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Almost normal subgroup

Prove that a subgroup which contains half of all elements is a normal subgroup.proof: subgroup normal subgroupA finite $p$-group has normal subgroup of index $p^2$Can we define the normal set without $G$ being a group?Embedding of Frobenius group as a normal subgroup?Determing whether a subgroup is normalIf a normal subgroup shares elements with a conjugacy class, then it contains it entirely?Local property of a subgroupSylow normal subgroupsAny Subgroup containing commutator subgroup is normal.

3

$begingroup$

Let $G$ be a group then $N$ a subgroup of $G$ is said to be normal subgroup of $G$ if $forall g in G$, $g^{-1}Ng = N$.

Is there any notion which is weaker than normal subgroup? I mean something like almost normal subgroup or nearly normal subgroup in which some of the elements only follow the normality condition.

group-theory

share|cite|improve this question

edited 29 mins ago

Ali Taghavi

237329

asked 1 hour ago

I_wil_break_wallI_wil_break_wall

535

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Groupprops lists some - groupprops.subwiki.org/wiki/Normal_subgroup#Weaker_properties
  $endgroup$
  – Eevee Trainer
  1 hour ago
  

  
  
  
  

add a comment | 

3

$begingroup$

Let $G$ be a group then $N$ a subgroup of $G$ is said to be normal subgroup of $G$ if $forall g in G$, $g^{-1}Ng = N$.

Is there any notion which is weaker than normal subgroup? I mean something like almost normal subgroup or nearly normal subgroup in which some of the elements only follow the normality condition.

group-theory

share|cite|improve this question

edited 29 mins ago

Ali Taghavi

237329

asked 1 hour ago

I_wil_break_wallI_wil_break_wall

535

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Groupprops lists some - groupprops.subwiki.org/wiki/Normal_subgroup#Weaker_properties
  $endgroup$
  – Eevee Trainer
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

Let $G$ be a group then $N$ a subgroup of $G$ is said to be normal subgroup of $G$ if $forall g in G$, $g^{-1}Ng = N$.

Is there any notion which is weaker than normal subgroup? I mean something like almost normal subgroup or nearly normal subgroup in which some of the elements only follow the normality condition.

group-theory

share|cite|improve this question

edited 29 mins ago

Ali Taghavi

237329

asked 1 hour ago

I_wil_break_wallI_wil_break_wall

535

$endgroup$

share|cite|improve this question

edited 29 mins ago

Ali Taghavi

237329

asked 1 hour ago

I_wil_break_wallI_wil_break_wall

535

share|cite|improve this question

edited 29 mins ago

Ali Taghavi

237329

asked 1 hour ago

I_wil_break_wallI_wil_break_wall

535

edited 29 mins ago

Ali Taghavi

237329

edited 29 mins ago

Ali Taghavi

237329

asked 1 hour ago

I_wil_break_wallI_wil_break_wall

535

asked 1 hour ago

I_wil_break_wallI_wil_break_wall

535

3 Answers
3

active

oldest

votes

3

$begingroup$

Consider the set of all conjugates of a subgroup $Hleq G$, defined by $C={gHg^{-1}mid gin G}$. If $H$ is normal, clearly $|C|=1$, so $|C|$ can be considered to be a measure of "how far away from being normal" a subgroup is.

share|cite|improve this answer

answered 49 mins ago

YiFanYiFan

4,0711627

$endgroup$

add a comment | 

2

$begingroup$

What about $N$ admits only a finite number oof conjugates ? For instance, when the group $G$ acts on a finite space, the stabilizer of a point has this property.

share|cite|improve this answer

answered 1 hour ago

ThomasThomas

4,062510

$endgroup$

add a comment | 

0

$begingroup$

In the contex of locally compact topological groups which are equiped with the Haar measure, one can consider the following:
"For almost all $gin G$ we have $g^{-1} Ng=N$"

In the context of abstract groups one can says "$cap_g g^{-1} N g$ has finite index in $N$.

