# Representation theory of a finite p-group over a field of characteristic p: dim of invariants =1 => dim of coinvariants = 1?

I am trying to understand the proof of Proposition 4 in S. Ullom, Integral normal bases in Galois extensions of local fields, Nagoya Math. J. Volume 39 (1970), 141-148. The PDF is available here: http://projecteuclid.org/euclid.nmj/1118798052

It appears that the following result is used, but I'm afraid that I don't quite see the proof. Let $k$ be a field of characteristic $p$ and let $G$ be a finite $p$-group. Let $W$ be a left $k[G]$-module such that $\dim_k W = |G|$. Suppose that $\dim_k W^{G} = 1$. Then $\dim_k W_G = 1$. Here $W^G$ denotes invariants and $W_G$ denotes coinvariants.

I have to admit that I know relatively little about representation theory in characteristic $p$. One idea would be to consider the dual representation $\hat{W}$, but I only got so far: $\dim_k W^{G} = 1 \implies W$ is indecomposable $\implies \hat{W}$ is indecomposable. But maybe this is not the right approach. If I could show that the Tate cohomology groups of $W$ vanish, then of course the desired result drops out, but I think this is rather strong medicine.

Is anyone able to give a proof of the above claim? I suspect the solution is fairly easy, but I just don't see it at the moment.

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The point is that you can inject $W$ into $k[G]$ and because of the equality of dimension this is an iso. This lets you to conclude that the coinvariants are $1$-dimensional.

The injection goes as follows:

$k[G]$ is a projective $k[G]$-module for trivial reasons, so its $k$-linear dual is injective, but $k[G]^*\cong k[G]$. So $k[G]$ is an injective $k[G]$-module.

Now $G$-invariants of $k[G]$ is a one dimensional vector space, which we may identify with $W^G$. Since $k[G]$ is injective, there exists a map $\phi: W\rightarrow k[G]$ which induces an injection on $W^{G}$. This means that $Ker \phi \cap W^G=(Ker \phi)^G=0$, but since $G$ is a $p$-group and we are in char $p$, we get that $Ker \phi=0$.

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Nice answer, modp ! –  Joël May 4 '12 at 16:31
A very nice answer - thanks! If I understand correctly: under the given hypotheses, by combining your argument with its `dual' we in fact have $\dim_k W^G=1 \Leftrightarrow W≅k[G] \Leftrightarrow dim_k W_G = 1$. –  Henri Johnston May 5 '12 at 17:32
Maybe it's useful to add some general perspective to the efficient answer given. The approach taken by modp is undoubtedly the most natural one, taking advantage of the extra assumption that $W$ has dimension equal to $|G|$. This assumption is essential, since without it a $k[G]$-module may well have a space of covariants of higher dimension when the space of invariants is 1-dimensional. (Concrete examples occur when you consider restrictions to a Sylow $p$-subgroup in a finite group of Lie type of various standard modules.)
It's a standard fact that any finite $p$-group has only the trivial simple module over a field of characteristic $p$ (unlike the radically more complicated situation in characteristic 0). In particular, any module with a 1-dimensional subspace of invariants is automatically indecomposable. Though the indecomposable $k[G]$-modules can be arbitrarily difficult to study in general, the special assumption that $\dim_k W = |G| = \dim_k k[G]$ then yields an isomorphic embedding of $W$ onto $k[G]$: For any finite group the group algebra over a field is both injective and projective as a left module. In the special case of a $p$-group, the group algebra is itself the projective cover (= injective hull) of the unique simple module. In particular, the socle (here the space of all invariants) is isomorphic to the head (here the space of all coinvariants). Indeed, the module $W$ is self-dual.