# What is a permutation of a Group?

I've become confused with a certain sentence of an abstract algebra text.

The author is talking about the Field extension $ℚ(x_1, x_2, ... , x_n)/ℚ$, where each $x_i$ is one of the roots of a polynomial of degree $n$.

He follows to explain that "the most important property of $ℚ(x_1, x_2, ... , x_n)$ is that it is symmetric with respect to $x_1, x_2, ... , x_n$, in the sense that any permutation $*$ of $x_1, x_2, ... , x_n$ extends to a bijection * of $ℚ(x_1, x_2, ... , x_n)$ defined by

$$* f(x_1, ... , x_n) = f(*x_1, ... , *x_n)$$

for each rational function $f$ of $x_1, ... , x_n$. Moreover, this bijection * obviously satisfies

$$*(f + g) = *f + *g,$$ $$*(fg) = (*f)(*g)$$

The issue is that I'm not really understanding what this permutation $*$ is, nor why it has the properties that the author claims it has.

I also don't get what he means by "symmetric". He was talking before about how each coefficient $a_i$ of the polynomial is a function of the roots. He called them "elementary symmetric functions". I guess -and might be completely wrong here- that the name has something to do with the fact that substituting a certain root $x_j$ with $x_k$ in any of these 'symmetric functions' wouldn't mess up the equation.

I don't know much about Group theory, so forgive me if I ask too many questions about the notation/terminology of the replies

I would really appreciate any help/thoughts!

What the author states is that for every permutation $\sigma$ of the set $\{x_1,x_2,\ldots,x_n\}$, there is a bijection of $\mathbb{Q}(x_1,x_2,\ldots,x_n)$ into itself thus defined: every rational function $f\in\mathbb{Q}(x_1,x_2,\ldots,x_n)$ is mapped int $\sigma f\in\mathbb{Q}(x_1,x_2,\ldots,x_n)$, where $\sigma f(x_1,x_2,\ldots,x_n)=f\bigl(\sigma(x_1),\sigma(x_2),\ldots,\sigma(x_n)\bigr)$.
The symmetric elementary polynomials are\begin{align}\sigma_1(x_1,x_2,\ldots,x_n)&=\sum_{k=1}^nx_k\\\sigma_2(x_1,x_2,\ldots,x_n)&=\sum_{i<j}x_ix_j\\&\cdots\\\sigma_n(x_1,x_2,\ldots,x_n)&=x_1x_2\ldots x_n\end{align}and if $x_1,x_2,\ldots,x_n$ are the roots of a monic polynomial $P(x)$, then$$P(x)=x^n-\sigma_1(x_1,x_2,\ldots,x_n)x^{n-1}+\sigma_2(x_1,x_2,\ldots,x_n)x^{n-2}-\cdots\pm\sigma_n(x_1,x_2,\ldots,x_n).$$