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 $G$

Question

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

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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 $ab \in NK$

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Answer 1

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 $N$.

Answer 2

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 $G$.