If $H$ and $K$ are normal subgroups with $H\cap K=\{e\}$, then $xy=yx$ for $x\in H, y\in K$. Claim : $H$ and $K$ are two normal subgroups of group $G$ such that $H\cap K = \{e\}$ then $ xy = yx $ for $x \in H$ and $y \in K$.
Proof : Let $x \in H$ and $y\in K$ then I need to prove that $xy =yx$. Let us assume that 
$$y^{-1}xy \neq x$$
$y^{-1}xy = z$ where $z \in H$
Question : I am not getting how to proceed further?
 A: Note $x^{-1}(y^{-1}xy) = (x^{-1}y^{-1}x)y$. The LHS is in $H$ since $H$ is normal and is a subgroup. The RHS is in $K$ since $K$ is normal and is a subgroup. So, the common value must be $e$.
A: Let $\pi:G\to G/H$ be the projection. Then $\pi(x)=\pi(x^{-1})=e$
so $\pi(xyx^{-1}y^{-1})=\pi(y)\pi(y^{-1})=e$, so $xyx^{-1}y^{-1}\in H$.
Similarly $xyx^{-1}y^{-1}\in K$....
A: You started out well. From
$$yxy^{-1}=x$$
notice that on the left you have a conjugation, which smells like a good idea to me, since one of the definitions of a normal subgroup is that it's closed under conjugation. So the left hand side is in $H$. Now continue:
$$(yxy^{-1})x^{-1}=e$$
The left hand side is a product of two elements of $H$ and so is in $H$. If we could show it's in $K$ as well, then we'd be done. But in fact, all we have to do is shuffle the parentheses around to get $y(xy^{-1}x^{-1})$ and see that the left hand side is indeed in $K$, for exactly the same reason it's in $H$. Now you can check that each of the steps in our reasoning is reversible to get back to the original equation you wanted to prove, $yx=xy$.
Another, maybe more direct way is to just calculate the "difference" between $xy$ and $yx$ and check that it's "zero" (identity). So we try to calculate $xy(yx)^{-1}$ and show it equals $e$. You should find that the reasoning is almost exactly the same.
