Proving a custom subgroup $HK$ is closed.

Question

Let $G$ be a group and let $H$ and $K$ be a normal subgroup and subgroup of $G$, respectively. Define $HK=\{hk \mid h\in H, k\in K\}$.

I want to show that $HK$ is closed under the group's operation.

I start by letting $h_1k_1, h_2k_2\in HK$ where $h_1, h_2\in H$ and $k_1, k_2 \in K$. Then, I want to show that $h_1k_1h_2k_2 \in HK$.

I thought I would use the fact that $H$ is a normal subgroup since our definition of normal is "$gH=Hg, \forall g\in G$". Yet, I realize that this is not as strong of a condition as I thought (I at first confused this with being abelian). Any suggestions as to a direction for this proof?

Answer 1

Meditate on the expression: $h_1 k_1 h_2 (k_1^{-1} k_1) k_2$.