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 $p$group 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 $p$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 $gh$ with $gN=hN$. Since $gN=hN$ and $(gN)^m=N$ it follows that $\{g,h\}^m\subseteq N$. Thus $(gh)^m\in FN$ 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 $(gh)^m$ is nilpotent and hence $gh$ is nilpotent. We conclude $J$ is nilpotent by another application of Wedderburn's theorem and hence contained in the radical of $FG$. In particular $g1$ 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 $g1$ is not nilpotent (else $g^{p^m}1=(g1)^{p^m}=0$ 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 $gh$ with $gP=hP$. 

