N is a normal subgroup of G if $aNa^{-1} \subset N $ for all $a ∈ G$. Prove that in that case, $aNa^{-1} = N $.

Question

I said if N is a normal subgroup of G when $aNa^{-1} \subset N $ aN = Na as N is a normal subgroup of $G$.

Therefore $aNa^{-1} = Naa^{-1} $ and $aNa^{-1} = N $.

I would like to go with this proof but it looks too simple for now?

Answer 1

Indeed since $N\lhd G$ then for all $g\in G$ we have $N^g\subseteq N$. Now do this for $g^{-1}\in G$ so we see that $N^{g^{-1}}\subseteq N$ and therefore $$(N^{g^{-1}})^g\subseteq N^g$$ But what is $(N^{g^{-1}})^g$ then?