Automorphisms and Galois Theory

Question

I have been given the polynomial $x^3-2$, I found the splitting field $E=\mathbb{Q}(\sqrt[3]{2},i\sqrt{3})$, then further deducing that $|G(E/\mathbb{Q})|=6$ and then that $G(E/\mathbb{Q}) \simeq S_3 \text{ or } C_6$. Further, by doing some calculations, I find that it is not commutative, thus $G(E/\mathbb{Q}) \simeq S_3$. However, I do not know how to find the order of these group elements, as I know by Lagrange's Theorem that the subgroups should consist of either one group of order 3 or three groups of order 2. My question is: how do I calculate this (the orders of these group elements)?

Answer 1

It is quite straightforward, actually. Could you try to write down the definition of each element in your automorphism group?

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Recall how you obtain those elements.$\newcommand{\rat}{\mathbb{Q}}$ First, any $\sigma\in G(E/\rat)$ is completely determined by its image on $2^{1/3}$ and $i\sqrt{3}$. Also, $\sigma$ must map an element in $E$ to one of its conjugates over $\rat$. Since there are $3$ conjugates of $2^{1/3}$ and $2$ conjugates of $i\sqrt{3}$ over $\rat$, there are $6$ possibilities of an automorphism of $E$ fixing $\rat$. We know that $|G(E/\rat)|=6$, so all those $6$ possibilities induced by the conjugates maps must be valid automorphisms.

To determine the order of, say, $\sigma\in G(E/\rat)$, simply look at how its powers act on the elements $2^{1/3}$ and $i\sqrt{3}$ (see the second sentence of the last paragraph). For example, if $\sigma$ is induced by $2^{1/3}\mapsto 2^{1/3}\exp(2\pi i/3)$ and $i\sqrt{3}\mapsto -i\sqrt{3}$, then we can construct the following multiplication table $$ \begin{array}{c|ccc} & \text{id} & \sigma & \sigma^2 \\ \hline \sqrt[3]{2} & \sqrt[3]{2} & \sqrt[3]{2}(\frac{1}{2}+i\frac{\sqrt{3}}{2}) & \sqrt[3]{2}\\ i\sqrt{3} & i\sqrt{3} & -i\sqrt{3} & i\sqrt{3} \end{array} $$ We see that $\sigma^2=\text{id}$ because they coincide on $2^{1/3}$ and $i\sqrt{3}$. Then $\sigma$ has order $2$. The calculation above is done as follows: $$ \begin{align*} \sigma^2(\sqrt[3]{2}) &= \sigma(\sqrt[3]{2}(\frac{1}{2}+i\frac{\sqrt{3}}{2})) \\ &= \sigma(\sqrt[3]{2})\sigma(\frac{1}{2}+i\frac{\sqrt{3}}{2}) \\ &= \sqrt[3]{2}(\frac{1}{2}+i\frac{\sqrt{3}}{2}) \sigma(\frac{1}{2}+i\frac{\sqrt{3}}{2}) \\ &= \sqrt[3]{2}(\frac{1}{2}+i\frac{\sqrt{3}}{2}) (\frac{1}{2}+\frac{1}{2}\sigma(i\sqrt{3}) )\\ &= \sqrt[3]{2}(\frac{1}{2}+i\frac{\sqrt{3}}{2}) (\frac{1}{2}-i\frac{\sqrt{3}}{2}) \\ &= \sqrt[3]{2}. \end{align*} $$ In this case, the calculation is rather complicated. If you use $E=\rat(\sqrt[3]{2},e^{2\pi i/3})$ for example, then it would become much simpler.