Maximal subgroups of $S_4$

Question

I have to prove that $$ \operatorname{Frat}(S_4):=\bigcap_{M\stackrel{\max}{\le} S_4}M=1 $$ but I don't know how to compute it since I don't know what are the maximal subgroups in $S_4$.

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EDIT: noting that $S_4\simeq G/Z$ where $G=GL_2(\mathbb F_3)$ and $Z=Z(G)$, I can show that $\operatorname{Frat}(S_4)$ is trivial computing $$ \bigcap_{\frac MZ\stackrel{\max}{\le}\frac GZ}M/Z $$ and showing that it is equal to $Z/Z$. And now the question become: what are the maximal subgroups of $GL_2(\mathbb F_3)$ which contain $Z$?

I know that all normal subgroups of $G$ which contain $Z$ are $Z$ itself $Q_8$ and $N=SL_2(\mathbb F_3)$ but I don't know if there is any relation between normal and maximal subgroups, or if in general these infos can help me.

SECOND EDIT: I thought that it would be sufficient to find two groups $A,B\le G$ s.t. $A\cap B=1$, $|A|=|B|=12$ and $Z\nleq A,B$.

In such a way $K:=ZA$ and $H:=ZB$ would be two subgroups of $G$ of order $24$ hence maximal, whose intersection is exactly $Z$.

How can I find these two groups $A$ and $B$?

Any hint is welcome.

Thanks!

Answer 1

Hints: If you know the subgroup lattice of $\;S_4\;$ this must be easy (and if you don't you can work it out now):

There are maximal subgroups of order $\;6,8,12\;$ , so the only non-trivial possibility is...and, in fact, it neither is, so...

Answer 2

DonAntonio's suggestion to calculate (and draw on paper) the complete lattice of subgroups of $S_4$ is a very good suggestion because (as you probably noticed from your string of exercises) $S_4$ is a very important group. However, it is not usually not necessary to find all of the subgroups in order to calculate the Frattini subgroup. In the case of $S_4$ it is particularly easy:

$S_4$ is a primitive permutation group, so $\Phi(S_4)=1$.

More explicitly, the for any $i \in \{1,2,3,4\}$ the subgroup $M_i$ of $S_4$ leaving $i$ in place is $S_{\{1,2,3,4\} \setminus\{i\}}$ and isomorphic to $S_3$. (A) It is easy to see $M_i$ is a maximal subgroup, for instance because it is transitive on $X_i=\{1,2,3,4\}\setminus\{i\}$ and so if you have $\pi$ outside of $M_i$, then $\pi(i) \in X_i$, so using $\sigma \in M_i$ we can make $\sigma(\pi(i))$ be anything, and conversely anything can be sent to $i$, so the subgroup $H$ generated by $M_i$ and $\pi$ is transitive, so $[H:M_i]=[G:M_i]=4$ so $G=H$. (B) It is also easy to the see the intersection of the $M_i$ is exactly the subgroup leaving $1$, $2$, $3$, and $4$ fixed, that is $\cap M_i = \{()\}$ is the trivial subgroup.

This proof works for any permutation group $G$ on $n$ points in which the stabilizer $M_1$ of the point $1$ has index $n$, and the stabilizer $M_{1,2}=M_1 \cap M_2$ of points $1$ and $2$ has index $n-1$ in $M_1$. Such groups $G$ are called “2-transitive groups.”

Answer 3

Hints: Here's a general fact. Suppose that $G$ is a finite group, and that $P$ is a Sylow $p$-subgroup of $G$ such any non-identity normal subgroup of $G$ contained in $N_{G}(P)$ is a $p$-group. Suppose also that $N_{G}(P)$ is not a maximal subgroup of $G.$ Then $[G:N_{G}(P)] = ab$ where $a>1,b >1$ are both integers congruent to $1$ (mod $p$). This should show you that all Sylow normalizers in $S_{4}$ are maximal subgroups.