SubGroups of $S_4$ of order $8$ and $12$. Describe by hand

Question

What's a "good" way of describing $S_4$ subgroups by hand? The following post gives a pretty detailed stategy but i can't figure out some tools that are used:

How to enumerate subgroups of each order of $S_4$ by hand

I think I'm not familiar with the use of the "Normalizer" that is used in this post, it may be useful to understand the strategy.

For subgroups of orders less than $8$, I can describe them mentally with intuition, but for subgroups of order $8$ and $12$, I realize I may not be using a good method as I can't describe them by hand.

More precisely, for the example of $A_4$ , the order $12$ subgroup of $S_4$, I understand the following reasoning: Prime factorization of $12$ is $2^2*3$, so, we know that this subgroup of order $12$ should (if it exists, what I can't prove except by exposing $A_4$) contain a $2$-sylow or a $3$-sylow, more precisely a subgroup of order $2^2$ or $3$. Since this point, I can't really figure out how to move forward.

I also tried to find a proper post that explains it but I never found explanation about the use of the normalizer,

Answer 1

To get a subgroup of order $12$, you might notice this is half of $24 = |S_4|$. You may also remember that elements of $S_4$ are either odd or even (and there's half of each). This might lead you to ask whether the set of all odd permutations, or the set of all even permutations, is a subgroup. If one of these is a subgroup, then we win. Of course, if you remember the definition of an even permutation, you'll find that the even elements do form a subgroup. This is $A_4$.

As for subgroups of order $8$, you might first realize that once you've found one subgroup of order $8$, you're done. This is a Sylow 2-subgroup, and so every subgroup of order $8$ is conjugate (thus isomorphic). So you might start thinking about what groups have order $8$ (there's only so many) and try to find one which is a subgroup of $S_4$.

After some thought, you'll probably land on the dihedral group $D_8$, which measures the symmetries of a square. Since a square has $4$ vertices, this feels plausible as something to do with $S_4$.

Since $D_8$ is defined geometrically, let's try to reason about it geometrically and show that it really is a subgroup of $S_4$, and then we'll be done.

Notice every element of $D_8$ defines a permutation of the vertices. For instance, the "rotate clockwise" action is the element $(4, 1, 2, 3)$ and the "flip along a vertical axis" element is $(4, 3)(1, 2)$. So then the permutations which arise as symmetries of this square form a subgroup isomorphic to $D_8$.

As an aside, this shows why these groups are all conjugate! Notice the only difference between the subgroups you get is the initial labeling of the vertices. So by relabeling the vertices (which corresponds to conjugation), you can move from one subgroup to another. You might want to think of this as a kind of "change of basis" for these symmetries.

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I hope this helps ^_^