Subalgebra of a group algebra
$begingroup$
Let $k$ be a field, $G$ a finite group, and $k[G]$ the group algebra.
Let $A$ be a subalgebra of $k[G]$. In general, $A$ is not the group algebra of some subgroup $H$ of $G$.
Question: Is there any criterion for when $A = k[H]$ for some subgroup $H$? Also, in that case, how do we read of the generating subgroup $H$? Will the situation become better/easier if I assume $A$ to be a sub-Hopf-algebra?
gr.group-theory rt.representation-theory noncommutative-algebra
asked Apr 18 at 20:19
StudentStudent
1394
$endgroup$
add a comment |
$begingroup$
Let $k$ be a field, $G$ a finite group, and $k[G]$ the group algebra.
Let $A$ be a subalgebra of $k[G]$. In general, $A$ is not the group algebra of some subgroup $H$ of $G$.
Question: Is there any criterion for when $A = k[H]$ for some subgroup $H$? Also, in that case, how do we read of the generating subgroup $H$? Will the situation become better/easier if I assume $A$ to be a sub-Hopf-algebra?
gr.group-theory rt.representation-theory noncommutative-algebra
asked Apr 18 at 20:19
StudentStudent
1394
$endgroup$
add a comment |
$begingroup$
Let $k$ be a field, $G$ a finite group, and $k[G]$ the group algebra.
Let $A$ be a subalgebra of $k[G]$. In general, $A$ is not the group algebra of some subgroup $H$ of $G$.
Question: Is there any criterion for when $A = k[H]$ for some subgroup $H$? Also, in that case, how do we read of the generating subgroup $H$? Will the situation become better/easier if I assume $A$ to be a sub-Hopf-algebra?
gr.group-theory rt.representation-theory noncommutative-algebra
asked Apr 18 at 20:19
StudentStudent
1394
$endgroup$
Let $k$ be a field, $G$ a finite group, and $k[G]$ the group algebra.
Let $A$ be a subalgebra of $k[G]$. In general, $A$ is not the group algebra of some subgroup $H$ of $G$.
Question: Is there any criterion for when $A = k[H]$ for some subgroup $H$? Also, in that case, how do we read of the generating subgroup $H$? Will the situation become better/easier if I assume $A$ to be a sub-Hopf-algebra?
gr.group-theory rt.representation-theory noncommutative-algebra
gr.group-theory rt.representation-theory noncommutative-algebra
asked Apr 18 at 20:19
StudentStudent
1394
asked Apr 18 at 20:19
StudentStudent
1394
asked Apr 18 at 20:19
StudentStudent
1394
asked Apr 18 at 20:19
StudentStudent
1394
asked Apr 18 at 20:19
StudentStudent
1394
1394
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The characteristic of the field is important here, when considering Hopf sub-algebras. The Cartier-Kostant-Milnor-Moore theorem says that a cocommutative Hopf algebra $H$ over an algebraically closed field $k$ of characteristic 0, is a semidirect product of a group algebra and an enveloping algebra of a Lie algebra. In particular, if $H$ is finite dimensional cocommutative Hopf algebra over $k$, then $H$ is isomorphic to a group algebra.
This theorem is not true over algebraically closed fields in positive characteristics, as there are restricted enveloping $p$-Lie algebras that are finite dimensional cocommutative Hopf algebras but not isomorphic to group algebras.
See the introduction of https://cel.archives-ouvertes.fr/cel-00374383/document (by Nicolas Andruskiewitsch) for further information.
answered Apr 19 at 8:18
Oeyvind SolbergOeyvind Solberg
4864
$endgroup$
1
$begingroup$
The CKMM theorem is very powerful! Is there a way to read off the group algebra part?
$endgroup$
– Student
Apr 19 at 12:24
$begingroup$
I have only used this when I start with something finite dimensional, and then, as the result states, the Lie algebra part is not there. I don't know how to you find the group algebra part when you start with something which is infinite dimensional.
