Suppose $G$ is a finite group. How can I prove that $\mathbb{C}[G]$ has a nonzero nilpotent element if and only if $G$ is not abelian?
Nilpotent elements in $\mathbb{C}[G]$
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group-theory
1 Answers
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It's because $\mathbb{C}[G]$ is isomorphic to a product of matrix algebras. Since $G$ is commutative iff $\mathbb{C}[G]$ is commutative and since the algebra of $n\times n$-matrices is commutative iff $n=1$, we see that $G$ is non-abelian iff at least one of the factors in the decomposition of $\mathbb{C}[G]$ is an$ n\times n$-matrix algebra with $n>1$. Such an algebra always contains a nilpotent element.