Difference between group algebra over a field and algebra over the same field?
$begingroup$
When is there a difference between the group algebra over a field and an algebra over the same field, that is generated by a multiplicative subgroup that is isomorphic to the group?
I think that for finite groups, the two notions should be the same, and that the difference occurs for some infinite groups?
abstract-algebra group-theory
edited Dec 15 '18 at 15:00
Brahadeesh
6,47942363
asked Oct 31 '18 at 1:43
compl11112222compl11112222
6
$endgroup$
|
show 1 more comment
$begingroup$
When is there a difference between the group algebra over a field and an algebra over the same field, that is generated by a multiplicative subgroup that is isomorphic to the group?
I think that for finite groups, the two notions should be the same, and that the difference occurs for some infinite groups?
abstract-algebra group-theory
edited Dec 15 '18 at 15:00
Brahadeesh
6,47942363
asked Oct 31 '18 at 1:43
compl11112222compl11112222
6
$endgroup$
$begingroup$
What do you mean by "an algebra over the same field, that is generated by the group"?
$endgroup$
– Eric Wofsey
Oct 31 '18 at 1:48
$begingroup$
Lets have a group G and a field F. I am wondering when is F[G] (the group algebra) different to the F-algebra generated by G?
$endgroup$
– compl11112222
Oct 31 '18 at 1:51
$begingroup$
Again, what do you mean by "the F-algebra generated by G"?
$endgroup$
– Eric Wofsey
Oct 31 '18 at 1:52
$begingroup$
I mean subalgebra of an F-algebra B generated by elements of G, where G is a subgroup of the B* ("linear span of G").
$endgroup$
– compl11112222
Oct 31 '18 at 2:00
3
$begingroup$
Please don't self-delete your post. That's unfair to the answerer who has spent time and effort in answering your question. We can (and will) undelete it.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 22:40
|
show 1 more comment
$begingroup$
When is there a difference between the group algebra over a field and an algebra over the same field, that is generated by a multiplicative subgroup that is isomorphic to the group?
I think that for finite groups, the two notions should be the same, and that the difference occurs for some infinite groups?
abstract-algebra group-theory
edited Dec 15 '18 at 15:00
Brahadeesh
6,47942363
asked Oct 31 '18 at 1:43
compl11112222compl11112222
6
$endgroup$
When is there a difference between the group algebra over a field and an algebra over the same field, that is generated by a multiplicative subgroup that is isomorphic to the group?
I think that for finite groups, the two notions should be the same, and that the difference occurs for some infinite groups?
abstract-algebra group-theory
abstract-algebra group-theory
edited Dec 15 '18 at 15:00
Brahadeesh
6,47942363
asked Oct 31 '18 at 1:43
compl11112222compl11112222
6
edited Dec 15 '18 at 15:00
Brahadeesh
6,47942363
asked Oct 31 '18 at 1:43
compl11112222compl11112222
6
edited Dec 15 '18 at 15:00
Brahadeesh
6,47942363
edited Dec 15 '18 at 15:00
Brahadeesh
6,47942363
edited Dec 15 '18 at 15:00
Brahadeesh
6,47942363
6,47942363
asked Oct 31 '18 at 1:43
compl11112222compl11112222
6
asked Oct 31 '18 at 1:43
compl11112222compl11112222
6
asked Oct 31 '18 at 1:43
compl11112222compl11112222
6
6
$begingroup$
What do you mean by "an algebra over the same field, that is generated by the group"?
$endgroup$
– Eric Wofsey
Oct 31 '18 at 1:48
$begingroup$
Lets have a group G and a field F. I am wondering when is F[G] (the group algebra) different to the F-algebra generated by G?
$endgroup$
– compl11112222
Oct 31 '18 at 1:51
$begingroup$
Again, what do you mean by "the F-algebra generated by G"?
$endgroup$
– Eric Wofsey
Oct 31 '18 at 1:52
$begingroup$
I mean subalgebra of an F-algebra B generated by elements of G, where G is a subgroup of the B* ("linear span of G").
