Theorem. Let $\varphi: G \rightarrow H$ be a surjective homomorphism (epimorphism) of the groups $G$ and $H$. $\varphi$ provides a bijection of the set of subgroups $U$ of $G$ with $\operatorname{ker}\varphi \subseteq U$ onto the set of subgroups of $H$ that preserves subgroup inclusion. The following applies to a subgroup $U$ of $G$ with $\operatorname{ker}\varphi \leq U$: It holds that $[G: U]=[H: \varphi[U]]$.
I proved that $U\leq G$ implies $\varphi[U]\leq H$. I am left with showing that $[G: U]=[H: \varphi[U]]$. A hint is:
Show that for $a, b \in G$ $$ a b^{-1} \in U \Leftrightarrow \varphi(a) \varphi(b)^{-1} \in \varphi[U]. $$
One way is easy: $$ \varphi(ab^{-1})=\varphi(a)\varphi(b^{-1})=\varphi(a)\varphi(b)^{-1}\in \varphi[U]. $$ but I dont get the other one and I do not get how the statement follows.
First lets prove that if $\varphi(a)\varphi(b)^{-1}\in \varphi(U)$, then $ab^{-1}\in U$.
Let $a$ and $b$ be elements of $G$ such that $\varphi(a)\varphi(b)^{-1}\in\varphi(U)$. Then $\varphi(ab^{-1})\in\varphi(U)$, so there exists $x\in U$ such that $\varphi(x)=\varphi(ab^{-1})$. That means that $(ab^{-1})x^{-1}\in\ker(\varphi)$. But we are assuming that $\ker(\varphi)\leq U$, so this means $(ab^{-1})x^{-1}\in U$. Since $x\in U$, it follows that $ab^{-1}\in U$, as desired.
Now, in light of this equivalence, why are the indices the same? We show there is a bijection between the cosets of $U$ in $G$ and the cosets of $\varphi(U)$ in $\varphi(G)=H$. Define the map that sends the coset $Ua$ to the coset $U\varphi(a)$. This map is well defined and one-to-one, since: $$\begin{align*} Ua=Ub&\iff ab^{-1}\in U\\ &\iff \varphi(a)\varphi(b)^{-1}\in\varphi(U)\\ &\iff \varphi(U)\varphi(a) = \varphi(U)\varphi(b)\end{align*}$$ And this assignment is surjective: given a coset $\varphi(U)h$, there exists $a\in G$ such that $\varphi(a)=h$, so $Ua$ maps to $\varphi(U)h$.
Thus, we have established a bijection between cosets of $U$ in $G$ and cosets of $\varphi(U)$ in $H$, proving that $[G:U]=[H:\varphi(U)]$.