If $H$ is permutable and $A$ is a subgroup of $H$, is $A$ then permutable?
$begingroup$
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
edited Dec 10 '18 at 17:37
the_fox
2,89021537
asked Dec 9 '18 at 22:12
H.koby H.koby
427
$endgroup$
|
show 6 more comments
$begingroup$
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
edited Dec 10 '18 at 17:37
the_fox
2,89021537
asked Dec 9 '18 at 22:12
H.koby H.koby
427
$endgroup$
$begingroup$
Well, $AH=H=HA$.
$endgroup$
– Berci
Dec 9 '18 at 22:20
1
$begingroup$
Anyway, it's not true in this form: consider $H=G$..
$endgroup$
– Berci
Dec 9 '18 at 22:22
$begingroup$
I've fixed it! Thank you!
$endgroup$
– H.koby
Dec 9 '18 at 22:32
$begingroup$
But I still don't get it... How I can use that when B is not H ?
$endgroup$
– H.koby
Dec 9 '18 at 22:34
1
$begingroup$
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.
$endgroup$
– the_fox
Dec 9 '18 at 23:16
|
show 6 more comments
$begingroup$
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
edited Dec 10 '18 at 17:37
the_fox
2,89021537
asked Dec 9 '18 at 22:12
H.koby H.koby
427
$endgroup$
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
group-theory finite-groups
edited Dec 10 '18 at 17:37
the_fox
2,89021537
asked Dec 9 '18 at 22:12
H.koby H.koby
427
edited Dec 10 '18 at 17:37
the_fox
2,89021537
asked Dec 9 '18 at 22:12
H.koby H.koby
427
edited Dec 10 '18 at 17:37
the_fox
2,89021537
edited Dec 10 '18 at 17:37
the_fox
2,89021537
edited Dec 10 '18 at 17:37
the_fox
2,89021537
2,89021537
asked Dec 9 '18 at 22:12
H.koby H.koby
427
asked Dec 9 '18 at 22:12
H.koby H.koby
427
asked Dec 9 '18 at 22:12
H.koby H.koby
427
427
$begingroup$
Well, $AH=H=HA$.
$endgroup$
– Berci
Dec 9 '18 at 22:20
1
$begingroup$
Anyway, it's not true in this form: consider $H=G$..
$endgroup$
– Berci
Dec 9 '18 at 22:22
$begingroup$
I've fixed it! Thank you!
$endgroup$
– H.koby
Dec 9 '18 at 22:32
$begingroup$
But I still don't get it... How I can use that when B is not H ?
$endgroup$
– H.koby
Dec 9 '18 at 22:34
1
$begingroup$
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.
$endgroup$
– the_fox
Dec 9 '18 at 23:16
|
show 6 more comments
$begingroup$
Well, $AH=H=HA$.
$endgroup$
– Berci
Dec 9 '18 at 22:20
1
$begingroup$
Anyway, it's not true in this form: consider $H=G$..
$endgroup$
– Berci
Dec 9 '18 at 22:22
$begingroup$
I've fixed it! Thank you!
$endgroup$
– H.koby
Dec 9 '18 at 22:32
$begingroup$
But I still don't get it... How I can use that when B is not H ?
$endgroup$
– H.koby
Dec 9 '18 at 22:34
1
$begingroup$
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.
$endgroup$
– the_fox
Dec 9 '18 at 23:16
$begingroup$
Well, $AH=H=HA$.
$endgroup$
– Berci
Dec 9 '18 at 22:20
$begingroup$
Well, $AH=H=HA$.
$endgroup$
– Berci
Dec 9 '18 at 22:20
1
1
$begingroup$
Anyway, it's not true in this form: consider $H=G$..
$endgroup$
– Berci
Dec 9 '18 at 22:22
$begingroup$
Anyway, it's not true in this form: consider $H=G$..
$endgroup$
– Berci
Dec 9 '18 at 22:22
$begingroup$
I've fixed it! Thank you!
$endgroup$
– H.koby
Dec 9 '18 at 22:32
$begingroup$
I've fixed it! Thank you!
$endgroup$
– H.koby
Dec 9 '18 at 22:32
$begingroup$
But I still don't get it... How I can use that when B is not H ?
$endgroup$
– H.koby
Dec 9 '18 at 22:34
$begingroup$
But I still don't get it... How I can use that when B is not H ?
$endgroup$
– H.koby
Dec 9 '18 at 22:34
1
1
$begingroup$
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.
$endgroup$
– the_fox
Dec 9 '18 at 23:16
$begingroup$
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.
$endgroup$
– the_fox
Dec 9 '18 at 23:16
|
show 6 more comments
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$begingroup$
Well, $AH=H=HA$.
$endgroup$
– Berci
Dec 9 '18 at 22:20
1
$begingroup$
Anyway, it's not true in this form: consider $H=G$..
$endgroup$
– Berci
Dec 9 '18 at 22:22
$begingroup$
I've fixed it! Thank you!
$endgroup$
– H.koby
Dec 9 '18 at 22:32
$begingroup$
But I still don't get it... How I can use that when B is not H ?
$endgroup$
– H.koby
Dec 9 '18 at 22:34
1
$begingroup$
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.
$endgroup$
– the_fox
Dec 9 '18 at 23:16