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I want to understand the Specht module and use it to find the irreducible representations of $S_n$. I know that the Specht modules are spanned by polytabloids, which are constructed using the permutations $\sigma \in S_n$ belonging to the same conjugacy class.

My understanding is that since the Specht modules are cyclic (generated by a finite number of polytabloids), all one has to do to figure out the corresponding irreducible representation of $S_n$ is observe how an arbitrary permutation $\sigma \in S_n$ acts on each of the generators of the Specht module. Is this correct?

Is there a way of knowing how many generators there are (and what these generators are) for a Specht module corresponding to a general cycle shape $\lambda$? Thank you.

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  • $\begingroup$ You know that irreps or S_n are in bijection with conjugacy classes so are in bijection with partitions and so are bijection with {Specht modules}. So the only thing you are left to show is that each Specht module is irreducible. This is standard material, but I did a (hopefully clearer than standard) writeup on this some time ago; see questions 15-16 from math.columbia.edu/~maithreya/MLQFTGMvol1.pdf (q 15 answers your question and q 16 proves the result that the answer cites) $\endgroup$ – Maithreya Sitaraman Aug 2 '18 at 18:29
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The dimension of a Specht module is given by the number of standard Young tableaux of that shape. There is an explicit basis for the module in terms of polytabloids. You can find details in books like Fulton's Young Tableaux or Sagan's The Symmetric Group. That number of standard Young tableaux is given by the celebrated hook length formula.

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