Assume that H and K are subgroups of a group G and that K is normal in G. Then
HK is a subgroup of G, where
HK = {hk | h e H, k e K}
Thanks for any help
First note that $\displaystyle e\in HK$ (easy to verify).
Let $\displaystyle h_1k_1, h_2k_2\in HK$, we seek to show that $\displaystyle (h_1k_1)(h_2k_2)^{-1}\in HK$
So $\displaystyle (h_1k_1)(h_2k_2)^{-1}=h_1k_1k_2^{-1}h_2^{-1}$.
Now, we see that $\displaystyle h_1k_1k_2^{-1}h_2^{-1}=(h_1h_2^{-1})(h_2 k_1k_2^{-1}h_2^{-1})$. Observe that $\displaystyle h_1h_2^{-1}\in H$ and since $\displaystyle K\lhd G$, $\displaystyle h_2(k_1k_2^{-1})h_2^{-1}\in K$.
Thus $\displaystyle (h_1k_1)(h_2k_2)^{-1}\in HK\implies HK\leq G$.
Does this make sense?