Decomposing $Sym^2\mathbb{C}[X]$ without using characters There is a nice and not too long way to decompose $Sym^2\mathbb{C}[X]$ (permutation representation of $S_5$ acting on $\{x_1,\cdots,x_5\}$) in irreducible without using characters?
I know a way using characters, since it is sufficient to know the characters of the standard and trivial representation (using $\chi_{sym^2V}(g) = \frac{1}{2}\chi_V(g^2)+\chi_V(g)^2)$ and then taking the scalar product with all irreducible characters of $S_5$ watching at the table characters of $S_5$.
I'm wondering whether there's a way without using characters (schur functions or symmetric polynomial together with Schur functors are allowed)
Any help or tip would be appreciated.
 A: Yes, this can be done as follows, and this argument generalizes to $S_n$ with no difficulty.
$S^2$ of a permutation representation is itself a permutation representation: namely it's the permutation representation on the set of $2$-element multisets of $X$, which you can think of as $S^2(X)$. This permutation representation decomposes into two orbits, the multisets $\{ x, x \}$ and the subsets $\{ x, y \}$ where $x \neq y$. So we get
$$S^2(\mathbb{C}[X]) \cong \mathbb{C}[X] \oplus \mathbb{C} \left[ {X \choose 2} \right]$$
where ${N \choose 2}$ is the set of $2$-element subsets. To finish from here use the result from this math.SE answer which tells us that each representation $\mathbb{C} \left[ {X \choose k} \right]$ embeds into the next one $\mathbb{C} \left[ {X \choose k+1} \right]$ with irreducible cokernel. This still uses characters but it does not involve knowing the character table.
The conclusion is that $S^2(\mathbb{C}[X])$ decomposes into two copies of $1$, two copies of a $4$-dimensional irreducible, and a $5$-dimensional irreducible.
