I supposed $n_3=4$ and $n_2=3$, and then
I made $G$ act by conjugation on $\text{Syl}_3 (G)$.
I want to show that $G\cong S_4$ (looking at all order 24 groups here I saw the only one that has $n_3=|\text{Syl}_3(G)|=4$ and $n_2=|\text{Syl}_2(G)|=3$ is $S_4$) so I want to show that the kernel of this conjugation action is trivial.
So this action defines a representation $\varphi:G\rightarrow S_4$. I was thinking of using the order of $G/\ker(\varphi)$ but this didn't solve anything. So I thought about this kernel as the set that normalizes all three Sylow $3-$subgroups, the intersection of the normalizers.
If we say $\text{Syl}_3 (G)=\left\{P_1,P_2,P_3\right\}$,then I had, by the Orbit-stabilizer theorem, that the order of the stabilizers is $6$.
I saw that in $S_4$ any two of these normalizers intersect in two elements, but when I intersect all of them the intersection is trivial. I want to prove that. So I was doing the following.
By Lagrange, the intersection of all, has order $1, 2, 3$ or $6$. It can't have order $3$, it'd be a normal Sylow $3-$subgroup and that's impossible. So it has order $1,2$ or $6$, and I want to discard $2,6$. How could I show that the normalizers are different (This would discard $6$)? And how could I show that the intersection of all of them has more than two elements?
I was thinking about saying that, if the intersection had order $2$, then there would be a normal subgroup of order $2$, and then a subgroup of order $6$ isomorphic to $\mathbb{Z}_2\rtimes_\phi \mathbb{Z}_3\cong \mathbb{Z}_6$. Could I get somewhere this way?
And how could I do the $2$ and $6$ parts?
Best Answer
This is how I see it. $G$ has 3 2-Sylow subgroups, and 4 3-Sylow subgroups. So $G$ acts on the 3-Sylow subgroups by conjugation, resulting in a map $\phi:G → S_4$. Let $K$ be the kernel of this map. Let $P$ be a 3-Sylow subgroup, and restrict $\phi$ to P; call the map $\phi_P$. Then the Sylow Permutation Theorem* states that the only fixed point of $\phi_P$ on the 4 conjugates is $P$ itself. This means that $\phi_P$ acts nontrivially on the other three 3-Sylow subgroups, implying that $K$ cannot contain an element of order 3, so that the order of $K$ is not divisible by 3, so it has to be 1, 2, 4, or 8.
Let $\psi: G → S_3$ be the action of $G$ on the 2-Sylow subgroups. Then $G$, order 24, acts on $S_3$ of order 6, implying a normal subgroup $V$ in $G$ of order 4. This subgroup then combines with a 3-Sylow subgroup, by semi-direct product, to form a subgroup $A$ of $G$ of order 12. There can be only one such subgroup of order 12, as there is no room for any more. So $A$ is of order 12 without normal 3-Sylow subgroups, so it is isomorphic to $A_4$ and $V$ is isomorphic to the Klein four-group. The 2-Sylow subgroups are of order 8, and each one of them must contain $V$ as a subgroup. Hence we have 1 identity element, 3 elements in $V$, 4 x 3 or 12 elements in the 2-Sylows outside of $V$, and 8 elements of order 3, accounting for all 24 elements of G. There is no more room for any more.
$K$ cannot be of order 8, because the subgroups of order 8 are not normal. $K$ cannot be of order 2, since in that case $K$ would be normal in $G$, so it combines with $P$ to form a subgroup of order 6 with subgroups of order 2 which are normal, hence is $Z_6$, containing elements of order 6, for which there is no room. Finally, $K$ can't be of order 4, since the normalizer of $P$ and its conjugates is of order 6, so the intersection of all these normalizers has to have order 1, 2, 3, or 6 and be contained in $K$, so it can't have order 4.
Therefore, $K$ is the trivial group, so that $\phi$ is an isomorphism between $G$ and $S_4$.
*The Sylow Permutation Theorem says that if a group $G$ has $n$ $p$-Sylow subgroups, then $G$ acts on $Syl_p(G)$ by conjugation, and the only fixed point of this action restricted to $P$ is $P$ itself. This occurs throughout the literature, mainly in proving that there are no simple groups of some specified order, but not as a theorem by itself.