Galois group permutation of roots

When considering the Galois group of the splitting field of the polynomial $x^3-2$, it is mentioned in my notes that $\sqrt[3]{2}$ can be mapped to $\sqrt[3]{2}$,$\sqrt[3]{2}\omega$ or $\sqrt[3]{2}\omega^2$, where $\omega$ is the cube root of unity. $\omega$ must be mapped to $\omega$ or $\omega^2$.

My question is why is this so? Sorry for the beginner question, but why can't $\sqrt[3]{2}$ be mapped to say $\omega$, or $\omega$ be mapped to say, 1?

Thank you very much for help.

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Hint : Can you prove a number can only be mapped to one of the other roots in its minimal polynomial? – dinoboy Oct 30 '12 at 14:54
You can find an explanation Example 7.2.14. – Reader Oct 30 '12 at 17:41
@Reader the link is dead, this seems like a very valuable post – frogeyedpeas Oct 18 '15 at 22:40

First recall the definition of an automorphism of a field $F$. A map $\sigma: F \rightarrow F$ is an automorphism if its a bijective homomorphism that is $\sigma(0)=0, \;\sigma(1)=1$ and for any $a,b \in F$ we have $\sigma(ab)=\sigma(a)\sigma(b)$ and $\sigma(a+b)=\sigma(a)+\sigma(b)$. In particular in your example $\omega$ cannot map to $1$ because $\omega \neq 1$.
In the general context we have the following result let $L/K$ be an algebraic extension and $\sigma$ a $K$-automorphism of $L$. By $K$-automorphism I mean that for any $\alpha \in K$ we have $\sigma(\alpha)=\alpha$ so $\sigma$ fixes $K$. If $x \in L$ then $\sigma(x)$ is a $K$-conjugate of $x$ that is, $\sigma(x)$ is a root of the minimum polynomial of $x$ over $K$. Let $p(t)=min_K(x,t)$ that is the minimum polynomial of $x$ then we have that
$$0=\sigma(p(x))=\sigma\left(\sum_{i=0}^n p_ix^i\right)=\sum_{i=0}^n \sigma(p_i)\sigma(x)^i=\sum_{i=0}^np_i\sigma(x)^i=p(\sigma(x)).$$ So $\sigma(x)$ is also a root of $p(t)$.
In the context of your question $\sqrt[3]{2}$ can't map to $\omega$ because their minimum polynomials over $\mathbb Q$ are different.