$endgroup$
– Oeyvind Solberg
Apr 20 at 7:35
add a comment |
$begingroup$
If $A$ is the group algebra of a subgroup, then $k[G]$ will be free as a module over $A$, and that may help to rule out some cases. This is not at all an "if and only if" statement, though.
Proposition 3.2.1 in Sweedler's book Hopf Algebras says that a cocommutative Hopf algebra is a group algebra if and only if it has a basis of group-like elements (those elements $x$ which satisfy $Delta(x) = x otimes x$). So if you have a sub-Hopf algebra, you can try to see if it has such a basis.
Ravenel says in Theorem 6.2.3 in Complex Cobordism and Stable Homotopy Groups of Spheres: "this is equivalent to the existence of a dual basis of idempotent elements ${y}$ satisfying $y_i^2=y_i$ and $y_iy_j=0$ for $i neq j$."
answered Apr 18 at 22:23
John PalmieriJohn Palmieri
2,35011726
$endgroup$
$begingroup$
Is there a characteristic assumption on the last part? Is there division by 2 in formula for y?
$endgroup$
– AHusain
Apr 18 at 22:29
$begingroup$
Great! Also, I think Sweedler provided a fairly nice answer already, since if A comes from some subgroup, then A is automatically a sub-Hopf-algebra!
$endgroup$
– Student
Apr 19 at 0:03
$begingroup$
For Ravenel's statement, I do not have access to the book for now. Would you mind pointing it out that in his content, is $A$ assumed to be a Hopf algebra?
$endgroup$
– Student
Apr 19 at 0:04
2
$begingroup$
Ravenel's book is available from his web page: web.math.rochester.edu/people/faculty/doug/mybooks/ravenel.pdf. He is certainly working in positive characteristic, but I don't know if this is necessary for his condition. He is certainly working with Hopf algebras, since otherwise there wouldn't be a multiplication on the dual.
$endgroup$
– John Palmieri
Apr 19 at 5:28
2
$begingroup$
@JoshuaGrochow: Ravenel uses his version to show that a particular Hopf algebra is a group algebra, so he found it useful.
$endgroup$
– John Palmieri
Apr 19 at 5:29
|
show 1 more comment
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The characteristic of the field is important here, when considering Hopf sub-algebras. The Cartier-Kostant-Milnor-Moore theorem says that a cocommutative Hopf algebra $H$ over an algebraically closed field $k$ of characteristic 0, is a semidirect product of a group algebra and an enveloping algebra of a Lie algebra. In particular, if $H$ is finite dimensional cocommutative Hopf algebra over $k$, then $H$ is isomorphic to a group algebra.
This theorem is not true over algebraically closed fields in positive characteristics, as there are restricted enveloping $p$-Lie algebras that are finite dimensional cocommutative Hopf algebras but not isomorphic to group algebras.
See the introduction of https://cel.archives-ouvertes.fr/cel-00374383/document (by Nicolas Andruskiewitsch) for further information.
answered Apr 19 at 8:18
Oeyvind SolbergOeyvind Solberg
4864
$endgroup$
1
$begingroup$
The CKMM theorem is very powerful! Is there a way to read off the group algebra part?
$endgroup$
– Student
Apr 19 at 12:24
$begingroup$
I have only used this when I start with something finite dimensional, and then, as the result states, the Lie algebra part is not there. I don't know how to you find the group algebra part when you start with something which is infinite dimensional.
$endgroup$
– Oeyvind Solberg
Apr 20 at 7:35
add a comment |
$begingroup$
The characteristic of the field is important here, when considering Hopf sub-algebras. The Cartier-Kostant-Milnor-Moore theorem says that a cocommutative Hopf algebra $H$ over an algebraically closed field $k$ of characteristic 0, is a semidirect product of a group algebra and an enveloping algebra of a Lie algebra. In particular, if $H$ is finite dimensional cocommutative Hopf algebra over $k$, then $H$ is isomorphic to a group algebra.
