TITLE: Finite groups with all irreducible representations one dimensional
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: 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.