Let $N$ and $K$ be subgroups of a group $G$ If $N$ is normal in $G$, prove that $NK = \{ n k : n \in K, k \in K \} $ is a subgroup of $G$ Let $N$ and $K$ be subgroups of a group $G$ 
part a) If $N$ is normal in $G$, prove that 
$$NK = \{ n k : n \in K, k \in K \} $$
is a subgroup of $G$
part b) if both $N$ and $K$ are normal subgroups of $G$ prove that $NK$ are normal 

Attmept 1  part a) if $a \in NK $ is $a^{-1} \in NK $ ?
leting $a^{-1}=n^{-1} k^{-1}$
$$
\begin{aligned}
a *a^{-1}&=nk *(n^{-1}k^{-1})
 \\      &=nk* k^{-1} n^{-1} &&  \text{since normal right??}
  \\      &=e
\end{aligned}$$
same argument needed for clusure 
$a=n_1 k_1 $ and $b =n_2 k_2$ is $ab \in NK??$
$$\begin{aligned}
ab= n_1 k_1 n_2 k_2 
  = n_1 n_2 k_1 k_2 && \text{ since normal?}
\end{aligned}  $$
so $ab \in NK$

 A: Here's how you do the product.  Let $a,b\in NK$.  Then, $a=n_1k_1$ and $b=n_2k_2$ for some $n_1,n_2\in N$ and $k_1,k_2\in K$.  We would like to show that $ab\in NK$.  Observe that $ab=n_1k_1n_2k_2$.
Approach 1: Since $N$ is normal, we know that for all $g\in G$, $gNg^{-1}=N$.  Now, by introducing $e=k_1^{-1}k_1$ we get
$$
ab=n_1k_1n_2k_2=n_1k_1n_2(k_1^{-1}k_1)k_2=n_1(k_1n_2k_1^{-1})k_1k_2.
$$ 
Since $N$ is normal, $k_1Nk_1^{-1}=N$, therefore, $k_1n_2k_1^{-1}$ equals $n_3$ for some $n_3\in N$.  Therefore, $ab=n_1n_3k_1k_2$.
Approach 2: Since $N$ is normal, we know that for all $g\in G$, $gN=Ng$.  In particular, $k_1N=Nk_1$, so there exists $n_3\in N$ so that $k_1n_2=n_3k_1$.  Therefore, $ab=n_1n_3k_1k_2$.
In both cases, $ab=(n_1n_3)(k_1k_2)$.  Since $n_1n_3\in N$ and $k_1k_2\in K$, by closure, $ab\in NK$.
Note here that $n_2$ and $k_1$ do not commute (as in the original post), but they almost commute (up to an element of $N$), i.e., $k_1n_2=n_3k_1$, $n_2$ might be different from $n_3$, but they are both in $N$.
A: For part (a), just use the following result:

For a group $G$, a subset $H$ of $G$ is a subgroup of $G$ if and only if $ab^{-1}\in H$ whenever $a,b\in H$.

To this end, suppose $a,b\in NK$, and write $a=n_1k_1$, $b=n_2k_2$. Then
$$ab^{-1}=(n_1k_1)(k_2^{-1}n_2^{-1})=n_1(k_1k_2^{-1}n_2^{-1}).$$
Since $N$ is normal, $k_1k_2^{-1}n_2^{-1}=n_3k_1k_2^{-1}$ for some $n_3\in N$, and thus
$$ab^{-1}=n_1n_3k_1k_2^{-1}\in NK.$$
Therefore $NK$ is a subgroup of $G$.
