, $H$ must have at least one element of order 4.

If $H$ has two elements of order 4 such their cycles are different, they will end up generating more than 8 elements. Thus, $H$ can have only one element of order 4. There are 3 different 4-cycles in $S_4$, and for each 4-cycle can be extended in only one way to form a subgroup of order 8 (I’m too lazy to write down everything, this needs some verification, but its mostly bashing numbers).

{ (), (0 1), (0 3 1 2), (0 3)(1 2), (0 2 1 3), (0 2)(1 3), (0 1)(2 3), (2 3) }

{ (), (0 1 3 2), (0 3), (0 3)(1 2), (0 2 3 1), (0 2)(1 3), (0 1)(2 3), (1 2) }

{ (), (0 1 2 3), (0 3 2 1), (0 3)(1 2), (0 2), (0 2)(1 3), (0 1)(2 3), (1 3) }

Note that all of these are isomorphic to $D_4$.

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Subgroups of order 6

No element of order 6. So must to isomorphic to $S_3$. Simple work from here. You will find 4 subgroups.