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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
See Emerton's answer to this question: mathoverflow.net/questions/12861/… –  Keenan Kidwell Jun 28 '11 at 18:30
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2 Answers

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
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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 I spanned by the elements g-h with gN=hN. Then $(g-h)^m$ belongs to the augmentation ideal of FN. Hence g-h is nilpotent. We conclude I is nilpotent and hence contained in the radical. In particular g-1 is in the radical for each g in N and so N is in the kernel of each irrep.

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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
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