Let $G$ be group and $ N \triangleleft G$ , $H < G$. Prove $NH < G$.

Question

Let $G$ be group and $ N \triangleleft G$ , $H < G$. Prove $NH < G$.

I am beginner in this so I am not sure if I proved it right.

Let $n_{1}h_{1}, n_{2}h_{2} \in NH$. We want to show that $ n_{1}h_{1}( n_{2}h_{2} )^{-1} \in NH$.

$ n_{1}h_{1}( n_{2}h_{2} )^{-1} = n_{1}h_{1}( h_{2}^{-1}n_{2}^{-1} ) = n_{1}(h_{1} h_{2}^{-1})n_{2}^{-1} $.

Because $H < G$, $h_{1} h_{2}^{-1} \in H$ and because ( I am not sure about this part) $ N \triangleleft G$ (then $\forall x \in G$, $ xN = Nx$ ) so $(h_{1} h_{2}^{-1})n_{2}^{-1} = n_{3}(h_{1} h_{2}^{-1})$ for some $n_{3} \in N$ and then

$n_{1}(h_{1} h_{2}^{-1})n_{2}^{-1} = n_{1}n_{3}(h_{1} h_{2}^{-1}) \in NH$.

Answer 1

$\forall n_1,n_2 \in N$, $\forall h_1,h_2 \in H$, we want to show that $ n_{1}h_{1}( n_{2}h_{2} )^{-1} \in NH$.

It's simpler to prove this directly: $$ n_1h_{1}(n_2h_2)^{-1} = n_1h_1(h_2^{-1}n_2^{-1}) = \left(n_{1}\underbrace{(h_{1} h_{2}^{-1})n_{2}^{-1}(h_{1} h_{2}^{-1})^{-1}}_{\in N}\right)(h_{1} h_{2}^{-1}) \in NH $$

Answer 2

It's perhaps easier if you do the separate checks:

1. $1\in N$ and $1\in H$, therefore $1=1\cdot1\in NH$;

2. if $n_1,n_2\in N$ and $h_1,h_2\in H$, then $(n_1h_1)(n_2h_2)=\bigl(n_1(h_1n_2h_1^{-1})\bigr)(h_1h_2)$

3. if $n\in N$ and $h\in H$, then $(nh)^{-1}=h^{-1}n^{-1}=(h^{-1}n^{-1}h)h^{-1}$

You may also want to prove that $NH=HN$:

• $nh=h(h^{-1}nh)$
• $hn=(hnh^{-1})h$

where the meaning of the symbols should be clear. This allows to skip 3.