Finite group, How can I construct solution step-by-step. 
  
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*I apologize for this easy question. But I'm trying to learn. That's why I need your help.
  

$\textbf{Example:}$ Show that group of order $36$ is not simple.
$\textbf{Solution:}$ Let $G$ be  group of order $36=2^23^2$. Then we have $n_3=1$ or $4$. Suppose $n_3=4$ Then the action of $G$ by conjugation on the set Syl$_3(G)$ of Sylow $3$-subgroups induces a homomorphism $$\phi:G\to S_4$$
 Assuming $G$ is simple, $\ker\phi=G$ or $\{1\}$
$\textbf{(i)}$ If  $\ker\phi=G$, then the conjugation by fixes all Sylow $3$-subgroups. This is a contradiction for the Sylow $3$-subgroups are all conjugate and we have assumed there is $4$ of them.
$\textbf{(ii)}$ If ker$\phi=\{1\}$, then $\phi$  is injective, which is a contradiction, for $\mid G\mid=36>24=\mid S_4\mid\blacksquare$

  
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*Ok. I must try to understand all steps. I need to more detail not hint above solution. I think it is more general.
  


Let Syl$_3(G)=\{A,B,C,D\}$. They are all $9$ orders.
Then the action
$\alpha_g:$ Syl$_3(G)\to$  Syl$_3(G)$ such that $\alpha_{g}(H)=gHg^{-1}$ for all $H$ in  Syl$_3(G)=\{A,B,C,D\}$. 
Consider the conjugation action $\phi :G\to Sym(G/A)\cong S_4$.( How can I show $Sym(G/A)\cong S_4$).
I see another approach as follow: 

  
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*$\ker\phi\subset A$, so $\mid \ker\phi\mid$ $\mid 9$
  
*$G/\ker\phi$ embeds into $S_4$; so $[G:\ker\phi]\mid 24$. That is $3\mid \ker\phi\mid$.
  

This tell us $\mid \ker\phi\mid=3$. $\ker\phi\unlhd G$.

  
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*How can I show $\ker\phi\subset A$. is this method general?
  
*For $\textbf{(i)}$ ;what is the mean of the conjugation by fixes all Sylow $3$-subgroups. 
  
*For $\textbf{(ii)}$; I cant see contradiction.


Your first solution is not quite right - with your count you are assuming that any pair of Sylow $3$-subgroups is intersecting only at the trivial element, which might not be true. The argument is frequently used when the order of a Sylow subgroup is prime, but here it is $9$.
The second solution is the one your are looking for. Note that it can never happen that ker$(\phi)=G$, since the kernel is contained in the normalizer of a Sylow $3$-subgroup, $N_G(P)$, where $P \in Syl_3(G)$. And $n_3=|G:N_G(P)|=4$.
