Could somebody give me a clue, related to the possible solution of the problem?
Let's denote a polynomial $f(x_{1}, x_{2}, x_{3}, x_{4}, x_{5})=x_{1} x_{2} x_{3} +x_{2} x_{3} x_{4} + x_{3} x_{4} x_{1} + x_{4} x_{1} x_{2}$. We consider a set of permutations $\omega$ such that $f_{\omega} (x_{1}, x_{2}, x_{3}, x_{4})=f$ (that actually tranform the polynomial into itself). I need to prove that the set of all these permutations creates a subgroup of $S_{4}$.
Actually, it's not a tough job to enumerate all the permutations needed. But the problem is to show that they actually create a subgroup of $S_{4}$. Could you give a clue, how to perform it in a better way?
Thank you in advance.
If two permutations fix a polynomial then so do their composition (apply them one after another and the polynomial remains fixed). If a permutation fixes a polynomial, then so does its inverse: first apply the permutation; the polynomial doesn't change. Then apply the inverse. Since this is an action, this is equivalent to applying the identity, which doesn't change the polynomial, and since we applied the inverse to the original polynomial the inverse also fixes the polynomial.