$endgroup$
– compl11112222
Oct 31 '18 at 2:00
3
$begingroup$
Please don't self-delete your post. That's unfair to the answerer who has spent time and effort in answering your question. We can (and will) undelete it.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 22:40
|
show 1 more comment
$begingroup$
What do you mean by "an algebra over the same field, that is generated by the group"?
$endgroup$
– Eric Wofsey
Oct 31 '18 at 1:48
$begingroup$
Lets have a group G and a field F. I am wondering when is F[G] (the group algebra) different to the F-algebra generated by G?
$endgroup$
– compl11112222
Oct 31 '18 at 1:51
$begingroup$
Again, what do you mean by "the F-algebra generated by G"?
$endgroup$
– Eric Wofsey
Oct 31 '18 at 1:52
$begingroup$
I mean subalgebra of an F-algebra B generated by elements of G, where G is a subgroup of the B* ("linear span of G").
$endgroup$
– compl11112222
Oct 31 '18 at 2:00
3
$begingroup$
Please don't self-delete your post. That's unfair to the answerer who has spent time and effort in answering your question. We can (and will) undelete it.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 22:40
$begingroup$
What do you mean by "an algebra over the same field, that is generated by the group"?
$endgroup$
– Eric Wofsey
Oct 31 '18 at 1:48
$begingroup$
What do you mean by "an algebra over the same field, that is generated by the group"?
$endgroup$
– Eric Wofsey
Oct 31 '18 at 1:48
$begingroup$
Lets have a group G and a field F. I am wondering when is F[G] (the group algebra) different to the F-algebra generated by G?
$endgroup$
– compl11112222
Oct 31 '18 at 1:51
$begingroup$
Lets have a group G and a field F. I am wondering when is F[G] (the group algebra) different to the F-algebra generated by G?
$endgroup$
– compl11112222
Oct 31 '18 at 1:51
$begingroup$
Again, what do you mean by "the F-algebra generated by G"?
$endgroup$
– Eric Wofsey
Oct 31 '18 at 1:52
$begingroup$
Again, what do you mean by "the F-algebra generated by G"?
$endgroup$
– Eric Wofsey
Oct 31 '18 at 1:52
$begingroup$
I mean subalgebra of an F-algebra B generated by elements of G, where G is a subgroup of the B* ("linear span of G").
$endgroup$
– compl11112222
Oct 31 '18 at 2:00
$begingroup$
I mean subalgebra of an F-algebra B generated by elements of G, where G is a subgroup of the B* ("linear span of G").
$endgroup$
– compl11112222
Oct 31 '18 at 2:00
3
3
$begingroup$
Please don't self-delete your post. That's unfair to the answerer who has spent time and effort in answering your question. We can (and will) undelete it.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 22:40
$begingroup$
Please don't self-delete your post. That's unfair to the answerer who has spent time and effort in answering your question. We can (and will) undelete it.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 22:40
|
show 1 more comment
1 Answer
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$begingroup$
An $F$-algebra $A$ generated by a multiplicative subgroup $G$ does not have to be isomorphic to the group algebra $F[G]$, regardless of any finiteness conditions. For a very simple example, taking $F=mathbb{Q}$, then $A=mathbb{Q}$ is generated by the group $G={1,-1}$ as a $mathbb{Q}$-algebra but is not isomorphic to $mathbb{Q}[G]$.
In order to conclude that $A$ is isomorphic to $F[G]$, you need to additionally know that $G$ is $F$-linearly independent as a subset of $A$. That means exactly that the canonical homomorphism $F[G]to A$ which is the identity on $G$ is an isomorphism, since $F[G]$ consists of formal linear combinations of elements of $G$.
answered Oct 31 '18 at 2:17
Eric WofseyEric Wofsey
188k14216346
$endgroup$
add a comment |
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1 Answer
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$begingroup$
An $F$-algebra $A$ generated by a multiplicative subgroup $G$ does not have to be isomorphic to the group algebra $F[G]$, regardless of any finiteness conditions. For a very simple example, taking $F=mathbb{Q}$, then $A=mathbb{Q}$ is generated by the group $G={1,-1}$ as a $mathbb{Q}$-algebra but is not isomorphic to $mathbb{Q}[G]$.
