Defining the normalizer, showing its a subgroup and $|H| = |G:N_G(H)|$ "Let G be a finite group. For a subgroup $H \subset G$ defing the normalizer $N_G(H) \subset G$. Show that the normalizer is a subgroup, that $H \unlhd N_G(H)$ and that the number of subgroups $H'$ conjugate to $H$ in $G$ is equal to the index of $|G:N_G(H)|$ of the normalizer".
For the normalizer, I have the definition as the biggest subgroup $\supset H$, such that$H \unlhd$ in it: $H \unlhd N_G(H)$. What does this exactly mean though? Is it basically the biggest normal subgroup in G?
Also, I don't understand how I would show the other stuff.
 A: Make the group $\,G\,$ act on the set $\,X:=\{K\;\;;\;\ K\leq G\}\,$ by conjugation. Thus, by the orbit-stabilizer theorem:
$$|\mathcal Orb(H)|=[G:Stab(H)]$$
but $\,\mathcal Orb(H)\,$ is just the set of all subgroups of $\,G\,$ conjugate to $\,H\,$ , and $\,Stab(H)\,$ is just $\,N_G(H)\,$, so...
A: Given a subgroup $H$, there are possibly a bunch of intermediate subgroups $K$ lying in between $H$ and $G$. $H$ has to be normal in at least one of these, since it's normal in itself. So we can keep going up the chain of subgroups until we arrive at a "largest" subgroup that $H$ is normal inside. Now, that subgroup need not be normal in $G$. It's also not always the largest normal subgroup of $G$ because the largest normal subgroup in $G$ is just $G$!. 
Now, in order to show the number of conjugates is equal to the index of the normalizer, I would start by writing out the conjugates of $H$: $\{H, g_1Hg_1^{-1}, \ldots, g_nHg_n^{-1}\}$. Now try and show that $\{N(H),  g_1N(H), \ldots, g_nN(H)\}$ are precisely the left cosets of $N(H)$. That is, you need to show no two cosets in that list are equal, and that every coset appears on that list. Now you just use the fact that $|G : N(H)|$ is by definition the number of cosets.
