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

If we have the polynomial:


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


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

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    $\begingroup$ Write variables to the 0th power too, and sort them. The answer should be clear then. $\endgroup$ – Jack Schmidt Jun 21 '12 at 17:33

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.

  • $\begingroup$ The $\tau$ argument isn't completely convincing. There is a subgroup strictly between $\langle \sigma\rangle$ and $\langle \sigma,\tau\rangle$, and I don't think you mention why it cannot be this group. You have the right answer, and I still think it is clear. (if anyone cares, it's a holomorph!) $\endgroup$ – Jack Schmidt Jun 21 '12 at 17:58
  • $\begingroup$ Dear @Jack Schmidt: but I mentioned that $\tau \sigma^i f \neq f$ and $\sigma^i \tau f\neq f$ so the stabilizer cannot contain an element involving $\tau.$ Isn't it enough? $\endgroup$ – Ehsan M. Kermani Jun 21 '12 at 18:06
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    $\begingroup$ Oddly, no. $\langle \tau \sigma^2 \tau, \sigma \rangle$ is a proper subgroup, and you didn't check whether $\tau \sigma^2 \tau f = f$ (or equivalently, that $\sigma^2 \tau f = \tau f$). $\endgroup$ – Jack Schmidt Jun 21 '12 at 18:12
  • $\begingroup$ indeed you're right! this is the only thing that must be examined by hand which I'm quite reluctant to do so. $\endgroup$ – Ehsan M. Kermani Jun 21 '12 at 21:29
  • $\begingroup$ Thanks I follow most of this. But how is it possible to know $S_4$ is generated by the above permutations? $\endgroup$ – Ben Davidson Jun 22 '12 at 13:39

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