10
$\begingroup$

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!

$\endgroup$
12
$\begingroup$

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

$\endgroup$
7
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ Because these are (normalizers of) non-normal Sylow subgroups. ;) $\endgroup$ – Dune Apr 3 '14 at 15:19
  • 1
    $\begingroup$ @Dune That's a bit circular, though, eh? ;) $\endgroup$ – Alexander Gruber Apr 3 '14 at 15:41
  • $\begingroup$ Sure. I just wanted to point out that these 3 examples belong to an infinite series of examples already mentioned. $\endgroup$ – Dune Apr 3 '14 at 20:16
5
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.