Prove that product of a nilpotent group and a supersolvable group is a supersolvable group 
Suppose that $G$ is a finite group, and $G = HK$, where $H,K \triangleleft G$, $H$ is a nilpotent group, $K$ is supersolvable group. Prove that $G$ is supersolvable group.  

Thanks.  
(The group $G$ is said supersolvable if it has normal series $G_{0}\triangleleft G_{1}\triangleleft\cdots\triangleleft G_{n}=G$ such that $G_{i}\triangleleft G$ and $G_{i}/G_{i-1}$ is cyclic for all $i = 0,1,2,...,n$.)  
PS. I knew that if $H$ and $K$ are supersolvable, normal subgroups of $G$, then $HK$ is not necessary supersolvable, but in this case it is supersolvable. 
 A: Start with the intersection of the groups in the lower central series of $H$ with $K$. The terms in this series are all normal in $G$. Since $K$ is supersolvable, we can refine this to a series in which all terms are normal in $K$ and the factor groups are cyclic. But because we have refined a series in which the factors are all centralized by $H$, the resulting terms are all normal in $H$ too. So we have found a series for $H \cap K$ with cyclic factor groups in which all terms are normal in $G$.
But $G/(H \cap K) \cong H/(H \cap K) \times K/(H \cap K)$ is a direct prodict of a nilpotent group and a supersolvable group, so it is supersolvable. Hence $G$ is supersolvable.
Added later: To see that the terms in the refinement are normalized by $H$,
note that $(\gamma_iH \cap K)/(\gamma_{i+1}H \cap K) \le Z(H/(\gamma_{i+1}H \cap K))$. So any subgroup of $(\gamma_iH \cap K)/(\gamma_{i+1}H \cap K)$ is normal in $H/(\gamma_{i+1}H \cap K)$. Hence any group $K_i$ with $\gamma_{i+1}H \cap K \le K_i \le \gamma_iH \cap K$ is normal in $H$.
