Let $K$ adn $N$ be subgroups of $G$. Assume that $N$ is normal. Then we know that $$ K / (K\cap N) \simeq KN/N. $$ I understand the proof using the map $k \mapsto kN$ which has kernel $K\cap N$. My question is, for KN/N to be a group, I would assume that $KN$ is a group as well. Is this true? In that case how can one show this?
I think one just needs to prove that $KN$ is closed. If $k,k'\in K$, and $n,n'\in N$, then I would need to show that $knk'n'\in KN$. I guess this uses the fact that $N$ is normal, but I can't figure out exactly how.
We need to show that it is a subgroup to do so we can show that it is closed under products and inverses. Suppose that $kn$ and $k'n'$ are in $KN$. Then, since $N$ is normal $(k')^{-1}nk' \in N$ and hence $knk'n'=(kk')(((k')^{-1}nk')n') \in KN$. Again since $N$ is normal $kn^{-1}k^{-1}$ is in $N$ and hence $(kn)^{-1} = k^{-1}kn^{-1}k^{-1} \in KN$.