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Let $N$ be a cyclic normal subgroup of a group $G$ and $H$ is any subgroup of $N$.

Prove $H$ is a normal subgroup of $G$


Guessinng that exists $a\in G$ where $\langle a\rangle=N$ of some order.

$H \subset N$, $ H$ is a subgroup of $N$.

$gH =Hg \equiv g h_1 =h_2 g \equiv g a^{k_1} =a^{k_2}g $.

*Not sure if on the right track just spit balling *

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marked as duplicate by Dietrich Burde, Michael Burr, Juniven, Namaste abstract-algebra Feb 28 '17 at 1:46

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Hints:

  • In a cyclic group of order $n$, for each $k$ dividing $n$, there is a unique subgroup of that order.

  • Let $g\in G$. $gHg^{-1}$ is a cyclic group of the same order as $H$ (and $H$ is a cyclic group as it is a subgroup of a cyclic group).

  • $gHg^{-1}$ is a subgroup of $N$ by the normality of $K$ ($gHg^{-1}$ is a subset of $N$ by normality of $N$ and $gHg^{-1}$ is a group, so its a subgroup of $K$).

Can you use these three facts to prove that $gHg^{-1}=H$?

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    $\begingroup$ ?typo... By K do you mean N? $\endgroup$ – DanielWainfleet Feb 27 '17 at 23:44
  • $\begingroup$ @user254665 Thanks $\endgroup$ – Michael Burr Feb 27 '17 at 23:48

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