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Let $\varphi:G\rightarrow H$ be a group homomorphism from group $G$ to group $H$. Show that, if $H$ is abelian, all subgroups of $G$ that contain $\mathrm{ker} (\varphi)$ are normal in $G$.

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Hint: $G / \ker \varphi$ is abelian. Hence $G' \le \ker \varphi$ where $G'$ is the commutator subgroup.


$G / \ker \varphi$ is isomorphic to a subgroup of $H$ by the first isomorphism theorem, hence abelian. By the familiar property of $G'$, we have $G' \le \ker \varphi$. It follows that any subgroup $K$ that contains $\ker \varphi$ also contains $G'$. Hence $K$ is normal by another familiar property of $G'$.

If you're not familiar with the property in the last statement, here is how to show it: Since $G/G'$ is abelian and $K/G'$ is a subgroup of $G/G'$, we have $K/G' \trianglelefteq G/G'$. We conclude that $K$ is normal by the fourth isomorphism theorem.

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    $\begingroup$ Why $G/kerφ$ is abelian? Can you give me more details? $\endgroup$
    – Misa
    Oct 28 '13 at 1:55
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    $\begingroup$ The first isomorphism theorem will help. $\endgroup$
    – B. Mackey
    Oct 28 '13 at 2:09
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    $\begingroup$ $G/ker\varphi\cong Im\varphi\subseteq H$ , $H$ is abelian so $G/ker\varphi$ is abelian. $\endgroup$
    – Misa
    Oct 28 '13 at 2:25
  • $\begingroup$ $G'\leq ker\varphi$, and what should we do next? $\endgroup$
    – Misa
    Oct 28 '13 at 2:26
  • $\begingroup$ @chuyenvien94 I've edited my answer. $\endgroup$ Oct 28 '13 at 10:17
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Notice that $\varphi$ induces a bijection between the subgroups of $G$ containing $\ker \varphi$ and the subgroups of $\varphi(G) \leq H$. Therefore, if $K$ is a subgroup of $G$ containing $\ker \varphi$, it is sufficient to show that $\varphi(gKg^{-1})=\varphi(K)$ for all $g \in G$; but the previous equality is clear since $H$ is abelian.

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Direct: Let $\ker(\phi) \leq N \leq G$. If $g \in G$, then $\phi(g N g^{-1})=\phi(g) \phi(N) \phi(g)^{-1} = \phi(N)$ shows $g N g^{-1} \subseteq \ker(\phi) N \subseteq N$.

(We only need that every subgroup of $H$ is normal. )

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  • $\begingroup$ Every subgroup of $H$ is normal because $H$ is abelian $\endgroup$
    – Misa
    Oct 29 '13 at 10:16
  • $\begingroup$ @Martin Brandenburg is this complete proof? $\endgroup$ Nov 9 '17 at 16:59
  • $\begingroup$ @Ayman Hourieh ? $\endgroup$ Nov 9 '17 at 17:11
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There is a general, easy to remember (and easy to prove) rule:

The preimage of a normal subgroup is normal.

Moreover, subgroups of $G$ containing $\ker(\varphi)$ are just preimages of subgroups of $H$. When $H$ is abelian, all its subgroups are normal and the claim follows from said rule.

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