# $S_n$ has only four (irred.) representations with degree $<n$ (for $n>6$)

I'm working on the following exercise:

For $n\ge 7$, $S_n$ has no irreducible representations of dimension $m$ with $2\le m\le n-2$.

There is a solution here but I'd like to follow the suggestion of Fulton and Harris (exercise 4.14) and prove this using the hook-length formula. Presumably the proof will proceed inductively, and I've shown the base case $n=7$, but it's the rest I'm struggling with.

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Claim: The only irreducible representations of $\mathcal{S}_d$ (over $\mathbf{C}$) which have dimension less than $d$ are $V_{(d)}$, $V_{(1,\ldots, 1)}$, $V_{(d{-}1, 1)}$ and $V_{(2,1,\ldots,1)}$ as well as the representations $V_{(2,2)}$ of $\mathcal{S}_4$ and $V_{(3,3)}$ & $V_{(2,2,2)}$ of $\mathcal{S}_6$.

F&H want you to show that for all $\lambda$ except those in the claim,$$\dim V_\lambda = \frac{d!}{\prod(\text{Hook lengths})}\geq\ d$$ by induction (removing the first column of $\lambda$). This is going to be slightly ugly, because it seems we have to say explicitly what happens when removing a column yields one of the seven exceptional diagrams listed in the claim.

First let's do the induction without worrying about those seven.

Proof: Assume the claim is true for all $d < k$. Now assume we have a partition $\lambda\vdash k$ (with $\lambda_1\geq\ldots\geq\lambda_n\geq 1$) and that $\lambda$ with one column removed (denoted $\hat\lambda$) is not one of the seven bad diagrams in the claim.

By induction, we're assuming that the hooks of $\hat\lambda=(\lambda_1-1,\ldots,\lambda_n-1)\vdash k-n$ satisfy$$\frac{(k-n)!}{\prod(\text{Hooks of }\hat\lambda)} \geq k-n$$ or equivalently$$(k-n-1)!\geq\prod(\text{Hooks of }\hat\lambda).$$

Starting from the top, the hook lengths of the first column of $\lambda$ are $\lambda_1-1+n$, $\lambda_2-2+n$, all the way to $\lambda_n$. So the product of the hooks of $\lambda$ is then$$\prod_{i=1}^n(\lambda_i+n-i)\cdot\prod(\text{Hooks of }\hat\lambda)$$so this part of the induction is complete if we show that this is at most $(k-1)!$.

Fact: $\lambda_i+n\leq k$. Why? There are $n$ boxes in the first col. of $\lambda$ and $\lambda_i$ boxes in the $i$th row of $\lambda$. So we double counted by the one box in the first col. and the $i$th row, but there must be another nonempty row in $\hat\lambda$ (otherwise $\hat\lambda$ would be one of the diagrams from the claim that we're putting off until later).

From the fact, we get $\lambda_i +n-i\leq k-i$, so$$\prod_{i=1}^n(\lambda_i+n-i)\ \leq\ (k-1)\cdots(k-n).$$But this combined with what we know about $\hat\lambda$ is gives us exactly$$\prod(\text{Hooks of }\lambda) = \prod_{i=1}^n(\lambda_i+n-i)\cdot\prod(\text{Hooks of }\hat\lambda)\leq(k-1)!$$This is what we wanted since we can rearrange this to$$k\leq\frac{k!}{\prod(\text{Hooks of }\lambda)}.$$

Now we have to worry about what happens if $\hat\lambda$ is one of the diagrams listed in the claim (since we couldn't use the inductive step on those). These are more tedious than anything since you can just explicitly compute the product of the hooks. For example, suppose $\hat\lambda$ is the partition corresponding to the trivial representation. Then we can take the dual of $\lambda$ and do the induction above unless $\hat\lambda$ is $(1,1)$. In that case, the product of the hooks is $2(n+1)(n)(n-2)!$ ($n$ is still the no. of boxes in the first column) so we want to know if$$\frac{(n-2)!}{2(n+1)(n)(n-2)!}\geq n-2.$$ This is true since this reduces to $n-1\geq 2$. The other cases can be handled similarly, but I don't think there is anything to be learned from writing it all out.

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