Let $G$ be a group and $A, B\leq G$. Suppose $H\trianglelefteq A$ and $K\trianglelefteq B$. Is it true that $HK\trianglelefteq AB$?
Notation: $\leq$ means subgroup and $\trianglelefteq$ means normal subgroup.
Thanks
Let $G$ be a group and $A, B\leq G$. Suppose $H\trianglelefteq A$ and $K\trianglelefteq B$. Is it true that $HK\trianglelefteq AB$?
Notation: $\leq$ means subgroup and $\trianglelefteq$ means normal subgroup.
Thanks
It is false. Take $H=\{1\}, A=G,$ and $K=B$. If the assumption was true we would have $K\lhd G$ but $K$ can be any subgroup.
No.
First of all, the set $AB$ is not in general a subgroup, consider eg $H = A$ and $K = B$ for easy examples.
Even if it is, the claim is false. Consider $G = B$, $H = A$. The claim fails since it is possible to find examples where $H$ and $K$ are subgroups of $G$, $K$ is normal in $G$, but $HK$ is not a normal subgroup.