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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.

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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 $ghg^{-1}\in H$?

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$f(ghg^{-1}) = f(h)$, we have that $ghg^{-1}kerf = hkerf$. My issue now is the presence of the $kerf$. Since they are sets I'm not sure if I can just eliminate directly since they appear both sides. Other thing I can see is that since we know that $hkerf$ is some element of H (H is a group so closed by operation), let $h^{'} = hkerf$. Then we have $ghg^{-1}kerf = h^{'}$. however the presence of $kerf$ also causes some issue. –  Friday Mar 11 '11 at 23:39
    
@MyNameIsDan: Rather: Let me write $N=\mathrm{ker}f$. You have $ghg^{-1}N = hN$. But remember what that means: it means that $h^{-1}(ghg^{-1})\in N$. And since $N\subseteq H$, then $h^{-1}(ghg^{-1})$ is in $H$. And the product of two elements of $H$ is always in $H$, so if you, say, multiply $h$ by this complicatedly-named-element we just discovered is in $H$, what do we get? –  Arturo Magidin Mar 12 '11 at 1:38
    
@MyNameIsDan: That is: you have to remember that if $G$ is a group and $K$ is a subgroup, then the equality $xK = yK$ means that $y^{-1}x\in K$ (in fact, is equivalent to that); that's what lets you go from statements about cosets to statements about elements. –  Arturo Magidin Mar 12 '11 at 1:39
    
@MyNameIsDan: Here is a supplement: Denoting $\mathrm{Ker(f)}$ by $N$, we have that G is the direct sum of $N$ and $J$, since $H$ contains $N$, it is a direct product of $N$ and a subgroup of $J$, which is of course normal in $G$. This argument, however, is aimed to give an intuition, instead of any rigor, which is why I didn't put it an answer. –  awllower Mar 12 '11 at 3:16
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@awllower: You could, but (i) I don't see that it will do you any good; and (ii) you don't have to use it at all. If you already know the isomorphism theorems, then this is a trivial consequence of the lattice isomorphism theorem; if you don't know them, then "exact sequences" is going to be well over your head. The simple argument I outline is more than enough. –  Arturo Magidin Mar 12 '11 at 3:56
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