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Let $G$ be a group and $H$ be a normal subgroup of $G$. Assume that $H$ is finite and cyclic. Show that every subgroup of $H$. is also normal in $G$.$$

I am trying to prove this problem, but absolutely cannot make any progress. I know that $H$ is generated by an element $a\in G$ with the order of $H$ being the order of $a$.

I also know that any subgroup $K$ of $H$ will be cyclic as well. I cannot seem to make any further progress from this.. any help would be great.

Thank you.

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3 Answers 3

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Let $a$ be a generator of $H$. Then any subgroup $K$ of $H$ is generated by $a^k$ for some $k$. For any $g \in G$, we have $ga^kg^{-1} = (gag^{-1})^k $ and $gag^{-1} \in H$ since $H$ is normal. Thus $gag^{-1} = a^l$ for some $l$. Thus $$ga^kg^{-1} = (gag^{-1})^k = (a^l)^k = (a^k)^l \in K$$ since $K$ is generated by $a^k$. It follows that $K$ is normal.

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  • $\begingroup$ Why is $ga^kg^-1 = (gag^-1)^k)$? $\endgroup$
    – Sank
    Commented Nov 3, 2016 at 20:37
  • $\begingroup$ $ga^2g^{-1}=gag^{-1}gag^{-1} = (gag^{-1})^2$. Use induction. $\endgroup$
    – user348749
    Commented Nov 3, 2016 at 23:54
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There is a generalization here. A subgroup $K$ of a group $H$ is called characteristic, if for any $\alpha \in Aut(H)$, $\alpha[K]=K$, and one writes $K$ char $H$. Note that $K$ is necessarily normal, since the inner automorphisms of $H$ fix $K$. Examples of characteristic subgroups are the commutator subgroup $H'$ and the center $Z(H)$.
Now, one can show in general that $K$ char $H \unlhd G$, implies $K \unlhd G$. So the thing you have to show is that all subgroups of a cyclic group are characteristic. And dear user282639, I leave that to you.

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As H is cyclic implies H is abelian and that implies every subgroup of H is normal in H but as H is normal in G which clearly implies every subgroup of H is normal in G.

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