I'm trying to show that $N_G(N_G(H)) = N_G(H)$ for some subgroup H < G. I also understand that the normalizer is described as the largest group which H is normal to, but I don't know how to use that information in a proof.
My attempt
I have shown that $N_G(H) \subseteq N_G(N_G(H))$ since $N_G(H)\lhd N_G(N_G(H)).$
I now want to show that $N_G(N_G(H)) \subseteq N_G(H)$, but I'm getting stuck. Here's what I have so far:
We need to show that if $$x \in N_G(N_G(H)) \Rightarrow x \in N_G(H).$$
i.e. $$xN_G(H)x^{-1} = N_G(H) \Rightarrow xhx^{-1} \in H$$
To show this, I considered taking an element $y \in N_G(H)$, but after that I'm lost. Can anyone provide some insight for how I should continue?
For the sake of completion, I figured that I would go ahead and place an answer to my question here.
The proposition stated in the question is false, unless (as Kenny Wong stated) H is a Sylow p-subgroup. The link to a proof of that is located here: Normalizer of the normalizer of the sylow $p$-subgroup.
To show that the the statement is false, examine the counterexample $D_8$, i.e. the set of symmetries on a square. (Sometimes, this is denoted as $D_4$).
$$G := D_8 = \{e,\ r,\ r^2,\ r^3,\ f,\ fr,\ fr^2,\ fr^3\}$$ where $r$ is a 90 degree rotation, and f is a flip. Let $H$ be $\{e,\ f\}$. Obviously, $H < G$.
$$N_G(H) = \{e,\ f,\ r^2,\ fr^2\}$$ $$\text{Then, }N_G(N_G(H)) = D_8.$$ $$\text{So, }N_G(H) \neq N_G(N_G(H)).$$