Let $(G, \star)$ and $(H, \blacksquare)$ be groups, and let $\varphi:G\rightarrow H$ be a group homomorphism. Let's also assume $\varphi$ is surjective. For any $h\in H$, we define $\varphi ^{-1}(h) = \{g\in G|\varphi (g)=h\}$.
How can I prove that for $h_1,h_2\in H$, $|\varphi ^{-1}(h_1)|=|\varphi ^{-1}(h_2)|$ ? That is, the number of elements of $G$ that $\varphi$ maps to $h_1$ is the same as to $h2$.
Step 1) Define $Ker_{\phi} =\{g\in G | \phi(g)= e_H \}$ this is the "kernel" of the map $\phi$
Step 2) For $h \in H$ since $\phi$ is surjective there exists $g_h \in G$ such that $\phi(g_h)=h$, now let $w \in G$ be "another" element in $G$ such that $\phi(w)=h$ we then have:
$$\phi((g_h)^{-1}w)=\phi(g_h^{-1})\phi(w)=\phi(g_h)^{-1}h=h^{-1} h = e_H$$
hence $(g_h)^{-1}w \in Ker_{\phi}$
Step 3) if $u,v \in G$ are such that $u^{-1} v \in Ker_{\phi}$ then $\phi(u)=\phi(u)e_H=\phi(u)\phi(u^{-1}v)=\phi(u(u^{-1}v))=\phi(v)$.
Step 4) let $h_1, h_2 \in H$ be given, there exists $g_1, g_2$ such that $\phi(g_1)=h_1, \phi(g_2)=h_2$ now define the function $F: \phi^{-1}(h_1) \rightarrow \phi^{-1}(h_2)$ as follows:
$$F(g)=(g_2)(g_1^{-1})g$$
with what was said above you can clearly see that $F$ is injective and surgective aswel, hence we must have: $$|\phi^{-1}(h_1)|=|\phi^{-1}(h_2)|$$