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How can I get the Wedderburn decomposition of the module $KG$, if I have the irreducible represenations of the group $G$? I am confused by a post here I found:

Why is the Wedderburn formula in this case wrong?

The last summand in the Wedderburn decomposition is $Mat(2,F_9)$, but the associated irreducible representation is $$< \begin{pmatrix} \ & \ & \ & 1 \\ \ & \ & 1 & \ \\ \ & 1 & \ & \ \\ 1 & \ & \ & \ \end{pmatrix} , \begin{pmatrix} \ & 1 & \ & \ \\ \ & \ & 1 & \ \\ \ & \ & \ & 1 \\ -1 & -1 & -1 & -1 \end{pmatrix} >. $$ I.E. this are matrices in $Mat(4,F_{3})$ (this is linked in the post). Why do I write than $Mat(2,F_9)$ as a summand?

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    Your general question is fairly hard to answer. If $K$ is a finite field whose characteristic is coprime to the order of $G$, then I can give you a simple answer in terms of the ordinary character table. If $K$ has characteristic 0, then the answer involves some occasionally hard number theory. If the characteristic of $K$ divides the order of $G$, then there is no Wedderburn decomposition, as such, but $KG$ mod its radical has a Wedderburn decomposition associated to the modular character table.2012-08-21
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    My answer below only seeks to help answer your confusion, so let me know if you need the fuller answer.2012-08-21

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