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Looking for nicer ways to work this out than having to check all permutations.

If we have the polynomial:

$p=x_1^2+x_2^2+x_3^2+x_4$

Then In order for $S_4$ to stabilize it it must leave $x_4$ uuntouched and we can move around the other variables so the stabilizer is just the number of permutations which leave $x_4$ which is quite straightforward. Is there a similar approach with the polynomial

$x_1x_2^2x_3^3+x_3x_4^2x_1^3+x_2x_3^2x_4^3+x_4x_1^2x_2^3$

When I say similar I just mean what's the smart way to solve this :-)

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    Write variables to the 0th power too, and sort them. The answer should be clear then.2012-06-21

1 Answers 1

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Let $f=x_1x_2^2x_3^3x_4^0+x_4x_1^2x_2^3x_3^0+x_3x_4^2x_1^3x_2^0+x_2x_3^2x_4^3x_1^0$ be the given symmetric polynomial i.e. it is invariant under the cyclic permutation $\sigma=(1\; 2\; 3\; 4)$ so is under the powers of $\sigma$ which has order $4.$ Now, $S_4$ is generated by $\tau=(1 \; 2)$ and $\sigma,$ but $f$ is not invariant under the action of $\tau$ i.e. $\tau f \neq f$ therefore, is not invariant under the elements of the form $\tau \sigma^i$ and $\sigma^i \tau.$ Since the stabilizer of $f$ in $S_4$ is the subgroup of $S_4,$ it must contain $\{1, \sigma, \sigma^2, \sigma^3\}$ and thus nothing more.

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    Dear @Ben Davidson: this is a basic result in group theory. For example, see theorem 2.6 http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/genset.pdf2012-06-22