Let $G$ be a group of order $4p^n$, where $p>2$ is prime and $n>0$. Show that $G$ is not simple. 
Let $G$ be a group of order $4p^n$, where $p>2$ is prime and $n>0$. Show that $G$ is not simple. (Hint: Consider the standard action of $G$ on $G/P$, where $P$ is a $p$-sylow subgroup.)

Let $P_k$ denote the k-sylow subgroup, and let $n_k$ denote the number of the conjugates. 
Case 1
Let $p \geq 5$. Then $p>4$, so $n_p \equiv 1 \mod p$ and $P_p \triangleleft G$. So $G$ is not simple. 
Case 2
Let $p=3$. 
a) Let $n=1$. Then $|G|=12$. If $P_3$ is normal, we are done. Otherwise, $G \cong A_{4}$ by Corollary 5.3.18, and we know that $A_4$ is not simple.
b) Let $n \geq 2$. By Theorem 1.96 of Advanced Modern Algebra (Rotman), we know that

If $H$ is a subgroup of finite index $n$ in a group $G$, then there exists a homomorphsm $\phi: G \rightarrow S_n$ with $ker \phi \subseteq H$. 

So we have a homomorphism $\phi: S \rightarrow S_4$ with $ker \phi \subseteq P_3$. Since $n \geq  2$, we know that $|G| \geq |S_4|$. So $\phi$ cannot be injective $\implies$ the kernel is nontrivial $\implies$ $G$ is not simple. 
Do you think my answer is correct? If it isn't, can you let me know where I went wrong? If it is correct, then I would appreciate it if somebody could possibly show me the link between the problem and the hint that was given...because I couldn't really see it. 
Thanks in advance 
 A: You are correct, though I have a couple of small suggestions. Firstly, in case 2.a. consider the fact that the Sylow $2$-subgroup has index $3$, so its core--the largest normal subgroup of $G$ contained in it--has index $6$. This is a quick and easy resolution of that case. Secondly, you should have a strict inequality $|G|>24=|S_4|$ to argue the fact that the kernel of your homomorphism is nontrivial.
You are, in fact, using the content of the hint! The way to construct that homomorphism is to have $G$ act on the four cosets of some Sylow $3$-subgroup. To be more specific, each element of $G$ corresponds, via this action, to a bijection on the set $G/P_3$. The group of bijections of a four element set is $S_4$.
A: To rewrite the proof using the suggested hint:
As $G$ acts on the set $G/P$ of size $4$, we have a homomorphism $\phi\colon G\to S_4$ (you used theorem 1.96 to arrive at this point). As the action is transitive, $\ker\phi\ne G$. Unless $|G|$ divides $24$, we have $\ker\phi\ne 0$ and we are done. Remains the case $|G|=12$. The only subgroup of index $2$ in $S_4$ is $A_4$ and is known to be not simple.
Or the other way around: If $G$ were simple then $G\cong\phi(G)$ would be a simple subgroup of $S_4$, hence of prime order.
