If $H$ is permutable and $A$ is a subgroup of $H$, is $A$ then permutable?












1














Let $G$ be a finite group, $Ale H$ where $H$ is a proper subgroup of $ G$.



How to show that if $H$ is permutable in $G$ (i.e. $HB = BH$ for all $Ble G$) then $A$ is permutable in $G$?



Well, I started by assuming the opposite (if $A$ is not permutable then $H$ is not permutable) and got two cases, when $B$ is $H$ and not, but I couldn't continue. Can you give me a hint, please?










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  • Well, $AH=H=HA$.
    – Berci
    Dec 9 '18 at 22:20






  • 1




    Anyway, it's not true in this form: consider $H=G$..
    – Berci
    Dec 9 '18 at 22:22










  • I've fixed it! Thank you!
    – H.koby
    Dec 9 '18 at 22:32










  • But I still don't get it... How I can use that when B is not H ?
    – H.koby
    Dec 9 '18 at 22:34








  • 1




    Here's how you can thank me properly: answer your own question by showing that $D_8$ is the smallest counterexample to your assertion. As a bonus, implement Derek's suggestion.
    – the_fox
    Dec 9 '18 at 23:16
















1














Let $G$ be a finite group, $Ale H$ where $H$ is a proper subgroup of $ G$.



How to show that if $H$ is permutable in $G$ (i.e. $HB = BH$ for all $Ble G$) then $A$ is permutable in $G$?



Well, I started by assuming the opposite (if $A$ is not permutable then $H$ is not permutable) and got two cases, when $B$ is $H$ and not, but I couldn't continue. Can you give me a hint, please?










share|cite|improve this question
























  • Well, $AH=H=HA$.
    – Berci
    Dec 9 '18 at 22:20






  • 1




    Anyway, it's not true in this form: consider $H=G$..
    – Berci
    Dec 9 '18 at 22:22










  • I've fixed it! Thank you!
    – H.koby
    Dec 9 '18 at 22:32










  • But I still don't get it... How I can use that when B is not H ?
    – H.koby
    Dec 9 '18 at 22:34








  • 1




    Here's how you can thank me properly: answer your own question by showing that $D_8$ is the smallest counterexample to your assertion. As a bonus, implement Derek's suggestion.
    – the_fox
    Dec 9 '18 at 23:16














1












1








1







Let $G$ be a finite group, $Ale H$ where $H$ is a proper subgroup of $ G$.



How to show that if $H$ is permutable in $G$ (i.e. $HB = BH$ for all $Ble G$) then $A$ is permutable in $G$?



Well, I started by assuming the opposite (if $A$ is not permutable then $H$ is not permutable) and got two cases, when $B$ is $H$ and not, but I couldn't continue. Can you give me a hint, please?










share|cite|improve this question















Let $G$ be a finite group, $Ale H$ where $H$ is a proper subgroup of $ G$.



How to show that if $H$ is permutable in $G$ (i.e. $HB = BH$ for all $Ble G$) then $A$ is permutable in $G$?



Well, I started by assuming the opposite (if $A$ is not permutable then $H$ is not permutable) and got two cases, when $B$ is $H$ and not, but I couldn't continue. Can you give me a hint, please?







group-theory finite-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 17:37









the_fox

2,43411431




2,43411431










asked Dec 9 '18 at 22:12









H.koby

427




427












  • Well, $AH=H=HA$.
    – Berci
    Dec 9 '18 at 22:20






  • 1




    Anyway, it's not true in this form: consider $H=G$..
    – Berci
    Dec 9 '18 at 22:22










  • I've fixed it! Thank you!
    – H.koby
    Dec 9 '18 at 22:32










  • But I still don't get it... How I can use that when B is not H ?
    – H.koby
    Dec 9 '18 at 22:34








  • 1




    Here's how you can thank me properly: answer your own question by showing that $D_8$ is the smallest counterexample to your assertion. As a bonus, implement Derek's suggestion.
    – the_fox
    Dec 9 '18 at 23:16


















  • Well, $AH=H=HA$.
    – Berci
    Dec 9 '18 at 22:20






  • 1




    Anyway, it's not true in this form: consider $H=G$..
    – Berci
    Dec 9 '18 at 22:22










  • I've fixed it! Thank you!
    – H.koby
    Dec 9 '18 at 22:32










  • But I still don't get it... How I can use that when B is not H ?
    – H.koby
    Dec 9 '18 at 22:34








  • 1




    Here's how you can thank me properly: answer your own question by showing that $D_8$ is the smallest counterexample to your assertion. As a bonus, implement Derek's suggestion.
    – the_fox
    Dec 9 '18 at 23:16
















Well, $AH=H=HA$.
– Berci
Dec 9 '18 at 22:20




Well, $AH=H=HA$.
– Berci
Dec 9 '18 at 22:20




1




1




Anyway, it's not true in this form: consider $H=G$..
– Berci
Dec 9 '18 at 22:22




Anyway, it's not true in this form: consider $H=G$..
– Berci
Dec 9 '18 at 22:22












I've fixed it! Thank you!
– H.koby
Dec 9 '18 at 22:32




I've fixed it! Thank you!
– H.koby
Dec 9 '18 at 22:32












But I still don't get it... How I can use that when B is not H ?
– H.koby
Dec 9 '18 at 22:34






But I still don't get it... How I can use that when B is not H ?
– H.koby
Dec 9 '18 at 22:34






1




1




Here's how you can thank me properly: answer your own question by showing that $D_8$ is the smallest counterexample to your assertion. As a bonus, implement Derek's suggestion.
– the_fox
Dec 9 '18 at 23:16




Here's how you can thank me properly: answer your own question by showing that $D_8$ is the smallest counterexample to your assertion. As a bonus, implement Derek's suggestion.
– the_fox
Dec 9 '18 at 23:16















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