quotient of normalizer and centralizer is cyclic group












1












$begingroup$


It's known that if $G/Z(G)$ is a cyclic group, then $G$ is Abelian. Since $G/Z(G)$ is just the special case $H=G$ in the $N/C$ theorem $C_G(H)triangleleft N_G(H)$. I wonder if the below statament is true:



If $H$ is a subgroup of $G$, if $N_G(H)/C_G(H)$ is a cyclic group, then $H$ is Abelian.



I tried using the proof in the original statement, but it didn't work out. If we set $N_G(H)/C_G(H)=langle aC_G(H)rangle$, how can i represent an element in $H$?










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$endgroup$








  • 2




    $begingroup$
    Your hypothesis implies that $H/Z(H)$ is cyclic.
    $endgroup$
    – Derek Holt
    Jan 7 at 21:41
















1












$begingroup$


It's known that if $G/Z(G)$ is a cyclic group, then $G$ is Abelian. Since $G/Z(G)$ is just the special case $H=G$ in the $N/C$ theorem $C_G(H)triangleleft N_G(H)$. I wonder if the below statament is true:



If $H$ is a subgroup of $G$, if $N_G(H)/C_G(H)$ is a cyclic group, then $H$ is Abelian.



I tried using the proof in the original statement, but it didn't work out. If we set $N_G(H)/C_G(H)=langle aC_G(H)rangle$, how can i represent an element in $H$?










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Your hypothesis implies that $H/Z(H)$ is cyclic.
    $endgroup$
    – Derek Holt
    Jan 7 at 21:41














1












1








1





$begingroup$


It's known that if $G/Z(G)$ is a cyclic group, then $G$ is Abelian. Since $G/Z(G)$ is just the special case $H=G$ in the $N/C$ theorem $C_G(H)triangleleft N_G(H)$. I wonder if the below statament is true:



If $H$ is a subgroup of $G$, if $N_G(H)/C_G(H)$ is a cyclic group, then $H$ is Abelian.



I tried using the proof in the original statement, but it didn't work out. If we set $N_G(H)/C_G(H)=langle aC_G(H)rangle$, how can i represent an element in $H$?










share|cite|improve this question











$endgroup$




It's known that if $G/Z(G)$ is a cyclic group, then $G$ is Abelian. Since $G/Z(G)$ is just the special case $H=G$ in the $N/C$ theorem $C_G(H)triangleleft N_G(H)$. I wonder if the below statament is true:



If $H$ is a subgroup of $G$, if $N_G(H)/C_G(H)$ is a cyclic group, then $H$ is Abelian.



I tried using the proof in the original statement, but it didn't work out. If we set $N_G(H)/C_G(H)=langle aC_G(H)rangle$, how can i represent an element in $H$?







abstract-algebra group-theory






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edited Jan 7 at 21:14









Bernard

123k741117




123k741117










asked Jan 7 at 21:04









IdeleIdele

1,020410




1,020410








  • 2




    $begingroup$
    Your hypothesis implies that $H/Z(H)$ is cyclic.
    $endgroup$
    – Derek Holt
    Jan 7 at 21:41














  • 2




    $begingroup$
    Your hypothesis implies that $H/Z(H)$ is cyclic.
    $endgroup$
    – Derek Holt
    Jan 7 at 21:41








2




2




$begingroup$
Your hypothesis implies that $H/Z(H)$ is cyclic.
$endgroup$
– Derek Holt
Jan 7 at 21:41




$begingroup$
Your hypothesis implies that $H/Z(H)$ is cyclic.
$endgroup$
– Derek Holt
Jan 7 at 21:41










1 Answer
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$begingroup$

Hint: $H cap C_G(H)=Z(H)$ and $H unlhd N_G(H)$






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    0












    $begingroup$

    Hint: $H cap C_G(H)=Z(H)$ and $H unlhd N_G(H)$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Hint: $H cap C_G(H)=Z(H)$ and $H unlhd N_G(H)$






      share|cite|improve this answer









      $endgroup$
















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        $begingroup$

        Hint: $H cap C_G(H)=Z(H)$ and $H unlhd N_G(H)$






        share|cite|improve this answer









        $endgroup$



        Hint: $H cap C_G(H)=Z(H)$ and $H unlhd N_G(H)$







        share|cite|improve this answer












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        answered Jan 7 at 22:35









        Nicky HeksterNicky Hekster

        29.1k63456




        29.1k63456






























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