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More precisely (the conditions can be tweaked to be more general as we may desire), let $G$ be a finite group and $K$ an algebraically closed field whose characteristic does not divide the order of the group. For $a\in G$, denote by $K^{[a]}$ the $K$-vector space spanned by the elements of the conjugacy class of $a$, with a representation $\rho:G\to GL(K^{[a]})$ extending the relations $\rho(g)x=gxg^{-1}$ for all elements $x\in[a]$ linearly to all of $K^{[a]}$. Then:

Conjecture. There is a collection of irreducible representations $V_{[a]}$ indexed by conjugacy classes of $G$, each distinct, such that $V_{[a]}$ is a direct factor of $K^{[a]}$ for each class $[a]$.

It may seem like a strange conjecture to make, but note that Specht modules follow this pattern and tabulate the irreps of the finite symmetric groups. (The motivation behind this question is the curiosity if the polytabloid construction method behind Specht modules could perhaps be plausibly generalized to arbitrary finite abstract groups.) Note that the number of conjugacy classes of $G$ is equal to the number of irreps, so this collection must be exhaustive. Is the conjecture true?

In the presence of an affirmative answer, one might wonder if such a bijection between the irreps and conjugacy classes is unique or can be made canonical, and in the event of a negative answer, under what (ideally, simple-to-state) conditions the conjecture would hold true.

I am disinclined to test even small cases because of the complexity of and my lack of understanding how to systematically, computationally go about decomposing the $K^{[a]}$ spaces into irreps, so I do confess I haven't done everything in my power to tackle the query, but I'll see what I can do tomorrow when I have time. (While I go to bed tonight, does anybody have advice or references on straightforward ways to check some small cases?)

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    *A New Approach to Representation Theory of Symmetric Groups* appears to generalize tableau methods to Coxeter groups, but I haven't looked at it very deeply.2012-10-19

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This is a nice question, but needs to be tweaked somehow. If $G$ is Abelian, then each $K^{[a]}$ just affords the trivial representation, and for any finite $G,$ the center $Z(G)$ will be in the kernel of the action of $G$ on each $K^{[a]}.$ More generally still, an irreducible character $\chi$ of $G$ occurs as a constituent of the character of one or more $K^{[a]}$ if and only if $\sum_{i=1}^{k} \chi(x_{i}) \neq 0,$ where $G$ has $k$ conjugacy classes with representatives $x_{1},\ldots x_{k}$ (there is a similar formula to determine the multiplicity of $\chi$ in the character afforded by an individual $K^{[a]}).$

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    @Tobias@ I meant complex irreducible character; since we are working over an algebraically closed field of characteristic coprime to the group order, even in the finite characteristic case, Brauer characters are genuine complex characters.2012-10-19