If $|G|=pn$, $|H|=p$, $p>n$ ,and $H\leq G$, then $H\triangleleft G$. Question: Assume that $G$ is a group and $H$ is a subgroup of $G$. If $|G|=pn$ and $|H|=p$ such that $p$ is a prime and $n$ is a positive integer less than $p$, then show that $H$ is a normal subgroup of $G$.
My process: For $a\in G$, and $Cl(a)=\{b\in G\mid \exists g\in G, gbg^{-1}=a\}$ we know that $|Cl(a)|=|G:C_G(a)|$. So, $|Cl(a)|=\dfrac{|G|}{|C_G(a)|}=\dfrac{pn}{|C_G(a)|}$. Also, $\forall a,b\in H$, $ab=ba$ since $H$ is cyclic and abelian. Therefore, $b\in C_G(a)$, in other words, $H\subseteq C_G(a)$ for all $a\in H$. By that, if $a\in H$, then $|Cl(a)|=\dfrac{pn}{|C_G(a)|}\leq \dfrac{pn}{p}=n$ because $|H|\leq |C_G(a)|$.
This is where I could not continue anymore. I thought that if I show that $|Cl(a)|$ divides $|H|$ for $a\in H$, then $|Cl(a)|=1$ since $n<p$, and $\forall a\in H$, $\theta_g(a)=a$ for all inner automorphism $\theta_g$, then $\theta_g(H)=H$.
 A: Assume that $H$ is not normal, then for some $g \in G$, $H^g \neq H$, hence $H \cap H^g $ is a proper subgroup of $H$. Since $|H|=p$, this implies $H \cap H^g=1$. But then
$$|G| \geq |HH^g|=\frac{|H| \cdot |H^g|}{|H \cap H^g|}=\frac{p \cdot p}{{1}}=p^2.$$
So $pn \geq p^2$, whence $n \geq p$, a contradiction.
As an obvious generalization (with the same line of proof or applying Sylow's theorems) one has the following.
Proposition Let $G$ be a finite group, $p$ a prime dividing its order. Let $P \in Syl_p(G)$ with $|G:P| \lt p$. Then $P \lhd G$.
A: $|G| =pn$  where $n<p.$
Now, $|H| = p$.
Let, if possible,
K be another subgroup of order $p$.
Since, $|G| =pn $ where $p>n$
By $\textit{Sylow theorem}$,
Number of sylow $p$ subgroup$ = 1+qp $ which must  divide $ n$.
Now, $1+qp|n $ and $n<p$
$\Rightarrow q=0 $,
So,  Sylow $p$ subgroup is unique.
Since, $p$ is prime and $ n<p$ then , $G$ has unique subgroup of order $p$.
$\mathcal{O}\left(gHg^{-1}\right) =p $
where $g \in \; G$.
Since, Subgroup is unique ,
hence, $gHg^{-1 }= H$
it is true $\forall \;g \in G$
Hence, $H \triangleleft\;G$ .
