Permutation on Polynomial is Group Action

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Theorem

Let $n \in \Z: n > 0$.

Let $F_n$ be the set of all polynomials in $n$ variables $x_1, x_2, \ldots, x_n$:

$F = \set {\map f {x_1, x_2, \ldots, x_n}: f \text{ is a polynomial in $n$ variables} }$

Let $S_n$ denote the symmetric group on $n$ letters.


Let $*: S_n \times F \to F$ be the mapping defined as:

$\forall \pi \in S_n, f \in F: \pi * \map f {x_1, x_2, \ldots, x_n} = \map f {x_{\map \pi 1}, x_{\map \pi 2}, \ldots, x_{\map \pi n} }$

Then $*$ is a group action.


Proof

Let $\pi, \rho \in S_n$.

Let $\pi * f$ be the permutation on the polynomial $f$ by $\pi$.

Let $e \in S_n$ be the identity of $S_n$.


From Symmetric Group is Group:

$e * f = f$

thus fulfilling Group Action Axiom $GA \, 1$.


Then we have that:

\(\displaystyle \paren {\pi \circ \rho} * f\) \(=\) \(\displaystyle \map \pi {\rho * f}\)
\(\displaystyle \) \(=\) \(\displaystyle \pi * \paren {\rho * f}\)

thus fulfilling Group Action Axiom $GA \, 2$

$\blacksquare$


Also see


Sources