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Find the group of permutations on $\{1, 2, 3, 4\}$ which leaves the symmetric polynomial $x_1 x_2+x_3x_4$ invariant.

What I know about this is as follows:

A polynomial $f(x_1, . . . , x_n)$ is invariant under $S_n$ if for all $\pi \in S_n$ $$f(\pi(x_1), . . . , \pi(x_n)) = f(x_1, . . . , x_n)$$ But here how will I find the permutation such that the polynomial is invariant.

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    The 2-cycle $(1,2)$ is in the group. Why? – 2017-01-13
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    Where did you get the question from? $x_1x_2+ x_3x_4$ is not a symmetric polynomial. – 2017-01-13
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    It is a symmetric polynomial. – 2017-01-13
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    By definition a [symmetric polynomial](https://en.wikipedia.org/wiki/Symmetric_polynomial) is a polynomial that is invariant under permutations of its variables. If $x_1x_2 + x_3x_4$ were a symmetric polynomial, then the answer to your question would be trivial. – 2017-01-13

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Well, if $\sigma\in G$, the group you want to find, then you have: $$\sigma(x_1)\sigma(x_2)+\sigma(x_3)\sigma(x_4) = x_1x_2+x_3x_4$$.

The key here is that the above identity should be thought of as an equality between two polynomials. Can you take it from here?

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    Then we can say that $\sigma$ is the identity element also. Except identity is there any? – 2017-01-13
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    no, there are more. look at $\sigma = (x_2,x_1, x_3, x_4)$, for example. – 2017-01-13
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    Even $(1 2), (3,4)$ this will also work. – 2017-01-13
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    yes, but you are looking for all such $\sigma $s that make it work. do you understand the question itself actually? – 2017-01-13
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    (1 2), (3,4), (1 2)(3 4), (1), I think these are the all. – 2017-01-13
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    @SachchidanandPrasad Keep going. $(1 3)(2 4)$ also works. Now multiply this by some of the ones you have, too. – 2017-01-13
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    @davidP why does it matter if I cannot have $1\to 3$? – 2017-01-13
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    @SachchidanandPrasad Sorry, I misread that as a cycle. – 2017-01-13
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    The permutation group acts on the indices not on the values of the polynomial: $$\sigma(x_1)\sigma(x_2)+\sigma(x_3)\sigma(x_4) = x_1x_2+x_3x_4$$ should read: $$x_{\sigma(1)}x_{\sigma(2)}+x_{\sigma(3)}x_{\sigma(4)} = x_1x_2+x_3x_4$$ – 2017-01-13