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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?

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I don't have enough rep to edit, but in the title Kernal should be Kernel –  David White Jun 28 '11 at 18:24
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See Emerton's answer to this question: mathoverflow.net/questions/12861/… –  Keenan Kidwell Jun 28 '11 at 18:30

2 Answers 2

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.

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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 $gP=hP$.

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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 at 15:49

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