Question

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?

Answer 1

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.

Answer 2

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.