Classification of groups $G$ of order 24. I supposed $n_3=4$ and $n_2=3$, I want to show that $G\cong S_4$. Then I made $G$ act by conjugation on $\text{Syl}_4 (G)=\{X_1,X_2,X_3,X_4\}$. This action defines a representation $\varphi:G\rightarrow S_4$. we have $\ker(\varphi)=\cap_{i=1}^{4}N_G(X_i)$ (the intersection of the normalizers), By Sylow theorem : $N_G(X_i)=6$ for all $i$. then $|\ker(\varphi)|\in\{1,2,3,6\}$ i want to show that $|\ker(\varphi)|=1$. By Sylow theorem : for all $i$ $N_G(X_i)$ has a single group of order 3 which is $X_i$ then
if $|\ker(\varphi)|=3$ then $X_1=X_2=X_3=X_4$ contradiction $(n_3=4) $
if $|\ker(\varphi)|=6$  then $X_1=X_2=X_3=X_4.$
Problem if $X$ and $Y$ two $3-$ Sylow  such that $X=Y$ then $N_G(X)=N_G(Y)$
 A: Since normal Sylow subgroups are unique, if $X$ and $Y$ are $3$-Sylows and $X\ne Y$, then $N(X) \ne N(Y)$ and $N(X) \cap N(Y)$ is a $3'$-group, and so has order $1$ or $2$ as each normalizer has order $6$.  If their intersection has order $1$ then the action of $G$ on its four $3$-Sylow subgroups is faithful and so since $|G|=24$, $G$ must be isomorphic to $\frak{S}_4$.  Now, suppose $|N(X)\cap N(Y)|=2$.  The kernel of the conjigation action must be contained in it, so we still have $G$ acting faithfully on its four $3$-Sylow subgroups, and thus isomorphic to $\frak{S}_4$, unless  $N(X)\cap N(Y)$ is a normal subgroup (call it $K$).  But if that were the case, then $G/K$ would have order $12$ and would have four $3$-Sylows since the $3$-Sylows would have distinct images in $G/K$.  ($K$ is central of order $2$, so we can't have $x\in yK$ for $x \ne y$ elements of order $3$.)  But that leaves room only for a unique $2$-Sylow in $G/K$, and its inverse image is clearly a unique $2$-Sylow in $G$, contradicting that there are three $2$-Sylows in $G$.
Note that this also shows that if a group of order $24$ with multiple $3$-Sylows is not $\frak{S}_4$, then it must be either $2\times \frak{A}_4$ or a double cover of $\frak{A}_4$, i.e., it must be $SL(2,3)$ or $2\times \frak{A}_4$.
Thus, except for abelian groups, $\frak{S}_4$, $2\times\frak{A}_4$, and $SL(2,3)$, all groups of order 24 are realized as semidirect products $3\rtimes H$, with $H$ of order $8$ acting nontrivially on the group of order $3$.
