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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 $S_4$.


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!

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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...

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  • $\begingroup$ Thanks! But... can you be less cryptical? :) $\endgroup$ – Joe Jun 7 '14 at 13:41
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    $\begingroup$ Of course I can...but I'm not sure I want to: have you already gotten the subgroups of $\;S_4\;$ ? It really isn't that hard even if you write down all those permutations as products of cycles. Are you already convinced that there're actually maximal subgroups of the above orders (say, can you construct subgroups with these orders)? As a final hint: "the only non-trivial posssibility" is if the wanted intersection has order two, but then this subgroup would be contained in the center of $\;S_4\;$ ... $\endgroup$ – DonAntonio Jun 7 '14 at 13:50
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    $\begingroup$ @Joe, I think you need to base your calculations on what you know about $\;S_4\;$, together with Sylow theorems and other stuff. For example, the only subgroup of order $\;12\;$ is the alternating one $\;A_4\;$ , which also has a normal (in the whole group) subgroup of order $\;4\;$ . Then you have the Sylow $\;2-$subgroups of order $\;8\;$ , and finally you have subgroups of order $\;6\;$ that are isomorphic to $\;S_3\;$ . All this is possible to deduce knowing a little about the elements of $\;S_4\;$ ...try it. And I can't see how that exercise requires to do less computations than whatever $\endgroup$ – DonAntonio Jun 7 '14 at 14:08
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    $\begingroup$ Ah, so now we're really close, @joe ! You now need to prove the following two things: (1) If a group has a normal subgroup of order two then that subgroup is contained in the group's center, or : $\;\left(N\lhd G\;\;and\;\;|N|=2\right)\implies N\subset Z(G)\;$ , and (2) $\;Z(S_4)=1\;$ . $\endgroup$ – DonAntonio Jun 7 '14 at 19:10
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    $\begingroup$ Ok, this is easy. $N=\{1,g\}\unlhd G$; hence since $g^x\in N\;\forall x\in G$, I have two possibilities: $g^x=1$ (but this holds iff $g=1$, contradiction) and $g^x=g\;\forall x\in G$ which is equivalent to $g\in Z(G)$ (that is the only valid). Hence $N\le Z(G)$. Then $Z(S_n)=1\;\forall n\ge4$ is well known. Thanks thanks thanks thanks. I'm studying for a group theory exam, my teacher is Andrea Lucchini, maybe you know him (a couples of theorem in Martin Isaac books got his name). He's very very good but very very hard! $\endgroup$ – Joe Jun 8 '14 at 8:58
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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.”

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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.

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