If $H$ and $K$ are subgroups of $G$, and $K$ is normal, then $HK$ is a subgroup of $G$

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$HK$ is the set of all products $hk$ where $h \in H$ and $k \in K$. Quotient groups are the next chapter in my textbook, so please avoid using them.

I figured out inverse-closure: $h^{-1}\in H$, $k^{-1}\in K$, $h^{-1}k^{-1}\in HK$

$h^{-1}k^{-1} = (kh)^{-1}$

RHS $= (hk^{'}h^{-1}h)^{-1}$ because $k$ is normal and $k = gk^{'}g^{-1}$ for any $g \in G$

RHS $=(hk^{'})^{-1}$ for every $h\in H$ and $k^{'}\in K$ possible.

Operation-closure is giving me trouble. I don't know how to prove every $hkh^{'}k^{'}\in HK$ I suspect this might be a bad place to start.

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Fill in details: for $\;h,h'\in H\;,\;\;k,k'\in K\;$ , we have

$$hkh'k'=\overbrace{h(kh'k^{-1})}^{\in H}\,\overbrace{kk'}^{\in K}\in HK$$