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Let $K$ be a field of $\operatorname{char}= p>0$ , let $G$ be finite group of order $p$, and $V$ is non zero $KG$-module.

How do I show that there exist non-zero $v\in V$ such that $gv=v $ for all $g\in G$, and how do I show that all the irreducible $KG$-modules are isomorphic, if that makes sense.

The first question makes sense for sure!

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    yes , sorry . i edited it.2012-04-24

2 Answers 2

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Here's mt_'s answer with more detail: Since we're in characteristic $p$, $(g-1)^p=g^p-1^p=g^p-1=1-1=0$. So $(g-1)$ is nilpotent in $KG$. Now pick $n$ to be the least integer such that $(g-1)^n=0$, and look at any nonzero element $v$ in the image of $(g-1)^{n-1}$. For such a $v$, $(g-1)v=0$, so $gv=v$.

Now that we've done the first part, let's look at any irreducible $KG$ module. We've already shown that it has an element which is fixed by $g$, and hence by $G$. But that must be the whole module, since if not, such an element would span a submodule (namely, this submodule would be the set of elements annihilated by $g-1$).

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Hint for the first part: Let $g \in G, g \neq 1$, so $g$ generates $G$. Show that $(g-1)^p=0$. It follows $(g-1)^p V=0$ but $V \neq 0$. There's therefore an $n \leq p$ such that $(g-1)^nV =\{ 0\}, (g-1)^{n-1}V \neq 0$. How does $g$ act on an element of $(g-1)^{n-1} V$?

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    What is the first symbol or phrase written that you don't understand? Do you not understand what it says or do you not understand why what it says is true?2012-04-24