Mixing
One dimensional representations of a simple group
$begingroup$
I keep seeing this following fact stated;
If $G$ is a simple group, then the only $1$ dimensional representation of $G$ is the trivial representation.
But I’ve not seen a proof and I can’t seem to find one online.
Could anyone please show me why this is true or provide me a reference to a proof?
Thank you!
abstract-algebra group-theory representation-theory
asked Jan 9 at 17:31
the manthe man
726715
$endgroup$
add a comment |
$begingroup$
I keep seeing this following fact stated;
If $G$ is a simple group, then the only $1$ dimensional representation of $G$ is the trivial representation.
But I’ve not seen a proof and I can’t seem to find one online.
Could anyone please show me why this is true or provide me a reference to a proof?
Thank you!
abstract-algebra group-theory representation-theory
asked Jan 9 at 17:31
the manthe man
726715
$endgroup$
2
$begingroup$
The simple group $Bbb Z/pBbb Z$ has $p$ nonisomorphic one-dimensional representations.
$endgroup$
– Lord Shark the Unknown
Jan 9 at 17:35
add a comment |
$begingroup$
I keep seeing this following fact stated;
If $G$ is a simple group, then the only $1$ dimensional representation of $G$ is the trivial representation.
But I’ve not seen a proof and I can’t seem to find one online.
Could anyone please show me why this is true or provide me a reference to a proof?
Thank you!
abstract-algebra group-theory representation-theory
asked Jan 9 at 17:31
the manthe man
726715
$endgroup$
I keep seeing this following fact stated;
If $G$ is a simple group, then the only $1$ dimensional representation of $G$ is the trivial representation.
But I’ve not seen a proof and I can’t seem to find one online.
Could anyone please show me why this is true or provide me a reference to a proof?
Thank you!
abstract-algebra group-theory representation-theory
abstract-algebra group-theory representation-theory
asked Jan 9 at 17:31
the manthe man
726715
asked Jan 9 at 17:31
the manthe man
726715
asked Jan 9 at 17:31
the manthe man
726715
asked Jan 9 at 17:31
the manthe man
726715
asked Jan 9 at 17:31
the manthe man
726715
726715
2
$begingroup$
The simple group $Bbb Z/pBbb Z$ has $p$ nonisomorphic one-dimensional representations.
$endgroup$
– Lord Shark the Unknown
Jan 9 at 17:35
add a comment |
2
$begingroup$
The simple group $Bbb Z/pBbb Z$ has $p$ nonisomorphic one-dimensional representations.
$endgroup$
– Lord Shark the Unknown
Jan 9 at 17:35
2
2
$begingroup$
The simple group $Bbb Z/pBbb Z$ has $p$ nonisomorphic one-dimensional representations.
$endgroup$
– Lord Shark the Unknown
Jan 9 at 17:35
$begingroup$
The simple group $Bbb Z/pBbb Z$ has $p$ nonisomorphic one-dimensional representations.
$endgroup$
– Lord Shark the Unknown
Jan 9 at 17:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A 1-dimensional representation of $G$ is a morphism from $G$ to the abelian group $mathbb C^*$; hence, it must factor through the quotient $G/[G;G]$ by the derived subgroup. Because $G$ is simple, the derived subgroup $[G,G]$ is either $G$ (so the morphism is trivial) or $(0)$ (so the group is simple abelian, and $G=mathbb Z/pmathbb Z$).
As Lord Shark the Unknown stated in the comment, this is all one can ask for, since $mathbb Z/pmathbb Z$ admits non trivial 1-dimensional representations.
answered Jan 9 at 17:40
Pierre PCPierre PC
267
$endgroup$
$begingroup$
Also worth noting that this works for all fields, not just the complex numbers.
$endgroup$
– Tobias Kildetoft
Jan 9 at 18:31
$begingroup$
Also also worth noting that if $G$ is non-abelian, then the statement is true.
$endgroup$
– David Hill
Jan 10 at 18:28
add a comment |
Your Answer
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
A 1-dimensional representation of $G$ is a morphism from $G$ to the abelian group $mathbb C^*$; hence, it must factor through the quotient $G/[G;G]$ by the derived subgroup. Because $G$ is simple, the derived subgroup $[G,G]$ is either $G$ (so the morphism is trivial) or $(0)$ (so the group is simple abelian, and $G=mathbb Z/pmathbb Z$).
