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Does every subgroup of an abelian group have to be abelian?

3

$begingroup$

My original problem is to show that E/L is an abelian extension over L and L/F is an abelian extension over F, given that E/F is an abelian extension over F and that L is a normal extension of F such that $Fsubseteq L subseteq E$.

So far I have proved that E is a normal extension of F, E is a normal extension of L, and L is a normal extension of F. I know that to prove abelian extension I must also prove that Gal(E/L) is an abelian group. I have shown that Gal(E/L) $subseteq$ Gal (E/F). In my mind it makes sense that I cannot lose commutativity therefore my subgroup must be Abelian too. How do I show this in a proof? Is it enough to show two elements in the subgroup must also exist in the larger group and that they must be commutative in the larger group? I feel like I know what needs to be done, just not how to phrase it.

abstract-algebra group-theory galois-theory abelian-groups

share|cite|improve this question

edited 4 hours ago

J. W. Tanner

5,1651520

asked 6 hours ago

MT mathMT math

253

$endgroup$

add a comment | 

3

$begingroup$

My original problem is to show that E/L is an abelian extension over L and L/F is an abelian extension over F, given that E/F is an abelian extension over F and that L is a normal extension of F such that $Fsubseteq L subseteq E$.

So far I have proved that E is a normal extension of F, E is a normal extension of L, and L is a normal extension of F. I know that to prove abelian extension I must also prove that Gal(E/L) is an abelian group. I have shown that Gal(E/L) $subseteq$ Gal (E/F). In my mind it makes sense that I cannot lose commutativity therefore my subgroup must be Abelian too. How do I show this in a proof? Is it enough to show two elements in the subgroup must also exist in the larger group and that they must be commutative in the larger group? I feel like I know what needs to be done, just not how to phrase it.

abstract-algebra group-theory galois-theory abelian-groups

share|cite|improve this question

edited 4 hours ago

J. W. Tanner

5,1651520

asked 6 hours ago

MT mathMT math

253

$endgroup$

add a comment | 

$begingroup$

My original problem is to show that E/L is an abelian extension over L and L/F is an abelian extension over F, given that E/F is an abelian extension over F and that L is a normal extension of F such that $Fsubseteq L subseteq E$.

So far I have proved that E is a normal extension of F, E is a normal extension of L, and L is a normal extension of F. I know that to prove abelian extension I must also prove that Gal(E/L) is an abelian group. I have shown that Gal(E/L) $subseteq$ Gal (E/F). In my mind it makes sense that I cannot lose commutativity therefore my subgroup must be Abelian too. How do I show this in a proof? Is it enough to show two elements in the subgroup must also exist in the larger group and that they must be commutative in the larger group? I feel like I know what needs to be done, just not how to phrase it.

abstract-algebra group-theory galois-theory abelian-groups

share|cite|improve this question

edited 4 hours ago

J. W. Tanner

5,1651520

asked 6 hours ago

MT mathMT math

253

$endgroup$

share|cite|improve this question

edited 4 hours ago

J. W. Tanner

5,1651520

asked 6 hours ago

MT mathMT math

253

share|cite|improve this question

edited 4 hours ago

J. W. Tanner

5,1651520

asked 6 hours ago

MT mathMT math

253

edited 4 hours ago

J. W. Tanner

5,1651520

edited 4 hours ago

J. W. Tanner

5,1651520

asked 6 hours ago

MT mathMT math

253

asked 6 hours ago

MT mathMT math

253

2 Answers
2

active

oldest

votes

4

$begingroup$

Huh, funny, we just went over this today in my algebra class.

Yes, subgroups of abelian groups are indeed abelian, and your thought process has the right idea.

