I gave a mapping A to C such that A is the set of left cosets in G described as $A$={$N(H), gN(H),...g_nN(H)$} for N(H) is the normalizer of H in G and C is the set of conjugates of H, $C$={$gHg^{-1}|g \in G$}. The mapping $\phi $ is $\phi (gN(H))=gHg^{-1}$. I need to show it is 1-1 and onto.
1-1 Part
Assume $\phi (g_1N(H))=\phi (g_2N(H))$. Hence $g_1Hg_1^{-1}=g_2Hg_2^{-1}$. We need to show $g_1N(H)=g_2N(H).$ Observe that $g_1Hg_1^{-1}=g_2Hg_2^{-1}$. So $g_2^{-1}g_1Hg_1^{-1}g_2=H$. So $g_2^{-1}g_1H(g_2^{-1}g_1)^{-1}$. Hence $g_2^{-1}g_1 \in N(H)$. Thus $g_2^{-1}g_1N(H)=N(H)$. So $g_1N(H)=g_2N(H).$ So $\phi $ is 1-1.
Onto
Let $y \in C$. Thus $y=g_1Hg_1^{-1}$ for some $g_1 \in G$. Therefore $y=\phi (x)$ for $x \in A$ where $x=g_1N(H)$. Thus $y=g_1Hg_1^{-1}=\phi(g_1N(H))=\phi (x)$. I have showed $\phi$ is well defined, thus $\phi ^{-1}(y)=x$. So $(\phi(y)) ^{-1}=x$. Is this right and if it is, what would the next step be?
How would this imply that [G:N(H)] is the number of conjugates of H in G.
Thank you
Well. Your "1-1" proof is spot on, but I think that your making the proof for "Onto" harder than it is. This is how I would do:
Let $y\in C$. Then there exists a $g_1$ so that $y=g_1Hg_1^{-1}$. Now choose the element $g_1N(H)\in A$. Since $\phi(g_1(N(H))=g_1Hg_1^{-1}$ and that $y\in C$ was arbitrarily choosen, we have shown that $\phi$ is "Onto".
Now.. your question about why $|G:N(H)|$ is the number of conjugates of $H$ in $G$. Since the map $\phi$ is bijective, we must have that $|A|=|C|$. Since $|C|$ is the number of conjugates of $H$, we just have to show that $|A|=|G:N(H)|$.
This can be done using Lagrange's Theorem. Since $N(H)$ is a subgroup of G, we know that $|N(H)|$ divides $|G|$ and that the number of left cosets of N(H) in G equals $\frac{|G|}{|N(H)|}$.