Counterexample for the normalizer being a normal subgroup Let $G$ be a group, and $H$ a subgroup.
$H$'s normalizer is defined: $N(H):=\{g\in G| gHg^{-1}=H \}$. 
Prove $N(H)$ is a normal subgroup of G, or give counterexample.
Intuitively it seems to me that this claim is wrong, however, I'm having trouble with finding a counterexample.
Thans in advance for any assistance! 
 A: An easy, concrete counterexample is any of the three subgroups of order $2$ in $S_3$.
A: For any finite group $G$ and any $p$-Sylow subgroup $P \leq G$ the following holds:

Every subgroup $U$ with $N_G(P) \leq U \leq G$ satisfies $N_G(U) = U$.

Especially every normalizer $N_G(P)$ provides a counterexample unless $P$ is normal in $G$.  Can you think of examples of Sylow subgroups which are not normal?
A: Let $G=\langle f,r : f^2 = r^8 = 1, rf =fr^7 \rangle$ be the dihedral group of order 16, and let $H=\langle f \rangle$. Then $N_G(H) = \langle f,r^4 \rangle$ and $N_G( N_G(H) ) = \langle f, r^2 \rangle$ and $N_G( N_G( N_G(H) ) ) = \langle f,r \rangle =G$.
In other words, $H$ is not normal, neither is its normalizer, but the normalizer of the normalizer is normal.
Using dihedral groups of order $2^n$ you can create arbitrarily long chains of non-normal normalizers that eventually end up being normal.
A: The normaliser of any proper nontrivial subgroup of a simple group would work too. Can you see why?
