$G$ is a group. $H\leq G$ and $K\trianglelefteq G$. Prove $HK\leq G$. I tried to use the following but $H$ would have needed to be a normal subgroup of $G$.
$$h_1k_1,h_2k_2\in HK\Rightarrow (h_1k_1)(h_2k_2)^{-1}\in HK.$$
How do you think I should prove it?

HINT: Your basic approach is fine. Note that since $K$ is normal in $G$, $Kg=gK$ for every $g\in G$. In particular, for any $k\in K$ and $g\in G$ there is a $k'\in K$ such that $kg=gk'$.
