# How to show the normaliser is a subgroup and that number of subgroups conjugate to H is equal to $|G : N_G(H)|$

I have to define the normalizer, show that its a subgroup, that $H$ is normal to the normalizer and that the number of subgroups conjugate to $H$ in $G$ are equal to the index $|G :N_G(H)|$ of the normalizer.

If we have a group $G$ and a subgroup $H$, then the normalizer, $N_G(H)$ is defined as the largest subgroup of $G$, which $H$ is normal in.

To show that is a subgroup, can I just say that by definition of a normalizer, it is a subgroup of $G$? Or can I just say the axioms and saying "this axiom holds as normalizer is a subgroup", which is basically the same thing?

Then to show the next bit, can I say that the Orbit Stabiliser theorem says $|G| = |Orb(x)| |G_x| \implies |Orb(h) = |G : G_h|$. The normalizer is the stabiliser of the group and the orbit is the subgroup $H$, so we get

$$H = |G : N_G(H)|$$

By Lagrange theorem, we know the RHS holds as we have already proves the normalizer is subgroup and so the proof is complete. We don't need to prove the equality as that comes from OS theorem which we assume to be proved.

• I think as you stated above, $N_G(H)=G_H$ and say since the normalizer is he stabilizer of the group so it is a subgroup of g, although, you can verify it by definition of a subgroup. – mrs Dec 31 '12 at 14:34
• In order to define $N_G(H)$ as the "largest" subgroup of $G$ in which $H$ is normal, you first need to define "largest", and then prove that there is such a subgroup and that it is unique. – Derek Holt Dec 31 '12 at 19:59

$a^{-1}Ha=b^{-1}Hb$ iff $(ab^{-1})^{-1}Hab^{-1}=H$ iff $ab^{-1}\in N_G(H)$ iff $a\in N_G(H)b$

Two distinct elements of $G$ will produce the same congugate of $H$ iff they belong to the same right coset of $N_G(H)$.

Thus, there are $|G:N_G(H)|$ different congugates of $H$.

• @Kaish: Note that epsilon proved that the normalizer is a subgroup just by your assumptions. :) – mrs Dec 31 '12 at 14:39
• @BabakSorouh So that proof proves the subgroup and the index bit? So I don't need to use OS theorem? – Kaish Dec 31 '12 at 14:45
• @Kaish: I again suggest you to read the answers here as Amr did for you math.stackexchange.com/a/265970/8581. Read Amr's and Matt's answeres. :-) – mrs Dec 31 '12 at 14:51

The second claim is also correct, at least I can say this for finite groups, I'm not shure if there is always a bijection for infinite groups. For finite $G$, $N_G (H)$ is the stabilizer of $H$ by the action of $G$ on the conjugacy classes of $H$, so your proof is correct.