Say $n \ge 5$. Let $P$ be the $(n-1)$ dimensional permutation module for $S_n$, i.e. the permutation representation on $\{(x_1, \dots, x_n) \in {\bf C}^n: \sum x_i = 0\}$. Prove that:
Is there a way to do this problem without using something "overpowered" such as Young tableaus?
These questions can be answered with a little bit of character theory. If you can work out the character of $P$, then the character of $\wedge^2 P$ is not so bad. For instance look at the bottom of the second answer of this question. Symmetric and exterior power of representation
Once you can compute the character of the relevant representations, finding their decompositions is not so difficult. If you have questions, we can discuss in the comments.
Write $G = S_n$. Recall that $$\mathbb{C}^n = P \oplus \text{span}(1, 1, \dots, 1).$$Let $\phi: \mathbb{C}^n \to P$ be the projection and $e_1, \dots, e_n$ the standard basis for $\mathbb{C}^n$. Therefore, $\phi$ is given explicitly as $$\phi(x_1, \dots, x_n) = (x_1 - \overline{x}, \dots, x_n - \overline{x}),$$where $$\overline{x} = {{\sum x_i}\over{n}}.$$Let $f_i = \phi(e_i)$.
The map $\Psi: \mathbb{C}^n \to P \otimes P$ defined by$$\Psi: \sum a_ie_i \mapsto \sum a_i(f_i \otimes f_i)$$is a $G$-homomorphism. Its image lies in $\text{Sym}^2P$, and it is injective.
Here is how to prove injectivity. Being a $G$-homomorphism, if $\Psi$ were not injective, its kernel would be a subrepresentation of $\mathbb{C}^n$; the only possibilities are $\text{span}(1, 1, \dots, 1)$ and $P$. In the first case, $$\Psi\left(\sum e_i\right) = 0 \implies \sum_i f_i \otimes f_i = 0$$and, in the second case, $$\Psi(e_1 - e_2) = 0 \implies f_1 \otimes f_1 = f_2 \otimes f_2.$$However, $f_1, \dots, f_{n-1}$ is a basis for $P$, and thus, $f_i \otimes f_j$ $(1 \le i,\, j \le n-1)$ is a basis for $P \otimes P$; this together with the fact that $$f_n = -f_1 - f_2 - \dots - f_{n-1}$$shows that neither possibility occurs.
Note also that $$\dim \text{Sym}^2P = {{n(n-1)}\over{2}} > n,$$and we have just shown that $\mathbb{C}^n$ is isomorphic to a subrepresentation of $\text{Sym}^2P$. Therefore, $\text{Sym}^2 P$ decomposes as a direct sum of $1$, $P$, and another representation $X$ $($not necessarily irreducible$)$. Consequently,$$P \otimes P \cong 1 \oplus P \oplus X \oplus \wedge^2 P.$$Note now that $\langle \chi_{P\otimes P}, \chi_{P\otimes P}\rangle \ge 4$ with equality if and only if $X$, $\wedge^2 P$ are irreducible. We shall show that equality holds; this implies that $X$ and $\wedge^2 P$ are both irreducible, as desired.
Let $\chi_P$ be the character of $P$, so that $\chi_P(g) = \text{Fix}(g) - 1$, where $\text{Fix}(g)$ denotes the number of fixed points of $g$ on $\{1, 2, \dots, n\}$. Then$$\langle \chi_{P \otimes P}, \chi_{P \otimes P}\rangle = {1\over{|G|}} \sum_{g \in G} (\text{Fix}(g) - 1)^4$$$$= {1\over{|G|}} \sum_{g \in G} (\text{Fix}(g)^4 - 4\text{Fix}(g)^3 + 6\text{Fix}(g)^2 - 4\text{Fix}(g) + 1).$$Now¹$${1\over{|G|}} \sum_{g \in G} \text{Fix}(g)^j = \text{# of }G\text{-orbits on }\{1, \dots, n\}^j.$$It remains to count the number of $S_n$-orbits on $\{1, \dots, n\}^j$ for each $1 \le j \le 4$. For example, with $j = 3$, there are five orbits: $\{(x, y, z)\text{ }|\text{ }x,\, y,\, z\text{ all distinct}\}$, $\{(x, x, y)\text{ }|\text{ }x \neq y\}$, $\{(x, y, x)\text{ }|\text{ }x \neq y\}$, $\{(y, x, x)\text{ }|\text{ }x \neq y\}$, and finally $(x, x, x)$. In all cases, the variables $x$, $y$, $z$ are understood to take values in $1, \dots, n$.
We shall abbreviate this reasoning by saying the following.
There is one orbit of type $(x, y, z)$, three orbits of type $(x, x, y)$, and one orbit of type $(x, x, x)$.
Reasoning similarly:
$j = 1$. One orbit.
$j = 2$. Two orbits: one of type $(x, y)$, one of type $(x, x)$.
$j = 3$. Five orbits $($as above$)$.
$j = 4$. Fifteen orbits: one of type $(x, y, z, w)$, six of type $(x, x, y, z)$, four of type $(x, x, x, y)$, three of type $(x, x, y, y)$, one of type $(x, x, x, x)$.
Thus,$$\langle\chi_{P \otimes P}, \chi_{P \otimes P}\rangle = 15 - 20 + 12 - 4 + 1 = 4,$$concluding the proof.
¹Indeed, if $G$ is any finite group acting on the set $X$, then $($Burnside's Lemma$)$$${1\over{|G|}} \sum_{g \in G} \text{Fix}(g)$$is the number of orbits of $G$ on $X$; apply this with $X = {1, \dots, n}^j$.
To prove Burnside's Lemma, note that $${1\over{|G|}} \sum_{g \in G} \text{Fix}(g) = \langle \chi_{\mathbb{C}X}, 1\rangle,$$ where $\mathbb{C}X$ is the permutation representation of $G$ associated to $X$. This inner product equals the dimension of the space of $G$-invariant vectors on $\mathbb{C}X$ $($why?$)$. But a function $f: X \to \mathbb{C}$ is $G$-invariant if and only if it is constant on each $G$-orbit; thus, the dimension of $G$-invariants equals the number of $G$-orbits.
