Problem
Show that if $|G|=p^3q$ and $G$ has no normal Sylow subgroups, then $G \cong \mathbb S_4$
The attempt at a solution
By the Sylow theorems we have:
-$n_p \equiv 1 (p), \space n_p|q$
-$n_q \equiv 1 (q), \space n_q|p^3$
Since $G$ has no normal subgroups, one can deduce from one of the Sylow theorems that $n_p,n_q \neq 1$, so the possibilities are $n_p=q, n_q \in \{p,p^2,p^3\}$
Since $|\mathbb S_4|=2^33$ and $\mathbb S_4$ has four $3-$Sylow subgroups, I should prove that $n_q=p^2$ and, specifically, that $p=2^3,q=3$. After that, I must somehow conclude $G \cong \mathbb S_4$
I would appreciate if someone could explain me how to show these things. Answers, hints, suggestions are welcome. Thanks in advance.
Best Answer
Extended hints: