Let $f \in \operatorname{Hom}(G,H)$ and $K = \ker(f)$. I am trying to show that there exists a bijection between the set of subgroups of $G$ containing $K$ and the set of subgroups of $H$ which are subgroups of $\operatorname{Im}(f)$.
Let $S = \{M < G \mid K \subseteq M\}$ and $R = \{J < H \mid J < \operatorname{Im}(f)\}$ In particular, I am trying to show that there exists a bijective function $\phi: S \rightarrow R$ defined by $\phi(M) = f(M)$.
$\underline{\text{Injectivity}}$: If $M_1, M_2 \in S$ s.t. $M_1 \neq M_2$ then $\exists r \in G \setminus \ker(f) $ and $r \in M_1 \setminus M_2$. We claim that $f(r) \notin f(M_2) \Rightarrow f(M_1) \neq f(M_2)$ thereby showing injectivity of $\phi$.
Suppose $f(r) = f(s^{-1})$ for some $s^{-1} \in M_2$.
Then $f(s^{-1})f(s) = f(r)f(s) = f(r \cdot s) = e \Rightarrow r \cdot s \in K \Rightarrow r s \in M_2$ since $K \subseteq M_2$. Then $s^{-1} \in M_2$ which implies that $(rs)s^{-1} = r \in M_2$ which is a contradiction since $r \in M_1 \ \backslash M_2$.
I am having troubles now showing surjectivity and would appreciate some help.
Thank you in advance.
Let $J$ be a subgroup of $H$ containing the image of $f$ and let $M$ be $f^{-1}(J)$. Then for any $m,n\in M$, $f(mn)=f(m)f(n)\in J$ because $J$ is a subgroup. Therefore, $M$ is a subgroup of $G$ and it only remains to show that it contains the kernel of $f$.
$k\in ker(f)\implies f(k)=e\in H=e\in J$ so so $k\in M$