As Lord Shark the Unknown stated in the comment, this is all one can ask for, since $mathbb Z/pmathbb Z$ admits non trivial 1-dimensional representations.
answered Jan 9 at 17:40
Pierre PCPierre PC
267
$endgroup$
$begingroup$
Also worth noting that this works for all fields, not just the complex numbers.
$endgroup$
– Tobias Kildetoft
Jan 9 at 18:31
$begingroup$
Also also worth noting that if $G$ is non-abelian, then the statement is true.
$endgroup$
– David Hill
Jan 10 at 18:28
add a comment |
$begingroup$
A 1-dimensional representation of $G$ is a morphism from $G$ to the abelian group $mathbb C^*$; hence, it must factor through the quotient $G/[G;G]$ by the derived subgroup. Because $G$ is simple, the derived subgroup $[G,G]$ is either $G$ (so the morphism is trivial) or $(0)$ (so the group is simple abelian, and $G=mathbb Z/pmathbb Z$).
As Lord Shark the Unknown stated in the comment, this is all one can ask for, since $mathbb Z/pmathbb Z$ admits non trivial 1-dimensional representations.
answered Jan 9 at 17:40
Pierre PCPierre PC
267
$endgroup$
$begingroup$
Also worth noting that this works for all fields, not just the complex numbers.
$endgroup$
– Tobias Kildetoft
Jan 9 at 18:31
$begingroup$
Also also worth noting that if $G$ is non-abelian, then the statement is true.
$endgroup$
– David Hill
Jan 10 at 18:28
add a comment |
$begingroup$
A 1-dimensional representation of $G$ is a morphism from $G$ to the abelian group $mathbb C^*$; hence, it must factor through the quotient $G/[G;G]$ by the derived subgroup. Because $G$ is simple, the derived subgroup $[G,G]$ is either $G$ (so the morphism is trivial) or $(0)$ (so the group is simple abelian, and $G=mathbb Z/pmathbb Z$).
As Lord Shark the Unknown stated in the comment, this is all one can ask for, since $mathbb Z/pmathbb Z$ admits non trivial 1-dimensional representations.
answered Jan 9 at 17:40
Pierre PCPierre PC
267
$endgroup$
A 1-dimensional representation of $G$ is a morphism from $G$ to the abelian group $mathbb C^*$; hence, it must factor through the quotient $G/[G;G]$ by the derived subgroup. Because $G$ is simple, the derived subgroup $[G,G]$ is either $G$ (so the morphism is trivial) or $(0)$ (so the group is simple abelian, and $G=mathbb Z/pmathbb Z$).
As Lord Shark the Unknown stated in the comment, this is all one can ask for, since $mathbb Z/pmathbb Z$ admits non trivial 1-dimensional representations.
answered Jan 9 at 17:40
Pierre PCPierre PC
267
edited Jan 9 at 17:49
answered Jan 9 at 17:40
Pierre PCPierre PC
267
answered Jan 9 at 17:40
Pierre PCPierre PC
267
answered Jan 9 at 17:40
Pierre PCPierre PC
267
267
$begingroup$
Also worth noting that this works for all fields, not just the complex numbers.
$endgroup$
– Tobias Kildetoft
Jan 9 at 18:31
$begingroup$
Also also worth noting that if $G$ is non-abelian, then the statement is true.
$endgroup$
– David Hill
Jan 10 at 18:28
add a comment |
$begingroup$
Also worth noting that this works for all fields, not just the complex numbers.
$endgroup$
– Tobias Kildetoft
Jan 9 at 18:31
$begingroup$
Also also worth noting that if $G$ is non-abelian, then the statement is true.
$endgroup$
– David Hill
Jan 10 at 18:28
$begingroup$
Also worth noting that this works for all fields, not just the complex numbers.
$endgroup$
– Tobias Kildetoft
Jan 9 at 18:31
$begingroup$
Also worth noting that this works for all fields, not just the complex numbers.
$endgroup$
– Tobias Kildetoft
Jan 9 at 18:31
$begingroup$
Also also worth noting that if $G$ is non-abelian, then the statement is true.
$endgroup$
– David Hill
Jan 10 at 18:28
$begingroup$
Also also worth noting that if $G$ is non-abelian, then the statement is true.
$endgroup$
– David Hill
Jan 10 at 18:28
add a comment |
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$begingroup$
The simple group $Bbb Z/pBbb Z$ has $p$ nonisomorphic one-dimensional representations.
$endgroup$
– Lord Shark the Unknown
Jan 9 at 17:35