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For $n\geq 7$, I would like to show that $S_n$ has no irredicuble representations of dimension $m$ for $2\leq m\leq n-2$.

The catch is that I am not allowed to use any "machinery" (evidently, this problem should be solvable having knowledge only of chapter 1 of Fulton and Harris).

At the moment, I don't really have any idea of how to approach this, especially without the use of character theory. Any suggestions would be greatly appreciated!

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    This is not true. The tensor product of the sign representation and the standard has dimension $n-1$, and is not isomorphic to the standard.2011-10-02
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    Thanks David. I'll edit the post to take this into account.2011-10-02
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    The case $n=7$ seems easy. Look at a 7-cycle. If it has a primitive seventh root as an eigenvalue, then it must have all six of them as eigenvalues, because it is conjugate to its non-trivial powers. That alone forces the dimension to be at least 6. OTOH if 1 is the sole eigenvalue, then all the 7-cycles would act trivially. Hence so would the subgroup that they generate, which is $A_7$... Alas, I cannot generalize this argument to higher $n$.2011-10-03
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    What about proving by induction that the only small irreducible reps of $S_n$ are the listed ones: two 1-dimensional, and two $(n-1)$-dimensional ones. So if $n\ge8$ is the smallest value with a counterexample, then the rep must remain irreducible when restricted to any of the $n$ conjugate 'point stabilizer' subgroups isomorphic to $S_{n-1}$. The restriction cannot be a sum of 1-dimensional reps, because then the copies of $A_{n-1}$ (and hence $A_n$) would act trivially. The remaining problem would be to show that neither of the $n-2$ reps of $S_{n-1}$ can be a restriction of a rep of $S_n$.2011-10-03
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    The result is true as stated for complex representations, but it is not true in general in other characteristics. For example, $S_n$ has a faithful irreducible representation of degree at most $n-2$ in characteristic $p$ whenever $p$ is a prime divisor of $n$ and $n >4.$ Take the obvious permutation module. Then $n$-long column vectors with zero sum form a submodule, but this submodule contains the span of the all-1 vector, which is invariant. I realise that characteristic 0 was the intent of the question, but it illustrates that the question is subtle.2011-12-20

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