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Can a finite group have 2D and 3D faithful irreducible representations?
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I am looking for finite groups that have a 2D (complex matrix) faithful irreducible representation and a 3D faithful irreducible representation. Up to order 1023 GAP found none. Other combinations of dimensions seem to occur. Are groups with 2D and 3D faithful irreps non-existing?
group-theory representation-theory
asked Jan 13 at 9:07
Joris VergeestJoris Vergeest
132
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I am looking for finite groups that have a 2D (complex matrix) faithful irreducible representation and a 3D faithful irreducible representation. Up to order 1023 GAP found none. Other combinations of dimensions seem to occur. Are groups with 2D and 3D faithful irreps non-existing?
group-theory representation-theory
asked Jan 13 at 9:07
Joris VergeestJoris Vergeest
132
$endgroup$
add a comment |
$begingroup$
I am looking for finite groups that have a 2D (complex matrix) faithful irreducible representation and a 3D faithful irreducible representation. Up to order 1023 GAP found none. Other combinations of dimensions seem to occur. Are groups with 2D and 3D faithful irreps non-existing?
group-theory representation-theory
asked Jan 13 at 9:07
Joris VergeestJoris Vergeest
132
$endgroup$
I am looking for finite groups that have a 2D (complex matrix) faithful irreducible representation and a 3D faithful irreducible representation. Up to order 1023 GAP found none. Other combinations of dimensions seem to occur. Are groups with 2D and 3D faithful irreps non-existing?
group-theory representation-theory
group-theory representation-theory
asked Jan 13 at 9:07
Joris VergeestJoris Vergeest
132
asked Jan 13 at 9:07
Joris VergeestJoris Vergeest
132
asked Jan 13 at 9:07
Joris VergeestJoris Vergeest
132
asked Jan 13 at 9:07
Joris VergeestJoris Vergeest
132
asked Jan 13 at 9:07
Joris VergeestJoris Vergeest
132
132
add a comment |
add a comment |
1 Answer
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Here is a proof that no finite group has faithful irreducible representations of degrees $2$ and $3$. It might be easier to do this using the classification, of finite subgroups of ${rm G}(2,{mathbb C})$, but I am more familiar with the subgroups of ${rm GL}(2,q)$ for finite $q$, so I will use that approach.
Let $G$ be a finite irreducible subgroup of ${rm GL}(2,{mathbb C})$. Let $p$ be a prime not dividing $|G|$. Then, by standard results in representation theory, the associated complex representation can be written over the finite field of order $q$ for some power $q$ of $p$.
The irreducible subgroups of ${rm GL}(2,q)$ are either imprimitive or semilinear, or they have normal subgroups isomorphic to ${rm SL}(2,3)$ or ${rm SL}(2,5)$.
The imprimitive and semilinear groups have a normal abelian subgroup of index $2$, and since all complex irreducible representations of abelian groups have degree $1$, the irreducible representations of $G$ have degree at most $2$.
On the other hand, ${rm SL}(2,3)$ or ${rm SL}(2,5)$ do not have faithful irreducible representations of degree $3$.
answered Jan 13 at 12:32
Derek HoltDerek Holt
53.6k53571
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1 Answer
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1 Answer
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$begingroup$
Here is a proof that no finite group has faithful irreducible representations of degrees $2$ and $3$. It might be easier to do this using the classification, of finite subgroups of ${rm G}(2,{mathbb C})$, but I am more familiar with the subgroups of ${rm GL}(2,q)$ for finite $q$, so I will use that approach.
Let $G$ be a finite irreducible subgroup of ${rm GL}(2,{mathbb C})$. Let $p$ be a prime not dividing $|G|$. Then, by standard results in representation theory, the associated complex representation can be written over the finite field of order $q$ for some power $q$ of $p$.
The irreducible subgroups of ${rm GL}(2,q)$ are either imprimitive or semilinear, or they have normal subgroups isomorphic to ${rm SL}(2,3)$ or ${rm SL}(2,5)$.
The imprimitive and semilinear groups have a normal abelian subgroup of index $2$, and since all complex irreducible representations of abelian groups have degree $1$, the irreducible representations of $G$ have degree at most $2$.
