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Timothy Chow
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Which finite groups have no irreducible represntationsrepresentations other than characters?

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Pablo
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A classical result states that all the irreducible representations of a finite group over $\mathbb{C}$ are characters if and only if $G$ is abelian. I would like to know what happens if we consider a field different from $\mathbb{C}$. Clearly as long as the field is algebraically closed and its characteristic does not divide $|G|$ the same conclusion holds.

Furthermore, I have proved the following claim:

Let $G$ be a finite group, and let $F$ be a field of characteristic not dividing $|G|$. Then all the irreducible representations of $G$ over $F$ are $1$-dimensional if and only if $G$ is an abelian group and $F$ has a primitive root of unity of order $\mathrm{exp}(G)$.

Is this a knowknown result? Is there a reference to it?

However, I am not sure what happens if the characteristic of $F$ divides $|G|$. For example, $S_3$ is a nonabelian group with all irreducible representations over $\mathbb{F}_3$ $1$-dimensional.

One verifies that all the irreducible representations are characters if and only if the image of $G$ in $\mathrm{GL}_{|G|}(F)$ under the representation coming from the action of $G$ on itself can be simultaneously triangulated. This implies that again $F$ must have a primitive root of certain orders, and that $G$ must be solvable to say the least (it is embedded in the subgroup of upper triangular invertible matrices).

In spite of that, I do not have a converse.

Is there a characterization (similar to those given above) of the cases in which a finite group has only 1-dimensional irreducible representations over a field $F$ of characteristic dividing $|G|$.

Hasn't this question been addressed before?

Both references and proofs are very much appreciated.

A classical result states that all the irreducible representations of a finite group over $\mathbb{C}$ are characters if and only if $G$ is abelian. I would like to know what happens if we consider a field different from $\mathbb{C}$. Clearly as long as the field is algebraically closed and its characteristic does not divide $|G|$ the same conclusion holds.

Furthermore, I have proved the following claim:

Let $G$ be a finite group, and let $F$ be a field of characteristic not dividing $|G|$. Then all the irreducible representations of $G$ over $F$ are $1$-dimensional if and only if $G$ is an abelian group and $F$ has a primitive root of unity of order $\mathrm{exp}(G)$.

Is this a know result? Is there a reference to it?

However, I am not sure what happens if the characteristic of $F$ divides $|G|$. For example, $S_3$ is a nonabelian group with all irreducible representations over $\mathbb{F}_3$ $1$-dimensional.

One verifies that all the irreducible representations are characters if and only if the image of $G$ in $\mathrm{GL}_{|G|}(F)$ under the representation coming from the action of $G$ on itself can be simultaneously triangulated. This implies that again $F$ must have a primitive root of certain orders, and that $G$ must be solvable to say the least (it is embedded in the subgroup of upper triangular invertible matrices).

In spite of that, I do not have a converse.

Is there a characterization (similar to those given above) of the cases in which a finite group has only 1-dimensional irreducible representations over a field $F$ of characteristic dividing $|G|$.

Hasn't this question been addressed before?

Both references and proofs are very much appreciated.

A classical result states that all the irreducible representations of a finite group over $\mathbb{C}$ are characters if and only if $G$ is abelian. I would like to know what happens if we consider a field different from $\mathbb{C}$. Clearly as long as the field is algebraically closed and its characteristic does not divide $|G|$ the same conclusion holds.

Furthermore, I have proved the following claim:

Let $G$ be a finite group, and let $F$ be a field of characteristic not dividing $|G|$. Then all the irreducible representations of $G$ over $F$ are $1$-dimensional if and only if $G$ is an abelian group and $F$ has a primitive root of unity of order $\mathrm{exp}(G)$.

Is this a known result? Is there a reference to it?

However, I am not sure what happens if the characteristic of $F$ divides $|G|$. For example, $S_3$ is a nonabelian group with all irreducible representations over $\mathbb{F}_3$ $1$-dimensional.

One verifies that all the irreducible representations are characters if and only if the image of $G$ in $\mathrm{GL}_{|G|}(F)$ under the representation coming from the action of $G$ on itself can be simultaneously triangulated. This implies that again $F$ must have a primitive root of certain orders, and that $G$ must be solvable to say the least (it is embedded in the subgroup of upper triangular invertible matrices).

In spite of that, I do not have a converse.

Is there a characterization (similar to those given above) of the cases in which a finite group has only 1-dimensional irreducible representations over a field $F$ of characteristic dividing $|G|$.

