Problem
Let $G$ be a group and $H,K$ subgroups of $G$, we define $HK=\{h.k : h \in H, k \in K\}$.
Prove that if $H$ or $K$ is normal, then $HK$ is a subgroup.
In order to show $HK$ is a subgroup, I have to verify the following three properties are satisfied:
$(i)$ $e \in HK$, this is almost immediate since$e=e.e$ and $e \in H, e \in K$ by the hypothesis $H,K$ are subgroups.
$(ii)$ $HK$ is closed under the operation of the group $G$, i.e., if $x,y \in HK$, then $x.y \in HK$.
$(iii)$ If $x \in HK$, then $x^{-1} \in HK$.
I am having some difficulty with $(ii)$ and $(iii)$:
In $(ii)$, if $x,y \in HK$, then $x=hk, y=h'k'$; $h,h' \in H, k,k' \in K$.
I want to show that $xy=(hk)(h'k') \in HK$. I don't know how to use the hypothesis say $H$ is normal.
The same goes for $(iii)$, suppose $x=hk \in HK$, how can I prove that $x^{-1}=k^{-1}h^{-1} \in HK$?
Any help would be appreciated.
Let's assume $H$ is normal -- the argument is analogous if $K$ is normal.
For $(ii)$, note that $$(hk)(h'k')=(hk)(h')(k^{-1}k)(k')=(h)(kh'k^{-1})(kk').$$ Do you see where to go from here, using that $H$ is normal in $G$?
For $(iii)$, try to come up with a similar trick ("multiply by $1$ in a clever way").
Another way to approach this problem would be to use the subgroup criteria: a nonempty set $A \subset G$ is a subgroup if and only if $a,b \in A \implies ab^{-1} \in A$. This will shorten the argument a bit, although the mechanics of the verification are the same.