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Suppose that there is a group $G$ and an odd prime $p$. Let $M$ be a $\mathbb{F}_p[G]$ module.

Is it true that $M$ cannot be irreducible but just indecomposible?

Thank you

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    I doubt the paper is wrong, independently of its age... The result I mentioned has been known for ages. You may be misunderstanding something o missing an hypothesis —telling us what paper it is that you are talking about would probably not hurt! In any case, please fix the question so that it asks what you intended to ask.2012-10-22

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It depends on $M$, of course.

For example, suppose $G=C_p$ is the ciclic group of order $p$, let $\sigma\in C_p$ be a generator, and let $M=\mathbb F_p\oplus\mathbb F_p$ be the $\mathbb F_p[G]$-module on which $\sigma$ acts as the matrix $\begin{pmatrix}1&1\\0&1\end{pmatrix}$. You should not have problems showing that this is a non-irreducible indecomposable module.

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It also depends on $p.$ If $F$ is a field of prime characteristic $p$ which does not divide the order of the finite group $G,$ then all indecomposable $FG$-modules are irreducible. On the other hand, if the prime $p = {\rm char}F$ divides $|G|,$ then there is always an indecomposable $FG$-module which is not irreducible ( for otherwise the group algebra $FG$ would be semisimple, and hence would have trivial Jacobson radical. But the sum of the group elements is a non-zero central nilpotent element of the center of $FG),$ so lies in the Jacobson radical of $FG.$