Let $G$ be a group of order $399$. Then by Sylow, it must have a normal subgroup of order $19$, denoted by $H$. Let $N, K$ be groups of order $7$ and $3$. Then $HN$ and $HK$ are groups of order $133$ and $57$. What is the possible order of their normalizer?
The normalizer can only be either themselves or the whole group by Lagrang
Best Answer
Without knowing more about $G$, both cases are possible for $HK$, but $HN$ is normal in $G$. Since $HN$ is of index 3 in $G$, it contains a normal subgroup (the normal core) of index $\leq 3!=6$ so must be $HN$ itself.
Consider the group $G=F_{21}\times C_{19}$ where $F_{21}$ is the nonabelian group of order 21. A subgroup $C_3\times C_{19}$ cannot be normal in $G$, or else the $C_3$ would be a characteristic subgroup of a normal subgroup, hence normal subgroup of $G$, contradicting $F_{21}$ nonabelian. So $N_G(HK)=HK$ in this case.
On the other hand, the abelian group of order 399 has $N_G(HK)=G$.