MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G$ be a finite group; let $F$ be a field of characteristic $p > 0$.

If I have an irreducible modular representation $\rho: G \to GL_n(F)$, does $\ker \rho$ contain all the normal $p$-subgroups of $G$? If so, how does one show this?

share|cite|improve this question
I don't have enough rep to edit, but in the title Kernal should be Kernel – David White Jun 28 '11 at 18:24
See Emerton's answer to this question:… – Keenan Kidwell Jun 28 '11 at 18:30

Yes, it does. Let $V$ be an irreducible $FG$ module where $F$ has characteristic $p >0$. Let $U$ be a normal $p$-subgroup of $G$. Then the fixed point space $V^{U}$ is non-zero (this is realy a separate argument by induction about a finite $p$-group acting on a finite-dimensional vector space over a field of characteristic $p$). But since $U$ is normal in $G$, the space $V^{U}$ is $G$-invariant. Since $G$ acts irreducibly on $V$ and $V^{U}$ is non-zero, we must have $V^{U} = V$, that is to say, $U$ acts trivially on $V$. Another (essentially equivalent) argument is to use Clifford's theorem, together with the fact that the only irreducible module in characteristic $p$ for a finite $p$-group is the trivial module.

share|cite|improve this answer
If F is a finite field, you can prove there is a fixd point by using that the number of fixed points for a p-group is equal to the size of the vector space mod p and observe that 0 is a fixed point. – Benjamin Steinberg Jun 28 '11 at 18:43

An alternative proof is as follows. If $G$ is a $p$-group then the augmentation ideal $I$ has basis elements $g-1$ with $g\in G$ and $g\neq 1$. Such an element is nilpotent since if $g$ has order $p^m$ then $(g-1)^{p^m}=0$. Thus $I$ has a basis of nilpotent elements. But an ideal of a finite dimensional algebra with a nilpotent basis is nilpotent by a theorem of Wedderburn. Thus $I$ is contained in the radical of $FG$ which implies $I$ is the radical since $FG/I=F$ is semisimple.

Next let $N$ be a normal $p$-subgroup of a group $G$. Say $N$ has index $m$. The natural map $FG\to F[G/N]$ has kernel $J$ spanned by the elements $g-h$ with $gN=hN$. Since $gN=hN$ and $(gN)^m=N$ it follows that $\{g,h\}^m\subseteq N$. Thus $(g-h)^m\in FN$ and also belongs to the augmentation ideal of $FG$. Since the augmentation of $FG$ restricts to the augmentation of $FN$ we conclude by the previous paragraph that $(g-h)^m$ is nilpotent and hence $g-h$ is nilpotent. We conclude $J$ is nilpotent by another application of Wedderburn's theorem and hence contained in the radical of $FG$. In particular $g-1$ is in the radical for each $g$ in $N$ and so $N$ is in the kernel of each irrep.

On the other hand, if order $g$ is not a $p$-power, then $g-1$ is not nilpotent (else $g^{p^m}-1=(g-1)^{p^m}=0$ for $m$ large enough). Thus no such element belongs to the radical. Therefore a normal subgroup is contained in the kernel of every irreducible representation over $F$ if and ony if it is a normal $p$-subgroup. In particular if $P$ is the unique maximal normal $p$-subgroup then the largest nilpotent Hopf ideal of $FG$ is the ideal spanned by $g-h$ with $gP=hP$.

share|cite|improve this answer
Ah, nice, thanks. – Dr Shello Jun 29 '11 at 14:38
Huh? Now that you edited your post, I don't understand it anymore... – darij grinberg Dec 28 '12 at 18:38
@darijgrinberg, I just saw your comment now. I hope it is now clear. – Benjamin Steinberg Jun 3 '14 at 15:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.