Showing this is pretty easy. Take an abelian group $G$ with subgroup $H$. Then we know that, for all $a,bin H$, $ab=ba$ since it must also hold in $G$ (as $a,b in G ge H$ and $G$ is given to be abelian).

share|cite|improve this answer

answered 6 hours ago

Eevee TrainerEevee Trainer

10.8k31843

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  can I use essentially the same reasoning to prove that L/F is an abelian extension as well?
  $endgroup$
  – MT math
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I believe so, yes.
  $endgroup$
  – Eevee Trainer
  4 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

If $G$ is an abelian group and $H$ is a subgroup, suppose $x, yin H$. Then in particular $x, yin G$, so $xy=yx$. Since $x, y$ were arbitrary, $H$ is abelian.

share|cite|improve this answer

answered 6 hours ago

Matt SamuelMatt Samuel

39.5k63870

$endgroup$

add a comment | 

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4

$begingroup$

Huh, funny, we just went over this today in my algebra class.

Yes, subgroups of abelian groups are indeed abelian, and your thought process has the right idea.

Showing this is pretty easy. Take an abelian group $G$ with subgroup $H$. Then we know that, for all $a,bin H$, $ab=ba$ since it must also hold in $G$ (as $a,b in G ge H$ and $G$ is given to be abelian).

share|cite|improve this answer

answered 6 hours ago

Eevee TrainerEevee Trainer

10.8k31843

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  can I use essentially the same reasoning to prove that L/F is an abelian extension as well?
  $endgroup$
  – MT math
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I believe so, yes.
  $endgroup$
  – Eevee Trainer
  4 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

Huh, funny, we just went over this today in my algebra class.

Yes, subgroups of abelian groups are indeed abelian, and your thought process has the right idea.

Showing this is pretty easy. Take an abelian group $G$ with subgroup $H$. Then we know that, for all $a,bin H$, $ab=ba$ since it must also hold in $G$ (as $a,b in G ge H$ and $G$ is given to be abelian).

share|cite|improve this answer

answered 6 hours ago

Eevee TrainerEevee Trainer

10.8k31843

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  can I use essentially the same reasoning to prove that L/F is an abelian extension as well?
  $endgroup$
  – MT math
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I believe so, yes.
  $endgroup$
  – Eevee Trainer
  4 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Huh, funny, we just went over this today in my algebra class.

Yes, subgroups of abelian groups are indeed abelian, and your thought process has the right idea.

Showing this is pretty easy. Take an abelian group $G$ with subgroup $H$. Then we know that, for all $a,bin H$, $ab=ba$ since it must also hold in $G$ (as $a,b in G ge H$ and $G$ is given to be abelian).

share|cite|improve this answer

answered 6 hours ago

Eevee TrainerEevee Trainer

10.8k31843

$endgroup$

share|cite|improve this answer

answered 6 hours ago

Eevee TrainerEevee Trainer

10.8k31843

answered 6 hours ago

Eevee TrainerEevee Trainer

10.8k31843

answered 6 hours ago

Eevee TrainerEevee Trainer

10.8k31843

2

$begingroup$

If $G$ is an abelian group and $H$ is a subgroup, suppose $x, yin H$. Then in particular $x, yin G$, so $xy=yx$. Since $x, y$ were arbitrary, $H$ is abelian.

share|cite|improve this answer

answered 6 hours ago

Matt SamuelMatt Samuel

39.5k63870

$endgroup$

add a comment | 

2

$begingroup$

If $G$ is an abelian group and $H$ is a subgroup, suppose $x, yin H$. Then in particular $x, yin G$, so $xy=yx$. Since $x, y$ were arbitrary, $H$ is abelian.

share|cite|improve this answer

answered 6 hours ago

Matt SamuelMatt Samuel

39.5k63870

$endgroup$

add a comment | 

$begingroup$

If $G$ is an abelian group and $H$ is a subgroup, suppose $x, yin H$. Then in particular $x, yin G$, so $xy=yx$. Since $x, y$ were arbitrary, $H$ is abelian.

share|cite|improve this answer

answered 6 hours ago

Matt SamuelMatt Samuel

39.5k63870

$endgroup$

share|cite|improve this answer

answered 6 hours ago

Matt SamuelMatt Samuel

39.5k63870

answered 6 hours ago

Matt SamuelMatt Samuel

39.5k63870

answered 6 hours ago

Matt SamuelMatt Samuel

39.5k63870