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Let G be an elementary abelian p-group and let V be a representation of G over an (algebraically closed) field of characteristic p. Suppose V has the following properties:

  1. Dim(V) = #P

  2. Dim(V^G)=1 (the dimension of the G-fixed subspace of V is 1).

Obviously the second property implies V is indecomposable. Does V in fact have to be the projective indecomposable?

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1 Answer 1

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Yes.

The second condition says that the socle $S=\operatorname{soc}(V)$ of $V$ (i.e., the largest semisimple submodule) is one-dimensional, and the injective hull of $V$ is the same as the injective hull of its socle, which is the regular representation $kG$. Since $V$ and $kG$ have the same dimension, they must be equal.

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  • $\begingroup$ Thanks! I figured it would either be straightforward or unknown. I'm glad it's the former. $\endgroup$
    – Jon Elmer
    Commented Aug 8, 2013 at 12:45

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