If $H$ is a subgroup of a group $G$ and $K$ is a normal subgroup of $G$ then show that $K$ is a normal subgroup of $HK$. 
If $H$ is a subgroup of a group $G$ and $K$ is a normal subgroup of $G$ then show that $K$ is a normal subgroup of $HK$.

Note $K$ is a normal subgroup of $HK$.
I thought I could just prove $K$ is a subset of $HK$ and vice versa so as to prove it is a normal subgroup but then how to do this I have no idea because I keep getting $hk$ element again while trying to do so.
I just tried to implement the steps to find $HK$ is a subgroup of $G$.
 A: I'll assume you've already shown that $HK$ is a subgroup of $G$.
First, it is clear that $K$ is a subgroup of $HK$ because $K$ is a group and $k=ek\in HK$ (as the identity $e$ is in the subgroup $H$).
To check normality, we note that for all $g\in G$ we have $gK=Kg$ because $K$ is a normal subgroup of $G$. But then if $g\in HK$, then we must also have $gK=Kg$ because $g\in HK\subset G$.
A: $K$ is a normal subgroup of $G$, i.e. $g^{-1}kg \in K$ for all $k \in K, g \in G$.
$HK=\left\{hk~|~h\in H, k \in K\right\}$ is a subgroup since $e = ee \in HK$, if $h_1k_1, h_2k_2 \in HK$ then $h_1k_1h_2k_2 = h_1(h_2h_2^{-1})k_1h_2k_2 = h_1h_2(h_2^{-1}k_1h_2)k_2 = h_1h_2k'_1k_2$ for some $k'_1\in K$ because $K$ is normal in $G$ and so $h_1k_1h_2k_2 = h_1h_2k'_1k_2 = (h_1h_2)(k'_1k_2) \in HK$, ie. $HK$ is closed under multiplication and $(hk)^{-1} = k^{-1}h^{-1} = (h^{-1}h) k^{-1}h^{-1} = h^{-1}(hk^{-1}h^{-1}) = h^{-1}k'\in HK$ for some $k' \in K$, i.e. $HK$ is closed under inverses.
We want to show $K$ is a normal subgroup of $HK$, i.e. $(hk)^{-1}k_1(hk) \in HK$ for all $k_1 \in K, h \in H, k \in K$. Expanding, we need $k^{-1}h^{-1}k_1hk \in HK$. You can use similar transformations as above to move the $k$ factors to the right-hand-side of the expression.
