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 $ab \in NK$
Best Answer
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 $N$.