# Representations of a cyclic group of order p over a field of characteristic p?

Let $p$ be a prime. My eventual goal is to prove that the only irreducible representation of a $p$-group over a field of characteristic $p$ is the trivial representation.

At the moment, I'm trying to prove a simpler claim: suppose $G$ is a (cyclic) group of order $p$, let $K$ be a field of characteristic $p$, and let $V$ be any representation of $G$ over $K$; if the dimension of $V$ (as a vector space over $K$) is greater than $1$, then $V$ must be a reducible representation.

Does anyone have any hints or suggestions on how to proceed? For example, where should I use the fact that $K$ has characteristic equal to $|G|$ (actually, I'm not even sure whether or not the above holds for arbitrary $K$) or the fact that $G$ is cyclic?

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Here are some hints:

1. How many elements does $V$ have? (I am assuming without loss of generality that it is finite-dimensional.)
2. What do you know about the sizes of orbits of the vectors in $V$ under the action of $G$?

If you can answer these questions, you will immediately be able to prove that the only irreducible representation is the trivial one.

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This is really cool! I think I see how to make this work when $K = \mathbb{F}_p$, but what if $K$ is not a finite field (so that $V$ contains infinitely many vectors)? –  Elliott Apr 20 '11 at 0:54
@Elliots: in that case, consider the action of the group on the vector space over the prime subfield generated by the orbit of any non zero vector. –  Mariano Suárez-Alvarez Apr 20 '11 at 1:03

In your special case, you can show that the group algebra of a cyclic group of order $p$ over a field $k$ of characteristic $p$ is isomorphic to $k[t]/(t^p)$. It follows that understanding $G$-modules is the same as understanding nilpotent endomorphisms of vector spaces of nilpotency index at most $p$: this can be done using Jordan canonical forms. Indeed, the Jordan form of the matrix of $t$ in a simple module has only one block, because the module is indescomposable, has only zero as eigenvalue because it is nilpotent, and you should have little trouble showing that that block must be of size $1$.

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Alternatively, show that the group algebra is local. It then follows that there is exactly one simple module.

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