This theorem is not true over algebraically closed fields in positive characteristics, as there are restricted enveloping $p$-Lie algebras that are finite dimensional cocommutative Hopf algebras but not isomorphic to group algebras.
See the introduction of https://cel.archives-ouvertes.fr/cel-00374383/document (by Nicolas Andruskiewitsch) for further information.
answered Apr 19 at 8:18
Oeyvind SolbergOeyvind Solberg
4864
$endgroup$
1
$begingroup$
The CKMM theorem is very powerful! Is there a way to read off the group algebra part?
$endgroup$
– Student
Apr 19 at 12:24
$begingroup$
I have only used this when I start with something finite dimensional, and then, as the result states, the Lie algebra part is not there. I don't know how to you find the group algebra part when you start with something which is infinite dimensional.
$endgroup$
– Oeyvind Solberg
Apr 20 at 7:35
add a comment |
$begingroup$
The characteristic of the field is important here, when considering Hopf sub-algebras. The Cartier-Kostant-Milnor-Moore theorem says that a cocommutative Hopf algebra $H$ over an algebraically closed field $k$ of characteristic 0, is a semidirect product of a group algebra and an enveloping algebra of a Lie algebra. In particular, if $H$ is finite dimensional cocommutative Hopf algebra over $k$, then $H$ is isomorphic to a group algebra.
This theorem is not true over algebraically closed fields in positive characteristics, as there are restricted enveloping $p$-Lie algebras that are finite dimensional cocommutative Hopf algebras but not isomorphic to group algebras.
See the introduction of https://cel.archives-ouvertes.fr/cel-00374383/document (by Nicolas Andruskiewitsch) for further information.
answered Apr 19 at 8:18
Oeyvind SolbergOeyvind Solberg
4864
$endgroup$
The characteristic of the field is important here, when considering Hopf sub-algebras. The Cartier-Kostant-Milnor-Moore theorem says that a cocommutative Hopf algebra $H$ over an algebraically closed field $k$ of characteristic 0, is a semidirect product of a group algebra and an enveloping algebra of a Lie algebra. In particular, if $H$ is finite dimensional cocommutative Hopf algebra over $k$, then $H$ is isomorphic to a group algebra.
This theorem is not true over algebraically closed fields in positive characteristics, as there are restricted enveloping $p$-Lie algebras that are finite dimensional cocommutative Hopf algebras but not isomorphic to group algebras.
See the introduction of https://cel.archives-ouvertes.fr/cel-00374383/document (by Nicolas Andruskiewitsch) for further information.
answered Apr 19 at 8:18
Oeyvind SolbergOeyvind Solberg
4864
answered Apr 19 at 8:18
Oeyvind SolbergOeyvind Solberg
4864
answered Apr 19 at 8:18
Oeyvind SolbergOeyvind Solberg
4864
answered Apr 19 at 8:18
Oeyvind SolbergOeyvind Solberg
4864
4864
1
$begingroup$
The CKMM theorem is very powerful! Is there a way to read off the group algebra part?
$endgroup$
– Student
Apr 19 at 12:24
$begingroup$
I have only used this when I start with something finite dimensional, and then, as the result states, the Lie algebra part is not there. I don't know how to you find the group algebra part when you start with something which is infinite dimensional.
$endgroup$
– Oeyvind Solberg
Apr 20 at 7:35
add a comment |
1
$begingroup$
The CKMM theorem is very powerful! Is there a way to read off the group algebra part?
$endgroup$
– Student
Apr 19 at 12:24
$begingroup$
I have only used this when I start with something finite dimensional, and then, as the result states, the Lie algebra part is not there. I don't know how to you find the group algebra part when you start with something which is infinite dimensional.
$endgroup$
– Oeyvind Solberg
Apr 20 at 7:35
1
1
$begingroup$
The CKMM theorem is very powerful! Is there a way to read off the group algebra part?
