Kernel of modular representation of a finite group 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?
 A: 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 $gP=hP$. 
A: 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.   
