# Groups acting on roots of polynomials

A passage from my textbook (Dummit and Foote) says

"Alternatively, since $G = Gal(\mathbb{Q}(\sqrt[3]{2}, \zeta_3)/\mathbb{Q})$ acts as permutations of the $3$ roots of $x^3 - 2$, $G$ is a subgroup of $S_3$, hence must be $S_3$ since it is of order $6$."

Why is $G$ a subgroup of $S_3$? I think I am missing something fundamental here.

Thanks!

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Because $G$ acts by permutation on the roots on $x^3-2$, faithfully, – Mariano Suárez-Alvarez Apr 12 '11 at 23:46
You said it: "$G$ acts as permutations of the 3 roots". – lhf Apr 12 '11 at 23:46

The elements of $G=\text{Gal}(\mathbb{Q}(\zeta_3,\sqrt[3]{2})/\mathbb{Q})$ can be identified with permutations of the three objects $a=\sqrt[3]{2}$, $b=\zeta_3\sqrt[3]{2}$, and $c=\zeta_3^2\sqrt[3]{2}$ as follows: An element $g\in G$ permutes these 3 objects by $$a\mapsto g(a)$$ $$b\mapsto g(b)$$ $$c\mapsto g(c)$$ and furthermore $g$ is uniquely determined by how it permutes these three objects.
Thus, each $g\in G$ can be associated with a permutation $\sigma_g\in S_3$ in such a way that the map $f:G\rightarrow S_3$ defined by $f(g)=\sigma_g$ is an injective homomorphism. We then identify $G$ with its image in $S_3$, i.e. we forget about the distinction between them - we can do this because $f$ is injective, so there is no "loss of information". This is the sense in which $G$ "is a subgroup" of $S_3$.