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Let $G$ be a finite group and let $\varphi:G \rightarrow F^{\times}$ be an homomorphism where $F$ is a field. $H$ is a subgroup of $G$ that contains $Ker(\varphi)$.

Prove that $H \lhd G$ and that $G/H$ is cyclic.

I have no idea how to prove this.

Thank you.

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You might first consider the case where $H = \ker \varphi$. Use the isomorphism theorems and the fact that a finite subgroup of $F^\times$ is cyclic (can $G$ assumed to be finite?) . –  marlu Jul 14 '12 at 10:04
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The claim is false as stated: for example take $G$ to be the multiplicative group of $\mathbb{Q}$, $\varphi$ to be the identity, and $H$ to be trivial. Perhaps you want $G$ to be a finite group? –  Chris Eagle Jul 14 '12 at 10:05
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2 Answers

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As was pointed out in the comments we have to assume that $G$ is finite. Let $K := \ker \varphi$. Then $K \leq H \leq G$ and $K$ is normal in $G$. By the isomorphism theorem, $G/K \cong \mathrm{im} \varphi$. This is a finite subgroup of $F^\times$, hence cyclic. Consider $H \hookrightarrow G \twoheadrightarrow G/K$. The kernel of this map is $H \cap K = K$, so $K$ is normal in $H$ and $H/K$ is a subgroup of $G/K$. Since $G/K$ is cyclic, we have $H/K \unlhd G/K$. Using this, we can show $H \unlhd G$: Let $x \in G$ and $h \in H$. Since $H/K \unlhd G/K$, there is an $h' \in H$ s.t. $x^{-1}hx \equiv h' \mod K$, i.e. $x^{-1}hx = h'k$ for some $k \in K$. The right side is an element of $HK = H$, hence $x^{-1}Hx \subseteq H$. This shows $H \unlhd G$. Now, by another isomorphism theorem, $G/H \cong (G/K)/(H/K)$. This is a quotient of a cyclic group, hence cyclic.

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Since $F^\times$ is abelian, the commutator subgroup $G'$ of $G$ is contained in $ker(\varphi)$. In general, it is easy to show that if $H$ is a subgroup of a group $G$ and $H \supseteq G'$, then $H$ must be normal.

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