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Given a finite group $G$, let $n$ be the smallest integer s.t. $G \subset S_n$ à la Cayley. I guess that if I want to construct the complex irreps (not just the character table) of $G$ then I could take the irreps of $S_n$ and restrict them to $G$. It seems plausible that after decomposition this might yield all the irreps of $G$, but I'm not entirely sure of this. In any event it seems inefficient, even if it works.

So my question is: what is the (is there a?) general technique for constructing all the inequivalent complex irreps of a finite group?

Maybe this is better suited to the underflow site, but since it's come up in actual work (albeit of the documentary sort) I'm posting here.

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    $\begingroup$ Just a comment, your suggestion for generating all irreps does work but as you say inefficient. $\endgroup$
    – Torsten Ekedahl
    Commented Nov 10, 2010 at 15:57
  • $\begingroup$ The $n$ is $|G|$, unless by à la Cayley you mean simply an injective morphism $G\to S_n$. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Nov 10, 2010 at 16:00
  • $\begingroup$ @Mariano: I do, but good point. $\endgroup$
    – Steve Huntsman
    Commented Nov 10, 2010 at 16:07
  • $\begingroup$ There is usually a better choice than $|G|$. You just have to find a subgroup $H<G$ such that the permutation representation $\mathbb{C}[G/H]$ is faithful. Usually, you will be able to find bigger groups than {1}. Still, as Torsten Ekedahl says, this method will be inefficient. $\endgroup$
    – Alex B.
    Commented Nov 10, 2010 at 16:08
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    $\begingroup$ One method of constructing representations is to induce irreducible representations from (maximal) subgroups. $\endgroup$
    – Bruce Westbury
    Commented Nov 10, 2010 at 16:11

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I think, the article by Vahid Dabbaghian-Abdoly, Journal of Symbolic Computation Volume 39, Issue 6, June 2005, Pages 671-688, entitled "An algorithm for constructing representations of finite groups", doi:10.1016/j.jsc.2005.01.002, and the references in the introduction give you what you are looking for. I think, this is the state of the art to this day. Basically, finding all the irreducible representations can be done in polynomial time.

I am not sure off the top of my head what the complexity of your suggested algorithm will be, but the bottleneck will likely be the fact that $S_n$ always has an irreducible representation of degree $n$. Generically, decomposing a representation of degree $|G|$ is much more work than you should have to do.

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Why representations? A better question is to find all irreducible characters or to fill the character table. You can do it efficiently for any given group.

After you have characters, you may want to construct its representations following MO

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    $\begingroup$ Which question is "better", depends on what you want to do. Also, my understanding of the question is that Steve already knows an algorithm for getting the representations from the character table, but is wondering about more efficient ways of doing it. That was not the focus of the question you have linked to. $\endgroup$
    – Alex B.
    Commented Nov 10, 2010 at 17:31
  • $\begingroup$ Violently disagree. If you have a finite group, a character is better than a representation. Most of the things can be done on the level of characters. You need a very subtle question (such that figuring out why two groups have the same charecter tables) to start worrying about representation itself. Nonwithstanding this, the link I give tells you how to extract any representation from the regular representation if this is what you are looking for. $\endgroup$
    – Bugs Bunny
    Commented Nov 10, 2010 at 20:54
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    $\begingroup$ @Bugs: Here is an example of a question which cannot be answered just by studying the character table. <i> What is the element in the top right corner of the matrix representing element $g$ (where $g$ is your favorite element of the group)?</i> On the other hand if one studies properties that are invariant under conjugation (both in the matrix group and in the original group itself), then you are probably right and one just needs the character table. $\endgroup$
    – user6976
    Commented Nov 11, 2010 at 12:46
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    $\begingroup$ The underlying context of my question is the (quantum) Fourier transform for nonabelian finite groups. "Strong Fourier sampling" requires one to obtain matrix elements. $\endgroup$
    – Steve Huntsman
    Commented Nov 11, 2010 at 23:36

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