If H and K are two normal subgroups of a group G .If the order of H and the order of K are relatively prime.prove that hk=kh My answer is : let  $h= gxg^{-1}$ in $H$ for $x$ in $H $
and    $k= gyg^{-1}$ in $K$ for $y$ in $K$
$hk =(gxg^{-1})(gyg^{-1})=g (xy) g^{-1}$
But $xy$ in $HK$. hence $HK$ is normal in $G$ i.e . $aHK =HKa$ 
$ahk=hka$
And w.r.t the order of $H=p$ and order of $K=q$ ....then order of $HK =pq$...

My favorite proof of this:
Consider $hkh^{-1}k^{-1}$: since $H \lhd G$, we have:
$hkh^{-1}k^{-1} = h(kh^{-1}k^{-1}) \in H$. Since $K \lhd G$, we have:
$hkh^{-1}k^{-1} = (hkh^{-1})k^{-1} \in K$.
Therefore $hkh^{-1}k^{-1} \in H \cap K$, and by Lagrange, the order of $hkh^{-1}k^{-1}$ divides $\gcd(|H|,|K|) = 1$.
Hence $hkh^{-1}k^{-1} = e$, and thus $hk = kh$.
