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.
Best Answer
$a^{-1}Ha=b^{-1}Hb$ 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 $H$.