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 $K \subseteq H$.
I have seen other solution, but I want someone to look at my solution and check if I have made any mistakes.
Consider cosets of sub group $H$, any element belonging to any cosets of $H$, other than H itself, must have order $s.p^k$, where $s$ divides m, since order of a coset element in quotient group divides order of that element in group. And thus all elements whose order is $p^k$ form must be within H.
Am I right, if not, what am I missing?
Why don't you look at the quotient $G/H$? This group has an order that is not divisible by $p$. Consider the image $KH/H (\cong K/(H \cap K))$. What can you say about this subgroup of $G/H$?