Prove that $k h = h k$ for $h \in H $ & $k \in K $

Question

I found the following statement in my lecture but I don't see exactly why it is true

We have $G $ a group and are given $H$ , $K$ two normal subgroups of $G$ such that :$\; H\cap{} K = \left\{e \right\}$

Then for all $\; h \in H $ and for all $\; k \in K$ we have :

$$hk=kh$$

Thanks in advance for your help .

Answer 1

Since both $H,K$ are normal subgroups, $$hkh^{-1}k^{-1} = h(kh^{-1}k^{-1}) \in H \\ hkh^{-1}k^{-1} = (hkh^{-1})k^{-1} \in K$$ Therefore $$hkh^{-1}k^{-1} \in H\cap K = \{e\},$$ so $hkh^{-1}k^{-1} = e$ or $hk=kh$.

Answer 2

We have $k^hk^{-1}=hkh^{-1}k^{-1}=h(h^{-1})^k$ in both $H$ and $K$ by normality.