In order to conclude that $A$ is isomorphic to $F[G]$, you need to additionally know that $G$ is $F$-linearly independent as a subset of $A$. That means exactly that the canonical homomorphism $F[G]to A$ which is the identity on $G$ is an isomorphism, since $F[G]$ consists of formal linear combinations of elements of $G$.
answered Oct 31 '18 at 2:17
Eric WofseyEric Wofsey
188k14216346
$endgroup$
add a comment |
$begingroup$
An $F$-algebra $A$ generated by a multiplicative subgroup $G$ does not have to be isomorphic to the group algebra $F[G]$, regardless of any finiteness conditions. For a very simple example, taking $F=mathbb{Q}$, then $A=mathbb{Q}$ is generated by the group $G={1,-1}$ as a $mathbb{Q}$-algebra but is not isomorphic to $mathbb{Q}[G]$.
In order to conclude that $A$ is isomorphic to $F[G]$, you need to additionally know that $G$ is $F$-linearly independent as a subset of $A$. That means exactly that the canonical homomorphism $F[G]to A$ which is the identity on $G$ is an isomorphism, since $F[G]$ consists of formal linear combinations of elements of $G$.
answered Oct 31 '18 at 2:17
Eric WofseyEric Wofsey
188k14216346
$endgroup$
add a comment |
$begingroup$
An $F$-algebra $A$ generated by a multiplicative subgroup $G$ does not have to be isomorphic to the group algebra $F[G]$, regardless of any finiteness conditions. For a very simple example, taking $F=mathbb{Q}$, then $A=mathbb{Q}$ is generated by the group $G={1,-1}$ as a $mathbb{Q}$-algebra but is not isomorphic to $mathbb{Q}[G]$.
In order to conclude that $A$ is isomorphic to $F[G]$, you need to additionally know that $G$ is $F$-linearly independent as a subset of $A$. That means exactly that the canonical homomorphism $F[G]to A$ which is the identity on $G$ is an isomorphism, since $F[G]$ consists of formal linear combinations of elements of $G$.
answered Oct 31 '18 at 2:17
Eric WofseyEric Wofsey
188k14216346
$endgroup$
An $F$-algebra $A$ generated by a multiplicative subgroup $G$ does not have to be isomorphic to the group algebra $F[G]$, regardless of any finiteness conditions. For a very simple example, taking $F=mathbb{Q}$, then $A=mathbb{Q}$ is generated by the group $G={1,-1}$ as a $mathbb{Q}$-algebra but is not isomorphic to $mathbb{Q}[G]$.
In order to conclude that $A$ is isomorphic to $F[G]$, you need to additionally know that $G$ is $F$-linearly independent as a subset of $A$. That means exactly that the canonical homomorphism $F[G]to A$ which is the identity on $G$ is an isomorphism, since $F[G]$ consists of formal linear combinations of elements of $G$.
answered Oct 31 '18 at 2:17
Eric WofseyEric Wofsey
188k14216346
answered Oct 31 '18 at 2:17
Eric WofseyEric Wofsey
188k14216346
answered Oct 31 '18 at 2:17
Eric WofseyEric Wofsey
188k14216346
answered Oct 31 '18 at 2:17
Eric WofseyEric Wofsey
188k14216346
188k14216346
add a comment |
add a comment |
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$begingroup$
What do you mean by "an algebra over the same field, that is generated by the group"?
$endgroup$
– Eric Wofsey
Oct 31 '18 at 1:48
$begingroup$
Lets have a group G and a field F. I am wondering when is F[G] (the group algebra) different to the F-algebra generated by G?
$endgroup$
– compl11112222
Oct 31 '18 at 1:51
$begingroup$
Again, what do you mean by "the F-algebra generated by G"?
$endgroup$
– Eric Wofsey
Oct 31 '18 at 1:52
$begingroup$
I mean subalgebra of an F-algebra B generated by elements of G, where G is a subgroup of the B* ("linear span of G").
$endgroup$
– compl11112222
Oct 31 '18 at 2:00
3
$begingroup$
Please don't self-delete your post. That's unfair to the answerer who has spent time and effort in answering your question. We can (and will) undelete it.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 22:40