Let $\Phi: G_1 \rightarrow G_2$ be a surjective homomorphism. Prove that there is 1-1 correspondence between subgroups that contain $K=\text{Ker}(\Phi)$ and subgroups in $G_2$
I tried proving directly that there is a bijection, given by:
$$K\leq A\rightarrow \Phi(A) $$
Injective: It is injective, because if $\Phi(A)=\Phi (B)$, we must have that for every $a\in A$, $\exists b\in B$ such that $\Phi(a)=\Phi(b)\Rightarrow \Phi(ab^{-1})=1\Rightarrow a=b k, $ where $k\in K$. Thus, $a\in B$, because $K\leq B$. This means that $A\subset B$. By symmetry, $A=B$.
Surjective: Let $H\leq G_2$. We must have $1\in H$, thus $K\leq \Phi^{-1}(H)$. $\Phi \Phi^{-1}(H)=H$, as pointed out by @Stinking Bishop in the comments.
Is there a nicer way to do this using the isomorphism theorem?
There is 1-1 correspondence between $\{ A\leq G_1 : K\leq A\}$ and subgroups of $\pi (G_1) = G_1 / K$.
surjective: as $\pi$ is surjective, for every subgroups $C \leq G_1 / K$ exists $\pi ^ {-1} (C) \geq K$.
injective: If $K \leq A \neq B \leq G_1$, then there is $a \in A$ (without loss of generality) s.t. $\forall b \in B$ $a\neq b$. Thus, $aK \neq bK$ $\forall b \in B$, then $\pi(A) = A/K \neq B/K = \pi(B)$.
$\pi(G_1) = G_1 / K \cong \Phi (G_1)$, due to isomorphism theorem, and there is 1-1 correspondence between their subgroups.