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?
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? 


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 nonzero (this is realy a separate argument by induction about a finite $p$group acting on a finitedimensional 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 nonzero, 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. 


An alternative proof is as follows. If G is a pgroup then the augmentation ideal I has basis elements $g1$ with $g\in G$ and $g\neq 1$. Such an element is nilpotent since if g has order $p^m$ then $(g1)^{p^m}=0$. 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 psubgroup of a group G. Say N has index m. The natural map $FG\to F[G/N]$ has kernel I spanned by the elements gh with gN=hN. Then $(gh)^m$ belongs to the augmentation ideal of FN. Hence gh is nilpotent. We conclude I is nilpotent and hence contained in the radical. In particular g1 is in the radical for each g in N and so N is in the kernel of each irrep. 

