Conjugating Subgroups... There is a corollary in my textbook that states,

The number of conjugates of $H \subset G$is $[G: N_G(H)]$, the index of the normalizer of H in G.

We know that $N_G(H)=\{x \in G| xHx^{-1} = H \}$ 
I'm a bit confused about what this is saying. So $[G: N_G(H)]$ is the number of disctinct elements of the form $gN_G(H)$ where $g \in G$, right? But I have some trouble understanding how that is related to the number of conjugates in H. So I was wondering if anybody could clarify it for me. 
Also, there is another corrolary that states,

Let H be a subgroup of G. THen $H \subset N_G(H)$. In particular, the number of conjugates of H in G divies [G:H].

From the previous corrolary, we know that the number of conjugates of H is $[G:N_G(H)]$. So how does that divide [G:H]. Again, I'm having some trouble understanding this corrolary too...
Thanks in advance 

Let $\{x_i\mid i\in I\}$ be a set of coset representatives for $N_G(H)$ in $G$.  For any $y\in G$, there are unique $i\in I$ and $x\in N_G(H)$ such that $y=g_ix$.  Then $yHy^{-1} = (g_ix)H(g_ix)^{-1} = g_ixHx^{-1}g_i^{-1} = g_iHg_i^{-1}$.  So every conjugate of $H$ is equal to some $g_iHg_i^{-1}$, and these are all distinct (left as easy exercise).
For the second part, use Geoff Robinson's comment (in general, if $H\leq K\leq G$, then $|G:H| = |G:K||K:H|$, which is an easy consequence of Lagrange's Theorem).
