Let $G$ be a group. If $H\leq G$ is a subgroup and $N\vartriangleleft G$, then $HN$ is a subgroup of $G$.

Question

Let $H$ be a subgroup of $G$ and $N$ a normal subgroup of $G$. Prove that $HN=\{hn\:|\:h \in H, n \in N\}$ is a subgroup.

Here what I have so far. It is not really much I understand what I need to show but I am stuck in one little thing which I can not by pass. Proof: I claim that $HN$ is not empty. Since $G$ is a group, $1_{g} \in G$. Since both $H$ and $N$ are subgroups of $G$, $1_{g} \in H$ and $1_{g} \in N$. Therefore $1_{g} \in HN$. I claim that $HN$ is closed. Let $h_{1}$$n_1$ and $h_{2}$$n_2$ $\in HN$. $(h_{1}n_1)(h_{2}n_2)=(h_{1}n_1h_{2}n_2)=h_{1}(n_1h_{2})n_2$. Now this is where my difficulty is I cannot swap $n_1$$h_{2}$ to get $h_{2}$$n_1$ so what do I do. Help!!!

Answer 1

Actually you don't need to swap $n_1h_1$ to get $h_2n_1$. When you have the expression

$$(h_1n_1)(h_2n_2)$$

just observe that this is equal to

$$(h_1h_2)(h_2^{-1}n_1h_2)(n_2)$$

By normality, $h_2^{-1}n_1h_2\in N$, then $(h_2^{-1}n_1h_2)(n_2)\in N$ and, since $h_1h_2\in H$, the element we considered lies in $HN$.

Answer 2

It is true that $h^{-1}nh=n'\in N\;\implies nh=hn'$ , because $\;N\lhd G\;$ , then

$$h_1n_1h_2n_2=h_1(n_1h_2)n_2=h_1(h_2n')n_2=(h_1h_2)(n'n_2)\in HN$$

Answer 3

You can calculate with the sets directly and don't have to calculate with elements.

Since $N$ is normal, we have $gN = Ng$ for all $g \in G$. It follows $HN=NH$.

Now if $H,N$ are subgroups of a group $G$ with $HN=NH$, then $HN$ is a subgroup. In fact, we have 1. $1=1 \cdot 1 \in HN$, 2. $HN \cdot HN = H(NH)N=H(HN)N=(HH)(NN)=HN$, and 3. $(HN)^{-1} = N^{-1} H^{-1} = NH = HN$.