$endgroup$
– Student
Apr 19 at 12:24
$begingroup$
The CKMM theorem is very powerful! Is there a way to read off the group algebra part?
$endgroup$
– Student
Apr 19 at 12:24
$begingroup$
I have only used this when I start with something finite dimensional, and then, as the result states, the Lie algebra part is not there. I don't know how to you find the group algebra part when you start with something which is infinite dimensional.
$endgroup$
– Oeyvind Solberg
Apr 20 at 7:35
$begingroup$
I have only used this when I start with something finite dimensional, and then, as the result states, the Lie algebra part is not there. I don't know how to you find the group algebra part when you start with something which is infinite dimensional.
$endgroup$
– Oeyvind Solberg
Apr 20 at 7:35
add a comment |
$begingroup$
If $A$ is the group algebra of a subgroup, then $k[G]$ will be free as a module over $A$, and that may help to rule out some cases. This is not at all an "if and only if" statement, though.
Proposition 3.2.1 in Sweedler's book Hopf Algebras says that a cocommutative Hopf algebra is a group algebra if and only if it has a basis of group-like elements (those elements $x$ which satisfy $Delta(x) = x otimes x$). So if you have a sub-Hopf algebra, you can try to see if it has such a basis.
Ravenel says in Theorem 6.2.3 in Complex Cobordism and Stable Homotopy Groups of Spheres: "this is equivalent to the existence of a dual basis of idempotent elements ${y}$ satisfying $y_i^2=y_i$ and $y_iy_j=0$ for $i neq j$."
answered Apr 18 at 22:23
John PalmieriJohn Palmieri
2,35011726
$endgroup$
$begingroup$
Is there a characteristic assumption on the last part? Is there division by 2 in formula for y?
$endgroup$
– AHusain
Apr 18 at 22:29
$begingroup$
Great! Also, I think Sweedler provided a fairly nice answer already, since if A comes from some subgroup, then A is automatically a sub-Hopf-algebra!
$endgroup$
– Student
Apr 19 at 0:03
$begingroup$
For Ravenel's statement, I do not have access to the book for now. Would you mind pointing it out that in his content, is $A$ assumed to be a Hopf algebra?
$endgroup$
– Student
Apr 19 at 0:04
2
$begingroup$
Ravenel's book is available from his web page: web.math.rochester.edu/people/faculty/doug/mybooks/ravenel.pdf. He is certainly working in positive characteristic, but I don't know if this is necessary for his condition. He is certainly working with Hopf algebras, since otherwise there wouldn't be a multiplication on the dual.
$endgroup$
– John Palmieri
Apr 19 at 5:28
2
$begingroup$
@JoshuaGrochow: Ravenel uses his version to show that a particular Hopf algebra is a group algebra, so he found it useful.
$endgroup$
– John Palmieri
Apr 19 at 5:29
|
show 1 more comment
$begingroup$
If $A$ is the group algebra of a subgroup, then $k[G]$ will be free as a module over $A$, and that may help to rule out some cases. This is not at all an "if and only if" statement, though.
Proposition 3.2.1 in Sweedler's book Hopf Algebras says that a cocommutative Hopf algebra is a group algebra if and only if it has a basis of group-like elements (those elements $x$ which satisfy $Delta(x) = x otimes x$). So if you have a sub-Hopf algebra, you can try to see if it has such a basis.
Ravenel says in Theorem 6.2.3 in Complex Cobordism and Stable Homotopy Groups of Spheres: "this is equivalent to the existence of a dual basis of idempotent elements ${y}$ satisfying $y_i^2=y_i$ and $y_iy_j=0$ for $i neq j$."
answered Apr 18 at 22:23
John PalmieriJohn Palmieri
2,35011726
$endgroup$
$begingroup$
Is there a characteristic assumption on the last part? Is there division by 2 in formula for y?
