Groups of order 24. 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?
 A: It's a bit harder to rule out $|K|=2$ (where $K$ is the kernel). In that case $G/K$ would have order $12$ and would also have $4$ Sylow $3$-subgroups, and so $G/K \cong A_4$. But $A_4$ has a normal Sylow $2$-subgroup, and hence so does $G$, so $n_2=1$ in this case. (Note that the group ${\rm SL}(2,3)$ has $n_3=4$, $n_2=1$. It is often hard to rule out cases for which there is a group with similar parameters.)
A: 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.
A: if stabilizer are same then you have an normal subgroup $N$ of order $6$. Let $H$ be subgroup of $N$ of order $3$. Notice that $H$ is uniqe in $N$ so $H$ must be normal in $G$ contradiction.
For other case, you have an normal subgroup of order $2$. Notice that any normal subgroup of order $2$ must be contained $Z(G)$ as $Z(G)$ is trivial contradiction.  
A: Offering a slightly different proof, similar to jimvb13's answer.
All references are from Dummit & Foote, Abstract Algebra, 3e.
Suppose $n_2=3, n_3=4$, and fix any of Sylow-2 groups called $P_2$, and any of Sylow-3 groups $P_3$.
Now, $|G|/|P_2| =3$, and the permutation of $P_2$ by left action gives a homomorphism $\varphi: G \mapsto S_3$.
Denote $\tilde{V} :=\mathrm {ker}\; \varphi \lhd G$.
Here, $\tilde{V} =4$ or $8$; indeed, for $\tilde{V} =12$ or $24$, the action would be not transitive.
But 8 is not possible either, in which case, $\tilde{V}$ as a Sylow-2 would be normal.
Thus $\tilde{V} =4$.
Now $\tilde{A} :=P_3 \tilde{V}_4$ is well defined (D&F, p.120, col.5), with $\tilde{A} =12$.
Since $|G|/|\tilde{A}| =2$, the smallest prime factor of 24, we see $\tilde{A} \lhd G$ (D&F, p.120, col.5).
It is possible to find a order-2 element $a \in G -\tilde{A}$.
In fact, even if all nonidentity elements in conjugates of $P_3$ are counted, we have only $3 \cdot 2 =6$ elements.
The rest $2$ elements has order to be multiple of 2.
So suppose $|\langle a \rangle| =2$, we see $G =\tilde{A} \rtimes \langle a \rangle$ (D&F, p.180, thm.12).
For brevity, call it $G =\tilde{A} \rtimes Z_2$.
We know that there are two nonisomorphic order-12 groups, $A_4$ and $S_3 \times Z_2$ (D&F, p.183).
Consider $(S_3 \times Z_2) \rtimes Z_2$, and I claim there is a contradiction.
Actually, for $P_3'$ a Sylow-3 subgroup of $S_3 \times Z_2$, we have $P_3'\; \mathrm{char}\; (S_3 \times Z_2) \lhd G$ (D&F, p.144), thus (recognized as subgroup of $G$) $P_3 \lhd G$ (see D&F, p.143), contrary to assumption.
Thus $\tilde{A} =A_4$ the alternating group, now $G =A_4 \rtimes Z_2$.
Notice the Klein group $V\; \mathrm{char}\; A_4 \lhd G$ and $V \lhd G$.
And if the product is direct, the only possibility of $P_2$ is recognized to be $P_2 =V \times Z_2$, but that clearly $P_2 \lhd G$.
Thus $G =A_4 \rtimes Z_2$ nontrivially, also we recognize $\tilde{V} =V$.
Consider $A_4 \leq S_4$, and pick any order-2 element in S_4 to form $A_4 \rtimes Z_2$.
We know that a conjugation within $S_4$ is just renumbering, and such conjugation exhaust all automorphisms $\mathrm{Aut} (A_4)$.
Considering the order, $A_4 \rtimes Z_2$ must be exactly $S_4$.
