Corollary 4.9. If $H$ is a subgroup of index $n$ in a group $G$ and no non trivial normal subgroup of $G$ is contained in $H$, then $G$ is isomorphic to a subgroup of $S_n$.
Corollary 4.10. If $H$ is a subgroup of a finite group $G$ of index $p$, where $p$ is the smallest prime dividing the order of $G$, then $H$ is normal in $G$.
If $|G|=pn$, with $p\gt n$, $p$ prime, and $H$ is a subgroup of order $p$, then $H$ is normal in $G$.
My attempt: By Lagrange theorem, $$\begin{align}|G| &= |H|[G:H]\\ &= p[G:H]\\ &=pn \\ \end{align}.$$ So $[G:H]=n$. If $n$ is prime, then $H\lhd G$ (by corollary 4.10). If $n$ is not prime. We claim $\exists \langle e\rangle \neq K\lhd G$ such that $K\subseteq H$. Assume towards contradiction, $\nexists$ such $K$. By corollary 4.9, $G\cong$ subgroup of $S_n$. By Lagrange theorem, $|G|=|\text{subgroup of }S_n|$ divides $|S_n|=n!$. That is $np\mid n!$. Which implies $p\mid (n-1)!$ and $p=i$ for some $1\leq i\leq n-1$. But $p\gt n$. Thus we reach contradiction. Hence $\exists \langle e\rangle \neq K\lhd G$ such that $K\subseteq H$. By Lagrange theorem, $|K|$ divides $|H|=p$. So $|K|=1$ or $p$. Since $K\neq \langle e\rangle$, we have $|K|=p$. Thus $K=H$ is normal in $G$. Is my proof correct?
I am uncertain about following step in my proof: Which implies $p\mid (n-1)!$ and $p=i$ for some $1\leq i\leq n-1$. Here is an alternative approach to above exercise which doesn’t use Sylow's theorem.