$endgroup$
– AHusain
Apr 18 at 22:29
$begingroup$
Great! Also, I think Sweedler provided a fairly nice answer already, since if A comes from some subgroup, then A is automatically a sub-Hopf-algebra!
$endgroup$
– Student
Apr 19 at 0:03
$begingroup$
For Ravenel's statement, I do not have access to the book for now. Would you mind pointing it out that in his content, is $A$ assumed to be a Hopf algebra?
$endgroup$
– Student
Apr 19 at 0:04
2
$begingroup$
Ravenel's book is available from his web page: web.math.rochester.edu/people/faculty/doug/mybooks/ravenel.pdf. He is certainly working in positive characteristic, but I don't know if this is necessary for his condition. He is certainly working with Hopf algebras, since otherwise there wouldn't be a multiplication on the dual.
$endgroup$
– John Palmieri
Apr 19 at 5:28
2
$begingroup$
@JoshuaGrochow: Ravenel uses his version to show that a particular Hopf algebra is a group algebra, so he found it useful.
$endgroup$
– John Palmieri
Apr 19 at 5:29
|
show 1 more comment
$begingroup$
If $A$ is the group algebra of a subgroup, then $k[G]$ will be free as a module over $A$, and that may help to rule out some cases. This is not at all an "if and only if" statement, though.
Proposition 3.2.1 in Sweedler's book Hopf Algebras says that a cocommutative Hopf algebra is a group algebra if and only if it has a basis of group-like elements (those elements $x$ which satisfy $Delta(x) = x otimes x$). So if you have a sub-Hopf algebra, you can try to see if it has such a basis.
Ravenel says in Theorem 6.2.3 in Complex Cobordism and Stable Homotopy Groups of Spheres: "this is equivalent to the existence of a dual basis of idempotent elements ${y}$ satisfying $y_i^2=y_i$ and $y_iy_j=0$ for $i neq j$."
answered Apr 18 at 22:23
John PalmieriJohn Palmieri
2,35011726
$endgroup$
If $A$ is the group algebra of a subgroup, then $k[G]$ will be free as a module over $A$, and that may help to rule out some cases. This is not at all an "if and only if" statement, though.
Proposition 3.2.1 in Sweedler's book Hopf Algebras says that a cocommutative Hopf algebra is a group algebra if and only if it has a basis of group-like elements (those elements $x$ which satisfy $Delta(x) = x otimes x$). So if you have a sub-Hopf algebra, you can try to see if it has such a basis.
Ravenel says in Theorem 6.2.3 in Complex Cobordism and Stable Homotopy Groups of Spheres: "this is equivalent to the existence of a dual basis of idempotent elements ${y}$ satisfying $y_i^2=y_i$ and $y_iy_j=0$ for $i neq j$."
answered Apr 18 at 22:23
John PalmieriJohn Palmieri
2,35011726
answered Apr 18 at 22:23
John PalmieriJohn Palmieri
2,35011726
answered Apr 18 at 22:23
John PalmieriJohn Palmieri
2,35011726
answered Apr 18 at 22:23
John PalmieriJohn Palmieri
2,35011726
2,35011726
$begingroup$
Is there a characteristic assumption on the last part? Is there division by 2 in formula for y?
$endgroup$
– AHusain
Apr 18 at 22:29
$begingroup$
Great! Also, I think Sweedler provided a fairly nice answer already, since if A comes from some subgroup, then A is automatically a sub-Hopf-algebra!
$endgroup$
– Student
Apr 19 at 0:03
$begingroup$
For Ravenel's statement, I do not have access to the book for now. Would you mind pointing it out that in his content, is $A$ assumed to be a Hopf algebra?
$endgroup$
– Student
Apr 19 at 0:04
2
$begingroup$
Ravenel's book is available from his web page: web.math.rochester.edu/people/faculty/doug/mybooks/ravenel.pdf. He is certainly working in positive characteristic, but I don't know if this is necessary for his condition. He is certainly working with Hopf algebras, since otherwise there wouldn't be a multiplication on the dual.
