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Let $k$ be a field of characteristic $p$ and $G$ a finite group. This question might be a dulicate of this question:

Which finite groups have no irreducible representations other than characters?

but I think it is still something that remained to explain.

If $G$ has all irreducible representations one dimensional then $G$ is an extension of a $p$-group (namely $G'$) by an abelian group as Ehud proved in his answer.

It still remains to show that this extension splits in order to deduce that $G$ is a semidirect product. Could you please explain why this extension should split?

I couldn't understand Pablo's the argument with the triangularizable matrices.

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    $\begingroup$ Read about the Schur-Zassenhaus theorem, which says that if $G$ is a finite group and $N \lhd G$ with ${\rm gcd}(|N|,[G:N]) =1,$ then $N$ is complemented in $G$. This was originally proved in the case that either $N$ or $G/N$ is solvable, which suffices for your situation ( given the Feit-Thompson theorem, it automatically follows that one of $N$ or $G/N$ is solvable when they have coprime orders, but that is not needed for your problem). $\endgroup$
    – Geoff Robinson
    Commented Jan 23, 2017 at 14:27
  • $\begingroup$ @ Geoff Robinson I know that Schur-Zassenhaus implies that N is complemented but I don't understand why in our case the commutator subgroup $G'$ should have order coprime to the order of $G/G'$. $\endgroup$
    – user103474
    Commented Jan 23, 2017 at 14:42
  • $\begingroup$ If it doesn't then maybe just replace $G'$ by the subgroup of $G$ corresponding to the Sylow $p$-subgroup of $G/G'$. $\endgroup$
    – Kevin Buzzard
    Commented Jan 23, 2017 at 15:09
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    $\begingroup$ It is the case that if all absolutely irreducible representations of $G$ are one-dimensional, then $G/O_{p}(G)$ is an Abelian group of order prime to $p$, and Schur-Zassenhaus may indeed be applied directly. To see this, note that $O_{p}(G)$ is in the kernel of all absolutely irreducible characteristic $p$ representations while on the other, if $K$ s the intrestection of the kernel of all such irreducible representations, then $K$ is a $p$-group and $G/K$ is Abelian, so $G$ has a normal Sylow $p$-subgroup, which is then contained in $K$ as it is a normal $p$-subgroup. $\endgroup$
    – Geoff Robinson
    Commented Jan 23, 2017 at 17:04
  • $\begingroup$ @Geoff Robinson Ok, now everything is clear. Thank you very much! $\endgroup$
    – user103474
    Commented Jan 23, 2017 at 17:46

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I decided to answer myself the question after Geoff Robinson's comments.

From Ehud's answer of the previous question it follows that $$ 1 \rightarrow G'\rightarrow G \rightarrow G/G'\rightarrow 1 $$ is an exact sequence where $G'$ is a p-group and $G/G'$ is abelian. On the other hand, if $G/K$ is abelian and $K$ is a p-group then it can been easily verified that a p-Sylow subgroup of $G$ containing $K$ is normal.

Thus as Kevin Buzzard suggests choosing a p-Sylow subgroup of $G$ containig $G'$ it follows that

$$ 1 \rightarrow P \rightarrow G \rightarrow G/P \rightarrow 1 $$

is an exact sequence where $P$ is a $p$-group and $G/P$ is abelian since $P$ contains the commutator subgroup $G'$.

Thus one can conclude that over any field of characteristic $p$ a finite group has all its irreducible representations one dimensional if and only if $G$ has a normal $p$-Sylow subgroup with abelian quotient.

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