Ring of Invariants of symmetric group The symmetric group $S_n$ acts on $\mathbb C^n$ by permuting the coordinates. In this case the ring of invariants is generated by elementary symmetric polynomials in n-variables. Now consider the regular representation of $S_n$, the basis of the vector space is indexed by the elements of $S_n$. Then what are the generators for the ring of invariants ? I guess the elementary symmetric polynomial in $n!$ variables generate the ring but I am not sure. 
 A: The elementary symmetric polynomial do not generate the ring if $n>2$, let $g_1,g_2$ be two distinct elements  of $S_n$, consider $G.(g_1,g_2)=\{gg_1,gg_2),g\in G\}$ the polynomial $\sum X_{gg_1}X_{gg_2}$ is invariant by $G$ but not invariant by $S_{n!}$.
A: The statement is obviously true for $n = 1$ and $n = 2$, but false for $n \geq 3$.
Let $\mathbb{C} S_n$ denote the regular representation of $S_n = \{\sigma_1, \dotsc, \sigma_k\}$, where $k = n!$. Then the ring of polynomials on $\mathbb{C} S_n$ is given as
$$
 A
 := \mathbb{C}[X_\sigma \mid \sigma \in S_n]
 = \mathbb{C}[X_{\sigma_1}, \dotsc, X_{\sigma_k}]
$$
and $S_n$ acts on $A$ via algebra automorphisms, which are given by
$$
 \tau . X_\sigma = X_{\tau\sigma}
 \quad
 \text{for all $\tau, \sigma \in S_n$}.
$$

Now consider the homogeneous polynomials of degree $2$ in $A$. The subspace $H$ of these polynomials has a basis given by $M := \{X_{\sigma_i} X_{\sigma_j} \mid 1 \leq i \leq j \leq k\}$ and both $H$ and $M$ are invariant under the actions of both $S_k$ and $S_n$.
A polynomial $P \in H$ with $P = \sum_{i \leq j} a_{ij} X_{\sigma_i} X_{\sigma_j}$ is symmetric if and only if the coefficients $a_{i,j}$ are constant on the $S_k$-orbits of $M$. Similarly $P$ is $S_n$-invariant if and only if the coefficients are constant on the $S_n$-orbits of $M$.
Because every $S_n$-orbit is contained in some $S_k$-orbit the above two criteria are equivalent if and only if the $S_k$- and $S_n$-orbits of $M$ coincide. We will determine these number of these two kinds of orbits to show that they are not equal (or at least estimate):

Notice that $M$ decomposes into  precisely two $S_k$-orbits, namely $\{X_{\sigma_i} X_{\sigma_i} \mid 1 \leq i \leq k\}$ and $\{X_{\sigma_i} X_{\sigma_j} \mid i < j\}$.
Next we estimate the number of $S_n$-orbits of $M$. First notice that
$$
 X_{\sigma_i} X_{\sigma_j} = \sigma_i.(X_1 X_{\sigma_i^{-1} \sigma_j})
 \quad
 \text{for all $1 \leq i,j \leq k$}
$$
So every orbit has a representant of the form $X_1 X_{\sigma_i} \in M$ with $1 \leq i \leq k$.
Notice that if $X_1 X_{\sigma_i} = \tau.(X_1 X_{\sigma_j}) = X_{\tau} X_{\tau \sigma_j}$ for some $1 \leq i,j \leq k$ then we have two possible cases: If $1 = \tau$ and $\sigma_i = \tau \sigma_j$ then $\sigma_i = \sigma_j$. If $1 = \tau \sigma_j$ and $\sigma_i = \tau$ then $\sigma_i = \sigma_j^{-1}$. So it directly follows that for the $s = \binom{n}{2}$ many transpositions $\tau_1, \dotsc, \tau_s \in S_n$ the representants $X_1 X_{\tau_1}, \dotsc X_1 X_{\tau_s}$ are not conjugated to each other. Therefore $M$ has at least $s$ orbits under the action of $S_n$. Because $n \geq 3$ it follows that $s > 2$.
So there are strictly more $S_n$ orbits than $S_k$-orbits, showing that the orbits of both actions do not conicide. So not every $S_n$-invariant polynomial is also symmetric.
