Let $K$ be an algebraically closed field of characteristic $0$ and $G$ a finite group. I know that the group ring $K[G]$ is a semisimple and so decomposes as a direct sum of $m$ simple components where $m$ is equal to the number of irreducible characters of $G$.

What is the exact relationship between these Wedderburn components and the (complete) set of irreducible characters? In particular how does this relate to the contragredient character of an irreducible character?

Many thanks for your help.

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    $\begingroup$ Well each component of the $K[G]$ decomposition is a finite-dimensional $K$-vector space, on which $G$ acts (via matrices). The associated character is the trace of these matrices. $\endgroup$ – user641 Sep 5 '13 at 20:52
  • $\begingroup$ @SteveD: If by "the associated character" you mean the irreducible character of $G$ that gave rise to this component, then this must be wrong. The sum of the dimensions of the components of $K[G]$ is obviously $|G|$, but the sums of the values at$~e$ of the irreducible characters is less; it is the sum of their squares that equals $|G|$. $\endgroup$ – Marc van Leeuwen Sep 6 '13 at 5:19
  • $\begingroup$ @MarcvanLeeuwen: There are multiple isomorphic components. For example, $\mathbb{C}[S_3]\cong V_1\oplus V_2\oplus V_3\oplus V_3$, where the $V_i$ are the components. Each copy of $V_3$ gives the same character. $\endgroup$ – user641 Sep 6 '13 at 6:13
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    $\begingroup$ @SteveD: The components in the question are summands of the algebra $K[G]$, and these are not as $K[G]$-module irreducible representations. Instead, viewed as $K[G]$-module they are isotypic components, and there are no more of them then there are distict irreducible characters (all copies of the same irreducuble are clumped together in one component). So while you are right as far as the decomposition of the regular representation into irreducibles is concerned, you are not talking about the same components as OP is. $\endgroup$ – Marc van Leeuwen Sep 6 '13 at 7:00
  • $\begingroup$ @MarcvanLeeuwen: Yes, I see now he is talking about the decomposition of $K[G]$ as a ring, sorry. $\endgroup$ – user641 Sep 6 '13 at 19:26

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