$endgroup$
– John Palmieri
Apr 19 at 5:28
2
$begingroup$
@JoshuaGrochow: Ravenel uses his version to show that a particular Hopf algebra is a group algebra, so he found it useful.
$endgroup$
– John Palmieri
Apr 19 at 5:29
|
show 1 more comment
$begingroup$
Is there a characteristic assumption on the last part? Is there division by 2 in formula for y?
$endgroup$
– AHusain
Apr 18 at 22:29
$begingroup$
Great! Also, I think Sweedler provided a fairly nice answer already, since if A comes from some subgroup, then A is automatically a sub-Hopf-algebra!
$endgroup$
– Student
Apr 19 at 0:03
$begingroup$
For Ravenel's statement, I do not have access to the book for now. Would you mind pointing it out that in his content, is $A$ assumed to be a Hopf algebra?
$endgroup$
– Student
Apr 19 at 0:04
2
$begingroup$
Ravenel's book is available from his web page: web.math.rochester.edu/people/faculty/doug/mybooks/ravenel.pdf. He is certainly working in positive characteristic, but I don't know if this is necessary for his condition. He is certainly working with Hopf algebras, since otherwise there wouldn't be a multiplication on the dual.
$endgroup$
– John Palmieri
Apr 19 at 5:28
2
$begingroup$
@JoshuaGrochow: Ravenel uses his version to show that a particular Hopf algebra is a group algebra, so he found it useful.
$endgroup$
– John Palmieri
Apr 19 at 5:29
$begingroup$
Is there a characteristic assumption on the last part? Is there division by 2 in formula for y?
$endgroup$
– AHusain
Apr 18 at 22:29
$begingroup$
Is there a characteristic assumption on the last part? Is there division by 2 in formula for y?
$endgroup$
– AHusain
Apr 18 at 22:29
$begingroup$
Great! Also, I think Sweedler provided a fairly nice answer already, since if A comes from some subgroup, then A is automatically a sub-Hopf-algebra!
$endgroup$
– Student
Apr 19 at 0:03
$begingroup$
Great! Also, I think Sweedler provided a fairly nice answer already, since if A comes from some subgroup, then A is automatically a sub-Hopf-algebra!
$endgroup$
– Student
Apr 19 at 0:03
$begingroup$
For Ravenel's statement, I do not have access to the book for now. Would you mind pointing it out that in his content, is $A$ assumed to be a Hopf algebra?
$endgroup$
– Student
Apr 19 at 0:04
$begingroup$
For Ravenel's statement, I do not have access to the book for now. Would you mind pointing it out that in his content, is $A$ assumed to be a Hopf algebra?
$endgroup$
– Student
Apr 19 at 0:04
2
2
$begingroup$
Ravenel's book is available from his web page: web.math.rochester.edu/people/faculty/doug/mybooks/ravenel.pdf. He is certainly working in positive characteristic, but I don't know if this is necessary for his condition. He is certainly working with Hopf algebras, since otherwise there wouldn't be a multiplication on the dual.
$endgroup$
– John Palmieri
Apr 19 at 5:28
$begingroup$
Ravenel's book is available from his web page: web.math.rochester.edu/people/faculty/doug/mybooks/ravenel.pdf. He is certainly working in positive characteristic, but I don't know if this is necessary for his condition. He is certainly working with Hopf algebras, since otherwise there wouldn't be a multiplication on the dual.
$endgroup$
– John Palmieri
Apr 19 at 5:28
2
2
$begingroup$
@JoshuaGrochow: Ravenel uses his version to show that a particular Hopf algebra is a group algebra, so he found it useful.
$endgroup$
– John Palmieri
Apr 19 at 5:29
$begingroup$
@JoshuaGrochow: Ravenel uses his version to show that a particular Hopf algebra is a group algebra, so he found it useful.
$endgroup$
– John Palmieri
Apr 19 at 5:29
|
show 1 more comment
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