Suppose $G$ is a finite group and $F$ is a field such that $\mathrm{char}\;F$ doesn't divide $|G|$. Suppose that $F$ is algebraically closed and $G$ is not abelian. How can I prove that $F[G]$ has infinitely many nilpotent elements?
Infinitely many nilpotent elements in $\mathbb{C}[G]$
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group-theory
representation-theory
finite-groups
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5If $x$ is nilpotent and $\lambda \in F$ then $\lambda x$ is also nilpotent. Since an algrebraically closed field is infinite, if there is one non zero nilpotent element there are infinitely many. – 2012-05-06
1 Answers
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It is a fundamental result in representation theory that the group algebra of a finite group over a field with characteristic not dividing the group order is semisimple and hence, by the Artin-Wedderburn Theorem, is isomorphic to a direct product of full matrix algebras over division rings. Furthermore, if the field $F$ is algebraically closed, then these matrix algebras are over $F$ itself. If the group is nonabelian, then at least one of these matrix algebras has dimension greater than 1, and hence contains infinitely many nonzero nilpotent elements.