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Given a polynomial $f(x_1, · · · , x_n) ∈ F[x_1, · · · , x_n]$ and a permutation $σ ∈ S_n$, I want to show that both $\prod_{σ∈Sn} {σ · f(x_1, · · · , x_n)}$ and $\sum_{σ∈Sn} {σ · f(x_1, · · · , x_n)}$ are symmetric polynomials.

I know a symmetric polynomial is a polynomial $F(X_1, …, X_n)$ in $n$ variables, such that if any of the variables are interchanged, one obtains the same polynomial. Or P is a symmetric polynomial if for any permutation $σ$ of the subscripts $1, 2, ..., n$ one has $F(X_{σ(1)},…, X_{σ(n)}) = P(X_1,…, X_n)$. But I am not sure how to proceed in the above example.

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  • $\begingroup$ You can't say "given a permutation $\sigma\in S_n$" and then sum over $\sigma\in S_n$. $\endgroup$
    – joriki
    Aug 23, 2018 at 5:49
  • $\begingroup$ Hint: for each $\tau \in S_n$, multiplying by $\tau$ on the left is a permutation of the group $S_n$. (the proof of Cayley's theorem). This is a common trick in group theory, averaging over a group. $\endgroup$
    – 4-ier
    Aug 23, 2018 at 6:12

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A polynomial $g$ is a symmetric polynomial if $\sigma g=g$ for all $\sigma \in S_n$. For your case we have $g=\sum\limits_\tau \tau(f)$. If $\tau$ runs over $S_n$ then the set $\sigma \circ \tau$ coincides with $S_n$.Then $$ \sigma(g)=\sum\limits_\tau \sigma(\tau(f))=\sum\limits_\tau (\sigma \circ \tau)(f)=\sum\limits_\tau \tau(f)=g, $$ so $g$ is an symmetric polynomial.

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  • $\begingroup$ Ah how about the case for product rather than sum? Is the proof the same? $\endgroup$
    – Homaniac
    Aug 23, 2018 at 9:10
  • $\begingroup$ yes, the proof is the same $\endgroup$
    – Leox
    Aug 23, 2018 at 9:28

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