TITLE: Factoring the group action on $\prod X_k^k $ gives another group?
QUESTION [0 upvotes]: Let's say you have a monomial symmetric polynomial, like the following
$$
m_{(1,2,3,4)}(X_1,X_2,\dots,X_{10})=X_1^1X_2^2X_3^3X_4^4 +X_2^1X_1^2X_3^3X_4^4 + \text{all permutations...}
$$
Then you can write this explicit example as
$$
\begin{eqnarray}
m_{(1,2,3,4)}&=&X_1^1X_2^2X_3^3X_4^4   +X_2^1X_1^2X_3^3X_4^4 + \text{all permutations...}\\
&=&\frac1N\left(\sum_{\rho\in S_{10}} \rho \right)X_1^1X_2^2X_3^3X_4^4, \tag{*}
\end{eqnarray}
$$
where the sum runs over all permutations $\rho$ of the symmetric group, i.e. $S_{10}$.
The factor $N=10!/6!$ is needed since $6!$ elements, those acting only on $X_5,\dots X_{10}$ and trivially the identity, leave $X_1^1X_2^2X_3^3X_4^4$ invariant. Let's call these elements $e\in E$ then we can write $(*)$ as
$$
m_{(1,2,3,4)}=\frac1N\left(\sum_{\rho\in S_{10}/E} \rho \right)\left(\sum_{e\in E} \rho_e \right) X_1^1X_2^2X_3^3X_4^4. 
$$
Is $E$ a group on it's own and how does this work for other symmetric polynomials like 
$$
m_{(1,2,2,4)}=X_1^1X_2^2X_3^2X_4^4  + \text{all permutations...?}
$$

REPLY [1 votes]: While thinking 'bout it I found this one: Fixed points and stabilizer subgroups

Given $g$ in $G$ and $x$ in $X$ with $g.x=x$, we say $x$ is a fixed point of $g$ and $g$ fixes $x$.
For every $x$ in $X$, we define the stabilizer subgroup of $x$ (also called the isotropy group) as the set of all elements in $G$ that fix $x$:
  $$
    G_x = \{g \in G \mid g.x = x\}.
$$
  This is a subgroup of $G$, though typically not a normal one.
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