Relation between the order of an element of a group and their character in a simple group
$begingroup$
Let $chi$ be the representation of a finite group $G$. Let $g in G$ be an element of order 2. If $G$ is a simple group but not cyclic of order 2, prove that $chi(g) equiv chi(1) mod 4$.
Proof that $chi(g) equiv chi(1) mod 2$ in any finite group can be found here:
Relation between the order of an element of a group and their character
but I'm having trouble connecting the two. Any help would be appreciated.
group-theory representation-theory characters simple-groups
asked Jan 25 at 11:08
OliverOliver
284
$endgroup$
add a comment |
$begingroup$
Let $chi$ be the representation of a finite group $G$. Let $g in G$ be an element of order 2. If $G$ is a simple group but not cyclic of order 2, prove that $chi(g) equiv chi(1) mod 4$.
Proof that $chi(g) equiv chi(1) mod 2$ in any finite group can be found here:
Relation between the order of an element of a group and their character
but I'm having trouble connecting the two. Any help would be appreciated.
group-theory representation-theory characters simple-groups
asked Jan 25 at 11:08
OliverOliver
284
$endgroup$
add a comment |
$begingroup$
Let $chi$ be the representation of a finite group $G$. Let $g in G$ be an element of order 2. If $G$ is a simple group but not cyclic of order 2, prove that $chi(g) equiv chi(1) mod 4$.
Proof that $chi(g) equiv chi(1) mod 2$ in any finite group can be found here:
Relation between the order of an element of a group and their character
but I'm having trouble connecting the two. Any help would be appreciated.
group-theory representation-theory characters simple-groups
asked Jan 25 at 11:08
OliverOliver
284
$endgroup$
Let $chi$ be the representation of a finite group $G$. Let $g in G$ be an element of order 2. If $G$ is a simple group but not cyclic of order 2, prove that $chi(g) equiv chi(1) mod 4$.
Proof that $chi(g) equiv chi(1) mod 2$ in any finite group can be found here:
Relation between the order of an element of a group and their character
but I'm having trouble connecting the two. Any help would be appreciated.
group-theory representation-theory characters simple-groups
group-theory representation-theory characters simple-groups
asked Jan 25 at 11:08
OliverOliver
284
asked Jan 25 at 11:08
OliverOliver
284
asked Jan 25 at 11:08
OliverOliver
284
asked Jan 25 at 11:08
OliverOliver
284
asked Jan 25 at 11:08
OliverOliver
284
284
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Let $rho$ be a representation affording $chi$. Then the eigenvalues of $rho(g)$ are $pm 1$. Suppose $m$ of them are $1$ and $n$ are $-1$. So $chi(g) = m-n$ and $chi(1) = m+n$.
If $chi(g) equiv 2 bmod chi(1)$, then $n$ is odd, and so $det(rho(g)) = -1$. But since the map $tau:G to {mathbb C}^*$ defined by $tau(g) = det(chi(g))$ is a group homomorphism with image a finite cyclic group, this would imply that $G$ had a normal subgroup of index $2$, contrary to hypothesis.
answered Jan 25 at 11:37
Derek HoltDerek Holt
54.2k53571
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add a comment |
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$begingroup$
Let $rho$ be a representation affording $chi$. Then the eigenvalues of $rho(g)$ are $pm 1$. Suppose $m$ of them are $1$ and $n$ are $-1$. So $chi(g) = m-n$ and $chi(1) = m+n$.
If $chi(g) equiv 2 bmod chi(1)$, then $n$ is odd, and so $det(rho(g)) = -1$. But since the map $tau:G to {mathbb C}^*$ defined by $tau(g) = det(chi(g))$ is a group homomorphism with image a finite cyclic group, this would imply that $G$ had a normal subgroup of index $2$, contrary to hypothesis.
answered Jan 25 at 11:37
Derek HoltDerek Holt
54.2k53571
$endgroup$
add a comment |
$begingroup$
Let $rho$ be a representation affording $chi$. Then the eigenvalues of $rho(g)$ are $pm 1$. Suppose $m$ of them are $1$ and $n$ are $-1$. So $chi(g) = m-n$ and $chi(1) = m+n$.
If $chi(g) equiv 2 bmod chi(1)$, then $n$ is odd, and so $det(rho(g)) = -1$. But since the map $tau:G to {mathbb C}^*$ defined by $tau(g) = det(chi(g))$ is a group homomorphism with image a finite cyclic group, this would imply that $G$ had a normal subgroup of index $2$, contrary to hypothesis.
answered Jan 25 at 11:37
Derek HoltDerek Holt
54.2k53571
$endgroup$
add a comment |
$begingroup$
Let $rho$ be a representation affording $chi$. Then the eigenvalues of $rho(g)$ are $pm 1$. Suppose $m$ of them are $1$ and $n$ are $-1$. So $chi(g) = m-n$ and $chi(1) = m+n$.
If $chi(g) equiv 2 bmod chi(1)$, then $n$ is odd, and so $det(rho(g)) = -1$. But since the map $tau:G to {mathbb C}^*$ defined by $tau(g) = det(chi(g))$ is a group homomorphism with image a finite cyclic group, this would imply that $G$ had a normal subgroup of index $2$, contrary to hypothesis.
answered Jan 25 at 11:37
Derek HoltDerek Holt
54.2k53571
$endgroup$
Let $rho$ be a representation affording $chi$. Then the eigenvalues of $rho(g)$ are $pm 1$. Suppose $m$ of them are $1$ and $n$ are $-1$. So $chi(g) = m-n$ and $chi(1) = m+n$.
If $chi(g) equiv 2 bmod chi(1)$, then $n$ is odd, and so $det(rho(g)) = -1$. But since the map $tau:G to {mathbb C}^*$ defined by $tau(g) = det(chi(g))$ is a group homomorphism with image a finite cyclic group, this would imply that $G$ had a normal subgroup of index $2$, contrary to hypothesis.
answered Jan 25 at 11:37
Derek HoltDerek Holt
54.2k53571
answered Jan 25 at 11:37
Derek HoltDerek Holt
54.2k53571
answered Jan 25 at 11:37
Derek HoltDerek Holt
54.2k53571
answered Jan 25 at 11:37
Derek HoltDerek Holt
54.2k53571
54.2k53571
add a comment |
add a comment |
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