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Does every subgroup of an abelian group have to be abelian?
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraA condition for a subgroup of a finitely generated free abelian group to have finite indexAdditive non-abelian group?Galois Group, Field Extension Prove AbelianCommutator subgroup is the minimal normal subgroup such that quotient group is abelianShowing that every subgroup of an abelian group is normalevery subgroup of the quaternion group is normalQuotient Group $D_2n/R$ is abelian where $R$ is the group of rotations?Proving a subgroup of a Galois group is normalNormal Subgroup of Galois GroupFaithful group action and Galois correspondence
$begingroup$
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
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
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
edited 4 hours ago
J. W. Tanner
5,1651520
asked 6 hours ago
MT mathMT math
253
$endgroup$
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
abstract-algebra group-theory galois-theory abelian-groups
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
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
edited 4 hours ago
J. W. Tanner
5,1651520
5,1651520
asked 6 hours ago
MT mathMT math
253
asked 6 hours ago
MT mathMT math
253
asked 6 hours ago
MT mathMT math
253
253
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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).
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 |
$begingroup$
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.
answered 6 hours ago
Matt SamuelMatt Samuel
39.5k63870
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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votes
$begingroup$
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).
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).
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).
answered 6 hours ago
Eevee TrainerEevee Trainer
10.8k31843
$endgroup$
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).
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
answered 6 hours ago
Eevee TrainerEevee Trainer
10.8k31843
10.8k31843
$begingroup$
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 |
$begingroup$
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
$begingroup$
can I use essentially the same reasoning to prove that L/F is an abelian extension as well?
$endgroup$
– MT math
6 hours ago
$begingroup$
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
$begingroup$
I believe so, yes.
$endgroup$
– Eevee Trainer
4 hours ago
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.
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.
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.
answered 6 hours ago
Matt SamuelMatt Samuel
39.5k63870
$endgroup$
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.
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
answered 6 hours ago
Matt SamuelMatt Samuel
39.5k63870
39.5k63870
add a comment |
add a comment |
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