Normal subgroup of a group

Question

Let $H$ be a subgroup of $G$ and $K$ be a normal Subgroup of $G$. I need to prove that $KH$ is a subgroup of G where, $KH=\{\text{$kh$ : $k \in K \wedge h \in H$}\}$.

Can somebody please help me in this proof?

Answer 1

$KH=\{\text{$kh$ : $k \in K \wedge h \in H$}\}$. Note that you can also imagine this as a right coset as : $KH=\{\text{$Kh$ : $h \in H$}\}$. But since K is a normal subgroup of G, so every left coset of K in G is same as the right coset of K in G. Hence Kh=hK. So $KH=\{\text{$Kh$ : $h \in H$}\}$ = {$hK$ : $h \in H$}= HK. And hence KH is a subgroup of G because of the result KH is a subgroup of G iff HK=KH. I hope you got my point.

Answer 2

Let $k,k'\in K$ and $h,h'\in H$. Let also $\tilde{h}:= h·h'\in H$

We must prove that $(kh)(k'h')\in KH$ and that $(kh)^{-1}\in KH$:

Then $khk'h' = khk'h^{-1}\tilde{h} = k(hk'h^{-1})\tilde{h}$ But $\tilde{k}:=hk'h^{-1}\in K$ since $K$ is a normal subgroup. So $khk'h' = k\tilde{k}\tilde{h} \in KH$

Now $(kh)^{-1} = h^{-1}k^{-1} = (h^{-1}k^{-1})(hh^{-1}) = (h^{-1}k^{-1}h)h^{-1}$ But $\hat{k}:=h^{-1}k^{-1}h \in K$ since $K$ is a normal subgroup. So $(kh)^{-1} = \hat{k}h^{-1}\in KH$