Proving $nk = kn$ 
Let $K$ and $N$ be normal subgroups of $G$ such that $K \cap N = \langle e \rangle$. Prove that $nk = kn$ for all $n \in N$ and for all $k \in K$.

Since $K$ and $N$ are normal, we have $nK = Kn$ and $kN = Nk$. However, on their own they don't imply $nk = kn$. How can I prove this?
 A: Suppose we have some $a=knk^{-1}n^{-1}. N$ is normal $\Rightarrow knk^{-1}\in N$. Also, $n^{-1}\in N \Rightarrow (knk^{-1})n^{-1}\in N$ since $N$ is closed. Further, $nkn^{-1}\in K$ since $K$ is normal. So, $a\in K \cap N \Rightarrow a\in\{e\} \Rightarrow a=e \Rightarrow knk^{-1}n^{-1}=e=kn(nk)^{-1} \Rightarrow nk=kn$.
A: Explain why $n^{-1}(k^{-1}nk)=(n^{-1}k^{-1}n)k\in K\cap N$ and you've got an answer.
A: In general, for normal $K$ and $N$,  $[K,N] \subseteq K \cap N$.
A: Since $N$ is normal, $knk^{−1}\in N$, and $k^{-1}nk\in N$ because $N$ is a group. Since $K$ is normal, $nk^{−1}n \in K$, which means by similar argument $n^{−1}kn \in K$. In either case, $n^{−1}k^{−1}nk \in N \cap K$ And since by hypothesis $N\cap K = \langle e \rangle$, it must follow that $n^{−1}k^{−1}nk$ is the only element of $N \cap K$ and that $k^{-1}nk=e$, which finally gives $kn=nk$, as desired. 
(Copying this from my comment response (now deleted) to someone else's answer because I cannot see my code from my original comment.)
