$|G|=p^n m$ , $p$ is prime and $\gcd(p,m)=1$. Let $H$ normal in $G$ of order $p^n$ and $K$ is a subgroup of $G$ of order $p^k$ then $K\subseteq H$ $|G|=p^n\ m$ where $p$ is prime and $\gcd(p,m)=1$ suppose that $H$ normal in $G$ of order $p^n$ and $K$ is subgroup of $G$ of order $p^k$ then show that $K\subseteq H$
this question is already asked here(
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 $p^n$.) but the accepted answer just contains hints and I am not able to prove above result by using that hints. Any help.
Here is My attempt:(without using hints in the accepted answer given in link) we are asked, not to use sylow theorems.
Given $\gcd(p,m)=1$ and so we have $\gcd(p^n, m)=1$. 
Since $H$ normal in $G$ and $K$ subgroup of $G$ so that $HK$ is subgroup of $G$ also, 
$|HK|=\frac{|H|\ |K|}{|H\cap K|}$ 
So that $|HK|=\frac{p^n\ p^k}{|H\cap K|}$
Since $H\cap K$ is subgroup of $H$ as well as of $K$. Hence by Lagrange's theorem we have, 
$|H\cap K|\ |\ p^n $ and $|H\cap K|\ |\ p^k$.
Case a: If $|H\cap K|=1$ then $|HK|= p^{k+n}$ but $|HK|\ |\ |G|=p^n\ m$ so we have $p\ |\ m$ which is contradiction as $\gcd(p,m)=1$
Case b: if $|H\cap K|<p^k$ then again we get $|HK|=p^{n+r}$ and so that we again get $p\ |\ m$ which is contradiction as $\gcd(p,m)=1$.
Case c:: if $|H\cap K|=p^k$ then as $H\cap K$ is subgroup of $K$ we have $H\cap K=K$ which gives $H\subseteq K$.
If $H\subset K$ then, this imply $|H|<|K|$ i.e. $p^n<p^k$ but $p^k\ |\ p^n\ m$ (as $|H|\ |\ |G|$). But this gives $p\ |\ m$ and we get again contradiction. So we must have $H$ is not proper subset of $K$.
If $H=K$ then $K\subseteq H$ holds.
Case d: if $|H\cap K|>p^k$ not possible because $|H\cap K|\mid |K|$ as $H\cap K$ is subgroup of $K$. 
So in my proof. None of the cases gives $K\subset H$ (i.e. $K$ proper subset of $H$). Is it possible? 
Is my proof is correct and sufficient?
Further, Is there is another method to prove this. Please Help 
 A: Your answer is almost correct and can be summarized in the following way (I am using your notations).
If you write $|H\cap K| = p^l$, you must have $|HK| = p^{n+k-l}||G| = p^n m$. This gives $l\geq k$ since $(p,m)=1
$.
Since $H\cap K \subseteq K$, you have $l\leq k$.
In the end you get $H\cap K = K$ by cardinality. It means that $K\subseteq H$.
A: Hint: look at $G/H$. The order of this quotient is not divisible by $p$. Now consider the canonical image of $K$. This is $KH/H$ and is isomorphic to $K/(K \cap H)$ and is a $p$-group. This yields $KH=H$, that is $K \subseteq H$.
A: An alternative: 
Suppose on the contrary that $K\not\subset H$. It means that $\exists x\in K$ such that $x\notin H$.  
We have $xH\in G/H$ and note that $xH$ is a non-identity element of factor group $G/H$, which has order $o(G/H)=m$.  So by Lagrange's theorem,we have $o(xH) | m$.
Also $x\in K\implies o(x)|p^k\implies o(xH)|p^k$. 
Using the above two relationships, we have $o(xH)|gcd (p^k,m)\implies o(xH)=1$ as $m,p$ are co-prime. 
And this contradicts our assumption that $xH$ is a non-identity element.
Therefore, our assumption that $K\not\subset H$ is wrong and hence by contradiction, $K\subset H$. This proves the result.
