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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?

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    $\begingroup$ $gng^{-1}=(gng^{-1}n^{-1})n$. $\endgroup$
    – Julien
    Jun 12, 2013 at 15:43

6 Answers 6

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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 .

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Hint: Since $H$ is abelian, $a^{-1}b^{-1}ab \in \operatorname{Ker}(f)$ for all $a, b \in G$.

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Hint: Every subgroup of an abelian group is normal.

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  • $\begingroup$ But $G$ is not abelian here. What am I missing? $\endgroup$
    – Brenin
    Jun 12, 2013 at 15:39
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    $\begingroup$ @atricolf the isomorphism theorems :-) subgroups containing the kernel are the same thing as subgroups of the image $\endgroup$ Jun 12, 2013 at 15:44
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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 :)

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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.

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$f:G\rightarrow~H$, 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.

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