# 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!

The normaliser of any proper nontrivial subgroup of a simple group would work too. Can you see why?

An easy, concrete counterexample is any of the three subgroups of order $2$ in $S_3$.

• Because these are (normalizers of) non-normal Sylow subgroups. ;) – Dune Apr 3 '14 at 15:19
• @Dune That's a bit circular, though, eh? ;) – Alexander Gruber Apr 3 '14 at 15:41
• Sure. I just wanted to point out that these 3 examples belong to an infinite series of examples already mentioned. – Dune Apr 3 '14 at 20:16

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?

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.