# Math Help - Urgent - normal subgroup proof

1. ## Urgent - normal subgroup proof

Stuck on the following question;

2. $i \Rightarrow ii$
$Ng=gN \Rightarrow g^{-1}Ng=N$ (multiply on the left by g inverse)

In particular this means
$g^{-1}Ng \subset N$

that is precisely the statement of ii

$ii \Rightarrow i$
Assume the hypothesis. For any g, it is clear that we have $g^{-1}Ng \subset N \Rightarrow Ng \subset gN$.

But then we also have
$(g^{-1})^{-1}Ng^{-1}\subset N \Rightarrow gNg^{-1} \subset N \Rightarrow gN \subset Ng$

Thus $gN=Ng$ as desired.