Every element of a subgroup $N$ normalizes $N$

Question

Let $N$ be a subgroup of a group $G$. By definition, an element $g\in G$ is said to normalize $N$ if $gNg^{-1}=N$. I am trying to show every element of $N$ normalizes $N$.

In other words, I want to show for all $n\in N$, $nNn^{-1}=N$.

My attempt:

Fix $n\in N$ and let $x\in nNn^{-1}$. Then $x=nhn^{-1}$ for some $h\in N$. Since $N$ is a group, $x\in N$. So $nNn^{-1}\subset N$. I am having a trouble showing the other inclusion. Let $y\in N$. I want to show that $y\in nNn^{-1}$ in order to show $N\subset nNn^{-1}$. Can someone help me figure this out?

Answer 1

Hint $$y=n(n^{-1}yn)n^{-1}\\ n^{-1}yn \in N$$

Answer 2

We can say even more: for every $n\in N$, we have $N=nN=Nn$ if $N$ is a subgroup and $n\in N$.

To prove $N\subseteq nN$, just consider $x=n(n^{-1}x)$ for $x\in N$.

We can use the same reasoning directly for $N\subseteq nNn^{-1}$: write $x\in N$ as $n(n^{-1}xn)n^{-1}$.