Homomorphisms and normal subgroups proof 
Let $f$ be a homomorphism mapping $G$ to $J$ groups. Assume that $J$ is abelian. Prove that if $H$ is a subgroup of $G$ and if $\mathrm{ker}(f)$ is a subset of $H$, then $H$ is normal in $G$.

I honestly have no idea how to start this proof. The fact that $J$ is abelian gives me that for all $a$ and $b$ in $G$, $f(ab) = f(ba)$; but this tells me nothing since $f$ is not injective. Furthermore, what exactly can I get from the fact that $\mathrm{ker}(f)$ is a subset of $H$? I understand the given information but I am having issues putting it all together as I see no possible relation!
My plan was to piece the information all together to somehow set things up for the normal subgroup test, however since I  see no connection with the given information my approach seems to be rather futile.
Another thought I just got was to prove that $f$ is injective then use the fact that $J$ is abelian to show that $ah = ha$ for all $a$ in $G$ and $h$ in $H$.
Thanks.
 A: You won't be able to prove that $f$ is injective, since there is absolutely no reason why it would be; and you can probably come up with examples of groups which satisfy the set-up but where $f$ is not injective. If you cannot, here's one: take $G=Q_8$, the quaternion group of order $8$. Let $J = C_2\times C_2$, the Klein $4$-group. Let $f\colon G\to J$ be the map that sends $1$ and $-1$ to $(0,0)$, $i$ and $-i$ to $(1,0)$; $j$ and $-j$ to $(0,1)$; and $k$ and $-k$ to $(1,1)$. Now take $H = \{1, -1, i, -i\}$. (Of course, $H$ is normal in $G$; but the point is that this set-up satisfies all the hypothesis, but $f$ is not injective; so you won't be able to prove, in general, that $f$ is injective).
If you already know the isomorphism theorems, you should remember that one of them says:

Theorem. Let $f\colon G\to K$ be a group homomorphism. The subgroups of $G$ that contain $\mathrm{ker}(f)$ are in one-to-one, inclusion preserving, normality preserving correspondence with the subgroups of $f(G)$. 

and that will solve your problem directly.
If you don't know the isomorphism theorems, start from what you need to start: let $h\in H$ and $g\in G$; you need to show that $ghg^{-1}\in H$, right?
Well, $f(ghg^{-1}) = f(g)f(h)f(g)^{-1} = f(g)f(g)^{-1}f(h) = f(h)$, since $J$ is abelian.
What do you know about two elements that map to the same thing under a group homomorphism (in terms of the kernel of course)? 
(Hint: $f(a) = f(b)$ if and only if $f(a)f(b)^{-1}=1$, if and only if $f(ab^{-1})=1$, if and only if ....)
What do you know about the kernel of $f$ in this case? Can you use these facts to conclude that $ghg^{-1}\in H$?
