TITLE: Let $|G|=p^nm$ where $p$ is a prime and $\gcd(p,m)=1$.Suppose that $H$ is a normal subgroup of $G$ of order $p^n$.
QUESTION [5 upvotes]: Let $|G|=p^nm$ where $p$ is a prime and $\gcd(p,m)=1$.Suppose that $H$ is a normal subgroup of $G$ of order $p^n$. IF $K$ is a subgroup of $G$ of order $p^k$, show that $K \subseteq H$.
Attempt: Given that $|G|=p^nm$ and $gcd(p,m)=1 => \gcd(p^n,m)=1$
$H \triangleleft G =>g^{-1}hg \subseteq H~~ \forall ~~g\in G, h\in H$
$G/H=\{gH ~~\forall~~g \in G \}$ and then $|G/H|=p^nm/p^n=m$
Now, $O(gH)~~|~~O(g)$ and $O(g)~~|~~p^nm$
$=> O(gH)~~|~~p^nm ..........(1)$
$O(gH)~~|~~m ....... (2) $ 
( By Lagrange's Theorem, $gH$ is an element in $G/H$ and $O(gH) ~~|~~|G/H|$ and $|G/H|=m$)
Since $\gcd(p^n,m)=1 => O(gH)= r $ s.t. $r$ divides $m$.
$=> g^r \in H$
If we show that $\exists h_1,h_2 \in H $ s.t. $k=h_1h_2$ then our result can be proved or if we prove that $K$ is a normal subgroup in $H$ and $|H : K|=1$, then also we can prove the result
How should I proceed ahead. Help will be appreciated.
Thank you.

REPLY [3 votes]: Let $g \in K, g \notin H$. Then $(gH)^m=H=(g^m)H \implies g^m∈H \implies (g^m)^{p^l}=e=(g^{p^l})^m$ for some $l$ But $g^{p^l} \in K$ As the order of K is $p^k$ the equality $(g^{p^l})^m=e$ is impossible. Proof is ended