Hasn't this question been addressed before?

Both references and proofs are very much appreciated.

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Pablo
  • 11.3k
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  • 68

A classical result states that all the irreducible representations of a finite group over $\mathbb{C}$ are characters if and only if $G$ is abelian. I would like to know what happens if we consider a field different from $\mathbb{C}$. Clearly as long as the field is algebraically closed and its characteristic does not divide $|G|$ the same conclusion holds.

Furthermore, I have proved the following claim:

Let $G$ be a finite group, and let $F$ be a field of characteristic not dividing $|G|$. Then all the irreducible representations of $G$ over $F$ are $1$-dimensional if and only if $G$ is an abelian group and $F$ has a primitive root of unity of order $\mathrm{exp}(G)$.

Is this a know result? Is there a reference to it?

However, I am not sure what happens if the characteristic of $F$ divides $|G|$. For example, $S_3$ is a nonabelian group with all irreducible representations over $\mathbb{F}_3$ $1$-dimensional.

One verifies that all the irreducible representations are characters if and only if the image of $G$ in $\mathrm{GL}_{|G|}(F)$ under the representation coming from the action of $G$ on itself can be simultaneously triangulated. This implies that again $F$ must have a primitive root of certain orders, and that $G$ must be solvable to say the least (it is embedded in the subgroup of upper triangular invertible matrices).

In spite of that, I do not have a converse.

Is there a characterization (similar to those given above) of the cases in which a finite group has only 1-dimensional irreducible representations over a field $F$ of characteristic dividing $|G|$.

Hasn't this question been addressed before?

Both references and proofs are very much appreciated.

A classical result states that all the irreducible representations of a finite group over $\mathbb{C}$ are characters if and only if $G$ is abelian. I would like to know what happens if we consider a field different from $\mathbb{C}$. Clearly as long as the field is algebraically closed and its characteristic does not divide $|G|$ the same conclusion holds.

Furthermore, I have proved the following claim:

Let $G$ be a finite group, and let $F$ be a field of characteristic not dividing $|G|$. Then all the irreducible representations of $G$ over $F$ are $1$-dimensional if and only if $G$ is an abelian group and $F$ has a primitive root of unity of order $\mathrm{exp}(G)$.

Is this a know result? Is there a reference to it?

However, I am not sure what happens if the characteristic of $F$ divides $|G|$. One verifies that all the irreducible representations are characters if and only if the image of $G$ in $\mathrm{GL}_{|G|}(F)$ under the representation coming from the action of $G$ on itself can be simultaneously triangulated. This implies that again $F$ must have a primitive root of certain orders, and that $G$ must be solvable to say the least (it is embedded in the subgroup of upper triangular invertible matrices).

In spite of that, I do not have a converse.

Is there a characterization (similar to those given above) of the cases in which a finite group has only 1-dimensional irreducible representations over a field $F$ of characteristic dividing $|G|$.

Hasn't this question been addressed before?

Both references and proofs are very much appreciated.

A classical result states that all the irreducible representations of a finite group over $\mathbb{C}$ are characters if and only if $G$ is abelian. I would like to know what happens if we consider a field different from $\mathbb{C}$. Clearly as long as the field is algebraically closed and its characteristic does not divide $|G|$ the same conclusion holds.

Furthermore, I have proved the following claim:

Let $G$ be a finite group, and let $F$ be a field of characteristic not dividing $|G|$. Then all the irreducible representations of $G$ over $F$ are $1$-dimensional if and only if $G$ is an abelian group and $F$ has a primitive root of unity of order $\mathrm{exp}(G)$.

Is this a know result? Is there a reference to it?

However, I am not sure what happens if the characteristic of $F$ divides $|G|$. For example, $S_3$ is a nonabelian group with all irreducible representations over $\mathbb{F}_3$ $1$-dimensional.

One verifies that all the irreducible representations are characters if and only if the image of $G$ in $\mathrm{GL}_{|G|}(F)$ under the representation coming from the action of $G$ on itself can be simultaneously triangulated. This implies that again $F$ must have a primitive root of certain orders, and that $G$ must be solvable to say the least (it is embedded in the subgroup of upper triangular invertible matrices).

In spite of that, I do not have a converse.

Is there a characterization (similar to those given above) of the cases in which a finite group has only 1-dimensional irreducible representations over a field $F$ of characteristic dividing $|G|$.

Hasn't this question been addressed before?

Both references and proofs are very much appreciated.

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Pablo
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