A subgroup containing a kernel of a group homomorphism into an abelian group is a normal subgroup. Let $ f \colon G \rightarrow H $ be a group homomorphism, where $ H $ is an abelian group.
If $\ker f \subset N $ for some $ N \subset G $, then $ N $ is a normal subgroup of $ G $.
I don't know how to start to prove the statement. Any hint?
 A: By the Lattice Theorem, subgroups of $G$ containing $\ker f$ are in one-to-one correspondence with subgroups of $G/\ker f$, and by the first isomorphism theorem,
$$G/\ker f \cong f(G)$$
So all subgroups of $G/\ker f$ are normal, and the corresponding subgroups are normal .
A: Hint: Since $H$ is abelian, $a^{-1}b^{-1}ab \in \operatorname{Ker}(f)$ for all $a, b \in G$.
A: Hint: Every subgroup of an abelian group is normal.
A: HINT n.3 
given $\phi$ homomorphism $G \to H$ for every $N$ normal in $H$, $\phi^{-1}(N)$ is normal in $G$.  Combine this with the other hints and the fact that for every $A \leq G$, $\phi(A) \leq H$ and you win :)
A: let $f:G_{1}\rightarrow~G_{2}$ be the group homomorphism, where $G_{2}$ is abelian.\
let $H_{1}$ be the subgroup of $G_{1}$ s.t. $ker(f)\subseteq~H_{1}$\
$f(h_{1}^{-1}g_{1}h_{1}g_{1}^{-1})=e_{2}$, where $g_{1}\in~G_{1}$ and $h_{1}\in~H_{1}$ and $e_{2}$ is the identity of $G_{2}$\
Since $G_{2}$ is abelian and $ker(f)\subseteq~H_{1}$ so we can easily have $g_{1}h_{1}g_{1}^{-1}\in~H_{1}$\
Hence $H_{1}$ is normal.
A: $f:G\rightarrow~H$,  by the first isomorphism theorem,
$$G/\ker f \cong f(G)$$
So you need to create a correspondance $G/N \rightarrow f(G)$ ,
now $f^{-1}(f(h'h))=f^{-1}(f(hh'))$ $f^{-1}(f(h')f(h))=f^{-1}(f(h)f(h'))$then $gN=Ng$ if ker(f) $\subset$ N, you can map element $e^H$ to ker(f) inside N for all elements.
