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 $\mathrm{exp}(G)$.***



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> Is this a known result? Is there a reference to it?
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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 $|G|$.
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Hasn't this question been addressed before?


Both references and proofs are very much appreciated.



Which finite groups have no irreducible representations other than characters?
A group satisfies this property if and only if it is an extension of a $p$ group by an abelian group. The reason is the following: Assume that indeed all the irreducible representations of $G$ are one dimensional. This means that all elements of the form $x-1$ where $x$ belongs to $[G,G]$ are in the Jacobson radical, because they act trivially on all irreducible representations. But since this is a finite dimensional algebra over a field, the Jacobson radical is nilpotent. This means that the element $x-1$ is nilpotent for every $x\in [G,G]$, and this can only happen if the order of $x$ is a power of $p$. Therefore, $[G,G]$ is a $p$-group, and the quotient $G/[G,G]$ is abelian. In particular, this implies that the group is solvable.


