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.

  • $\begingroup$ If the characteristic of the field divides the order of the group then it is a completely different game. It is no longer true that every representations splits as the direct sum of irreducible representations. For example, if $G$ is a finite $p$ group, and the field $k$ has characteristic $p$, then $kG$ only has one irreducible representation, namely $k$ itself with the trivial action. $\endgroup$
    – Ehud Meir
    Aug 4, 2015 at 12:42
  • $\begingroup$ @EhudMeir Dear Udi, I already know these things, and it is this different game that I want to play. Please note that your remark on trviality (which has an easy proof using orbits and actions) is a special case of my explanations above: the upper triangular matrices with $1$'s on the main diagonal form a $p$-Sylow subgroup of the general linear group, so every $p$-group is conjugate to a subgroup of it, and thus every representation of a $p$-group is simultaneously triangulizable showing that all the Jordan-Holder constituents are trivial as you claimed. $\endgroup$
    – Pablo
    Aug 4, 2015 at 12:52
  • $\begingroup$ Sorry for that. What I can say is that your example of $S_3$ generalizes: if you have a finite group $G$ which is a semidirect product of an abelian group $A$ with a $p$-group $P$ (where $A$ acts on $P$), and the ground field contains enough roots of unity ($exp(A)$ roots of unity) then all the irreducible representations of $G$ will be one dimensional. $\endgroup$
    – Ehud Meir
    Aug 4, 2015 at 13:00
  • $\begingroup$ @EhudMeir Does this follow from proposition 25 on page 62 subsection 8.2 (semidrect products by an abelian group) of Serre's Linear representations of finite groups? If you can prove your claim then we have an answer to the question. $\endgroup$
    – Pablo
    Aug 4, 2015 at 13:09
  • $\begingroup$ I do not have the book near me so it is hard to answer. However, I posted an answer. $\endgroup$
    – Ehud Meir
    Aug 4, 2015 at 13:17

1 Answer 1


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.

  • $\begingroup$ The fact that if all irreducible representations are $1$-dimensional then the group is a semidirect product of a $p$-group by an abelian group acting on it follows almost immediately from my observations in the question since simultaneous triangulation implies that the group is a subgroup of such a semidirect product (this is a feature of the subgroup of upper triangular matrices). What I am asking for is a converse - if a group is of this form, why should all the irreducible representations be characters. $\endgroup$
    – Pablo
    Aug 4, 2015 at 13:28
  • 2
    $\begingroup$ Sorry for the misunderstanding again. The converse is true due to the following reason: if you have such a semidirect product, and you have an irreducible representation $V$, then you can show that the subspace $V^P$ is a subrepresentation, nonzero, and therefore $P$ acts trivially on $V$. It is then just a representation of the quotient $A$, and all irreducible representations of $A$ are one dimensional. $\endgroup$
    – Ehud Meir
    Aug 4, 2015 at 13:33
  • $\begingroup$ That is fine because I think that Serr'es argument carries over to this case because he only needs that the abelian group be of order coprime to $p$. His argument is similar to the one you have written now. $\endgroup$
    – Pablo
    Aug 4, 2015 at 13:34
  • 2
    $\begingroup$ All this (in both cases) is well known. In particular, note that the characteristic $p$ case, any normal $p$-subgroup of $G$ acts trivially in each irreducible representation, which gives the converse you want. $\endgroup$ Aug 4, 2015 at 13:50
  • 2
    $\begingroup$ Yes, The Clifford theory was concerning the converse of the characteristic $p$ case, as I said. for the characteristic zero (or coprime to $|G|$) case, to prove the less obvious half, note that every irreducible is a direct summand of the regular module (Maschke's Theorem, and general theory of semisimple algebras, etc). If $G$ is Abelian group, take an element $x \in G$ of order $e = \exp(G)$. In some irreducible $F$-representation $\tau$ of $G$, $x$ has a primitive $e$-th root of unity $\omega$ as eigenvalue on extending scalars. If $\omega \not \in F$, $\tau$ has dimension greater than 1. $\endgroup$ Aug 4, 2015 at 16:12

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