On the other hand, ${rm SL}(2,3)$ or ${rm SL}(2,5)$ do not have faithful irreducible representations of degree $3$.
answered Jan 13 at 12:32
Derek HoltDerek Holt
53.6k53571
$endgroup$
add a comment |
$begingroup$
Here is a proof that no finite group has faithful irreducible representations of degrees $2$ and $3$. It might be easier to do this using the classification, of finite subgroups of ${rm G}(2,{mathbb C})$, but I am more familiar with the subgroups of ${rm GL}(2,q)$ for finite $q$, so I will use that approach.
Let $G$ be a finite irreducible subgroup of ${rm GL}(2,{mathbb C})$. Let $p$ be a prime not dividing $|G|$. Then, by standard results in representation theory, the associated complex representation can be written over the finite field of order $q$ for some power $q$ of $p$.
The irreducible subgroups of ${rm GL}(2,q)$ are either imprimitive or semilinear, or they have normal subgroups isomorphic to ${rm SL}(2,3)$ or ${rm SL}(2,5)$.
The imprimitive and semilinear groups have a normal abelian subgroup of index $2$, and since all complex irreducible representations of abelian groups have degree $1$, the irreducible representations of $G$ have degree at most $2$.
On the other hand, ${rm SL}(2,3)$ or ${rm SL}(2,5)$ do not have faithful irreducible representations of degree $3$.
answered Jan 13 at 12:32
Derek HoltDerek Holt
53.6k53571
$endgroup$
add a comment |
$begingroup$
Here is a proof that no finite group has faithful irreducible representations of degrees $2$ and $3$. It might be easier to do this using the classification, of finite subgroups of ${rm G}(2,{mathbb C})$, but I am more familiar with the subgroups of ${rm GL}(2,q)$ for finite $q$, so I will use that approach.
Let $G$ be a finite irreducible subgroup of ${rm GL}(2,{mathbb C})$. Let $p$ be a prime not dividing $|G|$. Then, by standard results in representation theory, the associated complex representation can be written over the finite field of order $q$ for some power $q$ of $p$.
The irreducible subgroups of ${rm GL}(2,q)$ are either imprimitive or semilinear, or they have normal subgroups isomorphic to ${rm SL}(2,3)$ or ${rm SL}(2,5)$.
The imprimitive and semilinear groups have a normal abelian subgroup of index $2$, and since all complex irreducible representations of abelian groups have degree $1$, the irreducible representations of $G$ have degree at most $2$.
On the other hand, ${rm SL}(2,3)$ or ${rm SL}(2,5)$ do not have faithful irreducible representations of degree $3$.
answered Jan 13 at 12:32
Derek HoltDerek Holt
53.6k53571
$endgroup$
Here is a proof that no finite group has faithful irreducible representations of degrees $2$ and $3$. It might be easier to do this using the classification, of finite subgroups of ${rm G}(2,{mathbb C})$, but I am more familiar with the subgroups of ${rm GL}(2,q)$ for finite $q$, so I will use that approach.
Let $G$ be a finite irreducible subgroup of ${rm GL}(2,{mathbb C})$. Let $p$ be a prime not dividing $|G|$. Then, by standard results in representation theory, the associated complex representation can be written over the finite field of order $q$ for some power $q$ of $p$.
The irreducible subgroups of ${rm GL}(2,q)$ are either imprimitive or semilinear, or they have normal subgroups isomorphic to ${rm SL}(2,3)$ or ${rm SL}(2,5)$.
The imprimitive and semilinear groups have a normal abelian subgroup of index $2$, and since all complex irreducible representations of abelian groups have degree $1$, the irreducible representations of $G$ have degree at most $2$.
On the other hand, ${rm SL}(2,3)$ or ${rm SL}(2,5)$ do not have faithful irreducible representations of degree $3$.
answered Jan 13 at 12:32
Derek HoltDerek Holt
53.6k53571
answered Jan 13 at 12:32
Derek HoltDerek Holt
53.6k53571
answered Jan 13 at 12:32
Derek HoltDerek Holt
53.6k53571
answered Jan 13 at 12:32
Derek HoltDerek Holt
53.6k53571
53.6k53571
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add a comment |
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