share|cite|improve this answer

answered 31 mins ago

Ali TaghaviAli Taghavi

237329

$endgroup$

add a comment | 

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3

$begingroup$

Consider the set of all conjugates of a subgroup $Hleq G$, defined by $C={gHg^{-1}mid gin G}$. If $H$ is normal, clearly $|C|=1$, so $|C|$ can be considered to be a measure of "how far away from being normal" a subgroup is.

share|cite|improve this answer

answered 49 mins ago

YiFanYiFan

4,0711627

$endgroup$

add a comment | 

3

$begingroup$

Consider the set of all conjugates of a subgroup $Hleq G$, defined by $C={gHg^{-1}mid gin G}$. If $H$ is normal, clearly $|C|=1$, so $|C|$ can be considered to be a measure of "how far away from being normal" a subgroup is.

share|cite|improve this answer

answered 49 mins ago

YiFanYiFan

4,0711627

$endgroup$

add a comment | 

$begingroup$

Consider the set of all conjugates of a subgroup $Hleq G$, defined by $C={gHg^{-1}mid gin G}$. If $H$ is normal, clearly $|C|=1$, so $|C|$ can be considered to be a measure of "how far away from being normal" a subgroup is.

share|cite|improve this answer

answered 49 mins ago

YiFanYiFan

4,0711627

$endgroup$

share|cite|improve this answer

answered 49 mins ago

YiFanYiFan

4,0711627

answered 49 mins ago

YiFanYiFan

4,0711627

answered 49 mins ago

YiFanYiFan

4,0711627

2

$begingroup$

What about $N$ admits only a finite number oof conjugates ? For instance, when the group $G$ acts on a finite space, the stabilizer of a point has this property.

share|cite|improve this answer

answered 1 hour ago

ThomasThomas

4,062510

$endgroup$

add a comment | 

2

$begingroup$

What about $N$ admits only a finite number oof conjugates ? For instance, when the group $G$ acts on a finite space, the stabilizer of a point has this property.

share|cite|improve this answer

answered 1 hour ago

ThomasThomas

4,062510

$endgroup$

add a comment | 

$begingroup$

What about $N$ admits only a finite number oof conjugates ? For instance, when the group $G$ acts on a finite space, the stabilizer of a point has this property.

share|cite|improve this answer

answered 1 hour ago

ThomasThomas

4,062510

$endgroup$

share|cite|improve this answer

answered 1 hour ago

ThomasThomas

4,062510

answered 1 hour ago

ThomasThomas

4,062510

answered 1 hour ago

ThomasThomas

4,062510

0

$begingroup$

In the contex of locally compact topological groups which are equiped with the Haar measure, one can consider the following:
"For almost all $gin G$ we have $g^{-1} Ng=N$"

In the context of abstract groups one can says "$cap_g g^{-1} N g$ has finite index in $N$.

share|cite|improve this answer

answered 31 mins ago

Ali TaghaviAli Taghavi

237329

$endgroup$

add a comment | 

0

$begingroup$

In the contex of locally compact topological groups which are equiped with the Haar measure, one can consider the following:
"For almost all $gin G$ we have $g^{-1} Ng=N$"

In the context of abstract groups one can says "$cap_g g^{-1} N g$ has finite index in $N$.

share|cite|improve this answer

answered 31 mins ago

Ali TaghaviAli Taghavi

237329

$endgroup$

add a comment | 

$begingroup$

In the contex of locally compact topological groups which are equiped with the Haar measure, one can consider the following:
"For almost all $gin G$ we have $g^{-1} Ng=N$"

In the context of abstract groups one can says "$cap_g g^{-1} N g$ has finite index in $N$.

share|cite|improve this answer

answered 31 mins ago

Ali TaghaviAli Taghavi

237329

$endgroup$

share|cite|improve this answer

answered 31 mins ago

Ali TaghaviAli Taghavi

237329

answered 31 mins ago

Ali TaghaviAli Taghavi

237329

answered 31 mins ago

Ali TaghaviAli Taghavi

237329