Hi, i wonder if there are interesting proofs about $S_n$ (group theoretic or not) using its character table. Using the MurnaghanNakayama rule you can for example prove that for $n>4$ $A_n$ is the only normal subgroup of $S_n$ because there are no nonlinear characters $x$ and $g$(not 1) in $S_n$ with $ x(g)=x(1)$, since $x(1)>x(g) $ . Do you know any other nontrivial theorems about $S_n$ with a proof using its charactertable ?
In an answer to an earlier question, I showed how to prove that the square root counting function $r_2: S_n\rightarrow \mathbb{N},\;g\mapsto \#\{h\in S_nh^2=g\}$ assumes its maximum at the identity, using the representation theory of $S_n$. Admittedly, you need to know slightly more than the character table. You need to be able to compute the FrobeniusSchur indicators of the characters, so you need to know how the conjugacy classes multiply. Alternatively, you just need to know that all representations are defined over $\mathbb{R}$, which you prove along to way to computing the character table anyway. In a comment to my answer, Richard Stanley remarks that, also using the representation theory of $S_n$, you can generalise this to the $k$th root counting function for any positive integer $k$. In an answer to the same question, Alon Amit remarks on possible generalisations to solving other polynomial equations in the elements of $S_n$. 


I'm not sure if you will consider this nontrivial, but from the character table you can very quickly show that the number of conjugacy classes of even permutations is always greater than or equal to the number of conjugacy classes of odd permutations. One just applies the general fact that the sum of the entries in any row of a character table (not weighted by the size of the conjugacy class) is a nonnegative integer (because the character $\chi ( \pi ) = \frac{\# S_n}{\# C_\pi}$, where $C_\pi$ is the set of conjugates of the permutation $\pi$, corresponds to an actual representation, namely, take a vector space with basis $\{ e_\pi \mid \pi \in S_n\}$ and define $g(e_\pi) = e_{g \pi g^{1}}$ for $g \in S_n$) to the sign character. 


The following is not strictly speaking something that can be read off from the character table. However, it is an elementary combinatorial identity about partitions which one can deduce from understanding the character theory of symmetric groups well enough, and looking at the character table does play a central role: For $\lambda \vdash n$ a partition of $n$ (i.e., $n = 1 \lambda_1 + 2 \lambda_2 + \cdots + n \lambda_n$) define $$ A(\lambda) = \prod_{i=1}^{n} n^{\lambda_n}, \qquad B(\lambda) = \prod_{i=1}^{n} (\lambda_n)! $$ Claim: $$\prod_{\lambda \vdash n} A(\lambda) = \prod_{\lambda \vdash n} B(\lambda) $$ The charactertheoretic proof proceeds as follows:
Putting together 2, 4, 5 we obtain $$ \det(M)^2 = \prod_{\lambda \vdash n} A(\lambda) B(\lambda) = \left(\prod_{\lambda \vdash n} B(\lambda)\right)^2 $$ and so the claimed identity. 


Because all the entries in the character table are integers and not just algebraic integers, you get that a proof that every permutation $\sigma$ of order $n$ is conjugate to all $\sigma^j$ for $j$ coprime to $n$. (Of course, one usually uses this in the opposite direction, to deduce that all entries are integers!) 

