Correspondence Theorem With The Kernel Problem:
Let $f \in Hom(G,H)$ and let $K = Ker(f)$.
(Do not use the Correspondence Theorem for the following questions)


*

*Show that the map $M \mapsto f(M)$ gives a bijection between the set of subgroups of $G$ containing $K$ and the set of subgroups of $Im(f) = f(G)$.

*Show that the bijection respects inclusions, indices, and normality (if $K < M_1, M_2 < G$, then $M_1 < M_2$ if and only if $f(M_1) < f(M_2)$, in which case $[M_2:M_1] = [f(M_2):f(M_1)]$ and $M_1 \trianglelefteq M_2$ if and only if $f(M_1) \trianglelefteq f(M_2)$).
Attempt:
I thought about using the First Isomorphism Theorem to produce an isomorphism between $G/K$ and $Im(f)$, and trying to associate each subgroup in $G$ to a coset in $G/K$ and therefore a bijection between the set of subgroups of $G$ and the set of subgroups of $G/K$, but I don't know how exactly to do this and how to make this idea rigorous.
For Part 2, I am very lost.
How would I approach this problem? Thank you.
 A: Consider the function $\varphi: M \mapsto f(M)$ from $Sub_K(G)$ to $Sub(Im(f))$. Giving that function a different name should help you.
$\varphi$ is surjective : let $H$ be a subgroup of $Im(f)$, then $f^{-1}(H) := \{ g \in G \big| f(g) \in H \}$ is clearly a subgroup of $G$ containing $K$, and $\varphi(f^{-1}(H)) = H$.
$\varphi$ is injective : let $H,H'$ be subgroups of $G$ containing $K$ such that $\varphi(H) = \varphi(H')$. Since the situation is symetrical it is enough to prove that $H \subseteq H'$ to ensure $H=H'$.
Let $h\in H$, since $\varphi(H) = \varphi(H')$, there is $h' \in H'$ such that $f(h)=f(h')$. But then $f(hh'^{-1})=e$ so that $hh'^{-1} \in K$, hence $hh'^{-1}$ belongs to $H'$, whence $h = hh'^{-1}\cdot h'$ belongs to $H'$ qed.
$\varphi$ respects indices : let $M_1 < M_2$ be subgroups of $G$ containing $K$. Write $M_2/M_1$ for the set of left cosets for $M_1$ in $M_2$.
$f$ induces a well defined map 
$$\begin{array}{clll} \bar{\varphi} : & M_2/M_1 & \mapsto f(M_2)/f(M_1) \\
\phantom{\varphi} & g\cdot M_1 & \mapsto  f(g)\cdot f(M_1) \end{array}$$
which is a bijection (just adapt the arguments that prove that $\varphi$ is a bijection to $\bar{\varphi}$ ) .
A: The proof is contained in Michael Artin's Algebra at the end of Chapter 2: Group Theory. 
