Prove that KN is a subgroup of G if N is a normal subgroup of G. Let $G$ be a group; $K$ and $N$ are subgroup of $G$. If $N$ is a normal subgroup of $G$, prove that $KN = \lbrace kn | k \in K, n \in N \rbrace$ is a subgroup too.
First step I tried is proving that $KN = NK$. Is it correct way?
Thanks for your time in advance.
 A: Yeah, your first step is correct.
Claim. $KN = NK$.
Proof.
Let $x \in NK$. Then, $x=nk$ for some $n \in N$ and $k \in K$. Note that $x=nk=k(k^{-1}nk)$. Since $k\in K$ and $k^{-1}nk \in N$ (since $N$ is a normal subgroup of $G$), we see that $x \in KN$. Hence, $NK \subseteq KN$.
Next, consider $y \in KN$. Then $y=kn=(knk^{-1})k \in NK$, for some $k \in K$ and $n \in N$. Thus, $KN \subseteq NK$. Hence, $KN = NK$.
Next, to prove that $KN$ is a subgroup of $G$, note that $e = e \cdot e \in KN$. If $a,b \in KN$, then $a = kn$ and $b=k_1n_1$ for some $k,k_1 \in K$ and $n,n_1 \in N$. Thus, $ab=k(nk_1)n_1=k(k_2n_2)n_1=(kk_2)(nn_2) \in KN$ 
since $KN = NK$, $nk_1 \in NK$, and $nk_1=n_2k_2$ for some $k_2 \in K$ and $n_2 \in N$. Finally, $b^{-1} = (k_1n_1)^{-1} = n_1^{-1}k_1^{-1} \in NK = KN$. Thus,
$(\forall a,b \in KN).ab^{-1}=(kn)(n_1^{-1}k_1^{-1}) \in KN$. Hence, $KN$ is a subgroup of $G$.
Hope it will help you.
A: Let's use the subgroup criterion:  closure under $x,y\to xy^{-1}$.  Given $x,y\in KN$, we have $x=k_1n_1, y=k_2n_2$.  Now $xy^{-1}=k_1n_1(k_2n_2)^{-1}=k_1n_1n_2^{-1}k_2^{-1}=k_1k_2^{-1}\hat n\in KN$.  I used normality of $N$ to get $\hat n$.
